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Physics 43 Homework Set 9 Chapter 40 Key

1. The wave function for an electron that is confined to x 0 nm is

a. Find the normalization constant.b. What is the probability of finding the electron in a 0.010 nm-wide region at x = 1.0 nm?

a)2 1Normalize: 1 0.559dx gives b nm

+

= =

b)1.005 1.005

2 2

0.995 0.995

(.559) exp( 2 / 3.2) 0.002286 .23%nm nm

nm nm

P dx x dx= = = = 17%

2. An electron in an infinite square well has a wave function that is given by ( )

=

L

x

Lx

2sin

2

for 0 x L and is zero otherwise.

a. Sketch the probability density function for this state.

b. Find the most probable position(s) of the electron in the well.

c. Find the expectation value of x.

d. Why or why not should the results of b and c be the same or different? That is, should the most

probable position(s) be the expectation value? (Hint: Use your sketch)

Solution in glass case.

3. A particle in an infinite square well has a wave function that is given by

( )

=

L

x

Lx

sin

21

for 0 x L and is zero otherwise. (a) Determine the probability of finding the particle between x = 0 and

x = L/3. (b) Use the result of this calculation and symmetry arguments to find the probability of finding

the particle between x = L/3 and x = 2L/3. Do not re-evaluate the integral. (c) What If? Compare the result

of part (a) with the classical probability.

(a) The probability is3 3 3

2 2

0 0 0

22 2 1 1sin cos

2 2

L L Lx x

P dx dx dxL L L L

= = =

3

0

21 1 1 2 1 3sin sin 0.196

2 3 2 3 3 4

Lxx

PL L

= = = =

.

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(b) The probability density is symmetric about2

Lx= .

Thus, the probability of finding the particle between2

3

Lx= and x L= is the same 0.196. Therefore, the

probability of finding it in the range2

3 3

L Lx is

( )1.00 2 0.196 0.609

P = = .

FIG. P41.21(b)

(c) Classically, the electron moves back and forth with constant speed between the walls,and the probability of finding the electron is the same for all points between the walls.Thus, the classical probability of finding the electron in any range equal to one-third of

the available space isclassical

1

3P = .

4. A particle in an infinitely deep square well has a wave function given by ( )

=

L

x

Lx

2sin

22

for 0 x L and zero otherwise. (a) Determine the expectation value of x. (b) Determine the probability of

finding the particle near L/2, by calculating the probability that the particle lies in the range 0.490L x

0.510L. (c) What If? Determine the probability of finding the particle near L/4, by calculating the

probability that the particle lies in the range 0.240L x 0.260L. (d) Argue that the result of part (a) does

not contradict the results of parts (b) and (c).

(a)

= =

2

0 0

2 42 2 1 1si n cos

2 2

L L

x xx x dx x dx

L L L L

= + =

2 2

2

00

4 4 41 1sin cos

2 16 2

L Lx x xx L L

xL L L L L

(b) Probability

= =

0.510 0.510

2

0.4900.490

2 42 1 1si n si n

4

L L

LL

x xLdx x

L L L L L

Probability ( )

= = 51

0.020 sin 2.04 sin 1.96 5.26 104

(c) Probability

=

0.260

2

0.240

41si n 3.99 10

4

L

L

xx

L L

(d) In the = 2n , it is more probable to find the particle either

near =4

Lx or =

3

4

Lx than at the center, where the

probability density is zero.

Nevertheless, the symmetry of the distribution means that

the average position is2

L.

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5. An electron is trapped somewhere in a molecule which is 7.13 nm long. Calculate theminimum kinetic energy of the electron.

3427

9

2 27 223

31

6.33 107.06 10 /

4 4 4 (7.13 10 )

(7.06 10 / )2.74 10 .0002

2 2(9.11 10 )

h h x Jsx p p x kgm s

x x m

p x kgm sK x J eV

m x kg

= =

= = = =

6. The wave function for a particle is

( )( )22 ax

ax

+=

for a > 0 and < x < +. Determine the probability that the particle is located somewhere between x = a

and x = +a.

P41.2 Probability ( )( )

2 1

2 2

1tan

aa a

aa a

a a xP x dx

a ax a

= = = +

( )1 11 1 1

tan 1 tan 14 4 2

P

= = =

7. The wave function of a particle is given by

( ) ( ) ( )kxBkxAx sincos +=

whereA, B, and k are constants. Show that is a solution of the Schrdinger equation, assuming the

particle is free (U = 0), and find the corresponding energy E of the particle.

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8. Consider a particle moving in a one-dimensional box for which the walls are at x= L/2 and x = L/2. (a)

Write the wave functions and probability densities for n = 1, n = 2, and n = 3. (b) Sketch the wave

functions and probability densities.

