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Chapter 11

Atomic Physics

_____________________________________________

11.0 Introduction

Early in twentieth century many prominent scientists doubted about the

existence atom needless to say how the electrons in an atom are arranged, what

are their motions, how atoms emit or absorb light, or even atoms are stable.

Today we can even pick up individual atoms and move them around.

In 1926 all these questions about existence of atom were answered with the

development of quantum physics. Its basic premise is that moving electrons,

protons, and particles of any kind are best viewed as matter waves, whose

motions are governed by Schrödinger’s wave equation. Although quantum

theory also applies to massive particles, there is no point to treat an automotive,

planet etc. as object with quantum theory. For such massive, slow-moving

objects, Newtonian physics and quantum physics yield same answer.

11.1 Some Properties of Atoms

Atoms are stable. Essentially all the atoms that form our tangible worlds have

existed without change for billions of years. Otherwise, we do not know what

the world is like if there is continuous changed into other form.

Atoms combine with each other to form stable molecules and stack up to

form rigid solids. An atom is mostly empty space but we can stand on the floor,

which made up of atoms without falling through it.

Atoms are put together systematically according to their chemical and

physical properties in the periodic table. Figure 11.1 shows examples of

repetitive properties of the element as a function of their positions in periodic

table. The figure also shows the plot of ionization energy of the element, which

is defined as the energy required to remove the most loosely bounded electron

from neutral atom is plotted as a function of the position in periodic table. It

shows remarkable similarity in chemical and physical properties of the elements

in each column show evidences that atoms are constructed according to

systematic rules.

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The elements are arranged in the period table in six horizontal periods.

Except for the first period, each period starts from the left with highly reactive

alkaline metal like lithium, sodium and etc and end with chemical inert noble

gas like neon, argon and etc. Quantum physics accounts for the chemical

properties of these elements. The numbers of elements in six horizontal periods

are 2, 8, 8, 18, 18, and 32.

Figure 11.1: A plot of ionization energies of the elements a function atomic number showing

periodic repetition of properties through six complete horizontal period of the

periodic table

Atoms emit and absorb light. Atom can exist only in discrete quantum states.

Each state has certain energy. An atom makes a transition from one state to

another state by emitting light or absorbing light. The light emitted or absorbed

as photon with energy governed by equation (11.1).

hv = Ehigh – Elow (11.1)

Atoms have angular momentum and magnetism. Figure 9.2 illustrates a

negative charged particle moving in circular orbit around a fixed center. The

orbiting particle has angular momentum L since the path is equivalent to tiny

current loop and magnetic dipole μ . They are perpendicular to the plane but in

opposite direction because the charge is negative.

Model shows in Fig. 11.2 is strictly the classical model. It does not

accurately represent an electron in atom. In quantum physics, the rigid orbit

model has been replaced by the probability model best visualized as a dot plot.

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Figure 11.2: A classical illustration of charged particle has angular momentum and magnetic

dipole moment

In 1915 well before the discovery of quantum physic, Albert Einstein and Dutch

physicist Wander Johannes de Haas carried out an experiment designed to show

that angular momentum and magnetic moment of individual atoms are coupled.

An iron cylinder is suspended using a thin fiber. Initially the magnetic dipole is

random position. Once the current is switched the solenoid around the iron

cylinder, the magnetic field B is created parallel to the axis of cylinder that

causes the magnetic dipole μ of iron aligned along the magnetic field. This

would cause a net angular momentum netL . To maintain zero angular momentum,

the cylinder begins to rotate around its axis to produce an angular

momentum rotL .

(a) (b)

Figure 11.3: Einstein-de Haas Experiment: (a) Initially no magnetic field and random

magnetic dipole and (b) with magnetic field, magnetic dipole lined up parallel

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11.2 Quantization of Angular Momentum

Every quantum state of an electron in an atom has an associated orbital angular

momentum and a corresponding orbital magnetic dipole moment.

The magnitude L of the orbital angular momentum L of an electron in an

atom is quantized, which can have a certain values. These values are governed

by equation (11.2).

)1( llL where l = 0, 1, 2, 3, …… n-1 (11.2)

in which l is the orbital quantum number and is the reduced Planck’s constant,

which is 2

h. l must be a value either zero or a positive value integral no greater

than (n-1), where n is the principal quantum number. Thus, for n = 4, l can have

0, 1, 2, and 3.

The components Lz of the angular momentum are also quantized and they

are given by

lL mz where ml = 0, 1, 2, ….. l (11.3)

Figure 11.4 shows the five quantized components Lz of the orbital angular

momentum for electron with orbital quantum number l = 2. These quantized

components correspond to magnetic quantum number ml=-2, m1=-1, m1= 0,

m1=1, and m1=2.

For every orbital angular momentum vector L shown in Fig. 11.4, there is

a vector pointing in the opposite direction representing the magnitude and

direction of the orbital magnetic dipole momentumorb

μ .

From Fig. 11.4, the minimum angle L between L and Lz is given by

6

2cos 1

z = 35.260 (11.4)

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Figure 11.4: The allowed values of Lz for an electron in a quantum state with orbital

quantum number l = 2

11.3 Magnetic Dipole Moment and Zeeman Effect

A magnetic dipole is associated with the orbital angular momentum L of an

electron. The magnetic dipole has an orbital magnetic dipole momentorb

μ , which

is related with angular momentum by equation (11.5).

Lq

μe

orbm2

(11.5)

The minus sign indicates that orb

μ is directed opposite L . L

μorb is called the

gyromagnetic ratio.

Owing to magnitude of L is quantized, the magnitude of orbμ is also

quantized following equation (11.6).

)1(m2 e

orb llq

μ (11.6)

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L and orb

μ cannot be measured directly. However, the components of them can

be measured along a given axis such as z-axis. The component z,orbμ of the

orbital magnetic dipole moment are quantized and given by

Bz,orb m μμ l (11.7)

ml is the orbital quantum number and B is the Bohr magneton, which is defined

as

T/J10x274.9m4

24

e

B

qh

μ (11.8)

me is the mass of electron.

