Solucionario serway cap 41

41Quantum Mechanics

CHAPTER OUTLINE

41.1 An Interpretation of Quantum Mechanics

41.2 The Quantum Particle under Boundary Conditions

41.3 The Schrödinger Equation41.4 A Particle in a Well of Finite

Height41.5 Tunneling Through a Potential

Energy Barrier41.6 Applications of Tunneling 41.7 The Simple Harmonic Oscillator

ANSWERS TO QUESTIONS

Q41.1 A particle’s wave function represents its state, contain-ing all the information there is about its location and motion. The squared absolute value of its wave function tells where we would classically think of the particle as spending most its time. Ψ 2 is the probability distribution function for the position of the particle.

*Q41.2 For the squared wave function to be the probability per length of fi nding the particle, we require

ψ ψ2 0 48

7 4

0 16=−

= =. .

nm nm nmand 0.4/ nnm

(i) Answer (e). (ii) Answer (e).

*Q41.3 (i) For a photon a and b are true, c false, d, e, f, and g true, h false, i and j true.

(ii) For an electron a is true, b false, c, d, e, f true, g false, h, i and j true.

Note that statements a, d, e, f, i, and j are true for both.

*Q41.4 We consider the quantity h2n2/8mL2. In (a) it is h21/8m

1(3 nm)2 = h2/72 m

1 nm2.

In (b) it is h24/8m1(3 nm)2 = h2/18 m

1 nm2.

In (c) it is h21/16m1(3 nm)2 = h2/144 m

1 nm2.

In (d) it is h21/8m1(6 nm)2 = h2/288 m

1 nm2.

In (e) it is 021/8m1(3 nm)2 = 0.

The ranking is then b > a > c > d > e.

Q41.5 The motion of the quantum particle does not consist of moving through successive points. The particle has no defi nite position. It can sometimes be found on one side of a node and sometimes on the other side, but never at the node itself. There is no contradiction here, for the quantum particle is moving as a wave. It is not a classical particle. In particular, the particle does not speed up to infi nite speed to cross the node.

463

Note: In chapters 39, 40, and 41 we use u to represent the speed of a particle with mass, reserving v for the speeds associated with reference frames, wave functions, and photons.

ISMV2_5104_ch41.indd 463ISMV2_5104_ch41.indd 463 7/2/07 5:36:56 PM7/2/07 5:36:56 PM

464 Chapter 41

Q41.6 Consider a particle bound to a restricted region of space. If its minimum energy were zero, then the particle could have zero momentum and zero uncertainty in its momentum. At the same time, the uncertainty in its position would not be infi nite, but equal to the width of the region. In such a case, the uncertainty product ∆ ∆x px would be zero, violating the uncertainty principle. This contradiction proves that the minimum energy of the particle is not zero.

*Q41.7 Compare Figures 41.4 and 41.7 in the text. In the square well with infi nitely high walls, the particle’s simplest wave function has strict nodes separated by the length L of the well. The

particle’s wavelength is 2L, its momentum h

L2, and its energy

p

m

h

mL

2 2

22 8= . Now in the well with

walls of only fi nite height, the wave function has nonzero amplitude at the walls. In this fi nite-depth well …

(i) The particle’s wavelength is longer, answer (a).

(ii) The particle’s momentum in its ground state is smaller, answer (b).

(iii) The particle has less energy, answer (b).

Q41.8 As Newton’s laws are the rules which a particle of large mass follows in its motion, so the Schrödinger equation describes the motion of a quantum particle, a particle of small or large mass. In particular, the states of atomic electrons are confi ned-wave states with wave functions that are solutions to the Schrödinger equation.

*Q41.9 Answer (b). The refl ected amplitude decreases as U decreases. The amplitude of the refl ected wave is proportional to the refl ection coeffi cient, R, which is 1− T , where T is the transmission coeffi cient as given in equation 41.22. As U decreases, C decreases as predicted by equation 41.23, T increases, and R decreases.

