msc_phychem.pdf

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UNIT-I

LESSON: 1 – SYMMETRY ELEMENTS AND SYMMETRY OPERATIONS CONTENTS 1.0. AIMS AND OBJECTIVES 1.1. INTRODUCTION 1.2. IDENTICAL CONFIGURATION 1.3. EQUIVALENT CONFIGURATION 1.4. SYMMETRY OPERATION 1.5. SYMMETRY ELEMENT 1.6. ROTATION AXIS OF SYMMETRY (Cn) 1.7. PLANE OF SYMMETRY OR MIRROR PLANE (σ) 1.8. CENTER OF SYMMETRY OR INVERSION CENTER (i) 1.9. ROTATION-REFLECTION AXIS N-FOLD (Sn) 1.10. IDENTITY (E) 1.11. INVERSE OPERATIONS 1.11.1. INVERSE OF s And i 1.11.2. INVERSE OF ROTATION [Cn

-1] 1.11.3. INVERSE OF Sn

1.12. LET US SUM UP 1.13. CHECK YOUR PROGRESS 1.14. LESSON - END ACTIVITIES 1.15. REFERENCES 1.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on symmetry elements and symmetry operations to the students. On successful completion of this lesson the student should have: * Understand the symmetry elements and symmetry operations. 1.1. INTRODUCTION Symmetry is a very fascinating phenomenon in nature. It is found in geometrical figures such as a cube, a sphere, an equilateral triangle, a rectangle, a square, a regular pentagon, a regular hexagon etc. Its importance was recognized by eminent Greek philosophers Pythagoras and Plato. 1.2. IDENTICAL CONFIGURATION An identical configuration is the one which is not only indistinguishable from the original one but also identical with it.

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1.3. EQUIVALENT CONFIGURATION An equivalent configuration is the one which cannot be distinguished from the original one but need not be identical with it. 1.4. SYMMETRY OPERATION A symmetry operation is a movement of the molecule such that the resulting configuration of the molecule is indistinguishable from the original. 1.5. SYMMETRY ELEMENT A symmetry element is a geometrical entity such as a line or a space or a point about which an operation of rotation or reflection or inversion is done. 1.6. ROTATION AXIS OF SYMMETRY (Cn) It is also called rotational axis, if a rotation around an axis by 360°/n results in a molecule indistinguishable from the original. Examples are water (C2) and ammonia (C3). A molecule can have more than one symmetry axis and the one with the highest number of n is called the principal axis and takes the z-axis in a Cartesian coordinate system. This axis of symmetry can be explained by taking the example of triangular planar boron trichloride molecule. In boron trichloride molecule an axis of symmetry is located perpendicular to the plane containing all the atoms. This is known as the C3 axis of symmetry. In general the symbol for proper axis of symmetry is Cn where n is known as the order of the axis. The order of the axis is given by the number of rotations by q, to get the identical configuration. n i s

alternatively given by the formula ÷ø

öçè

æ=

q

p2n , where q is the minimum angle of rotation to obtain

the equivalent configuration. n has non-zero positive integral values. Boron trichloride molecule has three C2 axes of symmetry in addition to the C3 axis (Fig.1.1). The C3 axis in this molecule is known as principal axis. In general, if there are Cn axes of different orders in a molecule, the axis with the highest order is referred to as the principle axis.

B

Cl

ClCl

C2

(a)



3

B

Cl

Cl

Cl

C3

(b) Fig.1.1. (a). The C2 axis of symmetry in BCl3 molecule. (b). The C3 principal axis in BCl3 molecule. 1.7. PLANE OF SYMMETRY OR MIRROR PLANE (σ) If reflection through a plane leaves an identical copy of the original molecule it has a plane of symmetry. Water has two of them: one in the plane of the molecule itself and one perpendicular to it. A symmetry plane parallel with the principal axis is dubbed vertical (σv) and one per perpendicular to it horizontal (σh). A third plane exists: if a symmetry plane bisects the angle between two n-fold axes that are perpendicular to the principal axis the plane is dihedral (δd). A plane can also be identified by its plane (xz),(yz) in the Cartesian coordination plane.

O

H

H

sv

C2

O

HH

C2

sv

Fig. 1.2. The reflection planes in water molecule. 1.8. CENTER OF SYMMETRY OR INVERSION CENTER (i) A molecule has a center of symmetry when for any atom in the molecule an identical atom can be found when it moves in a straight through this center an equal distance on the other side. An example is xenon tetrafluoride but not cisplatin even though both molecules are square planar. This is a point such that any line drawn through it meets the same atom at equal distances in opposite directions. All homo-nuclear diatomic molecules possess the centre of symmetry.



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Fig.1.3 lists the molecules with centre of symmetry. This element of symmetry is also called S2 axis. A particular symmetry element generates many symmetry operations. A Cn axis generates

a set of operations .,.....,,, 321 nnnnn CCCC The n

nC operation is equivalent to the identity operation.

H H C OO

C C

H

HH

H

N N

F

F

Fig.1. 3. Diagram showing molecules with centre of symmetry. 1.9. ROTATION-REFLECTION AXIS N-FOLD (Sn) It is also called improper rotational axis. Molecules with this symmetry element can have a 360°/n rotation around an axis followed by a reflection in a plane perpendicular to it without a net change. An example is tetrahedral silicon tetrafluoride with three S3 axes and the staggered conformation of ethane with S6 symmetry. It is the line about which a rotation by a specific angle followed by reflection in a plane perpendicular to the rotation axis is performed. Fig.1.4 shows the S6 axis in staggered from of ethane.

C

C

C

C

C

C

H

H

H

H

H

H H

H

H H

H

H

H

H

H

H

H

H

C6

s

Fig.1.4. The S6 axis in staggered form of ethane. 1.10. IDENTITY (E)



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This is a default symmetry element and every molecule has one. 1.11. INVERSE OPERATIONS Suppose for a molecule we carry out an operation P followed by Q such that Q returns all the atoms of the molecule to their original position then, Q is said to be the inverse of P. In such cases, QP = E =PQ Algebraically we can express Q = P-1, thus we can write P-1P = E = PP-1 because an operation and its inverse always commute. 1.11.1. INVERSE OF s And i In the case of inversion and reflection, the carrying out of these operations in succession leads to identity E, i.e. s.s = s2 = E and i.i= i2= E. Hence in these cases, these operations themselves are their own inverse that i = i-1 and s = s-1. 1.11.2. INVERSE OF ROTATION [Cn

-1] In the case of rotation, simply carrying out an operation for the second time does not give the original configuration. In Cn is the clockwise rotation by (2p/n)0 then Cn

-1 is an anticlockwise rotation by (2p/n)0 about Cn. Then, Cn

-1 Cn = E.

For example, in NH3 molecule ECC =-3

13 .

N

H1 H3H2

C3

N

H2 H1H3

C3

N

H1 H3H2

C3+

Clockwiserotation

Anti-clockwiserotation

C3-1

At the same time, for NH3, ECC =13

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The above way of carrying out the symmetry operations successively is algebraically represented as multiplication. If P and Q are two symmetry operations, PQ is a combined operation of carrying out Q first and then P. By conversion, first operation is written at right. For example, in the second case shown above for NH3 molecule, carrying out C2 first and then sv is represented as sv C2. 1.11.3. INVERSE OF Sn

As with Cn

-1, Sn-1 can be defined as rotation anticlockwise by (2p/n)0 followed by

reflection with perpendicular plane. We can prove for improper axis.



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( )( ) ECCSS hnhnnn == -- ss 1111

Since 11nn CC ss =

( ) ( ) ( ) ( )11111nhhnhnhnnn CCCCSS ssss --- ==

( ) ECCECCCC nnnnnhhn ==== --- 1111 ss

However, we can express Sn

-1interms of clockwise rotation about the same axis. We know that Sn

n= E when n is odd. Then

ESS nnn =- 11 (n-even) Thus, 11 -- = n

nn SS

ESS nn

n =- 112 (n-odd) Thus, 121 -- = nnn SS

1.12. LET US SUM UP In this lesson, we: Pointed out Ø Identical configuration Ø Equivalent configuration Ø Symmetry operation Ø Symmetry element Ø Rotation axis of symmetry (Cn) Ø Plane of symmetry or mirror plane (σ) Ø Center of symmetry or inversion center (i) Ø Rotation-reflection axis n-fold (Sn) Ø Identity (E) Ø Inverse operations Ø Inverse of s and i Ø Inverse of rotation [Cn

-1] Ø Inverse of Sn

1.13. CHECK YOUR PROGRESS 1. What is n improper axis of rotation? What are the operations generated byS5? How many of these are the distinct operations of S5? 2. Prove the following:

a) S2=I b) 326 CS = and c) ES n

n =



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1.14. LESSON - END ACTIVITIES 1. (a) Distinguish between (1). Symmetry element and symmetry operations. (2). Proper and improper rotation. (b) Show that C2(z) and s(xy) commute. 2. What is an inverse operation? Is this equivalent to any other combination of operations? Give an example. 1.15. REFERENCES 1. K.V. Raman, Group Theory and its applications to Chemistry, Tata McGraw-Hill Publishing Company limited, New Delhi. 2. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006- 2007. 3. V. Ramakrishnan, M.S Gopinathan, Group Theory in Chemistry, Vishal Publications.



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LESSON: 2 – GROUPS AND THEIR BASIC PROPERTIES

CONTENTS 2.0. AIMS AND OBJECTIVES 2.1. INTRODUCTION 2.2. GROUP 2.2.1. BASIC PROPERTIES OF A GROUP 2.2.2. ORDER OF GROUP 2.2.3. ABELIAN GROUP 2.2.4. NON-ABELIAN GROUP 2.2.5. ISOMORPHISM 2.3. SIMILARITY TRANSFORMATION AND CLASSES 2.3.1. SIMILARITY TRANSFORMATION 2.3.2. CLASS 2.4. GROUP MULTIPLICATION TABLE 2.4.1. IMPORTANT CHARACTERISTICS OF A GROUP MULTIPLICATION TABLE 2.5. SYMMETRY CLASSIFICATION OF MOLECULES INTO POINT GROUPS: 2.6. DIFFERENCE BETWEEN POINT GROUP AND SPACE GROUP 2.7. LET US SUM UP 2.8. CHECK YOUR PROGRESS 2.9. LESSON - END ACTIVITIES 2.10. REFERENCES

2.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on groups and their basic properties to the students. On successful completion of this lesson the student should have: * Understand the groups and their basic properties. 2.1. INTRODUCTION Having defined various symmetry operations in Lesson 1 we may now ask ourselves whether it is possible to classify the molecules into ‘Groups’ on the basis of the symmetry elements they posses. Is it possible to define certain symmetry groups so that all molecules belonging to a certain group have the same type of symmetry operations? Luckily, the answer is ‘Yes’. This means that we can give an accurate description of the symmetry of any molecule by knowing to which group it belongs.

2.2. GROUP A group is a complete set of members which are related to each other by certain rules. Each member may be called an element.



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2.2.1. BASIC PROPERTIES OF A GROUP Certain rules have to be satisfied by the elements so that they form a group. These rules are the following: 1. The product of any two elements and the square of any element must be elements of group (closure property). 2. There must be one element in the group which commutes with everyone of the elements and leaves it unchanged. 3. The associative law of multiplication should be valid. 4. For every element there must be a reciprocal (inverse) and this reciprocal is also an element of the group. RULE 1 If A and B are the element of the group and if AB = C, C must be a member of the group. Product AB means that we perform the operation B first and then operation A i.e., the sequence of operations is from right to left. It should be noted that the other product BA need not be same as AB. BA means doing A first and then performing the operation B later. Let BA = D. D must be a member of the group by rule 1. Usually AB BA and so C D. However, there may be some special elements A and B each that AB = BA. Then A and B are said to commute with each other or the multiplication of A and B is commutative. Such a group where any two elements commute is called an abelian group. H2O belongs to an abelian group. The four symmetry operations for H2O are E, C2z, sv(xz) and s’v(yz).The inter-relationships between these operations are given in the group multiplication table (Table 2.1). Table 2.1.

E C2z, sv(xz) s’v(yz) E C2z sv(xz) s’v(yz)

E C2z sv(xz) s’v(yz)

C2z

E s’v(yz)

sv(xz)

sv(xz) s’v(yz) E C2z

s’v(yz) sv(xz) C2z

E

Note that each member is its own inverse. The product of any two operations is found

among the four members.

( ) ECE v ===2'2

22 s

etcCCC vvvvvvv'

222'' ; sssssss ====

All these are noted from the table.



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RULE 2 Each group must necessarily have an element which commutes with every other element of the group and leaves it unchanged. Let A and B be the elements of the group. Let X be the element satisfying rule 2. i.e. XA = AX = A and also XB = BX = B. BA = BX2A; BX2 = B = BE, where we have set X2 = E (identity) It is clear BEn = B, n being any integer. This kind of element E which does not effect any change when multiplied with any element is a unique element and is called an identity operation E. For water, E, the identity operation satisfies this rule. It is so for all molecules. RULE 3 Associative law of multiplication must be valid. This means ABCD is the same as (AB) (CD), (A) (BCD) or (ABC) (D). ABC is the same as A(BC) or (AB)C. For example, we have for water

( )( ) EECEC vvvvvv === '''2

'2 ssssss

( ) 2'

2'

2'

2 CCECEC vvvvvv === ssssss

Multiplication simply means successive application of the symmetry operations in the order right to left. RULE 4 Inverse of an element A is denoted by A-1 (this does not mean 1/A). It is simply an element of the group such that A-1 A=E. In case of symmetry groups, A-1 is that element which

undoes or annuls the effect of A. For H2O we have, for example, C2C2 = E. Therefore 21

2 CC =-

i.e., C2 is its own inverse. This is true of all other elements for H2O. But this is not general. For

example, ECC ¹= 326 . Therefore 1

6-C is not C6.

2.2.2. ORDER OF GROUP The total number of elements present in a group is known as the order of the group. It is denoted by n. Example: 1. Water molecule belongs to C2v group of order 4 because it contains 4 elements namely E, C2z, sv and s’v. 2. Ammonia belongs to C3v group of order 6 as it contains 6 elements namely E, C3

1, C32,

sv(1), sv(2) and sv(3).



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2.2.3. ABELIAN GROUP A group is said to be abelian if for all pairs of elements of the group, the binary combination is commutative. That is AB = BC; BC = CB – and so on. Example: The elements of C2v point group E, C2z, sv and s’v form an abelian group as all the elements of this group commute with each other. 2.2.4. NON-ABELIAN GROUP A group is said to be non-abelian if the commutative law does not hold for the binary combinations of the elements of the group, i.e., AB BA. Example: The elements of C3v point group E, C3

1, C32, sv(1), sv(2) and sv(3) donot consecutive an

abelian group since the elements donot follow commutative law. 2.2.5. ISOMORPHISM Two groups are supposed to be isomorphic if they obey the following rules. 1. Both have same order and structure. 2. There is a one-to-one correspondence in all respects between the members of the two groups. If A1, B1, C 1, D1 and A2, B2, C 2, D2 are the members of the two isomorphic groups, A1 corresponds to A2, B1 corresponds to B2 and so on. 3. The relationship between the any two members of a group is exactly the same as the relationship between the corresponding members of the other group. Let us take the three groups listed below: (i) E, C2

(ii) E, i (iii) E, sh All the three are isomorphic groups.

22CE = ; 22 CEC =

2iE = ; iE = i

2hE s= ; hh E ss = etc.

2.3. SIMILARITY TRANSFORMATION AND CLASSES 2.3.1. SIMILARITY TRANSFORMATION Let A and X be the elements of a group and let us define B such that B = X-1 AX



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B is called the similarity transform of A by X, or A is said to be subjected to similarity transformation with respect to X. If A and B are related by a similarity transformation they are called conjugate elements. Take the NH3 molecule, for instance. Z axis is the C3axis.

N

Hb

HaHc

Z

N

Ha

HbHc

Z

N

Hb

HcHa

N

Hc

HbHa

N

Hb

HaHc

Z

s"'

s"'

120°

C3Z

C32Z

Fig. 2.1. Illustration of similarity transformation on NH3. There are three reflection planes. These are usually designated as follows: 1. Plane formed by z-axis and NHa bond: s’ or sa or sv’. 2. Plane formed by z-axis and NHb bond: s’’ or sb or sv’. 3. Plane formed by z-axis and NHc bond: s’’’ or sc or sv’. Let us prefer the designation s’, s’’ and s’’’. Let us perform a reflection (s’’’) with respect to the plane formed by NHc and z-axis. Let us perform s again. s’’’2 = E. Now let us find the similarity transform of C3 w.r.t. s’’’, i.e., (s’’’)-1C3s’’’ = ?

It is seen from the Fig. that (s’’’)-1C3s’’’ = 23C . Remember (s’’’) = (s’’’)-1. Thus C3 and

23C are conjugate elements.

The following rules about conjugate elements are notable: 1. Every element is conjugate of itself because every element is the similarity transforms of itself w.r.t. identity (E): E = E-1 and A = E-1 AE. 2. If A is the conjugate of B then B is the conjugate of A. This means that if A is the similarity transform of B by X, B is the similarity transform of A by X-1. We have A = X-1BX



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But (X-1)-1AX-1 = XA X-1 = X (X-1BX) X-1 = (XX-1) B (XX-1) = B (associative law) 3. If A is the conjugate of B and B is the conjugate of C, then A, B and C are mutually conjugate. 2.3.2. CLASSES A complete set of elements which are conjugate to one another is called a class of the group. Let us consider NH3. Set us the coordinate system in such a manner that ZNHa is in the yz plane. (Fig. 1) s’ is then syz. Without disturbing the NH3 molecule rotate the coordinate system by 120° w.r.t z axis, i.e., subject the coordinate system to C3. Now yz plane is ZNHb. syz is s’, s’’ and s’’’ are equivalent. s’ becomes same as that of s’’ if we change the coordinate system by a symmetry operation (C3) of the point group. s’ and s’’ are therefore in the same class. Example: Show that the three reflections of NH3 constitute a class. It is not difficult to show

that ECC =323 . . Hence, ( ) 1

323

-= CC .

Let as perform the similarity transformation of s’ by C3 in NH3.

"3

'233

'13 sss ==- CCCC

Thus s’ and s’’ are conjugate. Similarly we can show that s’, s’’ and s’’’ are mutually conjugate. Therefore s’, s’’ and s’’’ form a class. 2.4. GROUP MULTIPLICATION TABLE Every group is characterized by a multiplication table. The relationship between the elements of the binary combinations is reflected in the multiplication table. Consider a water molecule. It has four symmetry elements, viz., E, C2(z), sv(xz) and sv’(yz) (Fig.2.2).



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H

H

z

x

y

sv'(yz)

sv(xz)C2

O

Fig. 2.2. The Four symmetry elements of H2O molecules We can easily show that the product of any two symmetry elements is one of the four elements of the group. Thus, for instance, C2(z)sv(xz) = sv’(yz). Proceeding this way the symmetry operations of H2O can be listed in a group multiplication table (GMT) (Table 2.2).

E C2(z) sv(xz) sv’(yz)



C2(z) E sv’(yz) sv(xz)

sv(xz) sv’(yz) E C2(z)

sv’(yz) sv(xz) C2(z) E

Table 2.2. Group multiplication table of the symmetry operations of H2O molecule 2.4.1. IMPORTANT CHARACTERISTICS OF A GROUP MULTIPLICATION TABLE 1. It consists of h rows and h columns where h is the order of the group. 2. Each column and row is labeled with group element. 3. The entry in the table under a given column and along given row is the product of the elements which head that column and the row (multiplication rule is strictly followed). 4. At the intersection of the column labeled by Y and the row labeled by X, we found the element which is the product XY. 5. The following rearrangement theorem holds good for every ‘Group Multiplication Table’. “Each row and each column in the table lists each of the group elements once and only once. No two rows may be identical nor any two columns be identical. Thus each row and each column is a rearranged list of the group elements”. 2.5. SYMMETRY CLASSIFICATION OF MOLECULES INTO POINT GROUPS Molecules can be classified into point groups depending on the characteristic set of symmetry elements possessed by them. A molecular group is called a point group since all the elements of symmetry present in the molecule intersect at a common point and this point remains fixed under all the symmetry operations of the molecule. The symmetry groups of the molecules are denoted by specific symbols known as Schoenflies notations.



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Table 2.3. Some Molecular Point Groups Point Group

Symmetry Elements Examples

C1 E CHFClBr C2 E,C2 H2O2 C3 E,C3 C2H6

Cs E,sv NOCl

C2v E,C2,2sv H2O,CH2=O, pyridine C3v E,C3,3sv NH3,CHCl3,PH3

C¥v E,C¥,¥sv HCl, NO,CO

C2h E,C2, sh, i trans CHCl=CHCl D2h E,3C2, 3s, i CH2=CH2,naphthalene

D3h E,2C3, 3C2(^ to C3),3sv, sh, 2S3 BF3(trigonal planar)

D4h

E,C4, 4C2 (^ to C4),2sv, 2sd, sh, C2, S4(coincidence with C4), i

[PtCl4]2- ( s q u a r e planar)

D6h E,2C6,6C2(^ to C6),3sv, 3sd, sh, C2,2C3 ,2S6, 2S3,i C6H6 Td E, 4C3,3C2 ,3S4(coincidence with C2),6sd CH4

Oh

E,3C4,4C3,3S4 and 3C2 (both coincident with the C4 axes), 6C2 ,4S6, 3sh, 6sd

SF6

2.6. DIFFERENCE BETWEEN POINT GROUP AND SPACE GROUP Symmetry operations do not alter the energy of the molecule. Further in all the above operations the centre of the molecule is not altered as none of the operations involve a total translational movement of the molecule. Whatever happens to the molecule, the centre (point) is not changed. At least one point is fixed. Hence these are classified as ‘point group’ operations. In case of crystals operations such as ‘screw rotations’ and glide plane reflections can be additionally specified. Screw rotation involves a rotation with respect to an axis and then a translation in the direction of the same axis. Glide plane reflection is a reflection in a plane followed by a translation along a line in that plane. These are particular to crystals and the classification comes under what is known as space group. Note that here even the centre changes. Thus in short, in point group, there is at least one point (centre) which is not altered after all operations while in space group it is not possible to identify such a stationary point. 2.7. LET US SUM UP In this lesson, we: Pointed out Ø Group Ø Basic properties of a group Ø Order of group Ø Abelian group Ø Non-abelian group



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Ø Isomorphism Ø Similarity transformation and classes Ø Group multiplication table Ø Symmetry classification of molecules into point groups Ø Difference between point group and space group

2.8. CHECK YOUR PROGRESS 1. Explain why a set of numbers cannot form a group by the process of division. 2. Explain why the set of integers between 0 and µ do not form a group under the process of multiplication. 2.9. LESSON - END ACTIVITIES 1. Construct the multiplication table for the C3v point group to which NH3 molecule belongs. 2. Draw the structure of three distinct isomers of C2H2Cl2 and determine their point groups. Which of them is polar? 2.10. REFERENCES 1. K.V. Raman, Group Theory and its applications to Chemistry, Tata McGraw-Hill Publishing Company limited, New Delhi. 2. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006- 2007. 3. V. Ramakrishnan, M.S Gopinathan, Group Theory in Chemistry, Vishal Publications.



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UNIT-II

LESSON 3: REDUCIBLE AND IRREDUCIBLE REPRESENTATIONS CONTENTS 3.0. AIMS AND OBJECTIVES 3.1. INTRODUCTION 3.2. REDUCIBLE REPRESENTATION 3.3. IRREDUCIBLE REPRESENTATION 3.4. GRAND/GREAT ORTHOGONALITY THEOREM (G.O.T.) 3.5. CHARACTER TABLES FOR POINT GROUPS 3.6. CALCULATION OF CHARACTER VALUES OF REDUCIBLE REPRESENTATION PER UNSHIFTED ATOM FOR EACH TYPE OF SYMMETRY OPERATION 3.6.1. IDENTITY (E) 3.6.2. INVERSION AT THE CENTRE OF SYMMETRY (i) 3.6.3. REFLECTION IN A SYMMETRY PLANE (s) 3.6.4. PROPER ROTATION Cn’ 3.7. DETERMINATION OF TOTAL CARTESIAN REPRESENTATION T3N

3.8. DETERMINATION OF DIRECT SUM FROM TOTAL CARTESIAN REPRESENTATION 3.9. LET US SUM UP 3.10. CHECK YOUR PROGRESS 3.11. LESSON - END ACTIVITIES 3.12. REFERENCES 3.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on reducible and irreducible representations to the students. On successful completion of this lesson the student should have: * Understand the reducible and irreducible representations. 3.1. INTRODUCTION The set of matrices corresponding to the symmetry operations of a group is called its representation. Representations can be classified into (a) Reducible representations (reps) and (b) irreducible representations (irreps). 3.2. REDUCIBLE REPRESENTATION Let A, B, C… Be the matrices which form the representation of a group and let X be the similarity transformation matrix of this group such that



18

X-1AX = A’ ----- (1) X-1BX = B’ ----- (2) X-1CX = C’ ----- (3) Then, if X is the proper transformation matrix, we have

X-1AX = A' =

a1'

a2'

a3'

a4'

0

0

----- (4) The new matrix A’ is now blocked out along the diagonal into smaller matrices a1’, a2’, a3’, a4’, etc., with the off-diagonal elements equal to zero. Similarly, we have

X-1BX = B' =

b1'

b2'

b3'

b4'

0

0

----- (5) This is expressed by saying that the given sets of matrices form a reducible representation (rep). 3.3. IRREDUCIBLE REPRESENTATION If it is not possible to find a similarity transformation which will reduce the matrices A, B, C ... to block-diagonalized form, the representation is called an irreducible representation (irrep). 3.4. GRAND/GREAT ORTHOGONALITY THEOREM (G.O.T.) This is the most important theorem of group theory. It concerns the matrix elements which constitute the irreps of a point group. Mathematically it is stated as follows:

( ) ( ) ''

*

'' nnmmij

jiRjmni

ll

hRR

nmddd=GGå ----- (6)



19

Here Gi and Gj are the ith and jth irreps of a point group of order h with dimensions li and lj, respectively; Gi (R)mn is the mnth matrix element corresponding to the symmetry operation R belonging to the ith irrep and Gj (R)mn, is the complex conjugate of the m’n’th matrix element corresponding to the symmetry operation R belonging to the jth irrep. The ds are the well known ‘Kronecker deltas’ which have the following property:

îíì

¹

==

ji

jiij

,0

,1d ;

îíì

¹

==

',0

',1'

mm

mmmmd ;

îíì

¹

==

',0

',1'

nn

nnnnd ----- (7)

The summation is performed operations R of the molecule. If the matrix elements are real, then Gi (R) m’n’ = Gj (R)m’n’ ----- (8) The following three cases arise for the G.O.T. assuming that the matrix elements are real: 1. For two different irreps, i j, m=m’ and n=n’,

( ) ( )å =GGR

mnjmni RR 0 ----- (9)

2. For the same irrep i=j, m m’ and n n’,

( ) ( )å =GGR

nmjmni RR 0'' ----- (10)

3. For an irrep I and for m=m’, n=n’,

( )[ ]å =GR

imni lhR2 ----- (11)

In practice, we do not use the G.O.T. in the form given above but in a slightly different form involving the characters of the irreps. 3.5. CHARACTER TABLES FOR POINT GROUPS For practical purposes, it is sufficient to know only the characters of each symmetry class of a point group to which a molecule belongs. A character table lists the characters of all the symmetry classes for all the irreps of a group. Character tables for the C2v and C3v point groups are given in Tables 3.1 and 3.2. Table 3.1. Character table for C2v point group C2v I

E C2(z) sv (xz) II

sv’(yz) III

IV

A1 A2 B1 B2

1 1 1 1

1 1 -1 -1

1 -1 1 -1

1 -1 -1 1

z Rz x, Ry y, Rx

x2,y2,z2 xy xz yz

Table 3.2. Character table for C3v point group



20

C3v I

E 2C3

II 3sv

III IV

A1 A2 E

1 1 2

1 1 -1

1 -1 0

z Rz (x,y)(Rx,Ry)

x2+y2,z2

(x2-2,xy)(xz,yz)

The character tables can be obtained from the properties of the irreps given above. We can explain the character table by dividing it into four sections I, II, III, IV. Section I. In the top row, the Schoenflies symbol for the point group is given. This section also lists the Mulliken symbols for the different irreps. Symbols A and B are used to label one dimensional irreps and E and T to label two dimensional and three-dimensional irreps, respectively. The nomenclature E should not be confused with the identity operation. A is used when the character for the rotation about the principal axis is +1 and B when it is -1. In other words, A stands for symmetric and B for antisymmetric to such rotation. For a molecule having a centre of symmetry, the subscripts g and u are used to label the irreps that are respectively symmetric and antisymmetric to inversion through the centre of symmetry. Subscripts 1 and 2 are used to label the irreps that are resoectuvekt symmetric and antisymmetric to reflection in a vertical plane s v. The superscripts’ “are used it denote the irreps that are respectively symmetric and antisymmetric to reflection in a horizontal plane s h. Section II. This section gives the characters for all the symmetry operations of different irreps. The characters of the identity operation for the one-dimensional, two-dimensional and three- dimensional irreps are, respectively, 1, 2 and 3. Section III. It gives the transformation properties of the Cartesian coordinates x,y,z and rotations Rx, Ry, Rx about these axes. Section IV. It gives the transformation properties of the binary products of Cartesian coordinates xy,yz,zx etc. and the squares of the coordinates x2, y2, z2, x2+ y2, x2+ y2+z2 etc. The Cartesian coordinates, their squares and binary products, etc., listed in sections III and IV are referred to as the basis functions on which the symmetry operations operate. 3.6. CALCULATION OF CHARACTER VALUES OF REDUCIBLE REPRESENTATION PER UNSHIFTED ATOM FOR EACH TYPE OF SYMMETRY OPERATION The contribution to ( )Rc per unshifted atom for all symmetry operations R can be

worked out in the following manner. 3.6.1. IDENTITY (E) In this case, all three vectors remain unchanged for every unshifted atom as shown in Fig 3.1 where x’=x, y’=y and z’=z. The transformation matrix therefore, includes diagonal elements:



21

+1 0 0

0 +1 0

0 0 +1

Then ( )Ec per unshifted atom is +3.

y

z

x

y'

z'

x'

E

Fig. 3.1

3.6.2. INVERSION AT THE CENTRE OF SYMMETRY (i) Fig. 3.2 shows the effect for each unshifted atom, where x’ = -x, y’ = -y, and z’ = -z. Therefore, the matrix contains the following diagonal elements:

-1 0 0

0 -1 0

0 0 -1

Thus ( )ic per unshifted atom is -3.

y

z

x

y'

z'

x'i

Fig. 3.2



22

3.6.3. REFLECTION IN A SYMMETRY PLANE (s) The effect of any s on an unshifted atom is typically shown in Fig. 3.3, where x’ = x, y’ = y and z’ = z. The transformation matrix, therefore,

+1 0 0

0 -1 0

0 0 +1

Thus ( )sc per unshifted atom is +4.

y

z

x

y'

z'

x'

s

Fig. 3.3

3.6.4. PROPER ROTATION Cn’ Rotation is by (360/n)0, usually about z axis. For the unshifted atom, the result is as

shown in Fig 3.4, where q = (360/n)0. x’ =z, contributing +1 to ( )'nCc and x.y go to x’, y’

respectively.

x

y

zCn

z'

x' y' Fig. 3.4

3.7. DETERMINATION OF TOTAL CARTESIAN REPRESENTATION T3N The total Cartesian representation T3N can be derived from the contribution ( )Rc per

unshifted atom by simple arithmetic multiplication of ( )Rc with the number of unshifted atoms

for every symmetry operation. Hence the calculation of T3N involves two steps (i) to count the number of unshifted atoms for every symmetry operation. (ii) to calculate the contribution to



23

( )Rc for every unshifted atom in every type of symmetry operation. T 3N is also called the

reducible representation of the group for a particular transformation. ILLUSTRATIONS (i) For H20 molecule This molecule belongs to C2v group. The number of unshifted atom for each symmetry operation and the resultant T3N can be given as

C2v E C2 sxz syz

unshi f ted atoms 3 1 1 3

31-19T3N

O

H H (ii) For POCl3 molecule This molecule belongs to C3v point group. Since it is made of five atom, it will give

1515´ matrices. Using the method of unshifted atoms, the reducible representations can be worked out as given below

C2v E 2C3 3sv

unshi f ted atoms 5 2 3

3015T3N

P

O

Cl

ClCl

Rotation by C3

1 or C32 leaves P and O unshifted. Reflection by any sv leaves P, O and one Cl

unshifted. 3.8. DETERMINATION OF DIRECT SUM FROM TOTAL CARTESIAN REPRESENTATION The direct sum of total Cartesian representation T3N can be determined using the reduction formula.