The orbital angular moment and magnetic moment vector are opposite.

Thus, from equation (11.5), the z-component orbital magnetic dipole moment is

z

e

zm2

Lq

μ (11.9)

Substituting equation (11.3) into equation (11.9), one gets lL mz

e

zm2

mq

μ l (11.10)

Zeeman effect is the splitting of atomic energy levels and associated spectrum

lines when the atom is placed in magnetic field. An illustration is shown in Fig.

11.5.

(a) (b)

Figure 11.5: Illustration of Zeeman effect (a) with magnetic field and (b) with magnetic field

The potential energy V associated with the magnetic dipole orb is

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BV orb μ (11.11)

The interaction energy of the z-component of magnetic dipole is

Bm2

BVe

lz q

m where ml = 0, 1, 2, 3, …l (11.12)

In terms of Bohr magneton, which ise

Bm4

qh

μ , the interaction energy V is

BmV Bμl (11.13)

11.4 Electron Spin

Electron has intrinsic spin angular momentum S often called spin. Meaning it is

a basic characteristic of electron. The magnitude of S is quantized and depends

on a spin quantum number s, which is always 2

1 for electron, proton and

neutron. In addition the component S measured along any axis is quantized and

depends on the spin magnetic quantum number ms, which can have value 2

1 or

2

1 .

The magnitude S of the spin angular momentum S of any electron, whether

it is free or trapped, has the single value given by

)1( ssS (11.14)

where s is the spin quantum number that has value 2

1. Thus, the magnitude S of

the spin momentum S of any electron is )1( ssS 2

3 .

An electron has an intrinsic magnetic dipole, which associated with its spin

angular momentum S , whether the electron is confined to an atom or free. The

magnetic dipole has a spin magnetic dipole moments

μ , which follow equation

(11.15).

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Sq

μe

sm

(11.15)

The minus sign mean that s

μ is directly opposite of S . The z-component of the

spin magnetic dipole moment is equal to z

e

zm2

00232.2 Sq

μ , with a small

correction factor.

The magnitude of s

μ is also quantized following equation (11.16).

)1(me

s ss

qμ (11.16)

The components of spin angular momentum S and spin magnetic dipole

moment s

μ along any axis can be measured such as along z-axis. Thus, the

components Sz of the spin angular momentum are quantized and given by

sS mz (11.17)

where ms is the spin magnetic quantum number that can have two values either

2

1 for spin up or

2

1 for spin down i.e.

2

1SZ . The components z,sμ of the

spin magnetic dipole moment are also quantized and they are given by

BBz,s m2 μμμ s (11.18)

Figure 11.6 shows the two quantized components Sz of the spin angular

momentum for an electron and associated orientation for vector S . It also shows

the quantized components z,sμ of the spin magnetic dipole moment and the

associated orientation of sμ .

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Figure 11.6: The allowed values of Sz and and z of an electron

Example 11.1

Calculate the interaction energy for an electron in l = 0 state with no orbital

magnetic moment in magnetic field with magnitude of 2.00T and sketch a graph

to show the splitting of energy levels.

Solution

The interaction energy is equal to BV zμ ; z

e

zm2

00232.2 Sq

μ , and 2

1SZ .

Thus, the interaction energy is

B2

1

m200232.2V

e

q B

m200116.1

e

q.

em2

qis equal to Bohr

magneton.

The interaction energy is then equal to B00116.1V Bμ .

Substitute Bohr magneton value and magnetic field value, the interaction energy

is equal to B00116.1V Bμ 0.2x10x274.900116.1 24 = ±1.86x10-23

J or

±1.16x10-5

eV.

The sketch of energy splitting is as follow.

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11.5 Coulomb Force of Attraction

Two charged particles, which are also called point charges, have charge

magnitude q1 and q2 separated by a distance r. The electrostatic force or

Coulomb force F of attraction or repulsion between these two charged particles

follows equation

2

0

21

r4F

qq (11.19)

where 0 is the permittivity in vacuum, which has value 8.854x10-12

F/m.

Equation (11.18) is called Coulomb’s law, which name after Charles Augustin

Coulomb.

The potential energy of a charge body is equal to r4 0

21

qq. The minus

denotes that its frame of reference is at infinity.

11.6 Principle of Quantum Mechanic

The results of the photoelectric effect experiment performed as shown in Fig.

11.7 with incidence of the photon of various wavelengths on metal surface and

measured the energy of the emitted electron demonstrated the principle of

quantum mechanics. The energy in discrete packet called quanta called photon

was both postulated by Max Planck in 1900 regarding thermal radiation and

Einstein in 1905 regarding energy of light wave.

From the graph an equation can be deduced that is

KEmax = hv- q (11.20)

where e is the work function and the value of gradient of the plot is the value

of Planck's constant, which is 6.626x10-34

Js. KEmax is the maximum kinetic

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energy of electron. This equation is also known as Einstein photoelectric

equation.

Albert Einstein stated that energy of the incident photon hv must be equal

to the energy required to free electron, which is the work function plus the

maximum kinetic energy of the electron.

Figure 11.7: Plot of KEmax of emitted electron versus frequency v

11.7 Electron Wave-Particle Duality

Louis de Broglie, a French physicist, was recognized as the discoverer of the

wave nature of electron. The wave nature of electron is clearly shown in the

development of modern electronic instrumentations and experimental methods

at the turn of twentieth century such as electron microscope, where electron

behaves as wave rather than as particle of small sphere. In 1923 Arthur

Compton demonstrated the particle nature of electron, whereby the scattered

photon had larger wavelength than the wavelength ' of incidence x-ray,

which is termed Compton scattering that has relationship )cos1(cm

e

h

,

where is the scattering angle. Investigations related to the blackbody radiation

i.e., the characteristic radiation that a body emits when heated and the

photoelectric effect i.e., electron emission from matter by electromagnetic

radiation of certain energy revealed the quantum nature of light, whereas the

electron diffraction in crystals demonstrated the wave nature of particles.