A particle is confined in a rigid one-dimensional box of lengthL.Visualize:

Solve: (a)(x) is zero because it is physically impossible for the particle to be there because the box is rigid.

(b)The potential energy within the region L/2 xL/2 is U(x) =0 J. The Schrdinger equation in this region is

( )( ) ( )

22

2 2

2d x mE x x

dx

= =

where 22mE= .

(c) Two functions (x) that satisfy the above equation are sinx and cosx . A general solution to the Schrdingerequation in this region is

(x) =Asinx +Bcosx

whereAandBare constants to be determined by the boundary conditions and normalization.

(d)The wave function must be continuous at all points. =0 just outside the edges of the box. Continuity requires that also be zero atthe edges. The boundary conditions are (x=L/2) =0 and (x=L/2) =0.(e)The two boundary conditions are

( )2 sin cos sin cos 0

2 2 2 2

L L L LL A B A B

= + = + =

( )2 sin cos 02 2

L LL A B

= + =

These are two simultaneous equations. Unlike the boundary conditions in the particle in a box problem of Section 41.4,there are two distinct ways to satisfy these equations. The first way is to add the equations. This gives

2 cos 02

LB

=

B=0 (x) =Asinx

To finish satisfying the boundary conditions,

sin 02

L =

L=2, 4, 6, =2nwith n=1, 2, 3,

With this restriction on the values of , the wave function becomes ( ) ( )sin 2x A n x L = . Using the definition of

from part (b), the energy is

( )( )

2 2 22

2 2

22

2 8n

n hE n

mL mL

= =

n=1, 2, 3,

The second way is to subtract the second equation from the first. This gives

2 cos 02

LA

=

A=0 (x) =Bcosx

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To finish satisfying the boundary conditions,

cos 02

L =

L=, 3, 5, =(2n 1) n=1, 2, 3,

With this restriction on the values of , the wave function becomes ( ) ( )( )cos 2 1x B n x L = . Using the definition of, the energy is

( ) ( )2 2 2 2

2

2 22 1 2 1

2 8n

n hE nmL mL

= =

n=1, 2, 3,

Summarizing this information, the allowed energies and the corresponding wave functions are

( )

( )( )

( )( )

22

1 1 12

22

1 1 12

2 1cos 2 1 , 9 , 25 ,

8

2sin 2 4 , 16 , 36 ,

8

n

n

n x hB E n E E E

L mLx

n x hA E n E E E

L mL

= =

=

= =

whereE1=h2/8mL

2.

(f)The results are actually the same as the results for a particle located at 0 xL. That is, the energy levels are the sameand the shapes of the wave functions are the same. This has to be, because neither the particle nor the potential well have

changed. All that is different is ourchoice of coordinate system, and physically meaningful results cant depend on the

choice of a coordinate system. The new coordinate system forces us to use both sines and cosines, whereas before wecould use just sines, but the shapes of the wave functions in the boxhavent changed.

9. An electron is contained in a one-dimensional box of length 0.100 nm. (a) Draw an energy-level

diagram for the electron for levels up to n = 4. (b) Find the wavelengths of all photons that can be emitted

by the electron in making downward transitions that could eventually carry it from the n = 4 state to the n

= 1 state.(a) We can draw a diagram that parallels our treatment of standingmechanical waves. In each state, we measure the distance dfrom one node toanother (N to N), and base our solution upon that:

Since N to N2

d

= andh

p=

2

h hp

d= = .

Next,( )

( )

2342 2

2 2 31

6.626 10 J s1

2 8 8 9.11 10 kge e

p hK

m m d d

= = =

.

Evaluating,38 2

2

6.02 10 J mK

d

=

19 2

2

3.77 10 eV mK

d

= .

In state 1, 101.00 10 md = 1 37.7 eVK = .

In state 2, 115.00 10 md = 2 151 eVK = .

In state 3, 113.33 10 md = 3 339 eVK = .

In state 4, 112.50 10 md = 4 603 eVK = .

FIG. P41.5

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10. A particle is described by the wave function

(a) Determine the normalization constantA. (b) What is the probability that the particle will be found

between x = 0 and x = L/8 if its position is measured?

(a)

= 2

1dx becomes

= + = = 4 4

2 2 2 2

44

2 41cos si n 1

2 4 2 2

L L

LL

x x xL LA dx A A

L L L

or =2 4

A

L

and =2

A

L

.

(b) The probability of finding the particle between 0 and8

Lis

= = + =

8 8

2 2 2

0 0

2 1 1cos 0.409

4 2

L L

xdx A dx

L