24

ILLUSTRATION (1) POCl3 molecule The unshifted atoms and total Cartesian representation for this molecule which belongs to C3v point group is given by

C3v E 2C3 3sv

unshi f ted atoms 5 2 3

3015T3N

C3v E 2C3 3sv

1 1 1

-111

A1

A2

E 0-12

By applying reduction formula

( ) ( ) ( ) ( )[ ]

( ) ( ) ( ) ( )[ ]

( ) ( ) ( ) ( )[ ] 503310221516

1

113310211516

1

413310211516

1

2

1

=´´+-´´+´´=

=-´´+´´+´´=

=´´+´´+´´=

Ea

Aa

Aa

Therefore, the direct sum for total Cartesian representation is T3N = 4A1 + A2 + 5E (2) Reducible representation and direct sum for T3N for [PtCl4] 2-

[PtCl4] 2- belongs to D4h point group

Pt

Cl Cl

ClCl

D4h E 2C4C2 2C2' 2C2'' i 2S4 sh 2sv 2sd

unshif ted atoms

T3N

5 1 1 3 1 1 1 5 3 1

15 1 -1 -3 -1 -3 -1 5 3 1

Using the character table for D4h point group and reduction formula, it can be shown that



25

T3N = A1g + A2g + B1g + B2g + Eg + 2A2u + B2u + 3Eu 3.9. LET US SUM UP In this lesson, we: Pointed out

Ø Reducible Representation Ø Irreducible representation Ø Grand/great Orthogonality theorem (G.O.T.) Ø Character tables for point groups Ø Calculation of character values of reducible representation per unshifted atom for each

type of symmetry operation Ø Identity (e) Ø Inversion at the centre of symmetry (i) Ø Reflection in a symmetry plane (s) Ø Proper rotation Cn Ø Determination of total cartesian representation T3n Ø Determination of direct sum from total cartesian representation

3.10. CHECK YOUR PROGRESS 1. Define reducible and irreducible representation. 2. Construct the C2v character table. 3.11. LESSON – END ACTIVITIES 1. State and explain the Great Orthogonality Theorem. Use the conclusions obtained from the Orthogonality theorem to construct the character table for C2v group. 2. Using the Great Orthogonality theorem to construct the character table for C3v point group. 3.12. REFERENCES 1. K.V. Raman, Group Theory and its applications to Chemistry, Tata McGraw-Hill Publishing Company limited, New Delhi. 2. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006- 2007. 3. V. Ramakrishnan, M.S Gopinathan, Group Theory in Chemistry, Vishal Publications.



26

LESSON 4: GROUP THEORY AND VIBRATIONAL SPECTROSCOPY CONTENTS 4.0. AIMS AND OBJECTIVES 4.1. INTRODUCTION 4.2. GROUP THEORY AND NORMAL MODES OF VIBRATION OF POLYATOMIC MOLECULES 4.3. INFRA-RED ABSORPTION AND RAMAN SCATTERING SPECTROSCOPY 4.4. DETERMINATION OF SYMMETRY PROPERTIES OF VIBRATIONAL MODES 4.5. SYMMETRY SELECTION RULES FOR INFRA – RED RAMAN SPECTRA 4.6. MUTUAL EXCLUSION RULE 4.7. LET US SUM UP 4.8. CHECK YOUR PROGRESS 4.9. LESSON - END ACTIVITIES 4.10. REFERENCES 4.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on group theory and vibrational spectroscopy to the students. On successful completion of this lesson the student should have: * Understand the group theory and vibrational spectroscopy. 4.1. INTRODUCTION Group theory is a very powerful tool at the hands of a chemist, a theorist and a spectroscopist. It finds many applications the details of which are beyond the scope of this lesson. Some important applications of group theory are listed below. 1. Construction of hybrid orbitals. 2. Construction of SALCs (symmetry adapted linear combinations of atomic orbitals). SALCs are used in molecular orbital theory (MOT) of chemical bonding. 3. Determination of the irreducible to which the vibrational modes of molecules belong. 4. Determining which spectral transitions in infrared and Raman spectra are allowed or forbidden. 5. Determining the selection rules for transitions in carbonyl compounds and other chromophores. It is found that the former transitions are forbidden whereas the latter are allowed. 6. Determining which molecules are polar or nonpolar. 4.2. GROUP THEORY AND NORMAL MODES OF VIBRATION OF POLYATOMIC MOLECULES Group theory helps in two aspects of vibrational spectroscopy.



27

(i) Firstly helps in the classification of the normal modes of vibrations according to the irreducible representations of the point group of the molecule. (ii) Secondly, it helps in qualitatively finding out Raman and IR spectral activity of the fundamentals as well as overtone and combination bands. The number of fundamental modes of vibrations can be worked out in the following manner: Let us consider a molecule having N atoms. If we specify the position of each atom in space with the coordinates x, y and z, then there will be 3N coordinates (degrees of freedom) for the entire system. Since the molecule has translational, rotational and vibrational motion, these 3N coordinates (degrees of freedom) can be assigned to each type of motion as given below.

Degrees of freedom to describe the motion

Motion

Linear Non linear Translation 3 3 Rotation 2 3

Vibration 3N-5 3N-6

These degrees of freedom for vibrational motion are called the normal or fundamental modes of vibration. Depending on the type of molecule, these normal modes may be active either in IR or Raman or both. Example, the normal modes of vibration in H2O and CO2 molecules.

O

H H

O

H H

O

H H

C OO C OO C OO C OO

A knowledge of the symmetry of vibrational modes in molecules will be helpful in predicting whether these modes of vibrations will give rise to infrared or Raman spectrum or both these spectra. 4.3. INFRA-RED ABSORPTION AND RAMAN SCATTERING SPECTROSCOPY



28

In infra-red spectrum, the sample is irradiated with infra-red radiation leading to absorption of the radiation at frequencies corresponding to the absorptions give the vibrational frequencies of that molecule. Thus, infrared technique is a direct measurement of the vibrational frequencies which lie in infrared region. In Raman spectroscopy, the molecules are irradiated with uv or visible radiation causing perturbation of the molecule which in turn induces vibrational transitions. As a result, energy is taken up from or given out to the incident radiation, which is scattered at a shifted frequency. The differences in frequency between incident light and Raman scattered light correspond to vibrational frequencies. Further in Raman spectroscopy, plane-polarized light is used as incident light. The scattered light may be still polarized or depolarized. Hence, certain frequencies and hence certain vibrational modes may be found to give polarized Raman light and others give depolarized scattered light. Since in Raman spectroscopy a higher energy radiation than IR is used, the resulting data would consist of a series of infra-red absorption lines and a series of Raman scattered lines. These lines could be assigned to different vibrational modes only if the symmetry properties of these modes are clearly understood. 4.4. DETERMINATION OF SYMMETRY PROPERTIES OF VIBRATIONAL MODES The simple procedure for obtaining the representations of vibrational modes and hence understanding their symmetry properties involve the following steps: (i) Assign point group to the given molecule. (ii) Deduce the reducible representation T3N for all the symmetry operations of the point group using the relation

( ) ( )RUR iRxyz cc = ------ (1)

Where ( )Rxyzc = The character for the operation R in the reducible representation

RU = The number of unshifted atoms for the operation R

( )Ric = The character for operation R per unshifted atom.

(iii) The reducible representation T3N for the molecule is split into the various irreducible representations of the point group by using the standard reduction formula

( ) ( ) ( )piR

ppi RRgh

ap

ccå= 1 ----- (2)

(iv) The irreducible representations T3N thus obtained correspond to the translational, rotational and vibrational degrees of freedom. Thus T3N can be written as T3N = Tx + Ty + Tz + Rx + Ry + Rz +Tvib

Tvib = T3N- [Tx + Ty + Tz + Rx + Ry+ Rz]



29

ILLUSTRATIONS (a) Water molecule: (i) Water molecule belongs to C2v point group. (ii) The symmetry elements of this point group are E C2v s xz s yz

O

H H

C2

syz

sxz

(iii) The reducible representation T3N for this point group is

C2v E C2 sxz syz

31-19T3N

(iv) This reducible representation is decomposed into various irreducible representations using the standard reduction formula and by using the character table for this group.

C2v E C2 sxz syz

1A1

A2

B1

B2

1

1

1

1

1

-1

-1

1

-1

1

-1

1

-1

-1

1

Tz

Rz

Tx,Ry

Ty,Rx

( )[ ] 31.3.11.1.11.1.11.9.14

11 =++-+=Aa

2Aa = ( ) ( ) ( )[ ]1.3.11.1.11.1.11.9.14

1-+-+-+ = 1



30

( )( ) ( )[ ]

( )( ) ( )[ ] 31.3.11.1.11.1.11.9.14

1

21.3.11.1.11.1.11.9.14

1

2

1

=+-+--+=

=-++--+=

B

B

a

a

Thus T3N is given by

T3N = 3A1 + A2 + 2B1 + 3B2 (v) The sum of the irreducible representations of vibrational modes Tvib is related to T3N by the relation Tvib = T3N- [Tx + Ty + Tz + Rx + Ry+ Rz] Using the values of Tx, Ty, Tz , Rx, Ry and Rz in the character table for this groups , we get Tvib = (3A1 + A2 + 2B1 + 3B2) – (B1 + B2+ A1+ B2+ B1+ A1) Tvib = 2A1+ B2 Thus the three normal modes of vibration of water molecule belong to A1 and B2 representations. Of these, those vibrations which belong to A1 symmetry are called totally symmetric vibrations. SO2 molecule also has the same symmetry properties as water molecule and hence its sum of representations of vibrational modes Tvib is also Tvib = 2A1+ B2 (b) Ammonia Molecule (i) This belongs to C3v point group (ii) The various symmetry elements of this group are E C2v s xz s yz

N

H HH

C2

(iii) The reducible representation T3N using reduction formula, we get

E 2C3 3sv

2012T3N

(iv) By decomposing T3N using reduction formula, we get



31

T3N = 3A1 + A2 + 4E (v) A reference to the character for C3v group gives Tx + Ty + Tz = A1 + E Rx + Ry+ Rz = A2 + E Thus Tvib is obtained as Tvib = 2A1+ 2E Thus ammonia has four vibrational modes. (c ) BF3 molecule (i) This molecule belongs to D3h point group. (ii) The various symmetry operations present are E 2C3 3C2 s h 2S3 3 s v

B

F

F

F

(iii) The reducible representation T3N for BF3 molecule is

E 2C3 3C2 3sv

T3N 12 0 -2 4

sh 2S3

-2 2

(iv) This representation is decomposed by reduction formula to give T3N = A1’ + A2’ + 3E’+ 2A2”+E”. The symmetries of the translations (Tx, Ty, Tz) and rotations (Rx, Ry and Rz) as given by the character table for D3h are

Translation Rotation

(Tx,Ty) : E"

Tz : A2"

(Rx,Ry) : E"

Rz : A2"



32

Thus Tvib is given by Tvib = 2A1’+ 2E’+A2” (d) By a similar treatment the Tvib for the following molecules are obtained as Molecule Symmetry T3N Tvib SO2

POCl3

PtCl4

2- RuO4

C2v

C3v

D4h Td

3A1 + A2 + 2B1 + 3B2

4A1 + A2 +5E A1g + A2g + B1g + B2g+Eg +2A2u+B2u+3Eu

A1 + E +T1 + 3T2

2A1+ B2

3A1+ 3E A1g+ B1g+ B2g+A2u+B2u+2Eu

A1 + E +2T2

Knowing the symmetries of all the vibrational modes of a molecule, it is possible to predict which of them will be active in the infra-red and Raman spectra. To do this we must have knowledge of symmetry selection rules for two effects. 4.5. SYMMETRY SELECTION RULES FOR INFRA – RED RAMAN SPECTRA (A) SELECTION RULE FOR IR Consider a transition from the vibrational ground state of a molecule with wave function

0y to an excited vibrational state with wave function iy . The probability of such a transition Pi

occurring is given by

ò= tmyy dP ii 0

Where m = dipole moment of the molecule (a vector) dT =implies integration carried over all possible variables of the wave functions. The vector m can be split into three components mx, my, mz along the three Cartesian coordinates and one of the three integrals is given by

ò= tymy dP ikik 0 ( )zyxk ,,=

Symmetry selection rules:

(i) If one of the three integrals ò tymy dik0 is non-zero, then that vibrational mode is

inactive. This occurs when the product 10myy is totally symmetric, u, the character of this

direct product function is +1 for all symmetry opera5tions of the relevant point group.



33

(ii) If the integral ò tymy dik0 is zero, the probability of that transition is zero. Then it is

said to be forbidden in infra-red. How to find out symmetry of a particular vibration The integral consists of three product functions. Of these (i) yo, the ground state vibrational wave function, is always totally symmetric. (ii) The symmetry properties of mk and those of a translational vector along the same axis Tk are the same. (iii) The symmetry properties of y i are the same as those of vibrational mode i. Thus, if a vibrational mode has the same symmetry property as one of the translation vectors, Tx, Ty, Tz for that point group, then a transition from the ground state to that mode (in excited state) will be infra-red active. (B) RAMAN SELECTION RULE The probability of a vibrational transition occurring in Raman scattering is given by

ò= tayy dP ii 0

Where a = Polarizability of the molecule (a tenser) The Raman effect depends upon a molecular dipole induced in the molecule by the electromagnetic field of incident radiation. The induced dipole is proportional to the polarizability (a) of the molecule, which a measure of the ease with which the molecular electron distribution could be disterted. Since a, is a tensor, there are only six distinct components, viz, [For vibrational transitions ajk = akj (where j,k = x,y,z )] SELECTION RULE For a vibrational mode to be Raman active one of the six integral of the form

ò ¹ 00 tyay dijk ( )zyxkj ,,, =

As in the case of IR, the above integral will be non-zero only if ajk has the same symmetry as the mode described by the wave function y i. It is found that ajk has the same symmetry properties as x2, axy as xy and so on) Thus, a transition from the ground state to that mode would be Raman active, if a normal mode has the same symmetry as one of these binary combinations of x,y and z. Since the selection rules for infra – red and Raman spectra have different physical bases; there is no relationship between them. Thus an infra-red active mode might or might not be Raman active or conversely.



34

4.6. MUTUAL EXCLUSION RULE Consider a molecule which has a centre of symmetry. Point groups of molecules with this element of symmetry have two sets of irreducible representations. The representations which are symmetric with respect to inversion are called g representations. The representations which are antisymmetric to inversion are called u representations. Let us consider the inversion of a Cartesian coordinate x through the centre of inversion. The coordinate x becomes –x. Therefore all representations generated by x,y, or z must belong to a u representation. On the other hand, the product of two coordinates x and y does not change sign on inversion (-x.-y = xy). The product xy generates a g representation. All the other quadratic or binary coordinates also generate the g representation. From these rules we can conclude that in centrosymmetric molecules, the vibrational modes belonging to g symmetry species are Raman active and the modes belonging to u symmetry species are IR active. This rule is called the mutual exclusion rule. Another way of stating this rule is as follows: If a molecule has a centre symmetry, then any vibration that is active in the IR is inactive in the Raman and vice versa. Therefore, we can infer that a molecule has no centre of symmetry if the same vibration appears in both IR and Raman. Table1 lists the IR active and Raman active vibrational modes in some centrosymmetric and noncentrosymmetric molecules. CO2, C2H2 and N2F2 possess centre of symmetry. It is seen from Table1 that the IR active modes in these molecules are Raman inactive and vice versa. H2O, NH3, HCN and BF3 do not have centre of symmetry. In BF3 the vibrational mode with symmetry species E’ is IR active and Raman active. In H2O, NH3 and HCN, all the modes are IR active and Raman active. Table1 Molecule Point

Group Symmetry Species

IR Active Species

Raman Active Species

CO2 C2H2 N2F2 H2O NH3 BF3 HCN

D¥h

D¥h C2h

C2v C3v D3h C¥v

A1g,A1u,E1u 2A1g,E1g,E1u, A1u 3Ag,Au,2Bu

2A1,B2 2A1,2E A1’, 2E ’, A2” 2A1,E1

A1u,E1u E1u, A1u

Au,Bu

A1,B2 A1,E A2”, E ’ A1,E1

A1g

A1g,E1g

Ag

A1,B2 A1,E A1’, E ’,E” A1,E1

4.7. LET US SUM UP In this lesson, we: Pointed out Ø Group theory and normal modes of vibration of polyatomic molecules Ø Infra-red absorption and Raman scattering spectroscopy Ø Determination of symmetry properties of vibrational modes



35

Ø Symmetry selection rules for infra – red Raman spectra Ø Mutual exclusion rule

4.8. CHECK YOUR PROGRESS 1. How is mutual exclusion principle explained on the basis of symmetry of vibrational modes? 2. Taking water as example, illustrate how the various vibrational modes form basis for group representation? 4.9. LESSON – END ACTIVITIES 1. Explain the selection rules in IR and Raman spectroscopy from symmetry point of view. 2. Show that the normal modes form the basis for irreducible representations with a suitable example. 4.10. REFERENCES 1. K.V. Raman, Group Theory and its applications to Chemistry, Tata McGraw-Hill Publishing Company limited, New Delhi. 2. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006- 2007. 3. V. Ramakrishnan, M.S Gopinathan, Group Theory in Chemistry, Vishal Publications.



36

UNIT-III

LESSON 5: THE TIME-DEPENDENT AND TIME-INDEPENDENT SCHRÖDINGER EQUATIONS

CONTENTS 5.0. AIMS AND OBJECTIVES 5.1. INTRODUCTION 5.2. THE TIME-DEPENDENT SCHRÖDINGER EQUATION 5.2.1. ONE-DIMENSIONAL EQUATION FOR A FREE PARTICLE 5.2.2. OPERATORS FOR MOMENTUM AND ENERGY 5.2.3. EXTENSION TO THREE DIMENSIONS 5.2.4. INCLUSION OF FORCES 5.3. TIME-INDEPENDENT SCHRODINGER EQUATION 5.4. REQUIREMENTS OF THE ACCEPTABLE WAVE FUNCTION 5.5. BORN’S INTERPRETATION OF THE WAVE FUNCTION 5.6. LET US SUM UP 5.7. CHECK YOUR PROGRESS 5.8. LESSON - END ACTIVITIES 5.9. REFERENCES 5.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on the time-dependent and time- independent Schrödinger equations to the students. On successful completion of this lesson the student should have: * Understand the time-dependent and time- independent Schrödinger equations. 5.1. INTRODUCTION Classical mechanics applies only to macroscopic particles. For microscopic “particles” we require a new form of mechanics, which we will call quantum mechanics. Schrodinger formulated an important fundamental equation in 1926. Schrodinger argued that if micro-particles like electrons could behave like waves, the equation of wave motion could be successfully applied to them. 5.2. THE TIME-DEPENDENT SCHRÖDINGER EQUATION 5.2.1. ONE-DIMENSIONAL EQUATION FOR A FREE PARTICLE The wave function of a localized free particle is the one given in eqn.



37

( ) ( ) ( )[ ]dktkiikxkAtx ò¥

¥-

-=Y wexp, ----- (1)

For a free particle, the classical expression for energy is

m

pE x

2

2

= ----- (2)

úúúúúú

û

ù

êêêêêê

ë

é

=

=

==

mvp

vmp

m

vmmvE

x222

222

21

21

Replacing px by hk and E by wh we get

m

k

2

2h=w ----- (3)

úúúú

û

ù

êêêê

ë

é

=´=

==´=

upup

w

lpl

p

hh

phh

k

22

2

2

h

hQ

Substituting this value of w in Eq. (1)

( ) ( ) dktm

kkxikAtx ò

¥

¥-

úû

ùêë

é÷÷ø

öççè

æ-=Y

2exp,

2h ----- (4)

Differentiating ( )tx,Y with respect to t, we get

( )ò¥

¥-

úû

ùêë

é÷÷ø

öççè

æ-=

¶

Y¶dkt

m

kkxikAk

m

i

t 2exp

2

22 hh

----- (5)

Differentiating ( )tx,Y twice with respect to x, we get

( )ò¥

¥-

úû

ùêë

é÷÷ø

öççè

æ--=

¶

Y¶dkt

m

kkxikAk

x 2exp

22

2

2 h ----- (6)

Combining Eqs. (5) and (6), we have

( )

2

22

2

,

xmt

txi

¶

Y¶-=

¶

Y¶ hh ----- (7)



38

which is the one-dimensional Schrodinger equation for a free particle. 5.2.2. OPERATORS FOR MOMENTUM AND ENERGY To obtain the operators for momentum and energy, Eq. (7) may be written as

( ) ( )txx

ix

im

txt

i ,2

1, Y÷

ø

öçè

æ

¶

¶-÷

ø

öçè

æ

¶

¶-=Y÷

ø

öçè

æ

¶

¶hhh ----- (8)

From a comparison of Eqs. (2) and (8), it may be concluded that the energy ‘E’ and momentum ‘P’ can be considered as the differential operators.

t

iE¶

¶® h and

xiPx

¶

¶® h ----- (9)

operating on the wave function ( )tx,Y . Eq. (7) is obtained even if the operator for ‘P’ is taken as

xi

¶

¶h in place of

xi

¶

¶- h .

5.2.3. EXTENSION TO THREE DIMENSIONS The one-dimensional treatment given above can easily be extended to three dimensions. The three-dimensional wave packet can be written as

( ) ( ) ( )[ ] zyx dkdkdktiAt ò¥

¥-

-=Y wrkkr .exp, ----- (10)

Proceeding on similar lines as in the on-dimensional case we get the three-dimensional Schrodinger equation for a free particle as

( ) ( )tr

tr,

2

, 22

YÑ-=¶

Y¶

mti

hh ----- (11)

An analysis similar to the one made for one-dimensional system leads to the following operators for energy and momentum

t

iE¶

¶® h and Ñ® hiP ----- (12)

5.2.4. INCLUSION OF FORCES Modification of the free particle equation to a system moving in a potential V(r,t) can easily be done. The classical energy expression for such a system is given by

( )trVm

pE ,

2

2

+= ----- (13)



39

Schrodinger then made the right guess regarding the operators for r and t as rr ® and tt ® ----- (14) Replacing E,p,r and t in Eq. (13) by their operators and allowing the operator equation to operate on the wave function ( )t,rY , we get

( ) ( ) ( )ttV

mti ,,

2

, 22

rrtr

Yúû

ùêë

é+Ñ-=

¶

Y¶ hh ----- (15)

which is the time-dependent Schrodinger equation for a particle of mass ‘m’ moving in a potential V(r,t). The quantity in the square bracket in Eq. (15) is the operator for the Hamiltonian of the system. In general, it’s solution will be complex because of the presence of i in the equation. The equation cannot be relativistically invariant as it contains first derivative in time and second derivative in space coordinates. 5.3. TIME-INDEPENDENT SCHRODINGER EQUATION The time-dependent Schrodinger equation (15) describes the evolution of quantum systems using time-dependent wave function ( )t,rY .It completely neglects the time dependence

of the operators. If the Hamiltonian operator does not depend on time, the variables r and t of the wave function ( )t,rY can be separated into two functions ( )ry and ( )tf .

( ) ( ) ( )tt fy rr =Y , ----- (1)

Substituting this value of ( )t,rY in Eq. (15)

( ) ( ) ( )ttV

mti ,,

2

, 22

rrtr

Yúû

ùêë

é+Ñ-=

¶

Y¶ hh

and dividing throughout by ( )ry and ( )tf we get

( )

( )( )

( ) ( )rrr

Yúû

ùêë

é+Ñ-= V

mdt

td

ti 2

2

2

11 hh

y

f

f ----- (2)

The left side of this equation is a function of time and right side function of space coordinates. Since‘t’ and ‘r’ are independent variables, each side must be equal to a constant, say E. This gives rise to the equations

( )

( )h

iE

dt

td

t-=

f

f

1 ----- (3)



40

and

( ) ( ) ( )rrr yy EVm

=úû

ùêë

é+Ñ- 2

2

2

h ----- (4)

Solution of Eq. (3) is straightforward and is given by

( ) ÷ø

öçè

æ-=

h

iEtCt expf ----- (5)

where ‘c’ is a constant. The equation for ( )ry , Eq. (4), is the time- independent Schrodinger

equation or simply Schrodinger equation. Since ( )ry determines the amplitude of the wave function ( )t,ry , it is called the

amplitude equation. Equation (1) now takes the form

( ) ( ) ÷ø

öçè

æ-=Y

h

iEtt exp, rr y ----- (6)

The constant C is included in the normalization constant for ( )ry .

Significance of the separation constant, E, can be understood by differentiating ( )t,ry in

Eq. (1) with respect to time and multiplying by hi . Then

( ) ( )tEt

ti ,

,r

rY=

¶

Y¶h ----- (7)

Multiplying both sides of Eq. (7) by *Y from left and integrating over the space coordinates from ¥- to ¥ . We get

( ) Ett

i =Y÷ø

öçè

æ

¶

¶Yò

¥

¥-

,* rh ----- (8)

As the left side Eq. (8) is the expectation value of the energy operator, the constant ‘E’ is the energy of the system. The same can be understood from Eq. (4) as

( )rVm

+Ñ- 22

2

h

is the operator associated with the Hamiltonian of the system. 5.4. REQUIREMENTS OF THE ACCEPTABLE WAVE FUNCTION

The interpretation given to Y or 2Y imposes certain restrictions on acceptable values of Y .



41

A physical system is described by the probability density ( ) 2, trY and the normalization

integral equation.

( ) 1,

2

=Yò¥

¥-

tdtr

For the probability density to be unique and the total probability to be unity, the wave function must be finite and single valued at every point in space. The probability current density j, equation

( ) ( )YÑY-YÑY= **

2,

m

itrj

h

Another important parameter of the probability interpretation contains Y and YÑ . Hence Y has to be continuous and YÑ must be finite.

The Schrodinger equation has the term YÑ 2 .For YÑ 2 to exist YÑ must be continuous. For a wave function ( )t,rY to be acceptable, ( )t,rY and YÑ must be finite, single valued and

continuous at all points in space. Any function that is finite, single valued and continuous can be taken as an acceptable (well-behaved) wave function. 5.5. BORN’S INTERPRETATION OF THE WAVE FUNCTION The wave function ( )t,rY has no physical existence, since it can be complex. Also it

cannot be taken as a direct measurement of the probability as (r,t) since the probability is real and non-negative. However, ( )t,rY must in someway be an index of the presence of the particle at

(r,t). A universally accepted statistical interpretation was suggested by Born in 1926.He

interpreted the product of ( )t,rY and its complex conjugate *Y as the position probability

density P(r,t).

( ) ( ) ( ) ( ) 2* ,,,, trtrtrtrP Y=YY= ----- (1)

The quantity ( ) ty dtr2

, is the probability of finding the system at time‘t’ in the small

volume of element td surrounding the point r. When ( ) tdtr2

,Y is integrated over the entire

space one should get the total probability, which is unity. Therefore

( ) 1,

2

=Yò¥

¥-

tdtr ----- (2)

For Eq. (2) to be finite, ( )t,rY must tend to zero sufficiently rapidly as ±¥®r .Hence,

one can multiply ( )t,rY by a constant, say N, so that YN satisfies the condition in Eq, (2).Then



42

( ) 1,22

=Yò¥

¥-

tdtrN ----- (3)

The constant ‘N’ is called the normalization constant and Eq. (3) the normalization condition. Since the Schrodinger equation is a linear differential equation, YN is a solution of it. The wave functions for which the integral in Eq. (2) does not converge will be treated depending on the nature of the functions. 5.6. LET US SUM UP In this lesson, we: Pointed out Ø The time-dependent Schrödinger equation Ø One-dimensional equation for a free particle Ø Operators for momentum and energy Ø Extension to three dimensions Ø Inclusion of forces Ø Time-independent Schrodinger equation Ø Requirements of the acceptable wave function Ø Born’s interpretation of the wave function

5.7. CHECK YOUR PROGRESS 1. What are the requirements of an acceptable wave function? 2. Outline the Born’s interpretation of the wave function. 5.8. LESSON – END ACTIVITIES 1. Is the time dependent Schrodinger equation relativistically invariant? Explain. 2. Derive the time independent Schrodinger Equation. 5.9. REFERENCES 1. R.K. Prasad, Quantum Chemistry, Wiley Eastern Limited, New Delhi. 2. IRA N. Levin, Quantum Chemistry, Prentice-Hall of India private limited, New Delhi. 3. G. Aruldhas, Quantum mechanics, Prentice-Hall of India private limited, New Delhi. 4. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006- 2007.



43

LESSON 6: OPERATORS

CONTENTS 6.0. AIMS AND OBJECTIVES 6.1. INTRODUCTION 6.2. ALGEBRA OF OPERATORS 6.2.1. ADDITION AND SUBTRACTION 6.2.2. MULTIPLICATION 6.3. COMMUTATOR OPERATOR 6.4. LINEAR OPERATOR 6.5. EIGENVALUES AND EIGENFUNCTIONS 6.6. BASIC POSTULATES OF QUANTUM MECHANICS 6.6.1. POSTULATE-I 6.6.1.1. RULES FOR SETTING UP A QUANTUM MECHANICAL OPERATORS 6.6.1.2. SOME OPERATORS OF INTEREST 6.6.1.2.1. MOMENTUM OPERATOR 6.6.1.2.2. HAMILTONIAN OPERATOR 6.6.1.2.3. ANGULAR MOMENTUM OPERATOR 6.6.2. POSTULATE-II 6.6.2.1. SCHRODINGER EQUATION AS AN EIGENVALUE EQUATION 6.6.3. POSTULATE-III 6.6.4. POSTULATE-IV 6.6.4.1. STATIONARY STATES 6.7. AVERAGE OR EXPECTATION VALUES 6.7.1. UNCERTAINTIES IN POSITION AND MOMENTUM 6.8. LET US SUM UP 6.9. CHECK YOUR PROGRESS 6.10. LESSON - END ACTIVITIES 6.11. REFERENCES 6.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on operators to the students. On successful completion of this lesson the student should have: * Understand the operators. 6.1. INTRODUCTION An operator is a symbol for transforming a given mathematical function into another function. It has no physical meaning if written alone. For example, √ is an operator which in itself does not mean anything, but if a quantity or a number is put under it, it transforms that

quantity into its square root, another quantity. Similarly, dx

d is an operator which transforms a



44

function into its first derivative with respect to x; for example dx

d transform the function sin x

into the function cos x.

In general, if A denotes an operator which transforms the function f(x) into the function

g(x), then we write )()(ˆ xgxfA = . For example,

1. Let A be dx

d, and f(x) = ax2; then axax

dx

dxfA 2)()(ˆ 2 == , i.e., g(x) = 2ax.