Consider an electron particle moving with velocity v and mass m, its

momentum p shall be p = mv. The wave nature of electron is shown in

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Einstein's famous E = mc2 equation where E also equals to hv. Combination the

two equations it gives

m =2c

h (11.21)

The momentum p of the photon is p = mc. Substitute this relation into equation

(11.21), it gives

p = c

hv (11.22)

However, v

c then

p =

h (11.23)

which is the Louis de Broglie equation. The equation is important because it

shows the linkage between particle in terms of momentum and wave in terms of

wavelength. Rewrite equation (11.23) as de Broglie wavelength dBr, it becomes

vmr

dBr

h

p

h (11.24)

where mr is the relativistic mass, which is mr = m

v c

e

1 2 2 / and also equal to me

for the case of electron. If 1

2

2m Ve is the kinetic energy E of the electron moving

with velocity v, de Broglie wavelength dBr is Em2 e

dBr

h

p

h .

Example 11.2

What is the de Broglie wavelength dBr of an electron with kinetic energy of

120eV?

Solution

The de Broglie wavelength dBr follows equation

Em2 e

dBr

h

1931

-34

10x602.1x120x10x11.9x2

Js6.63x10

= 1.1x10-10

m.

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According to Louis de Broglie hypothesis, various particles such as electrons,

protons, or macroscopic object could exhibit wave characteristics in certain

circumstances. Whenever, the value of de Broglie wavelength of a given

particle or a macroscopic object is much smaller than the dimensions of the

apparatus i.e., its components, classical mechanics should be applied. For an

example, a ball having a mass of 0.1kg moving with the speed of 100ms-1

, the

de Broglie wavelength is 100x1.0

10x63.6 34

dBr

p

h= 6.63x10

-35m, which is very small

number as compared to the objects that ball interacts with. Thus, classical

mechanics is applied to the description of the motion of the ball. On the other

hand, if a particle, such as electron has a speed of about 8.0x103ms

-1. Its de

Broglie wavelength is 9.1x10-8

m, which is larger than the spacing between

atoms and the size of the atom. Thus, quantum mechanical description is

appropriately applied.

11.8 Heisenberg Uncertainty Principle

According to quantum-mechanical principles, there is inherent limitation to the

accuracy of a measurement of a quantum-mechanical entity. This limitation is

the result of (i) the particle-wave duality and (ii) inevitable interaction between

the instrument of observer and the entity observed.

Heisenberg uncertainty principle discovered by Werner Heisenberg in 1927.

Heisenberg uncertainty principle states that one cannot measure values with

arbitrary precision of certain conjugate quantities such as momentum and

position that share uncertainty relation. Mathematically the product of the

uncertainties of these two measurements follows equation (11.25).

2

xph

(11.25)

In quantum mechanics, a particle may be described in terms of a wave packet of

length x, which is the resultant of individual waves with different amplitudes

and nearly equal wavelengths. The position of the particle is defined so that the

probability of finding the particle at the center of the packet is highest. For

smaller x, the position of the particle is more accurately defined. However,

since the range of wavelengths in the packet now must be wider, there is

greater uncertainty in the momentum p. The particle’s uncertainties are related

by Heisenberg’s uncertainty principle which is shown in equation (11.25). This

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implies that on the atomic scale it is impossible to measure simultaneously the

precise position and momentum of a particle.

The description above can be interpreted as a continuous distribution of

wavelengths can produce localized wave packet, whereby each wave has

different momentum according to Louis de Broglie. Thus, there is a widespread

of wavelength contributing x in the wave packet. If the momentum of the

wave can be more precisely determined, then there is an increase in uncertainty

of its position x.

Example 11.3

An electron moving in x-direction with measured speed of 2.05x106m/s of

0.50%. What is the minimum uncertainty with which one can simultaneously

measure the position of the electron along x-direction?

Solution

The uncertainty equation is

2

xph

.

The delta momentum p of the electron is 0.5/100x9.11x10-31

x2.05x106 =

9.35x10-27

kgms-1

.

The uncertainty of the position x is equal to 6.63x10-34

/2/9.35x10-27

=

1.13x10-8

m = 11.3nm.

11.9 Schrödinger's Wave Equation

In 1926 Erwin Rudolf Josef Alexander Schrödinger provided a formulation

called wave mechanics, whereby it incorporated the principles of quanta

introduced by Plank and wave-particle duality principle by Louis de Broglie.

Based on de Broglie's principle, the motion electron in crystal is explained by

wave theory. This wave theory is described by Schrödinger's wave equation,

which is shown as one-dimensional equation (11.26).

2 2

22m

x t

xV x x t

x t

t

( , )( ) ( , )

( , )j (11.26)

where (x, t) is the wave function which is position and time dependent and j is

the imaginary constant equal to 1 .

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(x, t) can be separated using technique of separation variables into time-

dependent of wave function, position independent, or time-independent, portion

of the wave function, which shall be (x, t) = (x)(t), where (x) is a function

of position only and (t) is the function of time only. Equation (11.27) shall then

be equal to

2 2

22mt

x

xV x x t x

t

t

( )

( )( ) ( ) ( ) ( )

( )j (11.27)

or

2 2

22

1 1

m x

x

xV x

t

t

t

( )

( )( )

( )

( )j (11.28)

Left side of equation (11.28) is position dependent and right side is time

dependent and each side of this equation must be a constant, which can be

denoted by a variable constant . Thus, the time dependent portion can be

written as

j1

( )

( )

t

t

t (11.29)

The solution of equation (11.29) shall be ( ) exp /t t j . This is a form of

classical sinusoidal wave where = E = hv, the total energy of the particle. The

left hand side of equation (11.28) will be equal to

2 2

220

m

x

xE V x x

( )( ) ( ) (11.30)

where V(x) is the potential experienced by the particle.