*Q41.10 Answer (a). Because of the exponential tailing of the wave function within the barrier, the tun-neling current is more sensitive to the width of the barrier than to its height.

Q41.11 Consider the Heisenberg uncertainty principle. It implies that electrons initially moving at the same speed and accelerated by an electric fi eld through the same distance need not all have the same measured speed after being accelerated. Perhaps the philosopher could have said “it is necessary for the very existence of science that the same conditions always produce the same results within the uncertainty of the measurements.”

Q41.12 In quantum mechanics, particles are treated as wave functions, not classical particles. In classical mechanics, the kinetic energy is never negative. That implies that E U≥ . Treating the particle as a wave, the Schrödinger equation predicts that there is a nonzero probability that a particle can tunnel through a barrier—a region in which E U< .

*Q41.13 Answer (c). Other points see a wider potential-energy barrier and carry much less tunneling current.


Quantum Mechanics 465

SOLUTIONS TO PROBLEMS

Section 41.1 An Interpretation of Quantum Mechanics

P41.1 (a) ψ x Ae A x Aii x( ) = = ×( ) + ××( )5 00 10 10

10

5 10 5 1.

cos sin 0010 x A kx Ai kx( ) = ( ) + ( )cos sin goes

through a full cycle when x changes by l and when kx changes by 2π . Then kλ π= 2

where k = × =−5 00 10210 1. mπλ

. Then λ π=×( ) = × −2

5 00 101 26 1010

10mm

.. .

(b) ph= = × ⋅

×= ×

−

−−

λ6 626 10

1 26 105 27 10

34

1024.

..

J s

mkg m s⋅

(c) me = × −9 11 10 31. kg

Km u

m

p

me

e

= = =× ⋅( )× ×

−2 2 2 24 2

2 2

5 27 10

2 9 11 1

.

.

kg m s

001 52 10

1 52 10

1 60 103117

17

−−

−

( ) = × = ××kg

JJ

..

. −− =19 95 5J eV

eV.

P41.2 Probability P xa

x adx

a

aa

a

a

a

= ( ) =+( ) = ⎛⎝

⎞⎠⎛⎝⎞⎠

− −∫ ∫ψ

π π2

2 2

1tan−−

−

⎛⎝⎞⎠

1 x

a a

a

P = − −( )⎡⎣ ⎤⎦ = − −⎛⎝⎞⎠

⎡⎣⎢

⎤⎦⎥

− −11 1

1

4 41 1

π ππ π

tan tan == 1

2

Section 41.2 The Quantum Particle under Boundary Conditions

P41.3 E1192 00 3 20 10= = × −. .eV J

For the ground state, Eh

m Le1

2

28=

(a) Lh

m Ee

= = × =−

84 34 10 0 434

1

10. .m nm

(b) ∆E E Eh

m L

h

m Le e

= − =⎛⎝⎜

⎞⎠⎟−⎛⎝⎜

⎞⎠⎟=2 1

2

2

2

248 8

6 00. eeV

P41.4 For an electron wave to “fi t” into an infi nitely deep potential well, an integral number of half-wavelengths must equal the width of the well.

nλ2

1 00 10 9= × −. m so λ = × =−2 00 10 9.

n

h

p

(a) Since Kp

m

h

m

h

m

nn

e e e

= =( )

=×( ) = (−

2 2 2 2 2

9 22

2 2 2 2 100 377

λ. )) eV

For K ≈ 6 eV n = 4

(b) With n = 4, K = 6 03. eV

FIG. P41.4


466 Chapter 41

P41.5 (a) We can draw a diagram that parallels our treatment of standing mechanical waves. In each state, we measure the distance d from one node to another (N to N), and base our solution upon that:

Since dN to N =λ2

and λ = h

p

ph h

d= =λ 2

Next, Kp

m

h

m d de e

= = =× ⋅( )×

−2 2

2 2

34 2

2 8

1 6 626 10

8 9 11

.

.