2. Let A be a. (i.e., multiplication by a) and f(x) = x2 + c, then

acaxcxaxfA +=+= 22 ).()(ˆ , i.e., g(x) = ax2 +ac.

6.2. ALGEBRA OF OPERATORS Although operators do not have any physical meaning, they can be added, subtracted, multiplied and have some other properties. 6.2.1. ADDITION AND SUBTRACTION The addition or subtraction of operators yields new operators, the sum or the difference of operators being defined by,

( ) ( ) ( ) ( )xfBxfAxfBA ˆˆˆˆ ±=±

For example, let A be loge and B be dx

d, and ( )xf be ; then,

( ) ( ) ( )222 loglogˆˆ xdx

dxx

dx

dxfBA ee ±=÷

ø

öçè

æ±=±

( ) ( )xfBxfAxxeˆˆ2log ±=±=

6.2.2. MULTIPLICATION Multiplication of two operators means operations by the two operators one after the other,

the order of operation being from right to left; for example, ( )xfBA ˆˆ means that the function

( )xf is first operated on by B to yield a new function ( )xg which is then operated on by A to

yield the final function ( )xh ,

( ) ( )[ ] ( ) ( )xhxgAxfBAxfBA === ˆˆˆˆˆ

For example, let, A be 24x , B be dx

d,and ( ) 2axxf = , then,



45

( ) ( ) ( ) 42232 12344ˆˆ axaxxaxdx

dxxfBA =×=×=

The square of an operator means that the same operator is applied successively twice, i.e.,

( ) ( )xfAAxfA ˆˆˆ 2 = . For example,

Let dx

dA =ˆ and ( ) xxf sin= , then ( ) x

dx

dxfA sinˆ

2

2÷ø

öçè

æ=

Or ( ) ( ) xxdx

dx

dx

d

dx

dx

dx

dsincossinsin

2

-==úû

ùêë

é=÷

ø

öçè

æ

6.3. COMMUTATOR OPERATOR

For any two operators A and B , the difference ABBA ˆˆˆˆ - , which is simply denoted by

ABBA ˆˆˆˆ - or [A,B] is called “commutator operator”

If A and B commute then [A,B] = 0, where 0 is called the zero operator which means multiplying a function with zero.

In the earlier example, where, dx

dA =ˆ , 23ˆ xB = and ( ) xxf sin= ,the commutator is obtained as

follows

[ ] ( ) [ ] ( ) ( ) ( )xxfxxxxxxxxxfABBAxfBA 6sin6cos3cos3sin6ˆˆˆˆ, 22 ==-+=-=

or [ ] ( ) xxfBA 6, =

6.4. LINEAR OPERATOR An operator is said to be linear if its application on the sum of two functions gives the result which is equal to the sum of the operations on the two functions separately, i.e., if,

( ) ( )[ ] ( ) ( )xgAxfAxgxfA ˆˆˆ +=+

or ( ) ( )xcfACxcfA ˆˆ ×= , C = constant

Example:

(i) dx

dis a linear operator because ( )nm bxax

dx

d+ ( ) ( )nm bx

dx

dax

dx

d+=

(ii) , square root, is not a linear operator because ( ) ( ) ( ) ( )xgxfxgxf +¹+



46

6.5. EIGENVALUES AND EIGENFUNCTIONS

If an operator A operates on a well-behaved (i.e., finite, continuous and single-valued) function f to give the same function but multiplied by a constant, then the constant is called the

“eigenvalue” of the operator and the function f is called the “Eigenfunctions”. The equation is

called “eigenvalue equation”. For example, if the function axef -= is acted upon by the

operatordx

d, the result is ( ) ( )axax eae

dx

d -- -= .

Therefore, ( )a- is the eigenvalue and axe- is the eigenfunction of the operatordx

d.

6.6. BASIC POSTULATES OF QUANTUM MECHANICS The fundamental postulates are four which are stated and explained below. 6.6.1. POSTULATE-I Every physical property of a system (i.e. a particle or a system of particles) has a corresponding mathematical (quantum mechanical) operator. Physically measurable quantities of a particle are a position ( )x , momentum ( )p , kinetic

energy ( )T , potential energy ( )V , total energy ( )E , etc. The operators corresponding to these

quantities are given in table I. Table-I. Quantum Mechanical OPERATORS Corresponding to Various Physical Quantities Physical quantity Quantum mechanical operator

Position ( )x

Position ( )r

x-Component of momentum ( )xp

momentum ( )p

kinetic energy ( )T

x-Component of K.E. ( )Tx

x r

xi

h

¶

¶

p2

Ñi

h

p2

2

2

2

8Ñ-

m

h

p

2

2

2

2

8 xm

h

¶

¶-

p



47

Potential energy ( )V

Total energy ( )H

V

Vm

h+Ñ- 2

2

2

8p.

6.6.1.1. RULES FOR SETTING UP A QUANTUM MECHANICAL OPERATORS (i) Write down the expression for the physical quantity in classical terms, i.e., in terms of

Cartesian coordinates of position ( )zyx ,, and momenta ( )zyx ppp ,, .

(ii) Replace these coordinates and momenta by their corresponding operators (vide Table I above). (iii) Operator for a coordinate of position (say x) is multiplication by that variable x itself (i.e.x.).

(iv) Operators for a coordinate of momentum(say px) is xi

h

¶

¶

p2.

Take for example the kinetic energy (Tx) of a single particle moving in one direction(x).

m

pmvTx x

x22

12

2==

Therefore, the K.E. Operator

2

22

1ˆ ÷ø

öçè

æ

¶

¶=

xi

h

mxT

p

= 2

2

2

2

8 xm

h

¶

¶-

p ----- (1)

Similarly, potential energy operator ( ) ( )xVxV =ˆ . The total energy (H) is ( )xVm

px +2

2

;

hence, the total energy operator

( )xVdx

d

m

hH +-=

2

2

2

2

8ˆ

p ----- (2)



48

6.6.1.2. SOME OPERATORS OF INTEREST 6.6.1.2.1. MOMENTUM OPERATOR The kinetic, potential and total energies are derived from coordinates of momentum and position. but how is momentum operator to be derived? This is done by using a more fundamental property of electron wave itself. For an electron wave the wave function may be represented by the function, ( )lpy ixA 2exp ±= ----- (3)

Differentiating with respect to x,

( ) yl

plp

l

py iixA

i

dx

d 22exp

2±=±±=

But by de Broglie relationship,

xp

h=l

Therefore,

ypy

xph

i

dx

d 2±=

or,

dx

d

i

hpx

yy

P±=

2 ----- (4)

Buty is a function which on being removed from the equation reduces it to a differential

operator,

dx

d

i

hpx

P±=

2

) ----- (5)

The operator Px in the =x direction is dx

d

i

h

p2 and in the negative –x direction it is

dx

d

i

h

p2- .

6.6.1.2.2. HAMILTONIAN OPERATOR The operator corresponding to the total energy of a system, written as a sum of kinetic

and potential energies, is called Hamiltonian operator ÷ø

öçè

æ Ù

H . The total energy of a single particle

of mass m is

( ) ( ) Vpppm

zyxVm

pH zyx +++=+=

2222

2

1,,

2 ----- (6)



49

where V is written for V(x,y,z). The corresponding operator will, therefore, be

.2

1 222ÙÙÙÙÙ

+÷ø

öçè

æ++= Vppp

mH zyx

But

2

2

2

22

42 x

h

xi

hp x

¶

¶-=÷

ø

öçè

æ

¶

¶=

Ù

pp;

similarly,

2

2

2

22

4 y

hp y

¶

¶-=

Ù

p and

2

2

2

22

4 z

hp z

¶

¶-=

Ù

p

Thus,

ÙÙ

+÷÷ø

öççè

æ

¶

¶+

¶

¶+

¶

¶-= V

zyxm

hH

2

2

2

2

2

2

2

2

8p

Ù

+Ñ-= Vm

h 2

2

2

8p ----- (7)

For a system of n particles,

ÙÙ

+Ñ-= å Vim

hH

i i

2

2

2 1

8p ----- (7a)

where mi is the mass and Ñ i2 the Laplacian operator of the ith particle. 6.6.1.2.3. ANGULAR MOMENTUM OPERATOR The angular momentum (L) is a very important physical quantity for rotating systems. Classically, it is given by the vector product of position ( )r

rand linear momentum ( )p

r,

prLrrr

´= ----- (8)

If jirr

, and kr

are unit vectors along x, y and z coordinates respectively, we have

zkyjxirrrrr

++= ,

and

zyx pkpjpiprrrr

++=

Therefore,

( ) ( )zyx pkpjpizkyjxiLrrrrrrr

++´++=

( ) ( ) ( )xyzxyz ypxpkxpzpjzpypi -+-+-=rrr

Also, by definition,

zyx LkLjLiLrrrr

++= ----- (9)

where Lx, Ly and Lz represent the three components of L. Replacing px, p y and pz by their corresponding operators we obtain the operators for the three components of angular momentum.



50

Thus,

( )

( )

( )ïïïï

þ

ïïïï

ý

ü

÷÷ø

öççè

æ

¶

¶-

¶

¶=-=

÷ø

öçè

æ

¶

¶-

¶

¶=-=

÷÷ø

öççè

æ

¶

¶-

¶

¶=-=

xy

yx

i

hpypxL

zx

xz

i

hpxpzL

yz

zy

i

hpzpyL

xyz

zxy

yzx

p

p

p

2ˆˆˆ

2ˆˆˆ

2ˆˆˆ

----- (10)

6.6.2. POSTULATE-II The possible values of any physical quantity of a system (e.g. energy, angular momentum etc) are given by the eigenvalues a in the operator equation,

yy aA =)

----- (11)

where A)

i s the operator corresponding to that physical quantity and y is a well-behaved

eigenfunction. The eigenvalues and eigenfunctions for a system can be obtained by solving the operator equation. While the Eigenfunctions and the operator may be real or complex, the eigenvalues must be real because they represent observable physical quantities. Moreover, quantum mechanical operators of interest, e.g. momentum, energy, etc., are Hermitian operators which have been shown earlier to have real eigenvalues. Consider, for example, the operator equation for momentum px (in one dimension),

( ) ( ) ( )xpdx

xd

i

hxp xx y

y

py ==

2

) ----- (12)

This is an ordinary differential equation whose solution may be easily obtained as

( )h

xipAx xpy 2exp)( ±= ----- (13)

where A is a constant. Now, any value of px is allowed that keeps the function ( )xy well-

behaved, i.e., finite, single-valued and continuous. If we put kh

px =p2

, then the eigenfunction of

momentum operator becomes, ( )ikxA ±= expy ----- (14)

with eigenvalues

p2

khpx = ----- (15)



51

6.6.2.1. SCHRODINGER EQUATION AS AN EIGENVALUE EQUATION A particularly important eigenvalue equation is the equation for energy of a system. The operator for energy E (K.E+P.E) is the Hamiltonian operator. For a single particle in three dimensions, the Hamiltonian is

( )zyxVm

hH ,,ˆ

8ˆ 2

2

2

+Ñ-=p

----- (16)

The eigenvalue is,

( ) ( )zyxEzyxH ,,,,ˆ yy = or simply yy EH =ˆ

i.e.

yyyp

EVm

h=+Ñ- ˆ

82

2

2

----- (17)

This is the familiar Scharodinger equation. It is a second order partial differential equation which has to be solved in order to obtain expressions for the eigenfunctions,y , and the

observable quantity, energy (E), of the particle. Since energy is a very important observable property of a system, and can often be measured experimentally, a topic of major importance in the quantum chemistry is solving of Schrodinger equation. 6.6.3. POSTULATE-III The expected average (expectation) value of a physical quantity (M) of a system, whose state function is y , is given by

tyy

tyy

d

dMM

òò

=*

*)

----- (18)

where M

) is the operator for M.

It has to be noted that y is not necessarily an eigenfunction of M)

, though M)

has a set of

orthonormal eigenfunction ( )ij and y can be expressed as

å=i

iia jy

Hence,

tjj daMaMi

iii

ii ÷ø

öçè

æ÷ø

öçè

æ= åò å

)* , if y is normalized

åå ò=i j

jiji dMaa tjj)

**



52

But jjjM jlj =)

Therefore,

( )åå ò=i j

jjiji daaM tjlj **

( )åå ò=i j

jijji daa tjjl **

Since

0* =ò tjj dji for ji ¹ , ( ij and jj are orthogonal)

1= for ji = , ( ij and jj are normalized)

å=i

iii aaM l* or åi

iia l2

This expression means that each measurement of M must give one of the eigenvalues

il (postulate-II) and the average of many such measurements is equal to M . Further, ii aa * or 2

ia represents the fraction of the total number of measurements that give the eigenvalue il ,

andå = 0.12

ia .

Consider for example the expectation value of position, x, for a particle moving in one dimension, which is given by

òò+¥

¥-

+¥

¥-

== dxxdxxx yyyy **

because the operator for x is just x, for which the order of applying is immaterial. Similarly, the expected average value of momentum for the above system is,

ò+¥

¥-

= dxdx

d

i

hpx

yy

p*

2 ----- (19)

and K.E., ò+¥

¥-

-= dxdx

d

m

hTx 2

2

2

2

*8

yy

p ----- (20)

The limits of integration in each case will depend on the total length available for the particle’s motion. 6.6.4. POSTULATE-IV The time-dependent Schrodinger equation is given by,



53

t

ihH

¶

¶=

y

py

2

) ----- (21)

where H)

is the Hamiltonian operator, y is a function of position (x,y,z) as well

as time (t) and V is a function of position only. Since this a partial differential equation involving both position and time variables, it can be solved only by separating it into two differential equations, one containing position variable and the other containing time variable only. This can be done if we assume that, ( ) ( ) ( )tqtq jyyy ., == ----- (22)

where q is the collective symbol for the three coordinates of position x, y and z; dropping q and t for simplicity, yfy =

Then,

qq ¶

¶=

¶

¶ yj

y

or

2

2

2

2

qq ¶

¶=

¶

¶ yj

y ----- (23)

and

tt ¶

¶=

¶

¶ yy

y ----- (24)

Substituting equations (22),(23) and (24) into (21) and dividing the resulting equation by yj , we

get

t

ihV

qm

h

¶

¶=ú

û

ùêë

é+

¶

¶-

j

jpyy

y

p.

1

2

1.

8 2

2

2

2 ) ----- (25)

and

jj

pE

t

ih=

¶

¶

2 ----- (26)

Equation (25) is the familiar Schrodinger equation, also called the amplitude equation of the type

yy EH =)

in which the function y ,the wave function, is found by solving the equation

Equation (26) has a solution,

÷ø

öçè

æ-=

h

iEta

pf

2exp ----- (27)

So the time dependent wave function ( )tq,y is given by,



54

÷ø

öçè

æ-=Y

h

iEta

py

2exp ----- (28)

6.6.4.1. STATIONARY STATES States for which Y is given by (28) are called stationary states. A stationary state does not imply that the particle, or particles, of the system are at rest. It is stationary in the sense that the

probability density 2

Y and the energy are independent of time as long as the state of the system

is being described by the function Y . The probability density is given by,

*2

exp.2

exp ***2* yy

py

py aa

h

iEta

h

iEta =÷

ø

öçè

æ÷ø

öçè

æ-=Y=YY ------ (29)

which is independent of time. Thus, we see that the factor ÷ø

öçè

æ-

h

iEtp2exp is of no significance

and the essential part of the state function for a stationary state is the time-independent wave function y .

A knowledge of stationary states is of prime importance in understanding chemical problems. If the state of a system corresponds to one of these stationary states, every physical property of the system e.g. energy, charge density, dipole moment, etc., will be independent of time. Atomic and molecular spectra arise due to transitions between these stationary states brought about by the action of radiation. 6.7. AVERAGE OR EXPECTATION VALUES We can use the wave function to determine, in a state n, the average or expectation values of any physical quantity, e.g., position, momentum, energy, etc. Consider, for example, the position. Since the particle can be anywhere in the box, the result of a single measurement is never precise; but the average of a series of measurements has some meaning. This average can be computed by the formula,

dxL

xnx

Ldxxx

LLp

yy òò ==0

2

0

sin2

------ (30)

Let

;,,'

dun

Ldx

n

Luxor

L

xnu

pp

p==

when x = 0, u = 0 and when x = L, u = np ------ (31) Therefore,

dun

Lu

n

Lu

Lx

n

÷ø

öçè

æ= ò pp

p2

0

sin2



55

( )

uduun

Ln

ò=p

p 0

2

2sin

2 ------ (32)

Using the table of integrals

( )

p

p

n

uuuu

n

Lx

0

2

2 8

2cos

4

2sin

4

2úû

ùêë

é÷÷ø

öççè

æ--=

( )

( ) ( ) ( ) ( )úúû

ù

êêë

é

þýü

îíì

---þýü

îíì

--=8

100

8

2cos

4

2sin

4

22

2

pppp

p

nnnn

n

L

( )

( )24

22

2

Ln

n

LL =´=

p

p ------ (33)

Thus, quantum mechanical considerations show that the average value of the position of the particle in any state (n) lies in the middle of the box, which is the same as that obtained from classical point of view. This should not be surprising as the probability density curves (Fig. 1) indicate that the probability of finding the particle at a distance d to the left of the centre is exactly the same as that at the distance d to the right.

We can likewise calculate other quantities. A quantity of interest is 2x .

duuun

L

Lx

n

ò÷ø

öçè

æ=

p

p0

22

3

2 sin22

p

p

n

uuu

uu

n

L

L0

233

4

2cos2sin

8

1

46

2úû

ùêë

é-÷÷

ø

öççè

æ--÷

ø

öçè

æ=



56

( )úû

ùêë

é---÷÷

ø

öççè

æ--÷

ø

öçè

æ= 000

40

6

2 333pp

p

nn

n

L

L

( ) úû

ùêë

é-÷

ø

öçè

æ=

4

1

6

2 223p

pp

nn

n

L

L

úû

ùêë

é-=-=

2222

22

2

31

23 pp nn

LL ------ (34)

6.7.1. UNCERTAINTIES IN POSITION AND MOMENTUM Since the particle can be anywhere within the length L, the error (uncertainty) involved in any single measurement of position, Lx =D . the momentum lies between

;22 L

nhand

L

nh-+

so, the uncertainty,



57

,

,1,22 L

hpnfor

L

nh

L

nh

L

nhp xx ===÷

ø

öçè

æ--=D

and the product hpx x =DD . ------ (35)

which satisfy the Heisenberg uncertainty relation. 6.8. LET US SUM UP In this lesson, we: Pointed out Ø Algebra of operators Ø Addition and subtraction Ø Multiplication Ø Commutator operator Ø Linear operator Ø Eigenvalues and eigenfunctions Ø Basic postulates of quantum mechanics Ø Postulate-I Ø Rules for setting up a quantum mechanical operators Ø Some operators of interest Ø Momentum operator Ø Hamiltonian operator Ø Angular momentum operator Ø Postulate-II Ø Schrodinger equation as an eigenvalue equation Ø Postulate-III Ø Postulate-IV Ø Stationary states Ø Average or expectation values Ø Uncertainties in position and momentum

6.9. CHECK YOUR PROGRESS 1. Prove that the Hamiltonian Operator for the total energy of a system is Hermitian provided that the wave function is well behaved. 2. What are eigen functions and eigenvalues of an operator? 6.10. LESSON – END ACTIVITIES 1. Outline the different postulates of quantum mechanics. 2. Write notes on (a) Average or expectation values (b) Uncertainity in position and momentum



58

6.11. REFERENCES 1. R.K. Prasad, Quantum Chemistry, Wiley Eastern Limited, New Delhi. 2. IRA N. Levin, Quantum Chemistry, Prentice-Hall of India private limited, New Delhi. 3. G. Aruldhas, Quantum mechanics, Prentice-Hall of India private limited, New Delhi. 4. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, millennium edition, Vishal Publishing co.



59

UNIT-IV

LESSON 7: PARTICLE IN A ONE-DIMENSIONAL BOX

CONTENTS 7.0. AIMS AND OBJECTIVES 7.1. INTRODUCTION 7.2. PARTICLE IN ONE DIMENSIONAL BOX 7.3. NORMALIZATION OF y

7.4. ORTHOGONALITY OF THE WAVE FUNCTIONS 7.5. PARTICLE IN A THREE-DIMENSIONAL CUBICAL BOX 7.4.1. SEPARATION OF VARIABLES 7.4.2. DEGENERACY 7.6. ONE-DIMENSIONAL SIMPLE HARMONIC OSCILLATOR (S.H.O) 7.5.1. ENERGY OF S.H.O 7.7. THE RIGID ROTOR (OR) ROTATOR 7.8. LET US SUM UP 7.9. CHECK YOUR PROGRESS 7.10. LESSON - END ACTIVITIES 7.11. REFERENCES 7.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on particle in a one-dimensional box to the students. On successful completion of this lesson the student should have: * Understand the particle in a one-dimensional box. 7.1. INTRODUCTION Now consider the particle which is allowed to move in a limited space such a model, usually called “particle in a box” model, serves as the simplest case for the treatment of bound electrons in atoms and molecules. 7.2. PARTICLE IN A ONE-DIMENSIONAL BOX This is the simplest quantum mechanical problem. Here a particle of mass m is confined to move in a one-dimensional box of a length a, having infinitely high walls (Fig.1). It is



60

assumed, for the sake of simplicity, that the potential energy of the particle is zero everywhere inside the box, that is ( ) 0=xV ----- (1)

Thus, inside the box the Schrodinger equation, viz.,

( ) ( ) ( ) ( )xExxVdxdm

yy =úû

ùêë

é+- 22

2

2

h ----- (2)

takes the form ( ) ( )xEdxdm

yy =- 222

2

h ----- (3)

Our problem is solving this equation for energy E and the wave function ( )xy .

Mathematically, Eq.3 may be rewritten as

( ) 02 2

2

2

=+ yy

hmEdx

d ----- (4)

or as 02

2

2

=+ yy

kdx

d ----- (5)

where k2 ( )22 hmE= is a constant, independent of x.

Eq.5 is an ordinary second-order differential equation which has solution of the form kxBkxA sincos +=y ----- (6)



61

where A and b are constants. Before we proceed further, it should be pointed out that outside the box where ( ) ¥=xV , the Schrodinger equation (2) is

( ) ( ) ( )xExdxdm

yy =úû

ùêë

é¥+- 22

2

2

h ----- (7)

or ( ) 02

22

2

=¥-+ yy

Em

dx

d

h ----- (8)

This equation is satisfied if y is zero everywhere outside the box. This is another way of saying

that the particle cannot be found outside the box; it is confined within the box. This implies that y must be zero at the walls of the box, i.e., at x=0 and at x=a. Since the postulate of quantum

mechanics requires that y must be a continuous function of y , we are forced to conclude that in

Eq. 6, A must be zero. Thus, the solution of Eq. 5 is of the form kxB sin=y ----- (9)

Since 0=y at x=0 and at x=a, we have

0sin =kaB or 0sin =kaB or 0sin =ka so that ank p= ----- (10)

where n(=0,1,2,3,4.......,)is the quantum number. Hence, the allowed solutions of Eq. 4 are

( )axnBn pyy sin=º ; (n=1,2,3, ......) ----- (11)

Notice that the n=0 value of the quantum number, though allowed, is not acceptable since it implies that the wave function y is zero everywhere inside the box. This is not corre4ct since

the particle is taken to be inside the box, to begin with. From eqs.4 and 9, we have

( ) 22222 anmE p=h ----- (12)

or 2222222 82 mahnmanEE n === ph ; n=1,2,3,........ ( )p2h=h ----- (13)

Eq.13 gives the expression for the energy of the particle in a one-dimensional box. It should be committed to memory. Since energy depends upon the quantum number n, which can have any integral value, the energy level of the particle in the box are quantize. Before we discuss eq.13 further, let us determine the coefficient b in the wave function given by Eq. 11. This can be done by normalizing the wave function. 7.3. NORMALIZATION OF y



62

Since the total probability of finding the particle within the box is 1, therefore, according to Born’s interpretation of the wave function, the normalization of y requires that

ò =a

n dx0

21y ----- (14)

Substituting the value of ny from Eq.11, we have

( ) 1sin

2

0

=ò dxaxnBa

p ----- (15)

or ( ) 1sin0

22 =ò dxaxnBa

p ----- (16)

Since ( )qq 2cos12

1sin 2 -= , hence from Eqs. 14 and 16, we have

ò òò úû

ùêë

é÷ø

öçè

æ-=

a aa

n dxa

xndxBdx

0 00

22 2cos

2

1

2

1 py ----- (17)

[ ] 1022 =-aB ----- (18)

Hence, ( )2

1

2 aB = ----- (19)

Thus, the normalized wave functions for the particle in a one-dimensional box are

( ) ( ) ( )axnaxn py sin2 2

1

= ; n=1,2,3,....... ----- (20)

A few normalized wave functions for a particle in a one-dimensional box are given in Fig.2.



63

7.4. ORTHOGONALITY OF THE WAVE FUNCTIONS We have shown in lesson 6 that eigenfunctions of a Hermitian operator corresponding to different eigenvalues are necessarily orthogonal to each other, and that Hamiltonian is an Hermitian operator. For the problem of particle in a one-dimensional box, it can be demonstrated easily.

Let ,sin2

1L

x

L

py = and

L

x

L

py

2sin

22 =

be the two normalized eigenfunctions corresponding to the two eigenvalues E1 and E2 (characterized by quantum numbers 1 and 2 ). Then,

ò ò=L L

L

x

L

nx

Ldx

0 0

21

2sinsin

2 pyy

( ) ( )

dxL

x

L

x

L

L

òúúúú

û

ù

êêêê

ë

é +-

-

=0

2

12cos

12cos

2pp

( )

( )( )

( )L

L

xL

L

xL

L0

12sin

12

12sin

12

1úû

ùêë

é +

+-

-

-=

pp

( ) 0001

=-=L



64

7.5. PARTICLE IN A THREE-DIMENSIONAL CUBICAL BOX Let us now consider the motion of a particle of mass m confined to a three-dimensional

cubical box with edges of length a and volume equal to 3a . The potential is zero within the box and is infinite outside the box and at its boundaries. The time- independent Schrodinger equation for the particle is

( ) ( )zyxEzyxzyxm

,,,,2 2

2

2

2

2

22

yy =úû

ùêë

é÷÷ø

öççè

æ

¶

¶+

¶

¶+

¶

¶-

h ----- (21)

Eq. 21 may be rewritten as

02

2

2

2

2

2

2

=÷ø

öçè

æ+

¶

¶+

¶

¶+

¶

¶y

yyy

h

mE

zyx ----- (22)

7.5.1. SEPARATION OF VARIABLES Assuming that the wave function y is a product of three parts which separately depend on x,y, z,

we have ( ) ( ) ( ) ( )zZyYxXzyx =,,y ----- (23)

Hence, 2

2

2

2

x

XYZ

x ¶

¶=

¶

¶ y ----- (24)

2

2

2

2

y

YXZ

y ¶

¶=

¶

¶ y ----- (25)

2

2

2

2

z

ZXY

z ¶

¶=

¶

¶ y ----- (26)

Substituting Eqs.24-26 in Eq.22 and dividing throughout by ( )XYZm 22 h , we have

0111

2 2

2

2

2

2

22

=+úû

ùêë

é

¶

¶+

¶

¶+

¶

¶- E

z

Z

Zy

Y

Yx

X

Xm

h ----- (27)

We notice that the first term is a function of x only and is independent of y and z; the second term is a function of y only and is independent of x and z and the third term is a function of z only and is independent of x and y. The fourth term E is a constant. If energy E is written as the sum of three contributions associated with the three coordinates, then Eq. 27 can be separated into three equations. Thus, for instance, for motion along the X-axis (where X varies while Y and Z remain constant), the second and third terms on the left hand side of Eq.27 remain constant be Ex. Accordingly, we can write

xEx

X

Xm=÷÷

ø

öççè

æ

¶

¶-

2

22 1

2

h ----- (28)



65

yEy

Y

Ym=÷÷

ø

öççè

æ

¶

¶-

2

22 1

2

h ----- (29)

zEz

Z

Zm=÷÷

ø

öççè

æ

¶

¶-

2

22 1

2

h ----- (30)

where zyx EEEE ++= .

We see that each of the Eqs. 28-30 is of the form of Eq.3, the solution of which is given by Eq.20, i.e., we have

( ) ( ) ( )axnaxX x psin/2 2

1

= ; nx = 1,2,3... ----- (31)

and 2228mahnE xx = ----- (32)

Similar solutions exists for Y(y) and Z(z). Hence

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )aznaynaxnazZyYxXzyx zyx pppy sinsinsin8,, 2

13== ----- (33)

and ( )

2

2222

8ma

hnnnEEEE

zyx

zyx

++=++= ----- (34)

where nx, ny, nz =1,2,3,4,.......... 7.5.2. DEGENERACY It is seen from Eq.34 that the total energy depends upon the sum of the squares of three quantum numbers. It is evident that groups of different states, each specified by a unique set of quantum numbers, can have the same energy. In such a case, the energy level and the corresponding independent states are said to be degenerate. Consider, for instance, the energy level having energy = 14h2/8ma2. There are six combinations of nx,ny and nz which can give this value of energy: nx 1 1 2 3 2 3 ny 2 3 1 1 3 2 nz 3 2 3 2 1 1 This energy level is, therefore, 6-fold degenerate, i.e., its degeneracy is equal to 6. Proceeding in this manner we can calculate degeneracy of the energy levels of a particle in a three-dimensional cubical box. The results are shown in Fig. 3.



66

7.6. ONE-DIMENSIONAL SIMPLE HARMONIC OSCILLATOR (S.H.O) A diatomic vibrating molecule can be represented by a simple model, the so-called simple harmonic oscillator (S.H.O). The force acting on the molecule is given by f = -kx, where x is the displacement from the equilibrium position and k is a constant called the force constant. The potential energy V(x) of this molecule is given by

( ) 2

002

1kxdxkxdxfxV

xx

==-= òò ----- (35)

Eq. 35 is the equation of a parabola. Thus, if we plot potential energy of a particle executing simple harmonic oscillations as a function of displacement from the equilibrium position, we get a curve as shown in Fig. 4.



67

The vibrational frequency of the oscillator of mass m is given by

2

1

2

1÷ø

öçè

æ=

m

k

pu ----- (36)

It is more accurate to define the vibrational frequency as

2

1

2

1÷÷ø

öççè

æ=

mpu

k ----- (37)

where m is the reduced mass of the diatomic molecule defined as

21

111

mm+=

m ----- (38)

where m1 and m2 are the atomic masses of the two atoms. Using the potential energy given by Eq.35, for one dimensional S.H.O., the Schrödinger equation

yy EVm

=úû

ùêë

é+Ñ- 2

2

2

h is represented as

( ) ( )xExkxdx

d

myy =ú

û

ùêë

é+÷÷

ø

öççè

æ- 2

2

22

2

1

2

h ----- (39)

Mathematically,Eq. 39 may be written as

02

12 2

22

2

=÷ø

öçè

æ-+ y

ykxE

m

dx

d

h ----- (40)

Defining a and b as 2mE 2h=a and ( ) h21

mk=b , ----- (41)

Eq. 40 becomes ( ) 022

2

2

=-+ ybay

xdx

d ----- (42)

Defining a new variable x21

bx = , the above equation becomes ( Xi=x )

02

2

2

=÷÷ø

öççè

æ-+ yx

b

a

x

y

d

d ----- (43)

Eq. 42 has a solution of the form



68

( ) ( ) 2

2x

xfxy-

= e ----- (44)

Using this solution, Eq. 42 becomes

0122

2

=÷÷ø

öççè

æ-+- f

b

a

x

fx

x

f

d

d

d

d ----- (45)

Eq. 45 is identical in form to a well known second –order differential equation, called the Hermite equation,. viz.,

0222

2

=+- fx

fx

x

fn

d

d

d

d ----- (46)

The Hermite equation has solutions which depend upon the value of n. These solutions are called Hermite polynomials, Hn(x).i.e.,f (x) º Hn(x). The Hermite polynomial of degree n is defined as

( ) ( ) ( )ïî

ïíì

ïþ

ïýü

¶

¶-=

-

n

nn ee

xx

xx

2

2

1H n ----- (47)

A few Hermite polynomials are given below: H0(x) =1 H3(x) =8x3-12x H1(x) =2x H4(x) =16x4-48x2+12 H2(x) =4x2-2 The normalized wave functions of the one dimensional S.H.O. are then written as

( ) ( )xp

bxy x

nnn Hen

22

21

21

!2

-

úú

û

ù

êê

ë

é= Where n= 0,1,2,3,........ ----- (48)

The first three wave functions (with n= 0,1,2), the corresponding energy levels and the

probability functions 2

ny for the S.H.O., are shown in Fig. 5.