Combining equation (x, t) = (x)(t) and equation ( ) exp /t t j

shall yield

(x, t) = (x)(t) = t/exp)x( j (11.31)

In 1962, Max Born postulated that |(x, t)|2 is the probability of finding the

particle between position x and x+dx at a given time or |(x, t)|2 is the

probability density function. Thus,

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|(x, t)|2 = (x, t)

*(x, t) (11.32)

where *(x, t) is the complex conjugate function. Therefore,

*(x, t) =

*(x) exp ( / )j E t (11.33)

The product of the total wave function and its complex conjugate is given by

(x, t)*(x, t) = t)/E(exp)x(t)/E(exp)x( * jj- (11.34)

= (x)*(x)

Therefore, the probability density function |(x, t)|2 is |(x, t)|

2 = (x, t)

*(x, t)

= |(x)|2. This implies that probability density is independent of time.

Since |(x, t)|2 represents the probability density function, then for a

single particle ( )x x2

1-

d

and (x) must be finite and single valued.

We shall discuss a few applications of Schrödinger's wave equation. We

shall see how Schrödinger's wave equation is applied to derive the motion of

electron in free space, the wave solution for particle in infinite potential well,

tunneling of particle through a barrier, and harmonic oscillator.

11.9.1 Motion of Electron in Free Space

If there is no force acting on it, the potential energy V(x) is constant and the

total energy is such that E>V(x). For simplicity, the potential energy V(x) = 0

for all x then the time-independent wave equation (11.30) becomes

2

2 2

20

( )( )

x

x

m Ex

e

(11.35)

The solution of equation (11.35) is

Em2xexpB

Em2xexpA)x(

ee j-j (11.36)

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Taking into consideration of time-dependent portion of wave function, which

is ( ) exp /t E t j , the total solution for the wave function shall be

( , ) exp expx t A x m E Et B x m E Ete e

j - j

2 2 (11.37)

Equation (11.37) represents the traveling wave whereby the first term represents

the wave traveling in the +x direction and second term represents the wave

traveling in the -x direction.

The value of coefficients A and B shall be determined by the boundary

conditions. If +x direction is considered than the value of B shall be zero, then

the traveling-wave equation (11.37) shall be

( , ) expx t A x m E Ete

j

2 (11.38)

If we set k 2

, the wave vector k and wavelength

Em2 e

h , then equation

(11.38) shall become the traveling-wave equation as

txexpA)t,x( kj (11.39)

11.9.2 Wave Solution for a Particle in Infinite Potential Well

Let's discuss the wave solution for particle in infinite potential well. The

potential energy V(x) of the particle is a function of position as shown in Fig.

11.8.

If E is finite, then the wave function or (x) must be zero at both regions I

and II. The particle cannot penetrate these potential barriers. Thus, the

probability of finding it is zero at region I and III. The time-independent

Schrödinger's wave equation in region II where V(x) = 0 is

0)x(mE2

x

)x(22

2

(11.40)

A particular solution to this equation is

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xmE2

sinAxmE2

cosA)x(2221

(11.41)

One boundary condition is that the wave function (x) must be continuous so

that (x = 0) = (x = a) = 0. This will give rise A1 = 0 for x = 0 and (x = a) =

A2 sin amE2

2 = 0 for x = a. The equation is valid only if

a

mE22

n

, where n =

1, 2, 3,..., n, which is the quantum number. If n is negative value, it will give

rise to redundant solution and therefore, it is not considered.

Figure 11.8: Potential function of the particle in the infinite potential well

The coefficient A2 can be found from normalization boundary condition that is

( )x x2

1-

d

= ( ) ( )*x x x-

d

1 . If the wave solution (x) is real function then

(x) = *(x). Substitute into x

mE2sinA)x(

22

into ( ) ( )*x x x-

d

1 for

condition x = 0 to x = a, it becomes

1xxmE2

sinA2

22

2

d-

(11.42)

or

1xxa

sinA 22

2

dn

-

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The solution of the equation (11.43) will give rise to coefficient A2 equal to A2

= 2

a. Finally the time independent wave solution is given by

( ) sinxa

x

a

2 n, where n = 1, 2, 3, ....., n (11.43)

Thus, the probability density function is

a

xsin

a

2)x( 22 n

. Figure 11.9 shows

the probability density function plot for energy level n = 1, 2, and 3.

(a) (b) (c)

Figure 11.9: Probability density plot energy level n = 1, 2, and 3

For energy level n = 1, it is clearly shown that maximum probability density for

a/2, which at 50pm and minimum probability density functions are found at x =

0 or x = a. For energy level n = 3, the probability density function is

a

x3sin

a

2)x( 22 . For minimum probability density function,

a

x3 should be

equal to 0, , 2, and 3. This implies that minimum density function occurs at

x = 0, x = a/3, x = 2a/3, and x = a.

From the earlier analysis a

mE22

n

then the total energy E shall follow

equation (11.44).

2

222

n2ma

E

n

2

22

8ma

nh where n = 1, 2, 3, ...., n (11.44)

11.9.3 Wave Solution for Particle in Finite Potential Well

Let's discuss the wave solution for particle in finite potential well. The potential

energy V(x) of the particle is a function of position as shown in Fig. 11.10. If E

is finite then the wave functions or (x) is not zero at both regions I and II. The

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particle can penetrate these potential barriers. Thus, the probability of finding it

is not zero at region I and III. At region x < 0 and x > a, the potential energy is

equal to V0, while in between them the potential energy is equal to zero.

Figure 11.10: Potential function of the particle in the finite potential well

Let the wave functions associated in region I, II, and III to be I, II, and III

respectively, the one dimensional Schrödinger's wave equation for region I and

III is

)x(mE2

)x(mV2

x

)x(22

0

2

2

(11.45)

or )x()EV(m2

x

)x(2

0

2

2

.

In region II, the one dimensional Schrödinger's wave equation is

)x(mE2

x

)x(22

2

(11.46)

The wave function for region I and II are respectively equal to

)xkexp(B)xkexp(A)x( 00I (11.47)

and

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)xkexp(G)xkexp(F)x( 00III (11.48)

where 2

00

)EV(m2k

. From equation (11.47), which is

)xkexp(B)xkexp(A)x( 00I , as x approaching -, I approaches to zero.