69

7.6.1. ENERGY OF S.H.O The energy of the S.H.O is obtained by comparing Eqs. 45 and 46, hence we find that

12 += nb

a ----- (49)

Substituting for a and b from Eq. 41, we have

( )

( ) 12221

2

+= nmk

mEh

h ----- (50)

or ( ) 21

2

1mknE h÷

ø

öçè

æ+= ----- (51)

From Eq. 36, ( ) pn221

=mk ----- (52)

Combining Eqs. 51 and 52, we have

( )pn22

1h÷

ø

öçè

æ+=º nEE n

nhn ÷ø

öçè

æ+=

2

1;n=0,1,2,3,.....( p2h=h ) ----- (53)

The energy state with n=0 is the vibrational ground state with energy

nhE2

10 = ----- (54)



70

This energy is called the zero point energy of the oscillator. Classical mechanics predicts that the zero-point energy of the oscillator is zero whereas quantum mechanics predicts that the zero-point energy is non-zero. The occurrence of the zero-point energy is consistent with the Heisenberg uncertainty principle. 7.7. THE RIGID ROTOR (OR) ROTATOR A diatomic molecule rotating about an axis perpendicular to the intermolecular axis and passing through the center of gravity of the molecule constitutes an example of a rigid rotor, it being assumed that the internuclear distance does not change during rotation. The kinetic energy (K.E.) of the molecule is given by

K.E. º T ILI 22

1 22 == w ----- (55)

Where w is the angular velocity and I is the momentum of inertia of the rotating molecular. The angular momentum L =Iw. If no force acts on the rotor, we can set the potential energy V=0. Hence, the Hamiltonian is expressed as

IL 2 VTH 2=+= ----- (56)

The expression for L2 in spherical polar coordinates (r,q,f) is given by

úû

ùêë

é

¶

¶+÷

ø

öçè

æ

¶

¶

¶

¶-=

2

2

2

22

sin

1sin

sin

1

fqqq

qqhL ----- (57)

The Schrödinger equation yy EH =ˆ may thus be written as

yf

y

qq

yq

qqE

I=

úúû

ù

êêë

é

þýü

îíì

¶

¶+÷

ø

öçè

æ

¶

¶

¶

¶-

2

2

2

2

sin

1sin

sin

1

2

1h ----- (58)

The above equation may be written as

08

sin

1sin

sin

12

2

2

2

2=+

¶

¶+÷

ø

öçè

æ

¶

¶

¶

¶y

p

f

y

qq

yq

qqE

h

I ----- (59)

Eq. 59 contains two angular variables q and f. It can be solved by the method of separation of variables, i.e., we look for a solution of the form ( ) ( ) ( )fqfqy FQ=, ----- (60)

Substituting Eq. 60 into Eq. 59, we obtain

2

22

2

2 1sin

8sin

sin

ff

p

qq

q

q

¶

F¶

F-=+÷

ø

öçè

æ

¶

Q¶

¶

¶

Q h

IE ----- (61)



71

We can set both sides of Eq.61 equal to a constant, say m2, thereby obtaining two differential equations each in one variable. These equations are:

02

2

2

=F+¶

F¶m

f and ----- (62)

0sin

sinsin

12

2

=Q÷÷ø

öççè

æ-+÷

ø

öçè

æ

¶

¶

¶

¶

qb

q

yq

qq

m ----- (63)

Where 228 hIEpb =

Eq.62 has the solution

( ) ( )ff imN ±=F exp , 1-=i ----- (64)

where N is the normalization constant. This wave function is acceptable provided m is an integer. This condition arises because F must be single-valued. Thus, ( ) ( )pff 2+F=F ----- (65)

It follows, therefore, that exp (2pmi) =1 ----- (66)

Since xixe x sincos += (Euler’s relation) ----- (67) \ 12sin2cos =+ mim pp ----- (68) This can be true only if m=0, ± 1, ± 2, ± 3... etc. Let us now normalize the wave function F(f) to determine the normalization constant N.

ò =FF*p

f2

0

1d )20( pf ££ ----- (69)

Or ò =´p

ff f2

0

2 1deeN imim ----- (70)

Or ò =p

f2

0

2 1dN , i.e., N2 (2p) =1 so that

( ) 212

-= pN ----- (71)

Hence, the normalized wave functions become

( ) ( ) ( )fpf imm ±=F-

± exp221

; m=0, 1, 2, 3... ----- (72)

We shall not attempt to give a complete solution of Eq.63 but will merely state that if b=l(l+1) where l is the rotational quantum number, then this equation becomes a standard



72

mathematical equation whose solutions are known to be associated Legendre polynomials

( )qcosm

lP where l is either zero or a positive integer and l> m .

The normalized solutions are given by

( ) ( )( )( )

qqq cos!

!

2

1221

,

m

lml Pml

mll

úúû

ù

êêë

é

+

-´

+=Q=Q ± ----- (73)

The energy eigenvalues of the rigid rotor are obtained as follows:

( )18 22 +== llhIEpb ----- (74)

Thus,

( )

I

hllE

2

2

8

1

p

+= ; l = 0,1,2,3 ... ----- (75)

In spectroscopy it is customary to use the symbol J rather than l for the rotational quantum number so that the rotational energy levels are given by the expression,

( )

I

hJJEJ 2

2

8

1

p

+= ; J = 0, 1, 2, 3 ... ----- (76)

7.8. LET US SUM UP In this lesson, we: Pointed out Ø Particle in a one dimensional box Ø Normalization of y

Ø Orthogonality of the wave functions Ø Particle in a three-dimensional cubical box Ø Separation of variables Ø Degeneracy Ø One-dimensional simple harmonic oscillator (S.H.O) Ø Energy of S.H.O Ø The rigid rotor (or) rotator

7.9. CHECK YOUR PROGRESS 1. Derive an expression for the energy of a rigid rotor using the Schrodinger equation. 2. Calculate the degeneracy of the energy level with energy equal to i. 11(h2/8ma2) and ii 12(h2/8ma2) for a particle in a cubical box. 7.10. LESSON – END ACTIVITIES



73

1. Set up and solve the Schrodinger wave equation for a particle in a one dimensional box, with potential energy zero inside the box. Normalize the wave function. 2. Set up the Schrodinger wave equation for a simple harmonic oscillator, and solve it for the energy eigenvalues. 7.11. REFERENCES 1. R.K. Prasad, Quantum Chemistry, Wiley Eastern Limited, New Delhi. 2. IRA N. Levin, Quantum Chemistry, Prentice-Hall of India private limited, New Delhi. 3. G. Aruldhas, Quantum mechanics, Prentice-Hall of India private limited, New Delhi. 4. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, millennium edition, Vishal Publishing co.



74

UNIT-V

LESSON 8: THE SCHRODINGER EQUATION FOR HYDROGEN ATOM

CONTENTS 8.0. AIMS AND OBJECTIVES 8.1. INTRODUCTION 8.2. ANGULAR FUNCTIONS 8.3. PROBABILITY DENSITY AND RADIAL DISTRIBUTION FUNCTIONS 8.4. THE MOST PROBABLE DISTANCE OF THE H-ATOM (OR H-LIKE SPECIES) 1S ELECTRON 8.5. PHYSICAL INTERPRETATION OF THE HYDROGENIC ATOMIC ORBITALS 8.6. THE SCHRODINGER WAVE EQUATION FOR MULTI-ELECTRON ATOMS 8.6.1. TIME-INDEPENDENT PERTURBATION THEORY 8.6.2. APPLICATION OF FIRST-ORDER PERTURBATION THEORY TO HELIUM ATOM 8.7. VARIATION METHOD 8.8. LET US SUM UP 8.9. CHECK YOUR PROGRESS 8.10. LESSON - END ACTIVITIES 8.11. REFERENCES 8.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on the Schrödinger equation for hydrogen atom to the students. On successful completion of this lesson the student should have: * Understand the Schrödinger equation for hydrogen atom. 8.1. INTRODUCTION Hydrogen atom is the simplest of all atoms. It is a three-dimensional system and the Schrodinger equation for this system is

yy EH =ˆ ----- (1)

where ( )zyxVm

H ,,2

ˆ 22

+Ñ-=h

----- (2)

and where the Laplacian operator in Cartesian coordinates is given by



75

2

2

2

2

2

22

zyx ¶

¶+

¶

¶+

¶

¶=Ñ ----- (3)

The potential energy of interaction between the electron and the nucleus is given by

( ) rZerV 02 4pe-= ----- (4)

Since this attractive potential has spherical symmetry depending only upon r, it is convenient to express the Schrodinger equation in terms of polar coordinates (r,q,f), rather than Cartesian coordinates (x,y,z). The Cartesian coordinates are related to polar coordinates (see Fig.1) as follows: fq cossinrx =

fq sinsinry =

qcosrz = The Schrodinger equation for hydrogen atom in terms of polar coordinates becomes

04

2

sin

1sin

sin

11

0

2

2

2

222

2

2=÷÷

ø

öççè

æ++

¶

¶+÷

ø

öçè

æ

¶

¶

¶

¶+÷÷

ø

öççè

æ

¶

¶

¶

¶y

pe

m

f

y

qq

yq

qq

y

r

ZeE

rrr

r

rr h----- (5)

where m is the reduced mass of the electron and the nucleus, that is, ( )enne mmmm +=m .

Assuming that V(r) is a function of r only, the above wave equation can be solved by separating the variables: ( ) ( ) ( ) ( )fqfqy FQ= rRr ,, ----- (6)

where R(r) is the radial function which is a function of r only, and Q(q) and F(f) are angular functions. Substituting Eq. 6 into Eq. 5 then we get an equation which can be separated into three equations which are:



76

04

21

0

2

22

2

2=÷÷

ø

öççè

æ++-÷÷

ø

öççè

æR

r

ZeER

rdr

dRr

dr

d

r pe

mb

h ----- (7)

0sin

sinsin

12

2

=Q+Q

-÷ø

öçè

æ Qb

qqq

qq

m

d

d

d

d ----- (8)

or 02

2

2

=F+F

md

d

f -----(9)

where m and b are constants. The complete solution of Eq. 5 comprises the solutions of Eqs. 7,8 and 9. The energy values of hydrogen-like atoms are found to be given by

( ) ( )222

0

422

4

2

hn

eZEn

pe

mp-= ----- (10)

These energy levels are the same as those obtained from the Bohr theory. The normalized solutions of radial equation (9) are found to be

( )( ) ( )

[ ]( ) ( )rrr 12

3

3

0, 2exp

!2

!12 ++-÷

÷

ø

ö

çç

è

æ

+

---= l

lnl

ln Llnn

lnnaZrR ----- (11)

where ( )rnaZ 02=r and ( ) 2200 4 ea mpe h= ----- (12)

( )r12 ++l

lnL are the associated Laguerre polynomials defined as

( ) ( )úû

ùêë

é= -r

r

rrr

r ed

de

d

dL r

r

r

s

ssr ----- (13)

where s=2l+1 and r = n+l and n=1,2,3,4,...... . The number n is called the principal quantum number which determines the energies of different atomic orbitals and the distances of the electron from the nucleus. The probability of finding the electron between r and r+dr is given by the quantity R*(r)R(r)r2dr. 8.2. ANGULAR FUNCTIONS The solution of Eq. 8 gives

( )( )( )

( )( )qq cos

!2

!12 21

,

m

lml Pml

mll

úúû

ù

êêë

é

+

-+=Q ----- (14)



77

where ( )qcosm

lP are called associated Legendre polynomials defined as

( ) ( ) ( )ml

lmlm

l

m

ld

d

lP

+

+-

-=q

qqq

cos

1coscos1

!2

1cos

222 ----- (15)

In the above equation, the quantum number l (=0,1,2....) is the azimuthal quantum number. It determines the shape of the atomic orbitals. The solution of Eq. 9 gives

( ) ( )fp

f imm exp2

1=F ----- (16)

where p21 is the normalization constant and lm ±±±= ....,.........2,1,0 is the magnetic

quantum number. This name is given to this quantum number because in the presence of the external magnetic field, the states having different values of m have different energies. The phenomenon of removal of degeneracy of an energy state by application of external magnetic field is known as Zeeman effect, after P. Zeeman (1865-1943) who was the co-winner (with H.A Lorentz) (1853-1928) of the 1902 Physics Nobel Prize. These Dutch physicists were honored for their work on the influence of magnetism upon radiation phenomena. To sum up, the solution of the radial equation (7) gives quantum numbers n and l. The solution of the Q-equation (8) gives quantum numbers l and m, and the solution of the F-equation (9) gives quantum number m. In other words, the quantum numbers n, l, and m follow directly from the wave mechanical treatment. 8.3. PROBABILITY DENSITY AND RADIAL DISTRIBUTION FUNCTIONS As mentioned before, the value of y in the Schrodinger wave equation is important since

2y gives the probability of finding an electron of a given energy (i.e., of known quantum

numbers) at a given distance from the nucleus. Thus, by determining 2y at different distances

from the nucleus, it is possible to trace out a region of space around the nucleus where there is high probability of locating the electron. Each such region of space, as already pointed out, is called an orbital. The probability of locating the electron at different distances from the nucleus can be

represented graphically by plotting probability ( 2y ) against the distance (r) from the nucleus of

the atom. However, electron probability distribution is generally expressed in the form of radial probability distribution which means the probability of finding an electron within a small radial space around the nucleus. Let us suppose that the space around the nucleus is divided into a very large number of thin concentric shells of thickness dr at a distance r from the nucleus. The volume of a spherical shell between radii r and r + dr will be given by 4pr2dr. The probability of finding electron

within this spherical shell will be given by 4pr2 2y dr. The radial probability at a distance r from

the nucleus is thus given by the function, 4pr2 2y dr. In this function, while the probability factor 2y decreases, the volume factor 4pr2dr increases with increase in the value of r. The radial



78

probability distribution of the electron is thus obtained by plotting the function 4pr2 2y against

the distance r from the nucleus. The probability of finding an electron at zero distance from the nucleus is zero. The probability increases gradually as the distance increases, goes to a maximum and then begins to decrease. The peak of the curve gives the distance from the nucleus where the probability of finding the electron is maximum. This distance is called the radius of maximum probability.

In the case of hydrogen atom, the radius of maximum probability is°

A 0.529 . Bohr’s model restricts the electron to a definite orbit at a fixed distance from the nucleus. The wave mechanical model, however, gives merely the maximum probability of locating the electron at a given distance from the nucleus. In the case of hydrogen atom, for instance, according to Bohr’s calculations, the electron under ordinary conditions always stays at a

distance of °

A 0.529 from the nucleus. According to the wave mechanical model, however, the electron keeps on moving toward or away from the nucleus and the maximum probability of

locating it lies at a radius of °

A 0.529 from the nucleus. 8.4. THE MOST PROBABLE DISTANCE OF THE H-ATOM (OR H-LIKE SPECIES) 1S ELECTRON Consider the 1s orbital wave function of hydrogen atom given by

( ) 021

301 1 a

r

s ea-

= py ----- (1)

where r is the distance of the electron from the nucleus. The probability density for an electron in this orbital is given by

( ) 0

230

21 1 a

r

s ea-

= py ----- (2)

It may be noted that 21sy is independent of q and f and hence the electron distribution in

1s orbital is spherically symmetrical. Graphically the plot of 21sy against r is as shown by curve a

in Fig. 2.



79

There is also another procedure for describing the electron distribution in an atomic orbital. It consists in determining the probability of finding an electron in a spherical shell of thickness dr at a distance r from the nucleus. This is called radial probability. The radial probability distribution of electron is given by

( ) drdrrP s2

124 yp=

drea

r ar 02

30

2 14 -

÷÷ø

öççè

æ=

pp ----- (3)

The plot of P(r), i.e., of 21

24 syp against r is given by curve b in Fig.1. The maximum of

the curve is found by differentiating P(r) with respect to r and setting the derivative equal to zero, i.e.,

( )

0224

02

0

2

30

=÷÷ø

öççè

æ+-= - arer

a

r

adr

rdP ----- (4)

It readily follows from Eq. 4 that 0ar = . ( )002 ¹- areQ

Thus, we find that the maximum radial probability density occurs at a distance a0, the Bohr radius. We, therefore, conclude that the most probable distance of the 1s electron from the nucleus is precisely what has been predicted by the Bohr theory. As already mentioned, while the Bohr model restricts the electron to a definite orbit at a fixed distance from the nucleus, the wave mechanical model gives merely the maximum probability of locating the electron at a given distance from the nucleus.



80

8.5. PHYSICAL INTERPRETATION OF THE HYDROGENIC ATOMIC ORBITALS According to the Bohr theory, the angular momentum L of an electron in an orbital is given by ( )p2hnL = ; n=1, 2, 3, ..... ----- (5)

However, according to wave mechanics, the value of L is given by

( )[ ] ( )[ ] ( )p2112121

hllllL +=+= h ----- (6)

Thus , while according to the Bohr theory the ground state angular momentum is equal to h/2p since n=1(i.e., it is finite), according to wave mechanics its value is zero(Since l=0). Consider an electron of mass me moving about the nucleus at a distance r in an orbit with velocity u. The centrifugal force which keeps the electron away from the nucleus is given by

3

2

3

2222

rm

L

rm

rum

r

umF

ee

ee === ----- (7)

Where L= meur is the angular momentum of the electron. In terms of atomic units frequently used in quantum mechanics, we said me as well as e and h =1. Hence, in atomic units we can write Eq. 6 as

( )12 += llL ----- (8)

Substituting for L2 in Eq.7, we have

( ) 31 rllF += ----- (9)

The radial motion of an electron in an atom is controlled by the centrifugal force given by

Eq. 9 as well as by the coulombic force of attraction, viz., 22 rZe- (or 2rZ- in atomic units).

For the atom to be stable,

( )

01

23=-

+

r

Z

r

ll or ( ) Zllr 1+= ----- (10)

Eq.10 shows that , other things being equal, an electron in an orbital of high angular momentum tends to stay farther from the nucleus then an electron in a state of lower angular momentum. Fig. 3 shows the radial parts of the Hydrogenic wave functions for the principal quantum number n = 3. We see that the higher value of l, the less likely it is to be found near the nucleus. Thus, for a given value of n, the size of the orbit increases with increasing the azimuthal quantum number, l.



81

8.6. THE SCHRODINGER WAVE EQUATION FOR MULTI-ELECTRON ATOMS The Schrodinger wave equation cannot be solved exactly for atoms beyond hydrogen atom in the Periodic table. The troublesome term is the interelectron repulsion term in the Hamiltonian because of which the n-electron wave function cannot be split into n one-electron wave equations. Hence, methods have been developed for the approximate solution of the multi-electron Schrodinger wave equation. One of these methods is based on the time-independent perturbation theory and the other, called the variation method. Involves the selection of a trial wave function. 8.6.1. TIME-INDEPENDENT PERTURBATION THEORY The Schrodinger wave equation to be solved is

yy EH =ˆ ----- (1)

The Hamiltonian is decomposed into two parts as

'ˆˆˆ )0( HHH l+= ----- (2)

where )0(H is the unperturbed part and 'Hl is the perturbation where l is the perturbation parameter which measures the deviation of the problem of interest from the unperturbed system. It is further assumed that

)0(ˆ'ˆ HH <<l ----- (3) In general, l is set equal to unity which means that perturbation is fully applied.

Associated with )0(H are a set of eigenvalues )0(1E , )0(

2E , ........, )0(nE and the corresponding

eigenfunctions )0()0(2

)0(1 ,......., nyyy , i.e.,

)0()0()0(ˆnnn EH yy = ----- (4)



82

It is assumed that the eigenfunctions ny and the energy eigenvalues En of the total

Hamiltonian H can be expressed in the form of a power series of l :

( ) ( ) ( ) ( ) ....3321210 ++++= nnnnn ylyllyyy ----- (5a)

( ) ( ) ( ) ( ) ....3321210 ++++= nnnnn EEEEE lll ----- (5b)

The first term in Eq. 5a is the zeroth-order term, the second represents the first order correction; the third represents the second-order correction, etc., to the unperturbed zeroth-order

term. The eigen functions ( ) ( ) ,...., 21nn yy and the eigenvalues ( ) ( ) ,..., 21

nn EE are independent of l and ( ) ( ) ,...., 21

nn yy are so chosen that they are orthogonal to ( )0ny which is assumed to be normalized.

Omitted details, When Eq. 5a and 5b are solved, the results to the first-order in l are:

( ) ( )10nnn yyy +=

( )

( ) ( )

( ) ( )( )0

00

00

0'ˆ

mnm mn

mn

nEE

Hy

yyly å

¹ -

><+= ----- (6)

( ) ( )10nnn EEE +=

( ) ( ) ( ) ><+= 000 'ˆnnn HE yyl ----- (7)

where ( ) ( ) ( ) ( ) ( )00000 ˆˆnnnnn HHE >º=< yy ----- (8)

It can be further shown that

( ) ( ) ( ) >=< 001 'ˆnnn HE yy ----- (9)

( ) ( ) ( ) >=< 102 'ˆnnn HE yy ----- (10)

( ) ( ) ( ) >=< 203 'ˆnnn HE yy ----- (11)

and ( ) ( ) ( ) >=< -10 'ˆ nnn

nn HE yy ----- (12)

These equations show that the calculation of nth order energy requires a knowledge of (n-1)th order eigenfunctions. Since the contribution of second order, third order, etc., terms goes on successively decreasing, we are primarily interested in the first order correction. 8.6.2. APPLICATION OF FIRST-ORDER PERTURBATION THEORY TO HELIUM ATOM



83

We shall solve the Schrodinger wave equation for the ground state of helium atom using the first-order time- independent perturbation theory. In this case,

yy EH =ˆ ----- (13)

'ˆˆˆ )0( HHH l+= ----- (14)

where the unperturbed Hamiltonian, ( )0H , is given by

( ) ( ) ÷÷ø

öççè

æ+-Ñ+Ñ-=

210

222

21

20 11

42ˆ

rr

ZeH

pem

h ----- (15)

and the perturbation is the interelectron repulsion term :

1202 4'ˆ reH pe= ----- (16)

Here r1 and r2 are the distances of the two electrons from the helium nucleus of charge Ze, and r12 is the interelectron distance. We shall use atomic units, a.u. ( )141 0 ==»== pem meh , so

that

( ) ( ) ÷÷ø

öççè

æ+-Ñ+Ñ-=

21

22

21

0 22

2

1ˆrr

H ----- (17)

121'ˆ rH = ----- (18)

Since ( )0ˆ'ˆ HH << , it is pertinent to use perturbation theory.

Since ( )0H is the sum of two one-electron Hamiltonians, the unperturbed wave function ( ) ( )210 , rry can be written as the product of two Hydrogenic wave functions:

( ) ( ) ( ) ( ) ( ) ( )20

10

210 , rrrr yyy = ----- (19)

Where ( ) ( )ri0y is the wave function of the ith electron in a Hydrogenic atom with nuclear charge

= Ze. Thus,

( ) ( ) 0101

21

30

3

30

3

210 , aZraZr e

a

Ze

a

Zrr --

÷÷ø

öççè

æ´÷÷

ø

öççè

æ=

ppy

( ) ( )213 rrZeZ

+-= p (in a.u.) ----- (20)

The unperturbed ground state energy, ( )00E , is equal to the sum of the ground state energies

of two Hydrogenic atoms :

( ) ( ) ( ) 2220 22 ZZZEo -=-+-= ----- (21)



84

The first-order correction to the ground-state energy is

( ) ( ) ( ) >=< 0010 'ˆ yy HE

( ) ( ) ( ) ( ) 21210

210 ,'ˆ,* ttyy ddrrHrròò= ----- (22)

Substituting the value of ( ) ( )210 , rry from Eq. 20 into Eq. 22, we obtain

( ) ( )( ) 211221 12

2

61

0

tt

p

ddrrrZ

eZ

E+-

òò= ----- (23)

where the volume elements of the two electrons (in spherical polar coordinates are) :

11112

11 sin fqqt dddrrd =

212121212

212 sin fqqt dddrrd =

The evaluation of the interelectron repulsion integral (Eq.23) is rather tedious; it can be shown that it is given by

( ) ( )ZE 8510 = ----- (24)

Notice that ( )10E is positive, as was to be expected since the repulsion energy between two

electrons is always positive. Adding Eqs. 21 and 24, we have

( ) ( ) ZZEEE ÷ø

öçè

æ+-=+=

8

5210

000

÷ø

öçè

æ--= ZZ

8

52 ----- (25)

Re- incorporating the original units we find that

2

22

028

5

h

eZZE

m÷ø

öçè

æ--= ----- (25a)

Recalling that the ground state energy of hydrogen atom is 2

1- a.u. or 24 2hem- or -13.60eV,

we get for helium atom

( ) =÷ø

öçè

æ--= eVZZE 2.27

8

520 -2.75 a.u. = -74.80 eV ----- (26)



85

The experimental value is -2.904 a.u. or – 78.986 eV. The agreement between the theoretical and experimental values is not good. If the second-order and higher order contributions are included, the agreement improves but is still not very good. 8.7. VARIATION METHOD According to the variation method, also called the variation theorem, i f y is an

approximate wave function of a quantum mechanical system, an atom or a molecule, described

by the Hamiltonian H , then the energy eigenvalue of the system, approximately given by the integral

ò

ò=

><

><=

tyy

tyy

yy

yy

d

dHHE

*

ˆ*ˆ ----- (1)

or, if y is normalized (so that <y /y >=1), by the integral

0EE ³ ----- (2)

is an upper bound to the ground state energy, E0, of the system, i.e., We shall not give the proof of the variation theorem here. If the approximate wave function, also called the trial function, happens to be exact (true) wave function, then 0EE = ----- (3)

However, this is seldom the case since we have no idea of the exact wave function, to begin with. The application of the variation method involves the following steps: (i) Choose a trial wave function y dependent on variable parameters.

(ii) Evaluate the integral >< yy H

(iii) Minimise the above integral with respect to the variable parameters. (iv) The function y with the optimum value of the parameters is the best

approximation to the true wave function and the lowest value of >< yy H is the

nearest approximation to the true energy. 8.7.1. APPLICATION OF VARIATION METHOD TO HELIUM ATOM The Hamiltonian for He atom (in atomic units) is

( )1221

22

21

122

2

1ˆrrr

H +÷÷ø

öççè

æ+-Ñ+Ñ-= ----- (4)



86

where the symbols have their usual meanings. As a result of screening, each electron (1 and2) shields the other from the nucleus so that the two electrons do not ‘’see’’ the same nuclear charge. Here it is natural to use a wave function with nuclear charge Z less than 2. That is, Z is a variation parameter which we shall determine from the variational calculation. It is important to remember that the Hamiltonian in Eq. 4 contains the exact value of Z = 2.The vibrational parameter appears only in the trial wave function:

( ) 21

2

132

13

2,1ZrZr e

Ze

Zrr --

÷÷ø

öççè

æ´÷÷

ø

öççè

æ=

ppy ----- (5)

= ( ) ( )[ ]213 exp rrZZ +-p ----- (6)

>=< yy HE ˆ ----- (7)

Substituting the value of H from Eq. 4 in Eq. 7,

><+><-><->Ñ<->Ñ<-= yyyyyyyyyy1221

22

21

112

12

2

1

2

1

rrrE ----- (8)

The Laplacian operator in spherical polar coordinates (r,q,f) is given by

2

2

222

2

2

2

sin

1sin

sin

11

fqqq

qq ¶

¶+÷

ø

öçè

æ

¶

¶

¶

¶+÷

ø

öçè

æ

¶

¶

¶

¶=Ñ

rrrr

rr ----- (9)

Since the trial function (Eq. 5) does not depend upon angles q and f, differentiations with respect to q1, f1, q2 and f2 vanish so that for the first kinetic energy operator in Eq. 8 , we get

22222

22

11112

1

1

21

12

1

2321 sinsin

2

1

2

1211 fqqfqq

pyy dddrredddrre

rr

rre

Z ZrZrZr

òò--- ´ú

û

ùêë

é÷÷ø

öççè

æ

¶

¶

¶

¶-÷÷

ø

öççè

æ>=Ñ<-

----- (10) The functions of r2 are not affected by differentiation. Carrying out the integrations overall angles we obtain

( ) 2

0

22

21

21

1

21

12

10

2

2321

2

1

1

2

14

2

1drredrr

r

er

rre

Z ZrZr

Zr

òò¥

--¥

- ´úû

ùêë

é÷÷ø

öççè

æ

¶

¶

¶

¶-÷÷

ø

öççè

æ>=Ñ<- p

pyy ----- (11)

The second integral, evaluated using the standard integral, is equal to 2!/(2Z)3 so that Eq. 11 becomes



87

( )( )

( ) 12

1

10

3

23 11

2

!2

2

14 dreZr

re

ZZ ZrZr

úû

ùêë

é-

¶

¶÷ø

öçè

æ- -

¥-

ò

( ) ( ) 12

12

1

0

3 11 22 drerZZreZ ZrZr -¥

- +--= ò

( ) úû

ùêë

é+--= òò

¥-

¥-

0

122

0

12

13 1

1

122 drerdrerZZ ZrZr

= ( )( ) ( ) 22

!2

2

22

2

3

2

2

3 Z

Z

Z

Z

ZZ =

úúû

ù

êêë

é+-- ----- (12)

Similarly, the second kinetic energy operator term in Eq. 8 gives -2Z. The fifth term, the interelectron repulsion energy integral, is some what more tedious. So, we will not go into details about its evaluation, merely stating that it evaluates to (5/8)Z. Combining all these contributions Eq. 8 becomes

( ) ( ) ( ) ( )ZZZZZE 827852222 22 -=+-+= ----- (13)

Using the variation method, we minimize the energy with respect to Z, obtaining ,08272 =-=¶¶ ZZE where 69.11627 ==Z

Substituting this value of Z in Eq. 13, we obtain the ground state energy of helium atom: E0 = (27/16)2 - (27/8)(27/16) = -2.8476 a.u. = - (2.8476)(27.21 eV) = - 77.48 eV [ Q 1 a.u. = 27.21 eV ] ----- (14) This is much better than the first-order perturbation theory result, viz., - 74.80 eV, we obtained in Eq. 26, the experimental value being – 2.904 a. u., i.e., - 78.986 eV. in 1929, E.A. Hyleraas, using a more complicated wave function and the concept of the so-called electron correlation which stipulates that electrons try to avoid each other, obtained a result for the ground state energy of helium which is very close to the experimental value. 8.8. LET US SUM UP In this lesson, we: Pointed out Ø Angular functions Ø Probability density and radial distribution functions Ø The most probable distance of the H-atom (or H-like species) 1s electron Ø Physical interpretation of the hydrogenic atomic orbitals



88

Ø The Schrödinger wave equation for multi-electron atoms Ø Time-independent perturbation theory Ø Application of first-order perturbation theory to helium atom Ø Variation method

8.9. CHECK YOUR PROGRESS 1. Show that the following radial wave function of hydrogen atom is normalized.

( ) ( ) ( )0

23

00,1 exp2 ararR -=.