This shall mean constant A is equal to zero. As the result the wave function

becomes

)xkexp(B)x( 0I (11.49)

From equation (11.48), which is )xkexp(G)xkexp(F)x( 00III , as x

approaching , III approaches to zero. This shall mean constant G is equal to

zero. As the result the wave function becomes

)xkexp(F)x( 0III (11.50)

The wave function of region II is

)xkcos(D)xksin(C)x( 11II (11.51)

where 21

mE2k

.

At x = 0, )0kexp(B)0( 0I = )0kcos(D)0ksin(C)0( 11II , this implies that

constant B = D. This shall mean that wave function at region I is also equal to

)xkexp(D)x( 0I .

Differentiating wave equations of all regions, which are )xkexp(D)x( 0I ,

)xkcos(D)xksin(C)x( 11II , and )xkexp(F)x( 0III with respect to x, it yields

)xkexp(Dkdx

)x(00

I

(11.52)

)xksin(Dk)xkcos(Ckdx

)x(1111

II

(11.53)

)xkexp(Fkdx

)x(00

III

(11.54)

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At x = 0, dx

)0(

dx

)0( III

, this implies that Dk0 = Ck1. Substituting C = Dk0/k1

into wave function of region II, which is )xkcos(D)xksin(C)x( 11II , it

yields )xkcos(D)xksin(k

kD)x( 11

1

0II .

At x = a, II(a) = III(a) and dx

)a(

dx

)a( IIIII

. This shall mean that

)akcos(D)aksin(k

kD 11

1

0 )akexp(F 0 (11.55)

and

)aksin(Dk)akcos(Dk 1110 )akexp(Fk 00 (11.56)

With the finite well, the wave function is not zero outside the well. So with

xfinite

> a and from Heisenberg uncertainty principle pfinite

< xx , this suggests

that the average value of momentum is less in the finite well. Therefore, the

kinetic energy inside the finite well is less than inside the infinite well. In

addition, the number of allowed energy levels is finite, there is a possibility that

a finite well may be sufficiently narrow or sufficiently shallow that no energy

levels are allowed.

Let’s now attempt to obtain the energy levels in the finite well. Taking

equation (11.55) to divide with equation (11.56), it yields

)aksin(Dk)akcos(Dk

)akcos(D)aksin(k

kD

1110

11

1

0

)akexp(Fk

)akexp(F

00

0

(11.57)

This will yield )Lktan(kkkk2 1

2

0

2

110 . Substituting equation 2

0

2

0

)EV(m8k

h

and 2

2

1

mE8k

h

into this equation to eliminate k0 and k1, it yields a

transcendental equation for the energy level inside finite potential well, which is

11 Atomic Physics

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2

22

00

mEa8tan)VE2(EEV2

h (11.58)

The solution of equation (11.52) can be solved by numerical method. However,

one may use try and error method to fine the values of total energy by equating

left hand side of the equation with right hand side of the equation, which is

2

22

0

0 mEa8tan

VE2

EEV2

h. The results of energy of an electron calculated

using trigonometric function plot for V0 = 25eV and a = 0.5nm are shown in Fig.

11.11. The intersection of negative

2

22mEa8tan

hand negative

0

0

VE2

EEV2

are

the E values for level n = 1, n = 2, n = 3, while intersection of positive

2

22mEa8tan

hand positive

0

0

VE2

EEV2

are the E value for level n = 4, and n =

5.

Figure 11.11: Energy levels in finite potential well for a = 0.5nm and V0 = 25eV

11 Atomic Physics

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Using equation (11.44), which is 2

e

22

na8m

Enh

, the energies in infinite potential

well for a = 0.5nm are tabulated. Together with the energies of finite potential

well are shown in Fig. 11.12.

Energy Level Infinite Well (eV) Finite Well (eV)

1 1.504 1.123

2 6.015 4.461

3 13.533 9.905

4 24.059 17.162

5 37.593 24.782

6 54.133 Non quantized

Figure 11.12: Energies in infinite and finite potential wells for a = 0.5nm and V0 = 25eV

The wave functions of three energy levels of a finite potential well with a =

0.5nm and V0 = 25eV are shown in Fig. 11.13.

Figure 11.13: The wave functions of three energy levels of a finite potential well with a =

0.5nm and V0 = 25eV

11.9.4 Wave Solution for a Particle Tunneling

Figure 11.13 shows an electron of total energy E moving parallel along x-axis

and probability density function )x(2 of matter wave showing tunneling of

particle through barrier. The potential energy of the particle is zero except when

it is in the region 0 < x < a, where the potential energy has a constant value V0.

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Figure 11.14: An energy diagram showing potential energy barrier of V0 and the probability

density function )x(2 of electron wave showing tunneling of electron

through barrier

In classical theory, if the energy of particle E is less than the potential energy V0,

the particle will be bounded back at the boundary. In quantum mechanics, there

is a chance of leak through or tunnel through.

The Schrödinger's wave equation for left side and right of the barrier is

0)x(mE2

x

)x(22

2

and the solutions of this equation are

mE2xexpB

mE2xexpA)x(

j-jfor left side and

mE2xexpD

mE2xexpC)x(

j-jfor right side. Wave

function

mE2xexpB)x(

j-denotes the reflected back wave function. Wave

function

mE2xexpC)ax(

jdenotes the transmitted wave function, which

has lower and constant amplitude.

Within the barrier, the Schrödinger's wave equation is

0)x()EV(m2

x

)x(022

2

(11.59)

The solution of the wave function is

)xexp(F)xexp(E)a xand 0x( kk (11.60)

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where k is wave number and is equal to 2

0

2 )EV(m8

hk

.