2. Using the first order perturbation theory solve the Schrodinger wave equation for the ground state energy of Helium atom. 8.10. LESSON – END ACTIVITIES 1. Using the variation method solve the Schrodinger wave equation for the ground state energy of Helium atom. 2. Write the Schrödinger wave equation for Hydrogen atom in terms of polar coordinates. Separate the resultant equation in three equations using the technique of separation of variables. How do the quantum numbers n,l, and m emerge from the solution of the wave equation. 8.11. REFERENCES 1. R.K. Prasad, Quantum Chemistry, Wiley Eastern Limited, New Delhi. 2. IRA N. Levin, Quantum Chemistry, Prentice-Hall of India private limited, New Delhi. 3. G. Aruldhas, Quantum mechanics, Prentice-Hall of India private limited, New Delhi. 4. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, millennium edition, Vishal Publishing co.



89

UNIT-VI

LESSON 9: THERMODYNAMICS AND NON-IDEAL SYSTEMS: CHEMICAL POTENTIAL

CONTENTS 9.0. AIMS AND OBJECTIVES 9.1. INTRODUCTION: CONCEPT OF CHEMICAL POTENTIAL 9.2. GIBBS- DUHEM EQUATION 9.3. SOME IMPORTANT RESULTS 9.4. VARIATION OF CHEMICAL POTENTIAL WITH TEMPERATURE 9.5. VARIATION OF CHEMICAL POTENTIAL WITH PRESSURE 9.6. CHEMICAL POTENTIAL IN CASE OF A SYSTEM OF IDEAL GASES 9.7. APPLICATION OF THE CONCEPT OF CHEMICAL POTENTIAL 9.7.1. CLAPEYRON–CLAUSIUS EQUATION 9.7.2. INTEGRATED FORM OF CLAPEYRON-CLAUSIUS EQUATION FOR LIQUID Û VAPOR EQUILIBRIUM

9.7.3 . APPLICATION OF CLAPEYRON-CLAUSIUS EQUATION FOR LIQUID Û VAPOR EQUILIBRIA

9.7.4. CLAPEYRON-CLAUSIUS EQUATION FOR SOLID Û VAPOR EQUILIBRIUM 9.7.5. APPLICATION OF CLAPEYRON-CLAUSIUS EQUATION FOR SOLID Û LIQUID EQUILIBRIA 9.8. LET US SUM UP 9.9. CHECK YOUR PROGRESS 9.10. LESSON - END ACTIVITIES 9.11. REFERENCES 9.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on chemical potential to the students. On successful completion of this lesson the student should have: * Understand the concept of chemical potential. 9.1. INTRODUCTION: CONCEPT OF CHEMICAL POTENTIAL The thermodynamic properties, E, H, S, A and G are extensive properties because their values change with change in the mass (or) the numbers of moles of the system. In the derivation of various equations described earlier the change of was considered to be due to change in temperature and pressure only.



90

A tactic assumption was made that the systems under consideration was a closed system. (i.e.) there could be no change in the mass of the system. However, if the case of an open system containing two or more components, there can be change in the number of moles of various components as well. In that case, an extensive property like G must be a function not only of temperature and pressure but of the number of moles o the various components present in the system as well. Let T and P be the temperature and pressure respectively, of a system and let n1, n2, n5 …….ni be the respective number of moles of the constituents; 1, 2, 3,….i, then in view of what has been said above, the Gibbs free energy, G, must be a function of temperature, pressure and the numbers of moles of the various constituents (i.e.) ni)n3 n2, n1, P, (T, f G ¼¼= ------- (1)

Where ni n3n2n1 +¼¼+++ = total number moles = N (say) Then, for a small change in temperature, pressure and the numbers of moles of the components, the change in free energy dG will be given by the expression

i

ninnPTininnPT

i

ninPTiNTNP

dnn

Gdn

n

Gdn

n

GdP

P

GdT

T

GdG

.....2,1,,

2

...2,1,,2...2,,,,

..... úû

ùêë

é+ú

û

ùêë

é+ú

û

ùêë

é+ú

û

ùêë

é+ú

û

ùêë

é=

d

d

d

d

d

d

d

d

d

d

The quantity

innnPT

n

G

.......2

,1

,,2÷÷

ø

ö

çç

è

æ

¶

¶ is called partial molal free energy iG or

more often, chemical potential (µi) of the concerned component i. Thus,

iiG

njnnPTn

Gm==

÷÷

ø

ö

çç

è

æ

¶

¶

.......2

,1

,,2

------ (3)

The term chemical potential was first introduced by Gibbs. The physical significance of chemical potential easily follows from eqn (3). The chemical potential of a given substance is, evidently, the change in free energy of the system that results on the addition of one mole of that particular substance at a constant temperature and pressure, to such a large quantity of the system that is no appreciable change in the overall composition of the system. Eqn (2) may be written as

i

dni

dndndPNTP

GdT

NPT

GdG mmm

d

d

d

d+++ú

û

ùêë

é+ú

û

ùêë

é=

2211,,

------ (4)

Where µjµ2 µ1, ¼¼ are chemical potential of the components 1, 2 and j, respectively.



91

If temperature and pressure remains constant then,

dnµ .. dnµ dnµ (dG) jj2211 PT, ¼¼++= ------ (5)

If a system has a definite composition having n1, n2….ni moles of the constituents 1, 2 …j respectively, then on integrating eqn (5) we have

jj2211NP,T, µn ..µn µ n (G) ¼¼++= ------ (6)

From eqn (6) chemical potential may be taken as the contribution per mole of each particular constituent of the mixture to the total free energy of the system under conditions of constant temperature and pressure. It readily follows that for a total 1 mole of a pure substance, µG = (i.e.) free energy is

identical with chemical potential. 9.2. GIBBS- DUHEM EQUATION

We know that the eqn (6)

jj2211NP,T, µn ..µn µ n (G) ¼¼++= ------ (6)

Shows that the free energy of a system at constant temperature and pressure can be expressed as a sum of ‘nµ ‘terms for the individual components of the system. Diffentiating eqn (6), we obtain

dµn dnµ dµn dnµ dµn dnµ dG jjjj22221111 ++¼¼+++= (or)

)dµn ........dµn dµn()dnµ....... dnµ dn(µ dG jj2211jj2211 +++++= ------ (7)

But, according to eqn. (5) µjdnj ..……+ dn µ + dnµ = (dG) 2211P T, ------ (5)

The first term on right hand side of (7) is equal todG , at constant temperature and pressure. It follows therefore that at constant temperature and pressure, for a system of a definite composition 0 njdµj n2dµ2 n1dµ1 =¼¼++ ------ (8)

(Or) 0 dµn ii =å ------ (9)

This simple relationship is known as Gibbs – Duhem equation. For a system having only two components ((e.g.) a binary solution) the above equation reduces to 0 dµn dµn 2211 =+ ------- (10)



92

(Or)

2

1

21 dµ

n

n (-) dµ = ------ (11)

Eqn (11) shows that variation in chemical potential of one component affects the value for the other component as well. Thus, if dµ1 is positive, (i.e.) if µ1 increases then dµ2 must be negative, (i.e.) µ2 must decease and vice versa. 9.3. SOME IMPORTANT RESULTS In a special case when there is no change in the number of moles of the various

constituents of a system, that is, when the system is a closed one, then, dn ,dn ,dn j21 ¼¼ are all

zero. In such a case, eqn (4)

µjdnj ..……+ dn µ + dnµP

G

T

G dG 2211

,,

+÷ø

öçè

æ

¶

¶+÷

ø

öçè

æ

¶

¶=

NTNP

dT ------ (4)

Reduces to

NTNP

dT,, P

G

T

G dG ÷

ø

öçè

æ

¶

¶+÷

ø

öçè

æ

¶

¶= ------ (12)

It also follows from the expression G = H- T∆S

SdT TdS - PdV dE dG ++= ---- (A) PdV dE dq TdS +== ------ (B)

Combining eqns B &A we get, SdT VdP dG +=

That in a closed system SdT VdPdG = Hence, by equating coefficients of dT and dP in the above two equations, we get

S-=÷ø

öçè

æ

¶

¶

NP,T

G ------ (13)

and

V=÷÷ø

öççè

æ

¶

¶

NT,p

G ------ (14)

These results are important as they help us in deriving expressions for the variation of chemical

potential with temperature and pressure.

9.4. VARIATION OF CHEMICAL POTENTIAL WITH TEMPERATURE



93

The variation of chemical potential of any constituent (i) of a system with temperature can be derived by differentiating eqn (3) with respect to temperature and eqn (13) with respect to ni

iiG

njnnPTin

Gm==

÷÷

ø

ö

çç

è

æ

¶

¶

.......,,,2

1

------ (3)

S-=÷ø

öçè

æ

¶

¶

NP,T

G ------ (13)

The results are

NP

i

i TTn ,

2G ÷ø

öçè

æ

¶

¶=

¶¶

¶ m ------ (15)

and iS

jnnPTi

n

S

Ti

n-=

÷÷

ø

ö

çç

è

æ

¶

¶-=

¶¶

¶

,....1

,,

G 2 ------ (16)

Where Si , by definition, is the partial molal (or) molar entropy of the component i. It follows from equations (15) and (16) that

i

NP

i ST

-=÷ø

öçè

æ

¶

¶

,

m ------ (17)

Eqn (17) gives the variation of chemical potential µi of any constituent (i) with temperature. Since the entropy of a substance is always positive, hence, according to eqn (17), the chemical potential would decrease with increase in temperature. This is illustrated in fig (1) for a substance in solid, liquid and gaseous states. It is evident from this figure that at the melting point (Tm), the chemical potentials of the solid and liquid phases are the same. Similarly, at the boiling point (Tb); the chemical potentials of liquid and gaseous phases are the same. These observations are extremely useful in the phase rule studies.

Gas

Solid

Liquid

Temperature

Tm Tb

Chem

ical pote

ntial



94

Fig.1. Variation of chemical potential with temperature. 9.5. VARIATION OF CHEMICAL POTENTIAL WITH PRESSURE The variation of chemical potential of any constituent I with pressure may be derived by

differentiation eqn (3) with respect to pressure and eqn(14) with respect to in ,

iiG

njnnPTin

Gm==

÷÷

ø

ö

çç

è

æ

¶

¶

.......,,,2

1

------ (3)

W.r.t. P and

VP

G

NT

=÷ø

öçè

æ

¶

¶

,

------ (14)

W.r.t. n i

The results are

NT

i

i PnP

G

,

2

÷ø

öçè

æ

¶

¶=÷÷

ø

öççè

æ

¶¶

¶ m ------ (18) And

i

V

jnnPTi

n

V

Pi

n

G=

÷÷

ø

ö

çç

è

æ

¶

¶=

÷÷

ø

ö

çç

è

æ

¶¶

¶

.....1

,,

2 ------ (19)

Here iV by definition, is the partial molal (or) molar volume of the component i. It follows from

the eqns. (18) (19) that

i

NT

i VP

=÷ø

öçè

æ

¶

¶

,

m ------ (20)

Eqn. 20 gives the variation of chemical potential (µi) of any constituent i with pressure.

9.6. CHEMICAL POTENTIAL IN CASE OF A SYSTEM OF IDEAL GASES For a system of ideal gases, a faster development of equation

i

NT

i VP

-=÷ø

öçè

æ

¶

¶

,

m ----- (A) is also possible. In an ideal gas, nRTPV = . Consider now a system

consisting of a number of ideal gases, let n1, n2……, be the number of moles of each constituent present in the mixture. Then, in the ideal gas equation,n the total number of moles may be replaced by (n1 + n2 + ……). Hence,

( )P

RTnn

P

nRTV ....21 ++== ------ (1)



95

Differentiating equation (1) with respect to the in , at constant temperature and pressure,

we have

P

RTV

n

Vi

nnPTi

==÷÷ø

öççè

æ

¶

¶

....,,, 21

------ (2)

Substituting the value of iV ÷ø

öçè

æ=

P

RT equation (A), we have

P

RT

P NT

i =÷ø

öçè

æ

¶

¶

,

m ------ (3)

For a constant composition of the gases and at a constant temperature, equation (3) ay also be expressed in the form

PRTddPP

RTd i ln==m ------ (4)

Let ip be the partial pressure of the constituent ‘i’ present in the mixture, since, each

constituent behaves as an ideal gas, therefore,

RTnVp ii = ------ (5)

It follows from equation (5) and (1) that

Pn

np i

i ÷ø

öçè

æ= ------ (6)

Since in and n are constants, therefore, on taking logarithms and then differentiating, we get

Pdpd i lnln = ------ (7)

Substituting in equation (4) we have

ii pRTdd ln=m ------ (8)

On integrating equation (8), we get

( ) iPii pRT ln0 += mm ------ (9)

Where ( )Pi0m is the integration constant, the value of which depends upon the nature of the gas

and also on the temperature. It is evident from equation (9) that the “chemical potential of any constituent of a mixture of ideal gases is determined by its partial pressure in the mixture”. If the partial pressure of the constituent ‘i’ is unity, (i.e.), ip = 1, then

( )Pii0mm = ------ (10)



96

Thus, ( )Pi0m gives the chemical potential of the gaseous constituent ‘i’ when the partial

pressure of the constituent is unity, at a constant temperature. According to equation (5),

RTV

np i

i ÷ø

öçè

æ= ------ (11)

Now ÷ø

öçè

æ

V

ni represents molar concentration (i.e.), the number of moles per unit volume of

the constituent ‘i’ in the mixture. If this concentration is represented by iC , then equation (11)

gives

RTCp ii = ------ (12)

Introducing this value of ip in eqn. (9), we have

( ) ( )RTCRT iPii ln0 += mm

( ) i

tcons

Pi CRTRTRT lnln

tan

0 ++=44 344 21

m (Or)

( ) iCii CRT ln0 += mm ------ (13)

Where ( )Ci0m is a constant depending upon the nature of the gas and the temperature.

If 1=iC , then, ( )Cii0mm = ------ (14)

Thus ( )Ci0m represents the chemical potential of the constituent (i) when the concentration of the

constituent in the mixture is unity, at a constant temperature. Lastly, since ÷ø

öçè

æ

n

ni represents the

mole fraction ( )ix of the constituent ‘i’ in the mixture, equation (6) may be represented as

Pxp ii = ------ (15)

Substituting this value of ip in equation (9), we have

( ) ( )PxRT iPii ln0 += mm ------ (16)

( ) i

tcons

Pi xRTPRT lnln

tan

0 ++=44 344 21

m

( ) ixii xRT ln0 += mm ------ (17)

Where, one quantity ( )xi0m is also a constant which depends both on the temperature the total

pressure. If 1=ix then,

( )xii0mm = ------ (18)



97

Thus, the quantity ( )xi0m represents the chemical potential of the constituent ‘i’ when its mole

fraction, at a constant temperature and pressure is unity. 9.7. APPLICATION OF THE CONCEPT OF CHEMICAL POTENTIAL The concept of the chemical potential has been used in deriving a number of important generalizations such as the law of mass action the phase rule, the distribution law, the laws of osmotic pressure etc., 9.7.1. CLAPEYRON–CLAUSIUS EQUATION An equation of fundamental importance which finds extensive application in one component, two–phase systems, was derived by Clapeyron and independently by Clausius from the second law of thermodynamics and is generally knows as Clapeyron–Clausius equation. The two phases in equilibrium may be any of the following types: (i) Solid and liquid, S L at the melting point of the solid. (ii) Liquid and vapor, L V at the boiling point of the liquid. (iii)Solid and vapor, S V at the sublimation temperature of the solid (iv) One crystalline form and another crystalline form as, for example, Rhombic and monoclinic sulphur SR SM at the transition temperature of the allotropic forms Consider any two phase (say liquid and vapor) of one and the same substance in equilibrium with each other at a given temperature and pressure. It is possible to transfer any definite amount of the substance from one phase to another, in a thermodynamically reversible manner, (i.e.), infinitesimally slowly, the system remaining in a state of equilibrium all along. For example: By supplying heat infinitesimally slowly to the system, it is possible to change any desired amount of the substance from the liquid to vapor, phase at the same temperature and pressure. Similarly, by withdrawing heat infinitesimally slowly from the system, it is possible to change any desired amount of the substance from vapor to liquid phase with out change in temperature and pressure.

Since, the system remains in a state of equilibrium, the free energy change of either

process will be zero. We may conclude, therefore, that “equal amounts of a given substance must have exactly

the same free energy in the two phases at equilibrium with each other”. Consider in general, the change of a pure substance from phase ‘A’ to another phase ‘B’

in equilibrium with in at a given temperature and pressure. A B at constant temperature and pressure.

If GA is the free energy per mole of the substance in the initial phase ‘A’ and GB is the free energy per mole in the final phase ‘B’, then, since



98

GA = GB

Hence, there will be no free energy change. (i.e.,) G = GB – GA = 0

If the temperature of such a system is raised, say from T to T + dT, the pressure will also have to change. Say, from P to P + dP, in order to maintain the equilibrium. The relationship between dT and dP can be delivered from thermodynamics.

Let the free energy per mole of the substance in phase ‘A’ at the new temperature and pressure be GA + dGA and that in phase ‘B’ be GB + dGB. Since, the two phases are still in equilibrium, hence,

GA + dGA = GB + dGB ------ (1)

According to thermodynamics,

G = H – TS ------ (2) Since H = E + PV \ G = E + PV – TS ------ (3)

Upon differentiation,

dG = dE + PdV + VdP – TdS – SdT ------ (4) The first law equation for an infinitesimal change may be written as

dq = dE + dw ------ (5) If the work done dw is only due to expansion, then

dq = dE + PdV ------ (6) Now for a reversible process,

T

dqdS =

PdVdEdqTdS +== ------ (7)

Combining equations 4 and 7, we have

SdTVdPdG == ------ (8) The equation (8) gives change of free energy when a system under goes reversible a change of temperature dT and a change of pressure dP equation (8) for phase ‘A’ may be written as dTSdPVdG AAA -= ------ (9)



99

and for phase ‘B’ as

dTSdPVdG BBB -= ------ (10)

Since GA = GB, hence from equation (1), BA dGdG = ------ (11)

dTSdPVdTSdPV BBAA -=- ------ (12)

(or) AB

AB

VV

SS

dT

dP

-

-= ------ (13)

It may be noted that that since VA and VB are the molar volumes of the pure substance in the two phases ‘A’ and ‘B’ respectively, VB – V A represents the change in volume when one mole of the substance passes from the initial phase ‘A’ to the final phase ‘B’. It may be represented by VD . Similarly, SB – SA being the change in entropy for the same process, may be put as SD .

Hence V

S

dT

dP

D

D= ------ (14)

If ‘ q ’ is the heat exchanged reversibly per mole of the substance during the phase

transformation at temperatureT , then the change of entropy SD in this process is given by

T

qS =D

Hence VT

q

dT

dP

D= ------ (15)

Thus )( AB VVT

q

dT

dP

-= ------ (16)

This is Clapeyron-Clausius equation. This equation, evidently gives change in pressure dP which must accompany the change in temperature dT (or) vice verse, in the case of a system containing two phases of a pure substance in equilibrium with each other. Suppose the system consists of water in the two phases, viz., liquid and vapor in equilibrium with each other at the temperature T (i.e.) Water (liquid) water (vapor) The q = molar heat of vaporization, VHD

BV = volume of one mole of water in the vapor state, say gV

AV = Volume of one mole of water in the liquid state, say lV

Equation (16) therefore, taken the form



100

)( lg

V

VVT

H

dT

dP

-

D= ------ (17)

If the system consist of water its freezing point, then, the two phases in equilibrium will be water (solid) water (liquid) (ice) Equation (16) may then be written as

)( Sl

f

VVT

H

dT

dP

-

D= ------ (18)

Where fHD is the molar heat of fusion of ice.

9.7.2. INTEGRATED FORM OF CLAPEYRON-CLAUSIUS EQUATION FOR LIQUID Û VAPOR EQUILIBRIUM

The clapeyron- clausius equation

( )lg VVT

Hv

dT

dP

-

D= ------ (1)

as applied to liquid vapor equilibrium can be easily integrated. The molar volume of a substance in the vapor state is considerably greater than that in the

liquid state. In the case of water, for example the value of Vg at 100℃ is 18 × 1670 = 30060 ml while that of Vl is only a little more than 18 ml. thus Vg-Vl can be taken as vg without introducing any serious error. The Clapeyron equation (1) therefore, may be written as

gTV

Hv

dT

dP D= ------ (2)

Assuming that the gas law is applicable, (i.e.), RTPV = (Per mole)

Vg = P

RT ------ (3)

Hence, 2RT

HvP

RT

P

R

Hv

dT

dP D=´

D= ------ (4)

(or) 2T

dT

R

Hv

P

dP´

D= ------ (5)



101

(i.e.) 2

1

RT

Hv

dT

dP

P

D=´ ------ (6)

(or) 2

)(ln

RT

Hv

dT

Pd D= ------ (7)

Assuming that HvD remains constant over a small range of temperature, we have

2

lnT

dT

R

HvPd ò

D=ò ------ (8)

CTR

HvP +÷

ø

öçè

æD-=\

1ln ------ (9)

(or) CTR

HvP ¢+÷

ø

öçè

æD-=

1

303.2log ------ (10)

Where C and C¢ are integration constants. Equation (10) is, evidently, the equation of a straight

line. Hence, the plot Pln against ÷ø

öçè

æ

T

1should yield a straight line with slope

R

HvD- and

intercept = C. This enables evaluation of HvD . Equation (7) can also be integrated between limits of pressure P1 and P2 corresponding to temperature T1 and T2. Thus,

òòD

=2

1

2

2

1

)(lnT

T

P

PT

dT

R

HvPd ------ (11)

2

1

1ln

1

2

T

TTR

Hv

P

Púû

ùêë

éD-=\ ------ (12)

úû

ùêë

é-

D+=

21

11

TTR

Hv ------ (13)

úû

ùêë

é -D=

21

12

TT

TT

R

Hv ------ (14)

(or) úû

ùêë

é -D=

21

12

1

2log303.2TT

TT

R

Hv

P

P ------ (15)

The equation (15) is known as Clapeyron-Clausius equation. 9.7.3. APPLICATION OF CLAPEYRON-CLAUSIUS EQUATION FOR LIQUID Û VAPOR EQUILIBRIA



102

Equation (13) can be used for calculating the molar heat of vaporization, HvD liquid if we know the vapor pressures at two temperatures. Further, if HvD is known, vapor pressure at a desired temperature can be calculated from the knowledge of a single value of vapor pressure at a given temperature. It can also be used for calculating the effect of pressure on the boiling point of a liquid. A few examples are given below: 1. Calculation of molar heat of vaporization, HvD : The molar heat of vaporization of a liquid can be calculated if its vapor pressure at two different temperatures is known. Eg. 1: Vapor pressure of water at 95℃ and 100℃ are 634 and 760mm, respectively. Calculate

the molar heat of vaporization, HvD , of water between 95℃ and 100℃. Solution: Substituting the given data in eqn (13) we have

úû

ùêë

é -D=

-- )373)(368(

368373

314.8634

760ln

11 kk

kk

molJK

Hv

mm

mm = HvD = 41363 Jmol-1 ------ (12)

2. Effect of temperature on vapor pressure of a liquid: If vapor pressure of a liquid at one temperature is known, that another temperature can be calculated. Eg 2: The vapor pressure of water at 100℃ is 760mm. what will be the vapor p r e s s u r e a t

95℃ ? The heat of vaporization of water in this temperature range is 41.27 KJ per mole Solution: Substituting the given data in eqn (13) we have

úû

ùêë

é -´=

--

-

)373)(368(

368373

314.8

1027.41

760ln

11

132

kk

kk

molJK

Jmol

mm

p

∴ p2 = 634.3mm 3. Effect of pressure on boiling point: If boiling point of a liquid at one p r e s s u r e i s known, that at another pressure can be calculated. Eg 3: Ether boils at 33.5℃ at one atmosphere pressure. At what temperature will it boi l a t a pressure of 750mm, given that the heat of vaporization of ether is 369.86 J o u l e s p e r gram? Solution: Substituting the given data in eqn (13), we have

( )( )

úû

ùêë

é -=

--

--

))(5.306(

306

314.8

7486.369

760

750ln

2

211

11

Tk

JkT

molJK

gmolJg

mm

mm

T2 = 305.9K = 32.9℃



103

9.7.4. CLAPEYRON-CLAUSIUS EQUATION FOR SOLID Û VAPOR EQUILIBRIUM The clapeyron- clausius equation for solid Û vapor equilibrium may be put as

)( Sg

S

VVT

H

dT

dP

-

D= ------ (1)

Where SHD stands for the molar heat of sublimation of the substance. Since the molar

volume of a substance is the gaseous state as very much greater than that in the solid, Sg VV - can

be safely taken as gV .

Eqn (1) can thus be easily integrated, as before to give the following expression:

úû

ùêë

é -D=

12

12

1

2

303.2log

TT

TT

R

H

P

P S

9.7.5. APPLICATION OF CLAPEYRON-CLAUSIUS EQUATION FOR SOLID Û LIQUID EQUILIBRIA

The Clapeyron-Clausius equation (1) for solid Û liquid equilibrium cannot be integrated

easily since SV cannot be ignored in comparison with lV . Also the laws of liquid state are not as

simple as those for gaseous state. However, this equation can be used for calculating the effect of pressure on the freezing point of a liquid. This aspect will be discussed in chapter phase rule equilibria and phase rule. Eqn (1) can also be used for calculating heat of fusion from vapor pressure data obtained at different temperatures. Eg: 1

The vapor pressure of ice Û water system at 0.0075℃ is 4.58mm and at 0℃ i s 759.80mm of mercury. Calculate the molar heat of fusion of ice, given that the specific volumes of ice and water at 0℃ are 1.0907cc and 1.0001cc, respectively

Density of mercury at 0℃ = 13.6gm per cc. Solution:

dP = 759.8-4.58 = 755.22mm=75.52cm Hg

= 75.52´ 13.6´ 981 dynes cm-2 dT = 0.0075-0=0.00750 lV = 18 ´ 1.001

SV = 18 ´ 1.0907 cm3



104

Since vapor pressure increases on decreasing the temperature will have a negative sign. Substituting various values in eqn (1) we have

K0075.0

98160.1352.75 ´´ dynes 2-cm =

( ) 30907.10001.118273 --´

D

cm

H f

Hence, fHD = 5.98 ´ 1010 ergs mol-1 (1 dyn cm = 1 erg)

= 7

10

10184.4

1098.5

´

´

ergs

ergs1

1

-

-

cal

mol = 1429 cal mol-1

9.8. LET US SUM UP In this lesson, we: Pointed out Ø Concept of chemical potential Ø Gibbs- Duhem equation Ø Some important results Ø Variation of chemical potential with temperature Ø Variation of chemical potential with pressure Ø Chemical potential in case of a system of ideal gases Ø Application of the concept of chemical potential Ø Clapeyron–Clausius equation Ø Integrated form of clapeyron-clausius equation for liquid Û vapor equilibrium Ø Application of Clapeyron-Clausius equation for liquid Û vapor equilibria Ø Clapeyron-Clausius equation for solid Û vapor equilibrium Ø Application of Clapeyron-Clausius equation for solid Û liquid equilibria

9.9. CHECK YOUR PROGRESS 1. What is chemical potential? 2. How does chemical potential vary with temperature and pressure? Derive the Gibbs-Duhem equation. 9.10. LESSON – END ACTIVITIES 1. Derive Claperyon-Clausius equation in the form dP/dT= DHv/(TVg). Under what conditions can this equation be integrated and how? 2. What are the applications of Claperyon-Clausius equation? Explain with examples. 9.11. REFERENCES 1. J.C. Kuriacose and J. Rajaram, Thermodynamics, Shoban lal Nagin Chand & Co., Delhi. 2. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006-2007.



105

3. Gurdeep Raj, Advanced Physical Chemistry, Goel Publishing House, Meerut.



106

LESSON 10: THERMODYNAMICS AND NON-IDEAL SYSTEMS:

FUGACITY

CONTENTS 10.0. AIMS AND OBJECTIVES 10.1. INTRODUCTION: CONCEPT OF FUGACITY 10.2. FUGACITY AT LOW PRESSURES 10.3. DETERMINATION OF FUGACITY OF A GAS 10.4. CALCULATION OF FUGACITY AT LOW PRESSURE 10.5. FUGACITY OF A GAS IN A GASEOUS MIXTURE 10.6. FUGACITY OF A LIQUID COMPONENT IN A LIQUID MIXTURE 10.7. PHYSICAL SIGNIFICANCE OF FUGACITY 10.9. LET US SUM UP 10.10. CHECK YOUR PROGRESS 10.11. LESSON - END ACTIVITIES 10.12. REFERENCES 10.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on fugacity t o the students. On successful completion of this lesson the student should have: * Understand the concept of fugacity. 10.1. INTRODUCTION: CONCEPT OF FUGACITY Making use of the free energy function G Lewis introduced the concept of fugacity for representing the actual behavior of real gases which is distinctly different from the behavior of ideal gases. We know that variation of free energy with pressure at constant temperature is given by equation Viz.

VP

G

T

=÷ø

öçè

æ

d

d ------ (1)

This equation is applicable to all gases whether ideal or non- ideal. If one mole of a gas is under consideration, then V refers to molar volume. For an ideal gas, the above equation may be written as

( )P

dPnRTdG T =

úú

û

ù

êê

ë

é

=

=

P

RTV

RTPVQ

------ (2)

and for n moles as,



107

( ) ( )PnRTP

dPnRTdG T ln== ------ (3)

Integration of equation (3) yields,

PnRTGG ln+= * ------ (4)

Where *G , the integration constant, is the free energy of n moles of ideal gas at temperature, when the pressure P is unity. Equation (4), evidently, gives the free energy of an ideal gas at temperature T and pressure P . Integration of equation (3) between pressures 1P a n d 2P at constant temperatureT ,

yields,

1

2ln2

1P

PnRT

P

dPnRTG

P

P

==D ò ------ (5)

The corresponding equation for 1 mole of the gas would be

1

2lnP

PRTG =D ------ (6)

Equations (4) and (6) are not valid for real gases since V is not exactly equal toP

RT.

In order to make these simple equations applicable to real gas, Lewis introduced a new function f is called fugacity function. It takes the place of P in equation (3), which, for real

gases, may be expressed as

( )f

dfnRTdG T ln= ------ (7)

And equation (7) may be represented as

fnRTGG ln+= * ------ (8)

Where *G is the free energy of n moles of real gas at temperatureT , when its fugacity happens to be 1. Thus fugacity is a sort of fictitious pressure which is used in order to retain for real gases simple forms of equation which are applicable to ideal gases. Equation (8), evidently, gives the free energy of a real gas at temperature T and pressure P at which its fugacity can be taken as f .