Taking the exponential decrease function, the probability density function

)a x and 0x(2 is

)x2exp(E)a x0x( 22 k (11.61)

The transmission coefficient T is the measure of fractional amount of particle is

tunneled through. It is approximately equal to

)a2exp(V

E1

V

E16T

00

k

(11.62)

Since transmission coefficient is an exponential function, it depends on mass (in

this case is electron mass), thickness of the barrier a, and (V0 – E).

Tunneling is important in nuclear fusion. A fusion reaction can occur when

two nuclei tunnel through the barrier caused by their electrical repulsion and

approach each other closely enough for attractive nuclear force to cause them to

fuse. Fusion reaction occurs in the core of star including the Sun. Emission of

alpha particle from unstable nuclei is also through tunneling. An alpha particle

of energy E at the surface of a nucleus encounters a potential barrier that results

from the combined effect of the attractive nuclear force and electrical repulsion

of the remaining part of nucleus as shown in Fig. 11.15.

Figure 11.15: The potential energy function for an alpha particle interacting with nucleus of

radius R

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When the distance is such that rR, the radius of nucleus, the alpha particle

encounters square potential well, whereas rR, it encounters potential due to

electrical repulsive force.

11.9.4.1 Application of Particle Tunneling

One of the applications of electron tunneling is the scanning tunneling

microscope. Scanning tunneling microscope STM uses electron tunneling to

create image of surface down to the scale of individual atom. An extreme sharp

needle is brought very close to the surface within 1.0nm. When the needle is at a

positive potential with respect to the surface, electron can tunnel through the

surface potential energy barrier and reach the needle. The width of the barrier is

the distance between tip of needle and surface. The needle or probe scans across

the surface and at the same time move vertical upward or downward to maintain

constant current being registered due to tunneling. In this manner, the surface

topology of the sample can be constructed based on the motion of the probe.

Figure 11.16 illustrates the probe of a scanning tunneling microscope.

Figure 11.16: Probe of scanning tunneling microscope

Other examples of applications are tunneling diode used in microwave

application and Joseph junction. Joseph junction is consisting of two

superconductors separated by an oxide layer of 1.0 to 2.0nm thick. Electron

pairs in superconductors can tunnel through the barrier layer giving such a

device unusual circuit properties. Joseph junction is useful for establishing

precise voltage standard and measuring tiny magnetic field.

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Example 11.4

Suppose that the electron having energy E of 5.1eV approaches a barrier of

height V0 = 6.8eV and thickness a = 750pm. What is the probability that the

electron will be transmitted through the barrier?

Solution

The wave number is 2

0

2 )EV(m8

hk

=

234

19312

)10x63(6.

10x602.1x)1.58.6(10x11.9x8

=

6.67x109m

-1.

The transmission coefficient is

)a2exp(V

E1

V

E16T

00

k

= 3 )10x750x10x67.6x2exp( 129 = 3exp(-10.0) = 135x10

-6.

The probability is 135 electrons will tunnel through for every one million

electrons striking the barrier.

11.9.5 Wave Solution for Harmonic Oscillator

In Newtonian mechanic, a harmonic oscillator is a particle with mass m acted on

by a conservation force component Fx = - k’x. k’ is the force constant. The

corresponding potential energy function is 2

0 x2

1V 'k . The angular frequency is

m

k '

. The energy of the photon is hE . The harmonic oscillator has

the characteristic of Newtonian mechanics. Thus, the quantum-mechanical

analysis of the energy levels of harmonic oscillator would be multiples of

quantity shown in equation (11.63).

m

k '

(11.63)

The energy levels are based on Plank’s radiation law, which is 1e

c2)(

t/c

2

kh5

hI .

It has a good assumption that the energy levels are half integer multiple of .

Thus, equation (11.63) will be rewritten as

m

knn

'

2

1

2

1En n = 0, 1, 2, 3, 4,…, n (11.64)

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The Schrödinger's wave equation for the harmonic oscillator is

)x(Ex2

1m2

x

)x( 2'

22

2

k

(11.65)

The wave function approaches zero when x approaches infinity i.e. (x) 0 as

|| . The solution for the Ex2

1 2' k is

2/x'mexpC)x( 2k (11.66)

The constant C is chosen to normalize the function. Knowing that 1dx)x(2

.

Constant C can be found from integral table, which

a)xaexp( 22 . Thus, C

is found to be 4

8 '

mk. This implies that

2/x'mexp

m)x( 2

4

8 '

kk

.

The illustration of the energy levels of harmonic oscillator is shown in Fig.

11.17.

Figure 11.17: Energy levels of harmonic oscillator

As shown in Equation (11.64), the energy level difference between adjacent

energy level is .

Based on equation (11.64), the Schrödinger's wave equation for the

harmonic oscillator is

11 Atomic Physics

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)x(2

1x

2

1m2

x

)x( '2'

22

2

m

knk

(11.67)

The general formula for the normalized wave functions (x) is

2

'

4

'

n4

8 '

x2

mexpx

mH

!2

1m)x(

kk

n

k

n (11.68)

x

mH 4

'

n

kis the Hermite polynomial, after the French mathematician Charles

Hermite. The polynomial states Hn(y) is

22 xx e

dx

de1)y(H

n

nn

n (11.69)

For n = 0, 1, 2, 3, 4, and 5, the Hermite polynomials are 1)y(H 0 , y2)y(H 1 ,

2y4)y(H 2 2 , y12y8)y(H 3 3 , 12y48y16)y(H 24 4 , and

y120y160y32)y(H 35 5 respectively.

Thus, from equation (11.68), the first six wave functions for the harmonic

oscillator will be equal to

2

'

4

8 '

x2

mexp

m)0(

kk

2

'

42

'

4

8 '

x2

mexpx

m2

m)1(

kkk

2

'2

2

'

4

8 '

x2

mexp1x

m2

2

1m)2(

kkk

2

'

42

'34/3

2

'

4

8 '

x2

mexpx

m3x

m2

3

1m)3(

kkkk

2

'2

2

'4

2

'

4

8 '

x2

mexp3x

m12x

m4

24

1m)4(

kkkk

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x2

mexp

xm

30xm

40xm

8240

1k)5(

'

42

'34/3

2

'54/5

2

'

4

8 '

k

kkkm

(11.70)

From the above equations, the first four wave functions for the harmonic

oscillator are shown in Fig. 11.18. A is the amplitude of Newtonian oscillator.