Equation (7) on integration between fugacities 1f and 2f , at constant temperatureT ,

yields,

1

2lnf

fnRTG =D ------ (9)



108

The corresponding equation for one mole of the gas would be

1

2lnf

fRTG =D ------ (10)

As discussed above, equations (9) and (10) are applicable to real gases. Eg: 1 Calculate the free energy change accompanying the compression of 1 mole of a gas at 57

℃ from 25 to 200 atm. The fugacities of the gas at 57 ℃ may be taken as 23 and 91 atm, respectively, at pressure 25 and 200 atm. Solution:

1

2lnP

PRTG =D

= 1mole (8.314 11 -- molJK ) 330 K atm

atm

25

200ln = 5702.8 J

for more accurate value, we should use the equation involving fugacities. Thus,

1

2lnf

fRTG =D

= 1mole (8.314 11 -- molJK ) 330 K atm

atm

23

91ln = 3730.0 J

10.2. FUGACITY AT LOW PRESSURES

The ratioP

f, where P is the actual pressure, approaches unity when P approaches zero,

since in that case a real gas approximates to ideal behavior. The fugacity function, therefore, may be defined as

10lim

=®P

fP

it

Evidently, at low pressures, fugacity is equal to pressure. The two terms differ materially only at high pressures. 10.3. DETERMINATION OF FUGACITY OF A GAS

Equation fnRTGG ln+= * for one mole of a gas, may be put as

fnRTGG ln+= * ------ (1)



109

Differentiation of eqn (1) with respect to pressure at constant temperature and constant number of moles of the various constituents, (i.e.) in a closed systems, gives

÷ø

öçè

æ==÷

ø

öçè

æ

P

fRT

P

G

T d

d

d

d ln ------ (2)

since

VP

G

T

=÷ø

öçè

æ

d

d

it follows that

RT

V

P

f=÷

ø

öçè

æ

d

d ln ------ (3)

Thus, at definite temperature, eqn (3) may be written as VdPfRTd =)(ln ------ (4)

Since one mole of the gas is under consideration, ‘V’ is the molar volume of the gas. Knowing that for an ideal gas,

P

RTV = ------ (5)

The quantitya , defined as departure from ideal behavior at a given temperature, is given by

VP

RT-=a ------ (6)

Eqn (6) is multiplied by dP throughout, we get

VdPP

dPRTdP -=a ------ (7)

Combining equations (4) and (6), we have

dPP

dPRTfRTd a-=)(ln

Both sides dividing by RT (or)

RT

dPPdfd a-= ln)(ln ------ (8)



110

Integrating eqn (8) between pressures ‘O ’ and ‘ P ’, we have

ò-=P

dPRTP

f

0

1ln a ------ (9)

Now ‘a ’ is given by eqn (6), can be determined experimentally, at different pressures. These values of ‘a ’ are then plotted against corresponding pressures, as shown in fig. The area under the curve between pressure e=0 and any given pressure ‘ P ’, yields the value of

the integral dPP

aò0

, as illustrated by the shaded portion in fig.

+ve area

-ve area

pressure 'P'

posi

tiv

en

egat

ive

Plot of VP

RT-=a vs. P for the determination of fugacity of the gas

Incorporating this value in eqn (9) the fugacity ‘ f ’ can be evaluated at any given

pressure ‘ p ’of the gas.

Since ‘a ’ of the departure from ideal behavior, can be both positive as well as negative, the area under the curve (with respect to the pressure axis) can be both positive as well as negative. Thus, the fugacity of the gas can be both less than or more than the pressure. As is

evident from fig the area and hence the value of )(0

dPP

aò is positive at low pressure and negative

at high pressures. Hence, in accordance with eqn (9) fugacity ‘ f ’ of the gas would be less than

the pressure ‘ p ’ at low pressures and more than the pressure at very high pressures. This is

borne out by the data given in the table (1) for nitrogen gas at various pressures at 0℃. Fugacity of nitrogen gas at various pressures at 0℃

Pressure atmosphere Fugacity atmosphere



111

50 48.9 100 96.7 200 194.2 400 424.4 800 1191 1000 1834

Fugacity of hydrogen gas at various pressures at 0℃

Pressure atmosphere Fugacity atmosphere

25 25.4 50 51.5 100 106.1 200 225.8 500 685 1000 1899

It may be recalled that for hydrogen and helium PV is greater than RT for all pressures at ordinary temperatures. Hence fugacity in these gases always remains greater than the pressure.

This is borne out by the data in table (2) for fugacity of hydrogen gas at 0℃ .



112

10.4. CALCULATION OF FUGACITY AT LOW PRESSURE It has been found that the experimental value of a at low pressure assumes almost

constant value. Under such conditions, therefore, eqn

( )ò-=P

O

dPRTP

fa

1ln Becomes

RT

P

P

fa=ln ------ (1)

Now at low pressure, since gases tend to be ideal, pf = or

1=P

f ------ (2)

Making use of the fact that xln is approximately equal to 1-x when x approaches unity we have,

1ln -=P

f

P

f

Hence, P

f

P

fln1+= ------ (3)

RT

Pa-= 1 ------ (4)

RT

PV=

------ (5) Hence,

RT

VPf

2

= ------ (6)

This equation is useful in calculating fugacity at moderately low pressures. Eg: For a VanderWaals gas, express the fugacity as a function of V, T, R and the VanderWaals constants.



113

Solution:

From eqn ( ) VdpfRTd =ln --- (i) for one mole of the gas Integrating eqn (1)

( ) VdpfdRT òò =ln ------ (ii)

It is advisable to change the variable on the right hand side of the eqn (ii) by resorting to integration by parts. Accordingly,

VdpPVVdP òò -= ------ (iii)

Now, for one mole of the VanderWaals gas,

2V

a

bV

RTP -

-= ------ (iv)

(Where V is the molar volume) hence, from eqns (ii), (iii) and (iv),

CdVbVRTRTdVV

a

bV

RTPVfRT +---=úû

ùêë

é-

--= ò )ln(ln

2 ------ (v)

Where C is the integration constant.

To evaluate C, we recall that pf ® as op ® .

Also, as op ® at constant temperature a®VT , so that ( ) VbV ®- a n d

01

®V

.Hence,

( ) CPRTRTRTCVRTRTPRT +-=+-= /lnlnln

= CPRTRTRTRT ++- lnln

RTRTRTC -=\ ln 0=PQ ------ (vi) Substituting for C in eqn (v), we get

( ) RTRTRTV

abVRTPVfRT -+÷

ø

öçè

æ---= lnlnln ------ (vii)

( ) ÷÷

ø

öççè

æ÷ø

öçè

æ-

-+-=

V

a

bV

RTRTRTPV ln ------ (viii)

Rearranging the Vander-Waals equation, after expansion, neglecting the term 2/Vab and later

the term 2/Va as well, we obtain



114

V

a

bV

RTbRTPV -

-=- ------ (ix)

Hence, RTVabVRTbVbf /2)/(ln)/(ln --+-= this is the required expression for fugacity.

10.5. FUGACITY OF A GAS IN A GASEOUS MIXTURE

Remembering that for one mole of a pure substance, the free energy (G) is identical with the chemical potential m eqn )(ln)( fnRTddG T = c for one mole of

)(ln ii fRTdd =m ------ (1)

and eqn fnRTGG ln* += may be written as

iii fRT ln+=*

mm ------ (2)

Where im is the chemical potential of the gaseous component i at its unit fugacity.

10.6. FUGACITY OF A LIQUID COMPONENT IN A LIQUID MIXTURE

The eqn

iii fRT ln+=*

mm

is valid not only for the fugacity of a gas in gaseous mixture but also for the fugacity of a pure liquid in a liquid mixture. This easily follows from the following discussion. As is well known, in the case of phase equilibria, the chemical potential of any given component is the same in all phases. Thus if there are three phases a, b, and c containing a component i then at equilibrium,

)()()( cba iii mmm ==

Consider, for example, a liquid i in equilibrium with its vapor. The chemical potential of

the liquid, gili )()( mm =

Now according to the eqn fnRTGG ln* += chemical potential of a vapor and a gas

may be written as

igigi fRT ln)()( += *mm ------ (1)

Where as before gi )(mis the chemical potential of the vapor (when its fugacity is equal to

1. therefore, chemical potential of a liquid may, evidently, be written as

ilili fRT ln)()( += *mm ------ (2)



115

Where li )( *m stands for the chemical potential of the liquid when its fugacity is equal to

1. It follows from the above discussion that the fugacity of a pure liquid would be the same as that of its vapor in equilibrium with it, at a given temperature. 10.7. PHYSICAL SIGNIFICANCE OF FUGACITY In order to understand the physical significance of the term fugacity, consider a system consisting of liquid water in contact with its vapor water molecule in the liquid phase will have a tendency to escape into the vapor phase by evaporation, while those in the vapor phase will have a tendency to escape into the liquid phase by condensation. At equilibrium, the two escaping tendencies will be equal. It is now accepted that each substance in a given state has a tendency to escape from that state. This escaping tendency was termed as fugacity. 10.9. LET US SUM UP In this lesson, we: Pointed out Ø Concept of fugacity Ø Fugacity at low pressures Ø Determination of fugacity of a gas Ø Calculation of fugacity at low pressure Ø Fugacity of a gas in a gaseous mixture Ø Fugacity of a liquid component in a liquid mixture Ø Physical significance of fugacity

10.10. CHECK YOUR PROGRESS 1. Explain the term fugacity. How is it related to chemical potential? 2. Explain why the fugacity of Helium or Hydrogen is always more than the pressure.



116

10.11. LESSON – END ACTIVITIES 1. How can you determine the fugacity of gas using (i) Graphical method and (ii) From equations of state? 2. What is the physical significance of fugacity? How does fugacity vary with temperature? 10.12. REFERENCES 1. J.C. Kuriacose and J. Rajaram, Thermodynamics, Shoban lal Nagin Chand & Co., Delhi. 2. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006-2007. 3. Gurdeep Raj, Advanced Physical Chemistry, Goel Publishing House, Meerut.



117

LESSON 11: THERMODYNAMICS AND NON-IDEAL SYSTEMS: ACTIVITY

CONTENTS 11.0. AIMS AND OBJECTIVES 11.1. INTRODUCTION: CONCEPT OF ACTIVITY 11.2. ACTIVITY COEFFICIENT 11.3. TEMPERATURE COEFFICIENT 11.4. REFERENCE STATES OR STANDARD STATES 11.5. RATIONAL AND PRACTICAL APPROACHES 11.6. DETERMINATION OF ACTIVITY OF SOLVENT AND SOLUTE FROM COLLIGATIVE PROPERTIES 11.7. LET US SUM UP 11.8. CHECK YOUR PROGRESS 11.9. LESSON - END ACTIVITIES 11.10. REFERENCES 11.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on activity to the students. On successful completion of this lesson the student should have: * Understand the concept of activity. 11.1. INTRODUCTION: CONCEPT OF ACTIVITY It may be pointed out that since the absolute value of free energy (or) chemical potential is not known, it is impossible to evaluate of a substance. This difficulty has been overcome by referring all free energy and chemical potential measurements for any given substance to a standard reference point. Let im be the chemical potential of a substance ‘i’ in pure state and let

if be its fugacity

\Equation iii fRT ln+=*

mm

May then be put as

00

ln iii fRT+=*

mm ------ (1)

Let im be the chemical potential of the same substance in some other state. Then

iii fRT ln+=*

mm ------ (2)

The difference between chemical potential of a substance in any state and that in the pure state is given by



118

÷÷ø

öççè

æ=-

0

0ln

i

iii

f

fRTmm Or

÷÷ø

öççè

æ+=

0

0ln

i

iii

f

fRTmm ------ (3)

We may introduce here a new term, activity a and define it as

0f

fa = ------ (4)

or for a substance i , as

0

i

i

f

fa = ------ (5)

Activity of a substance in any given state is thus defined as the ratio of the fugacity of the substance in that state to the fugacity of the same substance in the pure state. The equation 3, therefore, reduces to

aRTii ln0

+= mm ------ (6)

Let a system consisting of one mole of a substance change from a state in which its chemical potential (or) free energy, both being identical since we are dealing with one mole of the substance is im , to another state in which its chemical potential is 2m , the change in chemical

potential mD , is then given by

( )10

20

12 lnln aRTaRT +-+=-=D mmmmm

or ÷÷ø

öççè

æ=D

1

2lna

aRTm ------ (7)

Comparing the above equation with that for an ideal gas, viz.

÷÷ø

öççè

æ=D=D

1

2lnP

PRTG m ------ (8)

It is evident that in the case o real gases, activity replaces pressure. Thus activity of a gas like fugacity, serves as a thermodynamic counter part of a gas pressure. In case of solutions it serves as a counter part of concentration of the solute in the given solution.

11.2. ACTIVITY COEFFICIENT For an ideal gas, activity is numerically equal to its pressure (i.e.) Pa = For real gases, however, activity is only proportional to its pressure (i.e.) Pa µ or Pa g= where g is known as

the activity coefficient Eg:



119

The activity of 2.5 moles of a substance changes from 0.05 to 0.35. What would be the change in its free energy at 27 ℃ ? Solution: The change of free energy for one mole of a substance is given by the relation

÷÷ø

öççè

æ=D

1

2lna

aRTG

\For n moles, ÷÷ø

öççè

æ=D

1

2lna

anRTG = (2.5 mol) (8.314 11 -- molJK ) (300K) ÷

ø

öçè

æ

05.0

25.0ln

= 12133.65 J 11.3. TEMPERATURE COEFFICIENT Temperature coefficient of a chemical substance is defined as the ratio of rate constants of a reaction at two different temperatures separated by 100C. The two temperatures are generally taken as 350 and 250C. Thus, the temperature coefficient is expressed as

Temperature coefficient = 025

035

k

k

In general the temperature coefficient is expressed as

Temperature coefficient = t

t

k

k 010+

Where kt is the specific rate of the reaction at t0C and kt + 10 is the specific rate of the same reaction at t+100. 11.4. REFERENCE STATES OR STANDARD STATES It may be emphasized again that there is no means of finding absolute values of free energy (G) or Chemical potential (m) of any substance. It is necessary, therefore, to make such measurements with reference to the value obtained for some convenient though arbitrary reference state called standard state. The standard state for a gas at any given temperature is defined as that state in which the fugacity of the gas is unity. Since a = f/f0, it follows that if f0 = 1, a = f. Evidently, the standard state for a gaseous component is such that fi = ai =1 The standard state for a liquid is the pure state of the liquid at one atmosphere pressure at any given temperature. In this state of the liquid, f = f0 and a = 1.The standard state for a liquid component I in a liquid solution is the pure state of the component such that ai = 1.



120

The standard state for a solid is the pure state of the solid at one atmosphere pressure at any given temperature. In this state, its activity a = 1. 11.5. RATIONAL AND PRACTICAL APPROACHES The concentration of a solution is usually expressed as mole fraction, molality or molarity. The standard state chosen depends on the concentration unit used. If the choice is mole fraction, it is referred to as the rational system, whereas if molality or molarity (practical units) are used it is called the practical system. (A) RATIONAL SYSTEM If the mole fraction of the solute, x2, is taken as the measure of the concentration of the solute, Henry’s law being applicable to solutes, f2 = K x2, where K is the Henry’s law constant. It is therefore desirable to choose the standard state for the solute in such a way that in a dilute solution, the activity becomes equal to the mole fraction of the solute. Thus it is desirable that

12

2 ®x

a as 02 ®x ----- (1)

The fugacity-mole fraction relationship over the entire range of mole fraction, 0 to 1, can be represented as the solid line in Fig1. For the very dilute solution as 02 ®x , it is seen that the

actual curve merges with the Henry’s law line.

Since 0222 ffa = , Eq. 1 can be written as



121

122

2

02

2

0=÷÷

ø

öççè

æ=÷÷

ø

öççè

æ

®® xf

fLim

x

aLim

xx xx ----- (2)

Since Henry’s law is also applicable to the solute, in the very dilute solution, Lim [f2/x2] for solid line = limiting slope = K. For Henry’s law line (dotted) 02 ®x .

Kx

fLim

xx=÷

÷

ø

ö

çç

è

æ

®2

'

2

0. Therefore

[ ] [ ] KxfLimxfLimxx xx

==®®

2

'

20

220

----- (3)

Since f2 = Kx2, eqn. (3) can be written as

12

2

02

=úû

ùêë

é

® Kx

fLimx

----- (4)

If equations 2 and 4 should hold good simultaneously f2

0 =K, the Henry’s law constant for this solute. From Fig. 1., it is seen that this state can be found by extrapolating the dotted line to a concentration x2 = 1. From Henry’s law f2’ =Kx2, it is seen that when x2 = 1, f2’ = k. This fugacity is taken as the standard fugacity for the solute. It should be noted that the standard fugacity (f2

0) is a hypothetical quantity and is not equal to the fugacity (f2’) of the pure solute. The standard state for the solute is chosen as the hypothetical liquid solution at the given temperature and 1 atm total pressure, in which the mole fraction of the solute is unity and yet it behaves ideally, obeying Henry’s law. If this law is obeyed over the entire range of composition (x2 = 0 to 1).

22

20

222 xK

KxKfffa ==== ----- (5)

Thus as 12 ®x , a2 becomes unity. The activity at any other concentration will be equal

to x2, the mole fraction of the solute. If the activity of the solute, a2 = f2/f2’ is plotted against x2, the curve obtained (Fig.2) is similar to the one in Fig.1, since each value of the fugacity is divided by the same constant, f2

0. For an ideal solution, the slope of the curve should be unity (Fig.2.). The activity of the pure solute, a2 is seen to be different from a2

0 . For any mole fraction, xj the activity coefficient gx is aj/xj. In Fig.2, Y=a2 and since the slope of the dotted line is unity, xj =X. Thus gx =Y/X. For a solution behaving ideally over the whole range of concentration, the activity will always be equal to its mole fraction. For non-ideal solutions, the standard state has no reality and it is preferable to define the standard state in terms of a reference state. It is seen from Fig.2 that the activity coefficient gx becomes unity as 02 ®x . It is thus possible to choose the infinitely

dilute solution as the reference state, such that as 02 ®x , 1®xn or 22 xa ® .



122

(B) PRACTICAL SYSTEM Molality is more widely used to express concentrations than mole fraction. In very dilute solutions, molality is proportional to mole fraction. Henry’s law is valid under these conditions (i.e.) f2 =Km2. If f2 is plotted against m2, the Henry’s law constant, K, can be obtained from the limiting slope of the curve (I) (Fig.3.). The choice of the standard fugacity should be such that as

1,02

22 ®ú

û

ùêë

é®

m

am or

12

02

2

02

2

0 22

=úû

ùêë

é=ú

û

ùêë

é

®® mf

fLim

m

aLim

mm ----- (1)

Under such limiting conditions, Henry’s law is valid (i.e.) f2 =Km2 or

12

2

02

=úû

ùêë

é

® Km

fLimm

----- (2)

For the same region ( )02 ®m of the solid line curve (Fig.3) equations 1 and 2 should

hold good simultaneously. This is true when f20 =K. In Fig.3., this condition is realized by

finding the fugacity corresponding to m2=1. The interpolation is done on the Henry’s law plot (dotted line) since this law is obeyed by ideal systems under all conditions. The standard state of the solute is the state, which has the fugacity that the solute in a solution of unit molality would



123

have, if Henry’s law is obeyed at this concentration. With increasing dilution, a solute approaches ideal behavior specified by the Henry’s law. However it is misleading to say that the standard state of the solute is the infinitely dilute solution. In such a case f2=0. If f2

0 =0, the activity at any finite concentration a2 = f2/f2

0 , would be infinite.

A curve similar to that in Fig.3 can be obtained by plotting activity, a2 of the solute against molality, m2 (Fig. 4). Since the mole fraction scale has limits of 0 and 1, the choice of x2 =1 as a standard state, is quite natural. Theoretically molality has no upper limit, but in practice the upper limit is the solubility of the substance.



124

The choice of standard state, m2

0 = 1 mole/kg is arbitrary. The standard state is the hypothetical 1 molal solution obtained by extrapolating the Henry’s law line to m2 =1. If the concentration of the solute is expressed in molarity (c) the standard state is chosen as the hypothetical state obtained when Henry’s law plot is extrapolated to c2 = 1 mol/l. 11.6. DETERMINATION OF ACTIVITY OF SOLVENT AND SOLUTE FROM COLLIGATIVE PROPERTIES VAPOUR PRESSURE MEASUREMENTS (A) SOLVENT The activity of a constituent, j, in solution is given by fj/fj

0, where fj is its fugacity in the solution and fj

0 is the value at the standard state (pure liquid at the same temperature as the solution and at 1 atm total pressure). If the vapour pressures are sufficiently low, ideal behaviour may be assumed; The fugacities may therefore be replaced by the respective partial pressures (i.e.)

0jjj ppa = ----- (1)

Since the effect of external pressure on activity is negligible, Eq. – can also be written as

jjj pPa = ----- (2)



125

Here pj is the partial vapour pressure of the solvent in equilibrium with the solution in which its activity is aj, and p is the vapour pressure of the pure solvent at the same temperature and 1atm pressure i.e. at the standard state. This method is useful in determining the activity of the solvent in aqueous solutions, mixtures of organic liquids etc. If the mole fraction or the concentration of the solvent is known, the activity coefficient of the solvent can be calculated. (B) SOLUTE If the solute is completely miscible with the solvent, the standard state is chosen as the pure liquid. The activity may be determined by using Eq.

jjj pPa =

If the solute is sufficiently volatile to permit the determination of its vapour pressure over the solution. For dilute solutions it is preferable to use the infinitely dilute solution as the reference state. The activity is proportional to its fugacity, the value of the proportionality constant depending on the standard state used or reference state used. If 1/k is the proportionality constant for the chosen reference state, a2 =f2/k. In dilute solutions, as [ ] 1,0 222 ®® xax or

kxkaf 222 == as 02 ®x

For a dilute solution f2 = p2, the partial pressure of the solute. Hence kpa 22 » ----- (1)

The value of k can be obtained by utilizing the fact that at high dilutions, the activity of the solute, a2* is equal to the mole fraction x2*. If p2* is the partial pressure under this condition, kpxa *** 222 == or ** 22 xpk = ----- (2)

The activity of the solute a2, in the given solution is given by Eq.1. Substituting for k,

÷÷ø

öççè

æ==

*

*

2

22

22

x

pp

k

pa ----- (3)

The activity coefficient ( )

2xg being a2/x2, from Eq. 3.

( )

÷÷ø

öççè

æ

÷÷ø

öççè

æ

==

*

*

2

2

2

2

22

2

x

p

x

p

x

axg ----- (4)

A graphical evaluation of k is possible, by plotting p2/x2against x2 and extrapolating it to x2 = 0. Once k is known, by using Eq. 3 the activity can be calculated. From Eq. 4 the activity coefficient can be calculated.



126

11.7. LET US SUM UP In this lesson, we: Pointed out Ø Concept of activity Ø Activity coefficient Ø Temperature coefficient Ø Reference states or standard states Ø Rational and practical approaches Ø Determination of activity of solvent and solute from colligative properties

11.8. CHECK YOUR PROGRESS 1. Define activity and activity coefficient. 2. What are the applications of activity concept to solutions? 11.9. LESSON – END ACTIVITIES 1. Explain briefly the rational and practical approaches of activity. 2. How can you determine (i) the activity of solvent from colligative properties (ii) the activity of solute. 11.10. REFERENCES 1. J.C. Kuriacose and J. Rajaram, Thermodynamics, Shoban lal Nagin Chand & Co., Delhi. 2. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006-2007. 3. Gurdeep Raj, Advanced Physical Chemistry, Goel Publishing House, Meerut. LESSON 12: THERMODYNAMICS AND NON-IDEAL SYSTEMS: THIRD

LAW OF THERMODYNAMICS

CONTENTS 12.0. AIMS AND OBJECTIVES 12.1. INTRODUCTION: THIRD LAW OF THERMODYNAMICS 12.3. NERNST HEAT THEOREM 12.4. CONSEQUENCES OF THE NERNST HEAT THEOREM 12.5. THIRD LAW OF THERMODYNAMICS: PLANK’S FORMULATION 12.6. STATEMENT OF LEWIS AND RANDALL 12.7. UNATTAINABILITY OF ABSOLUTE ZERO 12.8. THERMODYNAMIC QUANTITIES AT ABSOLUTE ZERO 12.9. APPLICATIONS OF THIRD LAW OF THERMODYNAMICS 12.10. THERMODYNAMIC PROBABILITY AND STATISTICAL THERMODYNAMICS 12.10.1. DERIVATION OF THE BOLTZMANN ENTROPY EQUATION



127

12.11. APPARENT EXCEPTIONS 12.12. LET US SUM UP 12.13. CHECK YOUR PROGRESS 12.14. LESSON - END ACTIVITIES 12.15. REFERENCES 12.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on third law of thermodynamics to the students. On successful completion of this lesson the student should have: * Understand the third law of thermodynamics. 12.1. INTRODUCTION: THIRD LAW OF THERMODYNAMICS

In the earlier development of chemical thermodynamics, the difficult task was to ascertain a quantitative relation between GD and HD in a chemical reaction and to find out GD from thermal data (i.e.), HD . A preliminary discussion will now be given which will deal with this problem thereby ultimately leading to the Nernst heat theorem and third law of thermodynamics. 12.3. NERNST HEAT THEOREM

Nernst postulate “for a process in a condensed system that the value o f ( )

T

G

dd

D

approaches zero asymptotically as the absolute zero is approached”. In other words it means that GD and HD curves meet each other at a short region above the absolute zero and run coinciding with each other upto absolute zero. This behavior is being shown by full lines and not by dotted lines (Fig. 1).

Fig. 1. Variation of GD and HD with change of temperature. The dotted lines curve reveals that the two values, GD and HD not only become equal to each other at absolute zero but their approach to each other becomes rapid and not gradual.



128

The mathematical expression of Nernst’s postulate may be expressed analytically as

( ) ( ) 000

=D=D ®®

HGT

Lt

T

Lt

T d

d ----- (1)

This mathematical expression expresses the Nernst heat theorem. 12.4. CONSEQUENCES OF THE NERNST HEAT THEOREM We know second law of thermodynamics that

( )

ST

G

P

D-=îíì

þýüD

d

d and

( )

P

P

CT

HD=

îíì

þýüD

d

d (Kirchoff’s equation)

When we set up the conditions demanded by the Nernst heat theorem, we get

000 =D=D ®® PLt

TLt

T CS ------ (2)

The significance of the equation (2) is that the entropy change tends to approach zero and the difference between the heat capacities of products and reactants also tends to approach zero as the temperature is lowered towards the absolute zero. This is the statement of Nernst heat theorem. Nernst heat theorem holds good only in the case of solids. Since no gas exists at 0K. Nernst heat theorem cannot apply to gases. It can also not be applied to liquids. 12.5. THIRD LAW OF THERMODYNAMICS: PLANK’S FORMULATION The absolute value of entropy of a pure solid (or) a pure liquid approaches zero at 0K.

00 =D® SLtT ------ (1)

If equation (1) is assumed correct, Nernst equation (2) follows immediately for

( ) 00

0 =÷ø

öçè

æ

D

D=D-

®

®T

GS

Lt

T

Lt

T

d ------ (2)

pure solids and liquids. Plank’s formulation is also consistent with the treatment of entropy

which is introduced in statistical mechanics. Plank statement asserts that KS0 is zero for only

pure solids and liquids whereas Nernst assumed that his theorem was applicable to condensed



129

phases, including solutions. According to plank “solutions at K0 have positive entropy equal to entropy of mixing”. 12.6. STATEMENT OF LEWIS AND RANDALL Lewis and Randall modified the statement of the third law of thermodynamics as follow. “Every substance has positive entropy but at the absolute zero the entropy may become zero, and it dose so become in the case of perfect crystalline substance”. From the point of view of a chemist it is the most practical tool for calculating free energy. Consider for example the transition S (rhombic, 0 K) S (monoclinic 0 K)

KS0D from heat capacity measurement is zero. Hence, both rhombic and monoclinic sulphur are

assigned zero entropy at 0 K. Thus no difficulty is involved for calculating absolute entropy of sulphur at a temperature T. 12.7. UNATTAINABILITY OF ABSOLUTE ZERO It is impossible to attain absolute zero in a finite series of operations. For example, if we consider entropy change at constant pressure

dTT

C

T

DQdS PP == ------ (1)

(or) ¥=÷ø

öçè

æ=÷

ø

öçè

æ

®® T

P

T

SLt

T

Lt

T 00 d

d ------ (2)

Equation (2) shows the entropy of any substance should tend towards infinity as T approaches absolute zero. Since this cannot happen clear no infinite series of processes lead to absolute zero. 12.8. THERMODYNAMIC QUANTITIES AT ABSOLUTE ZERO (i) Equivalence of G and H

Since 0S for any substance is zero, according to Lewis and Randall statement, is follows

that for any substance KKKK HTSHG 0000 =-= (or) KK HG 00 = ------ (1)



130

(ii) PCD in a chemical transformation

From Gibbs-Helmholtz equation

PT

GTHG ÷

ø

öçè

æ D+D=D

d

d ------ (1)

÷÷ø

öççè

æ÷ø

öçè

æ D=D-

PT

GS

d

dQ

T

HG

T

G

P

D-D=÷

ø

öçè

æ D

d

d

000

00

=D-D

=÷ø

öçè

æ D

®® T

HG

T

G KK

Lt

T

Lt

T d

d ------ (2)

000 =D-D KK HGQ

A close examination of (eqn. 2) shows that the result comes out to be an intermediate form 0

0 at

T=0. To resolve an indeterminate expression we may apply the mathematical rule of differentiating numerator and denominator with respect to independent variable ‘T’. Thus we have

00

=÷ø

öçè

æ D-÷

ø

öçè

æ D

® T

H

T

GLt

T d

d

d

d ------ (3)

(or)

PLt

T

Lt

T

Lt

T

CT

H

T

GD=÷

ø

öçè

æ D=÷

ø

öçè

æ D®

®®

0

00 d

d

d

d ------ (4)

It follows then

00 =D® PLt

T C

many investigations D have shown that PCD does approaches zero at T approaches zero.

(iii). Limiting value of Cp and Cv For a reversible temperature change in a substance at constant pressure,

T

dTC

T

DQdS

pp

p =÷÷ø

öççè

æ=

On integrating at constant pressure



131

òò +==T

pS

T

dTCSdS

0

0

Where 0S is integration constant. An examination of equation (2)

000

00

=D-D

=÷ø

öçè

æ D

®® T

HG

T

G KK

Lt

T

Lt

T d

d

Shows that if pC has a finite value at T = 0. Integral òT

p

T

dTC

0

will not give a finite value since

the denominator T approaches zero. Hence ‘S’ would not be finite. But according to Lewis and

Randall statements, ‘S’ must be finite at all temperature. Hence 00 =® PLt

T C following is an

analogous procedure, it can be shown that 00 =® PLt

T C .

12.9. APPLICATIONS OF THIRD LAW OF THERMODYNAMICS The third law of thermodynamics is used to calculate the absolute entropy of the solids, liquids and gases at different temperatures and the entropy changes of chemical reactions and other processes. Let us describe these one by one. 1. DETERMINATION OF ABSOLUTE ENTROPY OF SOLIDS It is useful in calculating the absolute entropies of pure substance at different temperature by using thermal data.

We knowT

dTCdS p= , this on integrating between temperature limits 0 and ‘T’ K

becomes as follows

TdCSST

pT ln0

0 ò=-

On setting up the condition demanded by the third law (i.e.)., 00 =S at T = 0. We get

TdCST

pT ln0ò= ------ (1)

Where TS = absolute entropy of the solid at temperature T. One can calculate the integrant in eqn.

(1) graphically by plotting pC against T (or) pC against Tln and the area of curve so obtained

gives the value of integrant (i.e.), TS between 0=T to any desired temperature (Fig. 2).



132

Fig. 2. Determination of absolute entropy

2. EVALUATION OF ABSOLUTE ENTROPIES OF LIQUIDS AND GASES In all these cases we generally start from the crystalline (solid) state of a given substance at absolute zero when its absolute entropy is taken as zero and then supplying heat to this solid; it can be converted into the required state of the substance at a given temperature. The sum of the entropy changes involving these conversions will give the value of absolute entropy of the specific substance at the given temperature. In all these calculations, the

absolute entropy at Ko0 has been taken to be zero. When we have to calculate the absolute entropy of a liquid, the following steps are involved. (1). First of all measurements are made on the solid form at the melting point. The entropy of the solid at this temperature ( )1SD is given as follows.

( )T

dTCS

s

T

p

m

ò=D0

1 ------ (1)

( )

spC is the heat capacity of solid.

(2). The integrant in eqn. (1) can be obtained graphically. Changing one solid into the liquid state at the melting point Tm. The entropy of this process (entropy of fusion) is given by

m

f

T

HS

D=D 2 ------ (2)

Where fHD is the molar heat of fusion of the substance.

(3). Heating the liquid from its melting point (Tm) to this boiling point (Tb). Te entropy involved in this case is given by



133

( )T

dTCS

bT

lpò=D0

3 ------ (3)

Where ( )

lpC is the heat capacity of the substance in the liquid state. This can be evaluated by

plotting ( )lpC vs Tln between temperature Tm and Tb and noting the area below the graph as

described before. (4). Changing the liquid into the gaseous state at the temperature Tb. The entropy involved here

4SD is the entropy of vaporization and is given by

b

v

T

HS

D=D 4 ------ (4)

Where vHD is the heat of vaporization per mole of the substance.

(5). Heating the gaseous from Tb to the required temperature (i.e.), Co25 (298.15 K). The entropy involved in this process, 5SD is given by

( )T

dTCS

gTb

pò=D15.298

5 ------ (5)

where ( )gpC is the heat capacity of the substance in the gaseous state at constant pressure. Thus

the absolute entropy of the gas at Co25 TS is equal to the sum of the all the entropies listed

above that is

( ) ( ) ( )T

dTC

T

H

T

dTC

T

H

T

dTCSSSSSS

gTb

p

b

v

T

lp

m

f

s

T

pT

bm

òòò +D

++D

+=D+D+D+D+D=15.298

00

54321

(3). CALCULATION OF FREE ENERGY CHANGES THE REACTIONS We know the simplified form of Gibbs-Helmholtz eqn. is STHG D-D=D ------ (1) If the reactants and products are in their standard states equation (1) may be written as follows

000 STHG D-D=D The standard entropy changes may be determined experimentally while the values of standard entropy changes may be obtained from the literature incorporating these values in eqn. (1) the standard free energy of a reaction may be easily calculated. Ex:

Calculate the standard free energy change 0GD of the following reaction

kJHgCOgOgCO 84.282);()(2

1)( 22 -=D®+ o



134

Solution:

In this case kJH 84.282-=D o = - 282840 J.

The value of 0SD is given by the expression.

treacproducts SSS tan000 -=D

The standard entropy of )(2 gCO is 213.80 11 -- molJK while the values for CO (g) and O

(g) are 197.90 and 205.01 11 -- molJK respectively.

Hence, [ ] [ ] [ ]{ }22

0000OCOCO SSSS +-=D

= 213.80 – (197.90 + 102.50) = - 86.60 11 -- molJK

000 STHG D-D=D

= - 282840 – (298 K) (- 86.60 1-JK ) = - 257033.2 J. 12.10. THERMODYNAMIC PROBABILITY AND STATISTICAL THERMODYNAMICS 12.10.1. DERIVATION OF THE BOLTZMANN ENTROPY EQUATION The Boltzmann entropy equation, viz., WkS ln= is probably the most famous equation in statistical thermodynamics. Its derivation is naturally simple and appealing. Before we derive it, we must distinguish between two kinds of probabilities, viz., mathematical probability and thermodynamic probability. Mathematical probability is a ratio of the number of case favorable to the occurrence of an event to the total number of equally probable cases. This probability always lies between 0 and 1. Thermodynamic probability is the number of microstates corresponding to a given macro state when we are dealing with the distribution of molecules amongst an extremely large number of energy levels. This probabilityW is a very large number, tending to be infinite. Boltzmann suggested that entropy can be related to the thermodynamic probability W as ( )WfS = ------ (1)

Consider two systems A and B whose entropies and microstates (also called complexions) are AS and BAB andWWS , , respectively, when the two systems are combined,

BA SSS += ------ (2)

By definition, the probability of a thermodynamic state is proportional to the number of complexions (macro states) required to achieve it. Mathematically, we know that the total probability is the product of the probabilities of the independent events. This is also true of the complexions. Hence, BA WWW ×= ------ (3)



135

and ( )BABA WWfSS =+ ------ (4)

This relationship suggests that entropies are additive and probabilities are multiplicative. This can be true only if S is a logarithmic function of W , (i.e.), WkS ln= ------ (5) This can be Boltzmann entropy equation. Eqn. (5) gives the quantitative definition of entropy of disorder. In equation (5) W is defined as the total number of different ways in which a given system in the specified thermodynamic state may be realized. 12.11. APPARENT EXCEPTIONS Entropies calculated using the third law is called thermal entropies. However, the statistical entropies, calculated by the method of statistical mechanics are more rigorous. It is found that the thermal entropies are somewhat smaller than the statistical entropies, the deviation

ranging from .1 to 4.8 11 -- molJK . We thus conclude that entropies of substance (such as H2, D2, CO, NO, N2O, H2O, etc.) are not zero at 0 K, as the third law formulates, but are finite. These entropies are called residual entropies. The existence of residual entropy in a crystal at 0 K is presumably due to the alterative arrangements of molecules in the solid. Such arrangements are Fig. for CO and N2O:

CO CO CO CO CO CO CO CO NNO NNO NNO NNO NNO NNO NNO NNO (a)

CO CO OC OC CO OC OC CO NNO NNO ONN ONN NNO ONN ONN NNO (b)

Fig. Alternative molecular arrangements (a) perfect crystal (b) actual crystal

Since both the arrangements are equally likely, from the Boltzmann entropy equation (5), WkS ln= , with ANW 2= (where AN is Avogadro’s number), we find

that 2ln2ln2ln RkNNkS AA ===

= (8.324 11 -- molJK ) (2.303) (0.3010) = 5.76 11 -- molJK Since the residual entropies are found experimentally to be less than this value, it is evident that the two alternative orientations of the CO and N2O molecules in the solid state at 0 K are not completely random. For H2 and D2, too, thermal entropies at 0 K are less than the corresponding statistical entropies. The calculation of the statistical entropy assumes that there exists an equilibrium

between ortho and para H2 at all temperatures. The mixSD of ortho H2 and para H2 is found to be



136

18.37 11 -- molJK in the vicinity of 0 K. When this value is added to the thermal entropy (calculated from heat capacity measurements), the arrangement with the statistical entropy is very good. 12.12. LET US SUM UP In this lesson, we: Pointed out Ø Third law of thermodynamics Ø Nernst heat theorem Ø Consequences of the Nernst heat theorem Ø Third law of thermodynamics: Plank’s formulation Ø Statement of Lewis and Randall Ø Unattainability of absolute zero Ø Thermodynamic quantities at absolute zero Ø Applications of third law of thermodynamics Ø Thermodynamic probability and statistical thermodynamics Ø Derivation of the Boltzmann entropy equation Ø Residual entropy

12.13. CHECK YOUR PROGRESS 1. State third law of thermodynamics. 2. Explain the relation between probability and third law. 12.14. LESSON – END ACTIVITIES 1. Explain the Nernst heat theorem. How does it lead to the enunciation of the third l aw of thermodynamics? 2. Write short notes on (a) Thermodynamic quantities at absolute zero (b) Statistical meaning of third law and its apparent exceptions. 12.15. REFERENCES 1. J.C. Kuriacose and J. Rajaram, Thermodynamics, Shoban lal Nagin Chand & Co., Delhi. 2. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006-2007. 3. Gurdeep Raj, Advanced Physical Chemistry, Goel Publishing House, Meerut.



137

UNIT-VII

LESSON 13: QUANTUM STATISTICS 13.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on quantum statistics to the students. On successful completion of this lesson the student should have: * Understand the quantum statistics. 13.1. INTRODUCTION Statistical thermodynamics is concerned with the calculation of quantities such as heat capacity, entropy, etc., in terms of atomic and molecular parameters. The discipline which deals with the computation of the microscopic properties of matters from the data on the microscopic properties of individual atoms or molecules is called statistical mechanicals or statistical thermodynamics. 13.2. PROBABILITY The word is commonly used to indicate the like hood of an event taking place. For example: - suppose a coin is flipped or tossed only two results are possible, i.e.., either the head or the tail will show up. The probability of the head showing up is ½ this does not mean that if a person tosses a coin 10 times or tosses10 coins. The chance of getting head is 5. The probability of any event occurring is given by Number of cases favoring a given occurrence C Probability = ------------------------------------------------------- = ------ Total number of equality possible cases r 13.3. TYPES OF STATISTICS Different physical situations encounter in nature are described by three types of statistics, Viz., the Maxwell–Boltzmann or (M-B) statistics, the Bose–Einstein or (B-E) statistics and the Fermi–Dirac or (F-D) statistics. The M–B statistics, developed long before the advent of quantum mechanics, is called classical statistics whereas; the B-E and F–D statistics are collectively called quantum statistics. The characteristics of the three types of statistics are summarized as follows.

1) In M-B statistics: -



138

The particles are assumed to be distinguishable and any number of particles may occupy the same energy level. Particles obeying M-B statistics are called boltzmannons or maxwellons.

2) In B-E statistics:-

The particles are indistinguishable and any number of particles may occupy a given energy level. This statistics is obeyed by particles having integrated spin, such as hydrogen H2 deuterium D2, nitrogen N2, Helium – 4 (He4) and photons. Particles obeying B-E statistics are called bosons.

3) In F-D statistics:-

The particles are in distinguishable but only one particle may occupy a given energy level. This statistics is obeyed by particles having half–integrated spin. Example: The protons, electrons, helium 3(He3) and nitric oxide (NO) particles obeying F-D statistics are called fermions.

13.4. THERMODYNAMIC PROBABILITY ( )W

Consider a system of ‘ N ’distinguishable particles occupying energy levels

......,, 210 eee etc., the total number of arrangements for playing ‘ 0n ’ particles in the ground state

energy level 10 ,ne particles in the first excited energy level 21 ,ne particles in the second excited

energy level 2e and so on, is known as the thermodynamic probability,W of the given macro

state. It is in general a very large number. Our problem is to determine ‘W ’ (i.e.) to determine how many microstates correspond to a given macro state. It can be shown that ‘W ’ is given by

!

!

!!.......!!

!

321 ij n

N

nnnn

NW

Õ== ------ (1)

Where inN å=

In eqn (1) ‘ N ’ is the total number of particles and the summation is over all the energy levels. It is possible to realize a given energy level in more than one way (i.e.) more than one quantum state has the same energy. When this happens, the energy levels is said to be

degenerate. Let ig be the degeneracy (or) multiplicity of the energy level ie . This means that if

there is one particle in the thi energy level, there are ig ways of distributing it. For two particles

in the thi level, there are 2ig possible distributions. Thus, for in particles in the thi level there

are in

ig possible distributions. Hence the thermodynamic probability for the system of N

particles is given by



139

´Õ=i i

n

in

ig

NW!

! Constant ------ (2)

13.5. BOLTZMANN EXPRESSION FOR ENTROPY Boltzmann suggested that entropy can be related to the thermodynamic probability W as ( )WfS = ------ (1)

Consider two systems A and B whose entropies and microstates (also called complexions) are AS and BAB andWWS , , respectively, when the two systems are combined,

BA SSS += ------ (2)

By definition, the probability of a thermodynamic state is proportional to the number of complexions (macro states) required to achieve it. Mathematically, we know that the total probability is the product of the probabilities of the independent events. This is also true of the complexions. Hence, BA WWW ×= ------ (3)

And ( )BABA WWfSS =+ ------ (4)

This relationship suggests that entropies are additive and probabilities are multiplicative. This can be true only if S is a logarithmic function ofW , (i.e.), WkS ln= ------ (5) This can be Boltzmann entropy equation. Eqn. (5) gives the quantitative definition of entropy of disorder. In equation (5) W is defined as the total number of different ways in which a given system in the specified thermodynamic state may be realized. 13.6. STIRLING’S APPROXIMATION The derivation Consider the figure where xln is plotted against x . In addition the figure also contains steps both above and below the smooth curve with the values equal to 321 ln,ln,ln etc. The area

under the smooth curve from 1=x to Nx = is given by eqn (1)

ò +-=N

l

NNNxdx 1lnln ------ (1)

Now, the same area can be approximated by adding up the areas of the rectangles formed by these steps. If we choose the rectangles whose tops lie above the curve, we find for the area

Nln.....lnln 32 +++

)........32ln( N´´´=



140

!ln N= ------ (2)

x

lnx

1 2 3 4 5 6

On the other hand if we choose the rectangles below the curve, we get )1ln(.....lnlnln 321 -++++ N

[ ])1(........321ln -´´´´= N

NNN ln!ln)!1ln( -=-= ------ (3)

úû

ùêë

é=

×-´´=-´´=-

N

N

N

NNNN

!)1.......(21)1.......(21)!1(

The correct value of the area lies between those given by eqn (2) and (3). Thus, 1ln +- NNN á !ln N ------ (4)

(Actual area) (Top rectangle) NN ln!ln - á 1ln +- NNN

(Bottom rectangle) (Actual area) (Or) NNNNN ln1ln!ln ++-á

1ln)1(!ln +-+= NNNN ------ (5)

Combining (4) and (5) we have

1ln)1(!ln1ln +-+áá+- NNNNNNN ------ (6)

Now, if ‘ N ’ is very large )10( 23» , we can neglect unity as compared with ‘ N ’ without

appreciably affecting the calculated value for the upper and the lower limits of !ln N with this approximation, the upper and the lower limit become identical and therefore, NNNN -= ln!ln



141

The check:

N !ln N NNN -ln Error

102 363.7 360.5 -0.8% 103 5912.1 5907.8 -0.07% 104 82,108.9 82,103.4 -0.007% 105 10,51,999 10,51,293 -0.0006%

Note: Each additional power of ‘W ’ in ‘ N ’ decreases the % error by roughly a factor of 10.

Therefore % error is entirely negligible for 2310»N . 13.7. STATE OF MAXIMUM THERMODYNAMIC PROBABILITY (Equilibrium State) The state of maximum thermodynamic probability is also the state of maximum absolute probability and it is also the equilibrium state.

!

!

in

NW

Õ= ------ (1)

Taking logarithms on both sides

inNW ln!lnln å-= ------ (2)

(Or) )ln(lnln iii nnnNNNW -å--= ------ (3)

We would like to adjust the in to make Wln (and henceW ) a maximum while keeping constant

the total number of particles

å= inN ------ (4)

Diffentiating the eqn (4)

å == 0idndN ------ (5)

(Since N is constant) Similarly,

å =--

= 0)ln(ln

iii

ii

nnndn

d

dn

Wd

)ln( iii

i

nnndn

d-å=



142

)11

(ln -+-= åi

iin

nn

Þidn Cross multiply

(Or)

å =-= 0.lnln ii dnnWd ------ (6)

Now the restriction given in eqn (5) has to be incorporated into eqn (6). This is done by using Lagrange’s multiplies ‘a ’. Multiplying eqn (5) by a we get,

å == 0idndN aa ------ (7)

(6)- (7)

å =+-=- 0)(lnln ii dnndNWd aa ------ (8)

[In getting the right hand side of eqn (8) we have subtracted term by term to right hand side of eqn (6)]

(Or) 0)(ln =+å ii dnn a

Since idn represent the variations in in and hence are not necessarily zero, eqn (8)

requires the coefficient to be equal to zero. (i.e.) 0)(ln =+ ain

(Or) aa -=-= enorn ii )(ln ------ (9)

But å= inN

å -- ==\ aa seeN

s

Neni ==\ -a

s

Nni = ------ (10)

Eqn (10) says the state and maximum thermodynamic probability is one with equal number of particles in each one of the ‘ s ’ aspects 13.8. MAXWELL-BOLTZMANN DISTRIBUTION LAW If we consider our investigation to a closed system of independent particles, it would meet the following two requirements



143

(i) The total number of particles is constant (i.e.)

å == inN Constant ------ (1)

(ii) The total energy ‘U ’ of the system is constant

å == iinU e Constant ------ (2)

The constancy of the total number of particles implies that

å ==i

idndN 0 ------ (3)

The constancy of the total energy implies that

å == 0iidndU e ------ (4)

It is well known that a system of ‘ N ’ distinguishable particles occupying energy levels of 1e , 2e etc. The total number of arrangements for planning no particles in the ground state

energy level 0e , 1n particles in the first excited level 1e and so on is known as thermodynamic

probability ‘W ’ is given by

!!.........!!

!

210 innnn

NW = ------ (5)

On taking logarithms on both sides of eqn (5), we get

......)!ln!ln!(ln!lnln 210 +++-= nnnNW

å-= !ln!ln inN ------ (6)

Stirling’s formula can be used for the factorial of large numbers (i.e.) NNNN -= ln!ln ------ (7)

iiii nnnn -= lnln ------ (8)

å åå å -=-= Nnnnnnn iiiiii lnln!ln ------ (9)

Combining eqns (7), (8), & (9) we get

å +--= NnnNNNW ii lnlnln

å-= ii nnNNW lnlnln ------ (10)

Differentiating and bearing in mind that ‘ N ’ is constant, we get



144

å å--= iiii ndndnnWdn lnlnln ------ (11)

Now,

å å å === 0ln i

i

iiii dn

n

dnnndn ------ (11a)

Hence at equilibrium,

å =-= 0lnln ii dnnWd ------ (12)

Eqn (12) gives the change in Wln which results when the number of particles in each energy level is varied. If our system were open, then in would vary without restriction and the

variation would be independent of one another. It would then be possible to solve eqn (12) by

setting each of the coefficients of the idn in eqn (12) equal to zero. However our system is not

open but closed and since ‘ N ’ is constant, the value of idn are not independent of one another,

as is seen from eqn (11a). Again, the energy of the system is constant too. How, then can we solve eqn (12) subject to the constraints of eqn (1) and (2)?. The desired solution is obtained by applying the method of Lagrange’s undetermined multipliers. Multiplying eqns (3) & (4) by the arbitrary constants a and b respectively and

subtracting the eqn (12) we get,

0)(lnln =++-=-- å iii dnndUdNWd beaba ------ (13)

As the variables ....., ii dnn¶ are independent of each other, so that the eqn (13) is to hold good,

then the summation must be zero

0ln =++ iin bea ------ (14)

0¹idnQ

)(ln iin bea +-=

(Or) )(

iei

nbea +-

=

(Or) ieei

nbea --= ------ (15)

This equation is known as Maxwell-Boltzmann distribution law. For complete generality, it is now necessary to make extension of this distribution law. In deriving eqn (15), assumption is made that each energy level is said to be non-degenerate. It is possible that there may be a number of quantum levels of almost identical energies, and for this, a statistical weight factor ‘ ig ’ is introduced for level ie . Hence the eqn (15) becomes as

ieei

gi

nbea --= ------ (16)



145

(or) KTi

eei

gi

n

e

a-

-= ------ (17)

Where KT

1=b

(or) KTi

eei

gi

n

e

a-

-å=å ------ (18)

å = NniQ

å-

-=\ KTi

eei

gN

e

a ------ (19)

Dividing eqn (17) by (19) we obtain

KTi

eei

g

KTi

eei

g

Ni

n

e

a

e

a

--

å

--

= ------ (20)

This is the general form of Maxwell–Boltzmann distribution law. 13.9. EVALUATION OF LAGRANGE’S UNDETERMINED MULTIPLIERS Evaluation ofa : We now proceed to determine a and b , we know that

å--

= iei

gNbea

------ (1)

Since å= inN

å-

=-

iei

g

Ne

bea ------ (2)

Defining a quantity ‘ q ’ called partition function, as

å-

= iei

gqbe

------ (3)



146

We obtain q

Ne =-a ------ (4)

Accordingly, the Boltzmann distribution law becomes

q

iei

Ng

in

be-

=

Partition function ‘ q ’ is a quantity of immense importance in the statistical

thermodynamics. We shall see presently that by evaluating the partition function for a system we can calculate the value of any thermodynamic function for that system. Evaluation of b :

The constant b , can be evaluated as follows

We know that [ ´Õ=!

!

in

in

ig

NW constant]

å -+= )!lnln(!lnln iii ngnNW ------ (1)

Applying Stirling’s approximation to Nln and !ln N , we have

å +-+-=i

iiiii nnngnNNN )lnln(ln

åå -+=i

iii

ii nngnNN lnlnln ------ (2)

Taking log of equation

q

iei

Ng

in

be-

= ------ (20)

We have

iii gqNn be-+-= lnlnlnln

Substituting this value into eqn (2), we get

å å -+--+=i i

iiiii gqNngnNNW )lnln(lnlnlnln be

å å+å-+-+=i

iiiiii ngnqNNNgnNN belnlnlnlnln ------ (3)

UqN b++= ln

Substituting this result into the Boltzmann equation



147

UKqNKWKS b+== lnln ------ (4)

From the combined statement of the first and the second laws of thermodynamics, we know that for a simple system, PdVTdSdU -= ------ (5) At constant volume (V = constant; 0=dV ) TdSdU = ------ (6)

TV

S

V

1=÷

ø

öçè

æ

¶

¶\ ------ (7)

Differentiating eqn (4) with respect to U at constantV , we get

VVV U

KUKU

S

q

NK

U

S÷ø

öçè

æ

¶

¶++÷

ø

öçè

æ

¶

¶=÷

ø

öçè

æ

¶

¶ bb

VV U

KUKUd

dq

q

NK÷ø

öçè

æ

¶

¶++÷

ø

öçè

æ

¶

¶=

bb

b

b ------ (8)

Also, from eqn

å-

=i

iei

gqbe

------ (8a)

N

Uq

d

dq-=

b ------ (9)

Substitution of eqn (9) in eqn (8) results in cancellation of the first and the last terms, giving

bKU

S

V

=÷ø

öçè

æ

¶

¶ ------ (10)

Comparing eqns (7) and (10), we find that

KT

1=b

Hence, from eqn (8a) the molecular partition function q becomes

å-

=i

KTei

gq

ie

------ (11)



148

And the Maxwell-Boltzmann distribution equation (eqn 2a) becomes

q

KTi

ei

Ng

in

e-

= ------ (12)

From eqn (12) we can easily obtain the ratio of the populations, (i.e.) the number of particles in any two energy levels ie and je . Thus,

KT

ji

e

jg

ig

jn

in

)( ee --

= ------ (13)

13.10. BOSE-EINSTEIN STATISTICS In Maxwell-Boltzmann distribution, particles are distinguished from one another. It means that if two particles interchange their positions or energy states, a new microstate or complex ion would arise. But in Bose Einstein statistics one considers the particles to be indistinguishable (i.e.) on interchanging. The two particles between two energy states, no new microstate or complex ion will arise. Let us consider four particles distributed between two cells x and y . (i.e.) three in ‘ x ’

and one in ‘ y ’. If these four particles are distinguishable as assumed in Boltzmann statistics.

Four complex ion or microstates would arise. On the other hand, if four particles are distinguishable as assumed in Bose Einstein statistics. Only one complex ion or microstate would arise.

abc abd acd bcd aaa

d c b a a

CellX

CellY

W = 4 W = 1

Suppose that cell is further divided into four compartments by portions, the sections are representing the energy states. The possible distributions for these are shown in fig. (2). There will be 20 possible distributions for three particles in cell‘ x ’ and four ways of distributions of one particle in cell‘ y ’. Therefore, the total possible distributions for three particles in a cell‘ x ’

and one particle in cell ‘ y ’will be 80204 =´ .

Let us now attempt a general expression of the microstates under such conditions.



149

Suppose there are in particles with energy 1e in which there are g states of energy. We

shall need )1( -g positions to place the in particles in ‘ g ’sections, each sections corresponding

to an energy state. It means that ‘ g ’ is the degeneracy of the level. Now the question arises how

to distribute in particles into ‘ g ’ sections without any restriction. Then the permutations of in

particles and )1( -g partitions simultaneously will be given by )!1( -+ gni . However this also

includes permutations of in particles amongst themselves and also )1( -g particles amongst

themselves because both these groups are internally indistinguishable. Hence the actual number in which in particles may be allocated in ‘ g ’ states is given by

aaa

aaaaaa

aaa

aaaa aa

aa

aa aa

aa

aa

aa

aa aa aa

aaa

aa

a

aa

aa

aa

aa

aa

aa

aa

aa

aa

aa

aa

Cell: X

Cell: Y

W = 20

W = 4

)!1(!

)!1(

-

-+

ii

i

gn

gn ------ (1)

As in the case of Maxwell-Boltzmann statistics, we assume that in the present case also the total number of particles is constant and the total energy of the system is also constant.

(i.e.) å == inN Constant ------ (2)

å == iinU e Constant ------ (3)

Thus, the thermodynamic probability ‘W ’ for the system of ‘ N ’ particles (i.e.) the

number of ways of distributing ‘ N ’ particles among various energy levels is given by



150

´-

-+Õ=

)!1(!

)!1(

ii

ii

gn

gnW Constant ------ (4)

Taking logarithm on both sides of eqn (4) we get

+----+= å )!1ln(!ln)!1(lnln iiii gngnW Constant ------ (5)

Here, too, since in a n d ig are very large numbers, we can invoke Stirling’s

approximation viz. xxxx -= ln!ln , neglecting unity Compared to in and ig , we get,

iiiiiiiiiiii gngnnngngngnW +-+-+-++= å lnlnln)(ln

iiiiiiiii ngnngngnW lnln)ln()(ln --++= å ------ (6)

Where we have set iiii gngn +=-+ 1 and ii gg =-1 . Since in is very large, it can be treated as

a continuous variable.

Differentiation of eqn (6) with respect to in and setting the differential equal to zero

gives for the most probable thermodynamic state of the system.

[ ] 0)ln(lnln =¶×+-å=¶ iiii ngnnW

(or) 0ln =¶úû

ùêë

é

+å i

ii

i ngn

n ------ (7)

From eqns (2) and (3)

å =¶=¶ 0inN ------ (8)

å =¶=¶ 0ii nU e ------ (9)

Applying the method of Lagrange’s undetermined multipliers )&( ba to eqns (8) and (9)

respectively we get

0ln =úû

ùêë

é++

+å ii

ii

i dngn

nbea

Each term of the summation is equal to zero

0ln =úû

ùêë

é++

+ii

ii

i dngn

nbea



151

Since the variations in¶ are independent of one another, )0( ¹¶ in it means that the

expression within the square bracket would be zero. Hence,

0ln =+++

i

ii

i

gn

nbea

i

i

i

n

gbea +=ú

û

ùêë

é+1ln

ie

in

ig bea +

=úú

û

ù

êê

ë

é+1

(or)

1)(

-+

=ie

ig

ni bea

This equation is known as Bose-Einstein law of distribution. 13.11. ENTROPY OF BOSE-EINSTEIN GAS Unlike in the case of Maxwell-Boltzmann statistics, in both Bose-Einstein and Fermi-Dirac statistics the indistinguishability of particles is assumed right in the beginning and the distribution laws described. Hence, no correction needs to be made at a latter stage. This is evident from the following derivation of the expression for the entropy of a Bose-Einstein gas. The thermodynamic probability of a Bose-Einstein system is

´-

-+Õ=

)!1(!

)!1(

ii

ii

gn

gnW Constant ------ (1)

Maximization of W leads to the Bose-Einstein distribution law

1)(

-+

=ie

ig

ni bea ------ (2)

Now, maxlnWKS = ------ (3)

Substituting eqn (1), after neglecting unity in comparison to ig , into eqn (3) gives

( )( )å --+=i

iiiiii ngngKS !ln!ln!ln

Application of Stirling’s approximation results in

( ) ( ) ( ){ }å +-+-+-++=i

iiiiiiiiiiii nnngggngngngKS lnlnln



152

or åþýü

îíì

÷÷ø

öççè

æ ++÷÷

ø

öççè

æ +=

i i

iii

i

iii

g

ngg

n

ngnKS lnln

åþýü

îíì

÷÷ø

öççè

æ++÷÷

ø

öççè

æ+=

i

ii

i

ii

g

ng

n

gnKS 1ln1ln

from equation (2) i

i

i

n

gbea +=÷÷

ø

öççè

æ+1ln

( )åþýü

îíì

÷÷ø

öççè

æ+++=\

i i

iiii

g

ngnKS 1lnbea

( )å

ïï

þ

ïï

ý

ü

ïï

î

ïï

í

ì

÷÷

ø

ö

çç

è

æ+++=

i

in

ig

igi

n

ig

iinKS 1lnbea ------ (4)

Let us now denote i

i

g

n as Z . Thus the argument of logarithm in equation (4) becomes

( ) zZ1

1+ . The function ( ) zZ1

1+ approaches, the base of natural logarithm, as its limit when Z

approaches zero. If ii Ng ññ , it follows that Z will be small. Accordingly we can say that

ei

ni

g

igi

n»

÷÷

ø

ö

çç

è

æ+1 , since for higher translational energy levels Z will be small. Thus,

( )å ++= ei

KnS i lnbea

or nkKnKS ++= bea ------ (5)

Equation (5) is identical to that obtained from Boltzmann statistics after correcting the latter for the indistinguishability. 13.12. FERMI-DIRAC STATISTICS In the Boltzmann or in the Bose-Einstein statistics, no restriction was made to the number of particles present in any energy state. But in applying Fermi-Dirac statistics to particles like electrons, the Pauli’s exclusion principle is taken into consideration; (i.e) two electrons (particles) in an atom cannot possess the same energy level. In simple words, it implies that not more than one particle can be assigned to a particular energy state.



153

In the example consider earlier (in Bose Einstein statistics), in which four particles were distributed in two cells, three in the cell ‘ x ’ and one in ‘ y ’, one will have a distribution as in

figure.

aa

aa

aa

aa

aa

aa

aa

aa

Cell: Y

W = 4

W = 4

Cell: X

When one applies the Fermi-Dirac statistics the total number of possible arrangements, with the restriction that one particle can only occupy a compartment is )44( ´ (i.e.) 16.

Consider that the in particles are distributed among the ig states )( ii gn á

where ig, as

before, is the degeneracy of the thi energy level. Imagine that the particles are distinguishable. This implies that the first particle may be placed in any one of the ig states and for each one of

these choice, the second particle may be placed in any one of the remaining )1( -ig states, and so

on. Thus the number of arrangements is given by the expression

Number of arrangements )!(

!

ii

i

ng

g

-=

Since, however the particles are indistinguishable, the above expression has to be divided by the possible number of permutations of in particles, viz, !in . Hence, the number of

arrangements of ‘ in ’ particles in the thi energy level is given by the expression[ ])!(!

!

iii

i

ngn

g

-

Thus, the thermodynamic probability ‘W ’for the system of ‘ N ’ particles (i.e.) the number of ways of distributing ‘ N ’ particles among the various energy level is given by

´-

Õ=

)!1(!