The results also shown that for x < -A and x> A are not forbidden region. The

maxima and minima for each function is n+1, an extra than the quantum number.

(a) (b)

(c) (d)

Figure 11.18: First wave functions for harmonic oscillator (a) n = 0, (b) n = 1, (c) n = 2, and

(d) n = 3

The probability density function 2(x) for n = 0, 1, 2, and 3 and its

corresponding Newtonian probability density function are shown in Fig. 11.19.

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(a) (b)

(c) (d)

Figure 11.19: Probability density function of harmonic oscillator for (a) n = 0, (b) n = 1, (c)

n = 2, and (d) n = 3

11.10 Atomic Model

In 1911, Ernest Rutherford suggested that atom is consisted of positively

charged nucleus with negatively charged electrons orbiting the nucleus at

constant speed. This was a great difference from the earlier model, which stated

that electrons were stuck into the spherical surface of the nucleus.

From the glow discharge experiment of hydrogen gas, Johannes Robert

Rydberg came out with an empirical formula, which stated that the line spectra

obeyed.

22

11R

1

Pjpj

(11.71)

where j<p and they are integrals. R c8

m32

0

0

4

h

q

is the Rydberg constant =

1.097x107m

-1.

Niels Bohr knew the results of hydrogen glow discharge and he thought

that the electron stayed at fixed stable orbit and did not radiate continuous range

of energies. He revised Rutherford hydrogen model and suggested that electron

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orbiting the nucleus in a circular orbit at constant velocity v. He came out with

three postulations. The first one states that there are certain orbits in which

electron is stable and does not radiate energy. The second one states that when

electron falls from orbit p to j, the energy, which it loses, is (Ep - Ej) and it is

radiated as a quantum of light, which follows equation (11.72).

Epj = hv (11.72)

where h is Planck's constant = 6.626x10-34

Js and v is the frequency of photon.

Frequency v can also be expressed as v =c

, where c is the speed of light in

vacuum which is 2.998x108ms

-1. Thus, the energy lost by the electron as it falls

from orbit p to orbit j shall be

Ep - Ej = hvpj =pj

c

h (11.73)

The total energy En of an electron orbiting in nth

orbit is equal to

En = n0

2

r8

q (11.74)

It comprises of kinetic energy n

2

0

2

er8

1Vm

2

1 q

and potential energy due to

coulomb force of attractionn

2

0 r4

1 q

, where the coulomb force of attraction is

2

n

2

0 r4

1 q

= m

v

re

n

2

and mv

re

n

2

is the centripetal force.

However, from Bohr's third postulation, which states that the angular

momentum is

mevrn = 2

nh= n (11.75)

where

2

h is the reduced Planck's constant.

The relation of rn and n can be obtained

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rn = 2

2

e

2

0

mn

q

h

(11.76)

For n =1, r1 = ao is called Bohr's radius and is equal to 0.529o

A . Thus, the radius

of n orbit rn 2

o

n A529.0r n . Substituting equation (11.76) into equation (11.74), it

yields equation (11.77).

222

0

e

4

n

1

8

mE

nh

q

(11.77)

where n is the subscript to denote energy level and eV6.138

m22

0

e

4

h

q. Thus,

energy of the electron at orbit n is E eVn 136

2

.

n.

For hydrogen atom whereby it has one electron occupying first orbit

closed to nucleus would have highest ionization energy. Ionization energy is

defined as energy required removing an electron from its stable state to infinity,

which is not in control of nucleus.

The shortcoming of Bohr's atomic model that it could not explain many

electron atom, was overcome by Arnold Johannes Wilhelm Sommerfeld

suggesting two more quantum numbers associated with the principal quantum

number n. They are: l denotes orbital quantum number for angular momentum

of electron and m is magnetic quantum number for behavior of electron in

magnetic field. The additional two quantum numbers meant that the orbit

denoted by n could be split into more orbits, which need not be circular. They

can be elliptical. The rough representation of an atom with circular and elliptical

orbits is shown in Fig. 11.20.

Figure 11.20: A representation of an atom with circular and elliptical orbits

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Later in 1925, George Eugene Uhlenbeck and Samuel Abraham Goudsmit

proposed a fourth quantum number called spin quantum number s, which is

responsible for the magnetic property of the element. Electron can have one of

the two different spin which either clockwise or anti-clockwise spin. Therefore,

the spin s can be s = + 1

2 or s =

1

2.

The summary of four quantum numbers is:

1. principal quantum number n;

2. orbital quantum number l;

3. magnetic quantum number ml;

4. spin quantum number s.

The maximum number of electron in an atom that can have the same principal

quantum number n is 2n2. l may be zero or a positive number such as l n -1. ml

l so ml can be negative, zero or positive integral.

l can also be shorthand written such that l = 0 means s; l = 1 means p, l = 2

means d and l = 3 means f.

Beside the way how to fill the quantum numbers, Friedrich Hund - Hund

rule - deduced from the measurement of magnetic moment of atom that same l

level is filled first by electron having the same value of spin i.e. 1

2 to be filled

first and followed by 1

2. This is also the reason why iron has magnetic

property because its outermost shells have four electrons having s = 1

2 and

they are not paired which termed spin pair. This gives rise to magnetic moment

of four Bohr magneton B, whereby one Bohr magneton e

Bm4

qh

= 9.27x10-

24Am

2.

Pauli exclusive principle states that no two electrons in atom can have the

same set of four quantum numbers i.e. a set of these four quantum numbers

describes an individual electron in a single atom.

As an illustration of how the four quantum numbers are applied, the

electronic configuration of chlorine that has atomic number Z = 17, is shown in

Fig. 11.21.