!

ii

i

gn

gW Constant ------ (1)

Taking logarithms of both sides of eqn (1) we have

[ ]++---= åi

iiii ngngW )!ln(!ln!lnln Constant ------ (2)

Applying stiriling’s formula to evaluate the factorial



154

NNNN -= ln!ln [ ]))ln()()ln(lnln iiiiiiiiiiii ngngngnnngggW --------å=

iiiiiiiiiiii ngngngnnnggg -+---+--å= )ln()(lnln

)ln()(lnln iiiiiiii ngngnngg ----å= ------ (3)

Differentiating eqn (3) with respect to in

iii

ii

iii

i

i dnngng

ngnn

nWd úû

ùêë

é--

---+-å= )ln(

)(

1)(ln

1ln ------ (4)

[ ]å --i

iiiii dnngn )ln(ln ------ (5)

The maximum value of ‘W ’ is obtained by equating Wd ln to zero

[ ] 0)ln(ln =--å iiii dnngn ------ (6)

But

å == 0idndN ------ (7)

å == 0idndN aa

å == 0ii dndU e ------ (8)

å == 0ii dndU beb

Multiplying eqn (7) by a and eqn (8) by b and adding to eqn (6) we get,

[ ] 0)ln(ln =++--å iiiii dnngn bea ------ (9)

Each term of the summation is equal to zero ( 0¹idnQ )

Since the variation in¶ are independent of one another, hence

[ ] 0)ln(ln =++--å iiiii dnngn bea ( 0¹idnQ )

[ ]iiii ngn bea ++-- )ln(ln = 0

i

ii

i

ng

nbea --=

-ln ; (or) ie

in

ig

in bea --

=-

(or) )(

ie

in

in

ig bea +

=úú

û

ù

êê

ë

é - ; (or) 1

)(+

+=

úú

û

ù

êê

ë

éie

in

ig bea



155

(or)

1)(

++

=ie

gi

n i

bea

This is known as Fermi-Dirac distribution. 13.13. ENTROPY OF FERMI-DIRAC GAS The thermodynamic probability of a Fermi-Dirac system is

( )

´-

= Õiii

i

nng

gW

!

!Constant ------ (1)

Maximization of W leads to the Fermi-Dirac distribution law

1)(

++

=ie

ig

ni bea ------ (2)

Now, maxlnWKS = ------ (3)

Substituting eqn (1), after neglecting unity in comparison to ig , into eqn (3) gives

( )( )å -+-=i

iiiiii nnggKS !ln!ln!ln

Application of Stirling’s approximation results in

( )( ) ( ) ( ){ }å +-++----=i

iiiiiiiiiiii nnnngngnggggKS lnlnln

or åþýü

îíì

÷÷ø

öççè

æ --÷÷

ø

öççè

æ -=

i i

iii

i

iii

g

ngg

n

ngnKS lnln

åþýü

îíì

÷÷ø

öççè

æ--÷÷

ø

öççè

æ-=

i

ii

i

ii

g

ng

n

gnKS 1ln1ln

from equation (2) i

i

i

n

gbea +=÷÷

ø

öççè

æ-1ln

( )åþýü

îíì

÷÷ø

öççè

æ--+=\

i i

i

i

iii

g

nn

gnKS 1lnbea



156

( )å

ïï

þ

ïï

ý

ü

ïï

î

ïï

í

ì

÷÷

ø

ö

çç

è

æ--+=

i

in

ig

igi

n

iinKS 1lnbea ------ (4)

Let us now denote i

i

g

n as Z . Thus the argument of logarithm in equation (4) becomes

( ) zZ1

1- . The function ( ) zZ1

1- approaches, the base of natural logarithm, as its limit when Z

approaches zero. If ii Ng ññ , it follows that Z will be small. Accordingly we can say that

e

in

ig

ig

in 1

1 »÷÷

ø

ö

çç

è

æ- , since for higher translational energy levels Z will be small. Thus,

å ÷÷ø

öççè

æ÷ø

öçè

æ-+=

eiKnS i

1lnbea

or nkKnKS -+= 'bea ------ (5)

Equation (5) is identical to that obtained from Boltzmann statistics after correcting the latter for the indistinguishability of particle. 13.14. LET US SUM UP In this lesson, we: Pointed out Ø Probability Ø Types of statistics Ø Thermodynamic probability ( )W

Ø Boltzmann expression for entropy Ø Stirling’s approximation Ø State of maximum thermodynamic probability (equilibrium state) Ø Maxwell-Boltzmann distribution law Ø Evaluation of Lagrange’s undetermined multipliers Ø Bose-Einstein statistics Ø Entropy of Bose-Einstein gas Ø Fermi-Dirac statistics Ø Entropy of Fermi-Dirac gas

13.15. CHECK YOUR PROGRESS 1. Derive the expression of Bose-Einstein statistics. 2. Derive Stirling’s approximation.



157

13.16. POINT FOR DISCUSSION 1. Maximizing the thermodynamic probability of a macro state and invoking Lagrange’s undetermined multipliers derive the expression for Maxwell-Boltzmann statistics. 2. Derive the expression for Fermi-Dirac statistics. 13.17. SOURCES 1. J.C. Kuriacose and J. Rajaram, Thermodynamics, Shoban lal Nagin Chand & Co., Delhi. 2. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006-2007. 3. Gurdeep Raj, Advanced Physical Chemistry, Goel Publishing House, Meerut.

UNIT-VIII

LESSON 14: PARTITION FUNCTION 14.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on partition function to the students. On successful completion of this lesson the student should have: * Understand the partition function.

14.1. INTRODUCTION: DEFINITION AND JUSTIFICATION OF NOMENCLATURE The partition function may be defined as the sum of the probability factors for different energy states or more conveniently it can be stated as the way in which the energy of a system partitioned among the molecules constituting the system.

It is expressed as kT

i

i

egqorQe

-S=)(

where gi is the statistical weight factor and is equal to the degree of degeneracy, (i.e.), the number of super- imposed energy levels k is the Boltzmann constant and equals to the ratio of the gas



158

constant R to the Avogadro’s number, eI is the energy of the quantum state in excess of the lower possible value and T is the temperature on Calvin scale. 14.2. MICROCANONICAL AND CANONICAL ENSEMBLES MICROCANONICAL ENSEMBLE It is a collection of a large number of essentially independent assemblies each of which possesses the same energy E, volume V and the number of systems N. For sake of simplicity, it is assumed that all the systems are of the same type. Such an ensemble in which the individual assemblies are separated by rigid and well insulated walls so that E,V and N for a particular assembly are not affected by the presence of other assemblies. CANONICAL ENSEMBLE It is defined as a collection of a large number of independent assemblies, having the same temperature T, volume V, and number of identical system, N. As all the assemblies possess the same temperature, T, it means that one could bring them in thermal contact with each other and also a large heat reservoir at the same temperature T. Thus, in canonical ensemble, systems can exchange energy but not particles. The canonical ensemble in which the individual assemblies are separated by rigid, impermeable but conducting walls. 14.3. RELATION BETWEEN THE TOTAL PARTITION FUNCTION OF A MOLECULE AND THE SEPARATE PARTITION FUNCTION

Consider a set of energy levels, represented by ie with degeneracies. The partition

function is defined by

å-

=i

KTei

gq

ie

------ (1)

If we substitute translational energies into this equation, we obtain the translational partition function transq . If we substitute rotational or vibrational energies, we get rotq or vibq .

Actually there is only one partition function for a molecule, namely, the one obtained by putting into the equation (1) all combinations of allowed energies and their corresponding degeneracies.Thus

å

-

×=

levelsallenergy

KTtottot

tot

egqe

------ (2)

Now,

elecvibrottrtot eeeee +++=

Most of the time zero (if all energies measured with respect to ground electronic level)

And elecvibrottrtot ggggg ×××=



159

KT

valueallenergy

elecvibrottr

elecvibrottr

eggggq)( eeee +++

-

å ×××=\

KTKTKTKT

elecvibrottrtot

elecvibrottr

eeeeggggqeeee

----

å ×××= ....

÷÷÷

ø

ö

ççç

è

æ

×÷÷÷

ø

ö

ççç

è

æ

×÷÷÷

ø

ö

ççç

è

æ

×÷÷÷

ø

ö

ççç

è

æ

×= åååå----

levelsallelec

KTelec

levelsallvib

KTvib

levelsallrot

KTrot

levelsalltra

KTtrtot

elecvibrottr

egegegegqeeee

.

Define

å-

×=

levelstra

KTtrtr

tr

egqe

å-

×=

levelsallrot

KTrotrot

rot

egqe

å-

×=

levelsallvib

KTvibvib

vib

egqe

å-

×=

levelsallelec

KTelecelec

elec

egqe

14.4. RELATION BETWEEN MOLECULAR PARTITION FUNCTION (q) AND CANONICAL PARTITION FUNCTION (Q) Consider a system of N non- interacting molecules. The Hamiltonian for such a system is the sum of separate term for the individual molecule and there will be no interaction terms.

121ˆ............ˆˆˆ

NHHHH +++= ------ (1)

The energy of such a system is the sum of the energies of all the N individual molecules

)()(2

)(1 .......... i

Nii

iE eee ++= ------ (2)



160

Where, )(1

ie is the energy of molecules 1 in a system with total energy ET . The canonical partition

function is defined as,

å-

×=i

KTi

i

egQe

------ (3)

Where the sum is over all the energy levels i of the system. Now

)()(

2

)(

1 .........i

gi

N

iiggg= ------ (4)

(2) and (4) in (3)

å÷÷

ø

ö

çç

è

æ +-

×ºi

KTi

N

ii

iN

ii

egggQ

)()(2

)(1 .....

)()(

2

)(

1 .....

eee

å åå ÷÷ø

öççè

æ÷÷ø

öççè

æ×÷÷

ø

öççè

æº ---

i i

iNi

Ni

ii

ii

KTeg

KTeg

KTegQ

)()(

)(2)(

2

)(1)(

1 .....eee

If all the molecules in the system are of the same kind, it is no larger necessary to distinguish their energy levels and \ equation (5) reduces to

úû

ùêë

é÷÷ø

öççè

æúû

ùêë

é÷÷ø

öççè

æ×ú

û

ùêë

é÷÷ø

öççè

æ= ååå ---

i

ii

i

ii

i

ii

KTeg

KTeg

KTegQ

)(1)(

2

)(1)(

2

)(1)(

1 .....eee

Or

N

i

ii

KTegQ

úúû

ù

êêë

é

÷÷

ø

ö

çç

è

æ= å -

)()(

1 .e

Now, since the molecular partition function is defined as

å ÷÷

ø

ö

çç

è

æ= -

i

ii

KTegq

)()(.

e Our last equation becomes

NqQ =

14.5. TRANSLATIONAL PARTITION FUNCTION ( transq )



161

For a particle of mass m, moving in an infinite three-dimensional box of sides, a, b and c, assuming that the potential is zero within the box, the energy levels obtained by the solution of the Schrödinger equation are given by the expression:

å×

-

×=

levelstrall

KTtrtr

tr

egqe

------ (1)

But

)(8

222

2

2

nznynxma

htr ++=e ------ (2)

Assuming the vessel to be cubical

2

22

2

22

2

22

888 ma

hnz

ma

hny

ma

hnxtr ++=e ------ (3)

(3) in (1)Þ

KTma

hnz

KTma

hny

levelstrall

KTma

hnx

ztrytrxtrtr eeegggq2

22

2

22

2

22

888,,,

--

×

-

×××= å

Or

×÷÷÷

ø

ö

ççç

è

æ

÷÷÷

ø

ö

ççç

è

æ

÷÷÷

ø

ö

ççç

è

æ

=-

×

-

×

-

×ååå KTma

hnz

levelsztrall

ztrKTma

hny

levelsytrall

ytrKTma

hnx

levelsxtrall

xtrtr egegegq2

22

2

22

2

22

8

,,

8

,,

8

,,

ztrytrxtrtr qqqq ,,, ××= ------ (4)

ztrytrxtr qqq ,,, ==

Let us evaluate xtrq ,

÷÷÷

ø

ö

ççç

è

æ

=-

×å KTma

hnx

levelsxtrall

xtrxtr egq2

22

8

,,, Where xtrg , =1

Since the translational energy levels for all practical purposes, are continuous, the summation in the above definition can be replaced by integration.



162

ò¥

-

=0

8, .

2

22

xKTma

hnx

xtr dneq

Standard integral= ò¥

- Õ=

0 2

1

2

1

2

2

a

e ax

Thus

2

1

8

2

1

,

2

2

2÷÷

ø

ö

çç

è

æ

Õ=

-KTma

h

xtr

e

q

2

1

2

2

,

8

2

1÷÷ø

öççè

æ Õ=

h

KTmaq xtr = a

h

mKT×÷

ø

öçè

æ Õ 2

1

24

8

ah

mKTq xtr ×÷

ø

öçè

æ Õ=

2

1

2,4

2 ------ (5)

Substituting (5) in (4) and remembering that all the individual comments are equal, we have

3

2

3

24

2a

h

mKTq tr ×÷

ø

öçè

æ Õ= (Or)

Vh

mKTq tr ×÷

ø

öçè

æ Õ=

2

3

24

2 RTPV = ;

P

RTV =

14.6. ROTATIONAL PARTITION FUNCTION ( rotq ) - DIATOMIC MOLECULES-

RIGID ROTORS The simplest system that undergoes rotational motion is a diatomic molecule. The rotational energy levels of a rigid diatomic rotor (i.e., a rotor whose inter-nuclear distance remains fixed during rotation), obtained by solving the Schrödinger wave equation are,



163

å¥

=

-

×=0J

KTJrot

J

egqe

------ (1)

Where

I

JJJ

2

)1( 2h+=e J = 0,1,2…..

and )12( += Jg J [degeneracy or statistical weight of the Jth level]

å¥

=

+-

+=\0

2

)1( 2

).12(J

IKT

JJ

rot eJqh

------ (2)

For quantum mechanical rigid rotor equation (2) has to be evaluated partially to get rotq .

If, however, we assume the rotation energy levels to be almost continuous (which will be true for large diatomic molecule at fairly high temperatures) the summation in equation (2) can be replaced by integration.

dJeJq IKT

JJ

rot ò¥

- +

+=0

2

2)1(

)12(h

------ (3)

Now, J (J+1) = J2 + J

( )[ ]

121

+=+

JdJ

JJd or ( )[ ] dJJJJd )12(1 +=+

\Equation (3) becomes,

)]1([).1(0

2

2

++ò¥

-

JJdJJeq IKTrot

h

Recall: òò¥

-¥

-

-=

00

1 axax ea

dxe

¥+-

-

úúû

ù

êêë

é -=

o

IKT

JJ

rot eIKT

q 2

)1(

2

2

2h

h



164

2

2

h

IKTq rot =

For homonuclear diatomic molecule expression for the partition function becomes,

2

2

hs

IKTq rot =

where s is called the symmetry number has a value of 2. 14.7. VIBRATIONAL PARTITION FUNCTION: DIATOMIC MOLECULES – HORMONIC OSCILLATOR For diatomic molecule vibrating as a simple harmonic oscillator (S.H.O), the vibrational energy levels, obtained by the solution of the Schrödinger wave equation are given by

nhVEvib2

1+= ------ (A)

where n is the vibrational frequency and V is the vibrational quantum number which has the values V = 0,1,2,3,....... The energy levels are non-degenerate, (i.e.) the degeneracy, g is unity.

iegq ii

be-S= ------- (B)

using equations A and B the vibrational partition function of the diatomic molecule is given by

å¥

=

-

×=0V

KTVvib

V

egqe

------ (1)

where, ne hVV )2

1( += V = 0,1,2…..

mpn k

2

1=

1=Vg :

å¥ +

-=0

)2

1(

.1KT

eqhV

vib

n

Let, xKT

h=

n



165

å¥

+-=0

)2

1( xVeq vib

Term by term,

.....25

23

2 +++=--- xx

x

eee

.....)1( 22 +++= --- xxx

eee

Since, x

x

xx

e

eee

-

-

--

-=+++

1.....)1(

22

We have, x

x

vibe

eq

-

-

-=

1

2

Or KT

h

KTh

vib

e

eq

n

n

-

-

-=

1

2

14.8. ELECTRONIC PARTITION FUNCTION Though it is possible in principle to solve the Schrödinger equation for the electronic states and energies of a molecule, it is more convenient to obtain this information from the spectroscopic data. For most of the molecules, the excited electronic energy levels lie so for above the ground state compared with kT (a typical excited state value being greater than

JeV 191032 -´» ) that all the molecules may be considered to be in the ground state at ordinary temperatures. Thus, contributions to the electronic partition function arising from excited electronic states may be neglected. The electronic partition function is given by

å-

ºevelsnicenergylallelectro

KTeleele

elec

egqe

KTele

KTeleele

eleele

egegq1,0,

1,0,

ee--

+= ------ (1)

Taking the lowest level as our reference level that is,



166

00, =elee we have, KT

eleeleele

ele

eggq1,

1,0, 1e

-+= ------ (2)

There is no general formula for the elee . The series is added term by term using spectroscopically

observed electronic energies.

For nearly all diatomic molecules, elee 1 is very much greater than KT at room

temperature and therefore all terms in equation (2) after the first term contribute negligibly (unless the temperature is K000,10ñ ). Therefore,

0,eleele gq = ------ (3)

For most diatomic molecules the ground electronic level is non-degenerate (ie)

0,eleele gq = = 1 an important exception is O2. For O2 , 30, =eleg due to spin

degeneracy. Another exception is NO, which has an odd number of electron 20, =eleg ,

due to the two possible orientations of the spin of the unpaired electron. 14.9. LET US SUM UP In this lesson, we: Pointed out Ø Introduction Ø Justification of nomenclature Ø Microcanonical and canonical ensembles Ø Relation between the total partition function of a molecule and the separate partition

function Ø Relation between molecular partition function (q) and canonical partition function (q)

Ø Translational partition function ( transq )

Ø Rotational partition function ( rotq ) - diatomic molecules-rigid rotors

Ø Vibrational partition function: diatomic molecules – hormonic oscillator Ø Electronic partition function

14.10. CHECK YOUR PROGRESS 1. Derive an expression for the molecular translational partition function of an ideal gas. 2. The fundamental vibrational frequency of F2 is 2.76´1013HZ. Calculate the vibrational partition function of F2 at 250c? 14.11. POINT FOR DISCUSSION 1. Derive an expression for the molecular rotational partition function of an ideal gas. 2. Derive an expression for the molecular vibrational and electronic partition function of an ideal gas.



167

14.12. SOURCES 1. J.C. Kuriacose and J. Rajaram, Thermodynamics, Shoban lal Nagin Chand & Co., Delhi. 2. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006-2007. 3. Gurdeep Raj, Advanced Physical Chemistry, Goel Publishing House, Meerut.



168

LESSON 15: HEAT CAPACITY OF SOLIDS

CONTENTS 13.0. AIMS AND OBJECTIVES 13.1. INTRODUCTION 13.2. PROBABILITY 13.3. TYPES OF STATISTICS 13.4. THERMODYNAMIC PROBABILITY ( )W

13.5. BOLTZMANN EXPRESSION FOR ENTROPY 13.6. STIRLING’S APPROXIMATION 13.7. STATE OF MAXIMUM THERMODYNAMIC PROBABILITY (Equilibrium State) 13.8. MAXWELL-BOLTZMANN DISTRIBUTION LAW 13.9. EVALUATION OF LAGRANGE’S UNDETERMINED MULTIPLIERS 13.10. BOSE-EINSTEIN STATISTICS 13.11. ENTROPY OF BOSE-EINSTEIN GAS 13.12. FERMI-DIRAC STATISTICS 13.13. ENTROPY OF FERMI-DIRAC GAS 13.14. LET US SUM UP 13.15. CHECK YOUR PROGRESS 13.16. LESSON - END ACTIVITIES 13.17. REFERENCES 15.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on heat capacity of solids to the students. On successful completion of this lesson the student should have: * Understand the heat capacity of solids. 15.1. INTRODUCTION In 1819, Dulong and Petit found that at constant pressure, the molar heat capacity at constant volume of most of the solid elements at room temperature was given by

116 --» molcalKCV

In order to rectify this problem, two theories of heat capacities were developed. The first by Albert Einstein in 1907 and the second (which is a modification of the Einstein theory) by peter Debye in 1912. 15.2. THE EINSTEIN THEORY OF HEAT CAPACITY



169

Einstein made the following assumptions in consulting his theory of heat capacities of mono atomic crystals. 1. The atoms in a crystal lattice undergo small oscillations (vibrations) about their equilibrium configurations. In fact, an ideal crystal can be considered as a system of ‘ N ’ non- interacting particles (i.e.) atoms. 2. Each atom vibrates independently of the others and has three independent vibrational degrees of freedom. Thus the crystal may be treated as a system of N3 i n d e p e n d e n t a n d distinguishable harmonic oscillations. 3. There are no electronic, translational or rotational modes of motions in a mono atomic crystal. Using assumptions 2 & 3, the molar vibrational partition function of the crystal can be written as

NvibVib qQ 3)(= ------ (1)

From eqn Nq

N

KTi

Q =

úúúú

û

ù

êêêê

ë

é-

å=

e

exp

Q=Molar partition function; q = Molecular partition function

(or)

÷÷÷÷

ø

ö

çççç

è

æ Q-

--Q

== T

E

eNT

ENqNQ VibVib 1ln3

2

3ln3ln ------ (2)

Where E)(- is the characteristics Einstein temperature for the vibration.

EQ =K

hn

Where n is the vibrational frequency of the oscillator. Thus the internal energy of an ideal Einstein crystal is given by

1

3

2

3

,

ln2

-

Q

Q+=

÷÷

ø

ö

çç

è

æ=

T

E

e

ENKNh

NVTVib

QKTU n

d

d

(Or)

1

30

-

Q

Q

=-

T

E

e

TE

RTUU ------ (4)



170

Where 0U is the zero point-energy

÷ø

öçè

æ== QERNh

2

3

2

3n . Hence the molar heat capacity

2

1

23

÷÷÷÷

ø

ö

çççç

è

æ

-

Q

Q

÷ø

öçè

æ Q=

T

E

e

T

E

e

T

ER

VC ------ (5)

Experimentally it is found that at temperature approaching zero, VC approaches zero and

in the limit of high temperatures, VC approaches the Dulong-Petit value of

R3 (i.e.) )6( 11 -- MolCalK .Einstein theory predicts these limiting values VC of quite successfully.

Thus as,

0¾®¾T T

E

eT

E

e

Q

=-

Q

1 ------ (6) So that

0¾®¾T T

E

eT

ER

VC

Q

÷ø

öçè

æ Q=

23 =0 ------ (7)

Again as ¥¾®¾T ,

÷ø

öçè

æ Q+

»Q

T

ET

E

e1

So that

0lim¾¾ ®¾ itT R

T

E

T

E

e

T

ER

VC 3

11

2

32

»

÷÷

ø

ö

çç

è

æ-

Q+

Q

÷÷

ø

ö

çç

è

æ Q» ------ (8)

The above results are illustrated in Fig.1 for a number of metallic and non-metallic crystals.



171

Fig.1. Temperature-dependence of CV for monotomic solids

Einstein theory however is not successful in predicting the VC values in the lower and

intermediate temperature ranges. The values predicted by it are lower than those actually observed. 15.3. DEBYE THEORY OF HEAT CAPACITY According to Debye’s theory the crystal is treated as a macromolecule containing ‘ L ’ atoms, for which there are )63( -L vibrational modes. Since L36áá ,this assembly can be taken

to have L3 vibrational modes. The frequencies of such vibrational range from 0n (ground state)

to a maximum value maxn .

This limit arises when the wavelength of the oscillations is of the same order as the inter atomic distances. Debye considered the solid to be a continuous elastic medium. The vibrations of the atoms could be regarded as the elastic waves propagated through the medium. The vibrational motion of the atoms can be considered to be similar to sound waves propagated through this elastic medium. The vibrational energy is quantized and in order to calculate the total energy of vibration it is necessary to know the number of vibrations at each value of the allowed frequency (energy). This number, a function of the frequency (i.e.) )(nf is

known to be proportional to 2n (from the elastic theory of solids).

2)( nan ¢=f ------ (1)



172

In eqn (1)a ’ is the constant of proportionality. Considering a narrow range of

frequency nd , the number of vibrations is nn df )( . The frequency ranges from 0n (equal to

zero) to maxn (Debye cut off frequency) and the energy levels of the vibrating solid are closely

spaced. Hence the total modes of vibration in the crystal can be evaluated by integration and equated to L3 . The total vibrational energy of the crystal is equal to the average energy of an oscillation of frequency ‘n ’ multiplied by the number of oscillations at that frequency. This must be

summed up or integrated over all the ranges of allowed frequencies 0n and maxn .

ò =max

0

3)(

n

n

nn Ldf

ò =¢max

0

32

n

n

nna Ld ------ (2)

Evaluating this integral,

3max

9

na

L=¢ ------ (3)

3max

9)(

nn

Lf = ------ (4)

QL

d3

nn = , L

L3

32 =¢

nna ,

3max

9

na

L=¢

ñáU has already been calculated (Einstein heat theory). Since 0n is equal to zero.

ò ñá=max

0

)( nn dfUU ------ (5)

ò-

=max

0

3max

29

1n

nnnn

dL

e

h

KT

h ------ (6)

SincedT

dUCV = , on differentiating eqn (6) with respect to temperature,

ò-

÷ø

öçè

æ

==max

0 3max

2

)1(

9

n

nnn

n

KT

hV

e

KTdLKT

h

dT

d

dT

dUC ------ (7)



173

nn

n

n

n

n

nd

KT

h

e

KT

h

e

KT

hR 2max

02

1

2

3max

9ò

÷÷÷÷

ø

ö

çççç

è

æ

-

÷ø

öçè

æ ------ (8)

The quantity K

hn has the dimensions o temperature and when maxnn = , the temperature

corresponding to KT

h maxnis called the Debye characteristic temperature Dq .Since

KT

hnis

dimensionless, it can be replaced by u ,

TKT

hU Dqn

== maxmax

Since K

hD

nq = ; u

h

KT

KT

hu ×=== n

n

dUh

KTd ×=n

And so

duuh

KTdU

h

KTu

h

KTd 2

3

2

2

2÷ø

öçè

æ=×÷

ø

öçè

æ=nn

Equation (8) can thus be written as

duuh

KT

e

ue

uh

KT

RC

u

u

V2

3

2

2

3

3 )1(

9÷ø

öçè

æ

-÷ø

öçè

æ=

(or)

duuh

KTU

ue

ueu

D

T

KT

hR 2

3

02)1(

233

9max

÷ø

öçè

æò

-÷÷

ø

ö

çç

è

æ÷ø

öçè

æ=

q

(or)

ò-

÷ø

öçè

æ÷ø

öçè

æ÷÷

ø

ö

çç

è

æ=

T

ue

duueu

h

KT

KT

h

D

TR

Dq

q 02)1(

333

94

(or)



174

ò-

÷÷

ø

ö

çç

è

æ=

T

ue

duueu

D

TR

Dq

q 02)1(

3

94

------ (9)

The high and low temperature limits of )(VCV can be obtained

22

24

4

.......!2

.......!2

1

2)1(÷÷ø

öççè

æ++

÷÷ø

öççè

æ+++

=- u

u

uuu

ue

duueu

Since u is small this reduces to 2

4

u

uor 2u

If DT q>> , ò÷÷

ø

ö

çç

è

æ»

Tduu

D

TRVC

D

V

q

q 0

3

9)( 2

Q32

2÷ø

öçè

æ=÷

ø

öçè

æ=

TKT

hu Dqn

TKT

hdU Dqn

==

\

3

3

13

9)(÷÷

ø

ö

çç

è

æ

÷÷

ø

ö

çç

è

æ=

TD

D

TRVCV

q

q

RVCV 3)( »

The Debye formula has the same high temperature limit as the Einstein equation. At very

low temperature ¥¾®¾T

Dqand the value of the integral become

15

4 4p and is independent of T .

The heat capacity varies under these conditions with

3

÷÷

ø

ö

çç

è

æ

D

T

q. The Debye equation for very low

temperatures is thus 3TCV a ¢= (Debye-T-cubed law). This law is used to calculate VC at very

low temperatures and to extrapolate the experimental results from the lowest accessible



175

temperature to the absolute zero. Since )(VCV is a function of

D

T

q only, a plot of VC against

D

T

qfor all solids should fall on a curve (Fig. 2).

Fig. 2. The Debye curve for atomic heat capacity

From Fig. 2 it is seen that )(VCV attains the classical value of R3 when

D

T

q i s

approximately one, (i.e.) when D

q is approximately 300. At temperatures near 300 K Dulong

and Petit’s law holds good for carbon, D

q is 1860 K and for other light elements also D

q being

large,

D

T

q is small and )(VCV value falls on the ascending portion of the curve (i.e.) the value is

less than R3 . A higher value for VC is reduced only at high temperatures.

15.4. PLANCK DISTRIBUTION LAW FOR BLACK BODY RADIATION In the derivation of Planck’s radiation formula, we will consider a small cavity in an opaque material as a theoretical model of a black body. Suppose the cavity has volume ‘V ’and it is full of electromagnetic radiation at temperature T .An electromagnetic wave of frequency ‘n ’

and velocity ‘C ’ consists of particles called photons of energy nh , momentum c

hn .



176

The cavity in an oppose material is assumed to be having a large number of photons, each one of them is indistinguishable from the other. The assembly of photons in the cavity may be compared to the assembly of gas molecules in vessels. Therefore, the assembly of photons is supposed to constitute a gas known as photon gas. The photon gas possess the following properties 1. Photons are particles of rest mass zero 2. Unlike the gas molecules the photons may be created and destroyed. In this case we have 0¹å idn . This is strictly in accordance with the Bose Einstein statistics b e c a u s e t h e

number of photons is not fixed in the cavity due to their absorption and re-emission by the walls of the cavity. 3. There are two possible spin states which are associated with the photons of spin 1. In other words the photon possess two allowed modes, the number of allowed states or modes per unit volume for the photons with momentum lying in the range P to dP will lie

within a cell of volume dPP 24p .Therefore, the total number of eigen states will be given by

V

h

dPPdPpg

3

24.2)(

p= ------ (1)

For a photon,

c

hhP

n

l== ------ (2)

Differentiating equation (2), we get

c

hddP

n= ------ (3)

On substituting equations (3) in (1), we get

c

hd

h

VPdg

npnn ×=

3

28)()(

c

hd

ch

Vh nnp×=

23

228

c

hP

n=Q ,

2

222

c

hP

n=\

nnp

dc

V3

28= ------ (4)

Equation (4) represents the total number of Eigen states which are lying in the frequency range n and nn d+ . Substituting eqn (4) in the Bose Einstein distribution law, we get



177

1)(

-+

=ie

ig

ni bea

1

8

1

)()(3

2

-

¶=

-

=

KTi

e

cV

KTi

e

dgdn

e

nnp

e

nn

(or)

1

83

2

-

¶=

KTi

e

cV

V

dn

e

nnp ------ (5)

The left hand side of the equation (5) represents the number of photons per unit volume. When it is multiplied by nh , the energy of photon gives density nndE in the specified range. On substituting nhE = , we get

1

83

3

-

¶=

KT

h

e

c

hdE

n

nnpnn ------ (6)

Equation (6) is the Planck distribution law for black body radiation. nndE is also known as the spectral energy density at the frequency n . 15.5. LET US SUM UP In this lesson, we: Pointed out Ø Introduction Ø The Einstein theory of heat capacity Ø Debye theory of heat capacity Ø Planck distribution law for black body radiation

15.6. CHECK YOUR PROGRESS 1. For Al, QE = 240K, Calculate Cv for Al on the Einstein model at (a) 50K and (b) 240K 2. For I2 at 10K, Cv = 4.02JK-1 mol-1. Calculate (a) QD and (b) Cv of I2 at 12K. 15.7. LESSON – END ACTIVITIES



178

1. Discuss the salient features of the Einstein theory of the heat capacity of monoatomic crystals. How did Debye modified it? Show the results of the Einstein and the Debye theories on a plot and comment briefly. 2. Explain briefly the heat capacities of solids. 15.8. REFERENCES 1. J.C. Kuriacose and J. Rajaram, Thermodynamics, Shoban lal Nagin Chand & Co., Delhi. 2. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006-2007. 3. Gurdeep Raj, Advanced Physical Chemistry, Goel Publishing House, Meerut.



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