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Atomic

Number Z Electron Number

Quantum Number

n l ml s

17

1 1 0

0 + 1/2

2 0 - 1/2

3

2

0 0 + 1/2

4 0 - 1/2

5

1

-1 + 1/2

6 0 + 1/2

7 +1 + 1/2

8 -1 - 1/2

9 0 - 1/2

10 +1 - 1/2

11

3

0 0 + 1/2

12 0 - 1/2

13

1

-1 + 1/2

14 0 + 1/2

15 +1 + 1/2

16 -1 - 1/2

17 0 - 1/2

Figure 11.21: Electronic configuration of chlorine

11.10.1 Electron Probability Distribution

Instead treating the electron as a point particle moving in a precise circle like

the one shown in Fig. 11.20, Schrödinger’s equation gives a probability

distribution surrounding the nucleus. The probability of finding an electron in a

small volume dV is dV2

. The integration over all space should be unity. i.e.

1dv2

. This shall mean 100% probability of finding electron somewhere in

the universe. The volume of the shell with inner radius r and other radius r+dr is

approximately equal to drr4dV 2 . Thus, the probability P(r) of finding an

electron within the radial range dr is equal to

drr4dV)r(P222 (11.78)

Figure 11.22 shows the probability P(r) of finding an electron for several

hydrogen atom wave functions. The figures show that for each function, the

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number of maxima is (n-1). Take an example n = 3, the number of maxima is

two, which is shown in 3p and n = 4, the number of maxima is three, which is

shown in 4p.

(a) l = 0 (b) l = 1 (c) l = 2 and l = 3

Figure 11.22: Radial probability distribution function P(r) for several hydrogen wave

functions plotted as functions of ration r/a

The graph with l = n-1, there is only one maximum located at r = n2a, where a is

Bohr’s radius. Thus, the largest possible l for each n like 1s, 2p, 3d, and 4f, P(r)

has a single maximum at n2a. Figure 11.22 shows the radial probability

distribution function 22r4)r(P , which indicates the relative probability of

finding an electron within a thin spherical shell of radius.

Figure 11.23 and Fig. 11.24 show three dimensional probability

distribution function 2

, which indicate the relative probability of finding an

electron within a small box at a given position. The darker cloud indicates high

value of 2

. Figure 11.23 shows the cross sections of spherically symmetric

probability clouds for three lowest s sub-shells which 2

depend only on the

radial coordinate r.

Figure 11.23: Three dimensional probability distribution function

2 for spherically

symmetric 1s, 2s, and 3s hydrogen atom wave functions

11 Atomic Physics

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Figure 11.24 show cross sections of the clouds for other electron for which 2

depends on both r and . In any stationary state if hydrogen atom, 2

is

independent of .

Figure 11.24: Cross sections of three dimensional probability distribution function

2 for a

few quantum states of the hydrogen atom. Mentally rotate each drawing along

z-axis to get three dimensional representation of 2

Tutorials

11.1. How many l value associated with n = 3?

11.2. How many ml number associated with l = 3?

11.3. For orbital quantum number with l = 3.

(a). What is the magnitude of the orbital angular momentum in a state

with l = 3?

(b). What is the magnitude of its largest projection on an imposed z-axis?

(c). What is the magnitude of its minimum projection on an imposed z-

axis?

11.4. An atom in a state with l = 1 emits a photon with wavelength 600nm as it

decays to a state with l = 0. If the atom is placed in a magnetic field with

magnitude B = 2.00T, determine the shifts in the energy levels and in the

11 Atomic Physics

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wavelength resulting from the interaction of the magnetic field and

atom’s orbital magnetic moment.

11.5. Calculate the value of two equal charges if they repel one another with

force of 0.1N when situated 50cm apart in vacuum.

11.6. The work function for cesium is 2.14eV.

(a). What is the maximum kinetic energy of the electron ejected from the

surface of cesium by light of wavelength = 546nm?

(b). What is the maximum speed of the electron?

11.7. A gain of sand with mass of 1.0x10-3

g appears to be at rest on smooth

surface. We locate its position to within 0.010mm. What velocity limit is

implied by our measurement of its position?

11.8. An electron revolving an orbit n around the nucleus has total energy equal

to En = n0

2

r8

q. Prove this formula using the formula of Coulomb force

of attraction between the electron and nucleus in which it follows

expression 2

n0

2

r4

q.

11.9. Calculate the de Broglie wavelength or an electron of kinetic energy

1.0eV and 100eV.

11.10. The equation is 1xxa

sinA 22

2

dn

-

. Evaluate for the value of constant A2

for limit x = 0 to x = a.

11.11. A ground state electron is trapped in a one-dimensional infinite potential

well with a = 100pm.

(a). What is the probability of that an electron can be detected in left one-

third of the well?

(b). What is the probability that the electron can be detected in middle

one-third of the well?

11.12. Calculate the energies and radii of second, third, and fourth orbit of

hydrogen atom.

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11.13. What is the maximum number of electron that occupied in a shell with

principal quantum number 3?

11.14. Reference to question 11.13, state the electronic configuration of this

shell.

11.15. The home-base level of Lyman series of hydrogen atom is n = 1. What is

the least energetic photon emitted in this series?

11.16. A 2.0eV electron encounters a barrier with height 5.0eV. What is the

probability that it will tunnel through the barrier of width a is 1.0nm?

11.17. A sodium atom of mass 3.82x10-26

kg vibrates with simple harmonic

motion in crystal. The potential energy increases by 0.0075eV when the

atom is displaced by 0.014nm from its equilibrium position.

(a). Find the angular frequency according to Newtonian mechanics.

(b). Find the spacing of adjacent energy levels.

(c). If the atom emits a photon during a transition from one vibrational

level to the next lower level, what is the wavelength of the emitted

photon?

11.18. Iron, transition element has atomic number Z = 26 and its electron

configuration is 1s2 2s2 2p6 3s2 3p6 4s2 3d6. Using four quantum

numbers to tabulate the electronic configuration of iron. From the results

discuss and comment why iron has magnetic property.