J s

110 31−( )⎡

⎣⎢⎢

⎤

⎦⎥⎥kg

Evaluating, Kd

= × ⋅−6 02 10 38

2

. J m2

Kd

= × ⋅−3 77 10 19

2

. eV m2

In state 1, d = × −1 00 10 10. m K1 37 7= . eV

In state 2, d = × −5 00 10 11. m K2 151= eV

In state 3, d = × −3 33 10 11. m K3 339= eV

In state 4, d = × −2 50 10 11. m K4 603= eV

(b) When the electron falls from state 2 to state 1, it puts out energy

E hfhc= − = = =151 37 7 113eV eV eV.λ

into emitting a photon of wavelength

λ = =× ⋅( ) ×( )

(

−hc

E

6 626 10 10

113

34 8. J s 3.00 m s

eV)) ×( ) =−1 60 1011 019.

.J eV

nm

The wavelengths of the other spectral lines we fi nd similarly:

Transition 4 3→ 4 2→ 4 1→ 3 2→ 3 1→ 2 1→

E eV( ) 264 452 565 188 302 113

λ nm( ) 4.71 2.75 2.20 6.60 4.12 11.0

*P41.6 For the bead’s energy we have both (1/2)mu2 and h2n2/8mL2. Then

n mumL

h

muL

h= =1

2

8 222

2 note that this expression can be thought of as 2L L

dNNλ=

Evaluating, n =×

−2 0 005( . kg)(10 m) 0.2 m

3.156 10 s (6.

10

7 6626 10 J s)10

× ⋅= ×−34

129 56.

P41.7 ∆Ehc h

m L

h

m Le e

= =⎛⎝⎜

⎞⎠⎟

−⎡⎣ ⎤⎦ =λ

2

22 2

2

282 1

3

8

Lh

m ce

= = × =−3

87 93 10 0 79310λ. .m nm

FIG. P41.5



P41.8 ∆Ehc h

m L

h

m Le e

= =⎛⎝⎜

⎞⎠⎟

−⎡⎣ ⎤⎦ =λ

2

22 2

2

282 1

3

8

so Lh

m ce

= 3

8

λ

P41.9 The confi ned proton can be described in the same way as a standing wave on a string. At level 1, the node-to-node distanceof the standing wave is 1 00 10 14. × − m, so the wavelength istwice this distance:

h

p= × −2 00 10 14. m

The proton’s kinetic energy is

K mup

m

h

m= = = =

× ⋅( )−1

2 2 2

6 626 10

2 1 672

2 2

2

34 2

λ.

.

J s

××( ) ×( )= ×

− −

−

10 2 00 10

3 29 10

27 14 2

13

kg m

J

1.6

.

.

00 10 J eVMeV

×=−19 2 05.

In the fi rst excited state, level 2, the node-to-node distance is half as long as in state 1. The momentum is two times larger and the energy is four times larger: K = 8 22. MeV.

The proton has mass, has charge, moves slowly compared to light in a standing wave state, and stays inside the nucleus. When it falls from level 2 to level 1, its energy change is

2 05 8 22 6 16. . .MeV MeV MeV− = −

Therefore, we know that a photon (a traveling wave with no mass and no charge) is emitted at the

speed of light, and that it has an energy of +6 16. MeV .

Its frequency is fE

h= =

×( ) ×( )×

−

−

6 16 10 1 60 10

6 626 10

6 19. .

.

eV J eV334

211 49 10J s

Hz⋅

= ×.

And its wavelength is λ = = ××

= ×−−c

f

3 00 10

1 49 102 02 10

8

21 113.

..

m s

sm

This is a gamma ray , according to the electromagnetic spectrum chart in Chapter 34.

P41.10 The ground state energy of a particle (mass m) in a 1-dimensional box of width L is E

h

mL1

2

28= .

(a) For a proton m = ×( )−1 67 10 27. kg in a 0.200-nm wide box:

E1

34 2

27

6 626 10

8 1 67 10 2 00 1=

× ⋅( )×( ) ×

−

−

.

. .

J s

kg 008 22 10 5 13 10

10 222 3

−− −

( ) = × = ×m

J eV. .

(b) For an electron m = ×( )−9 11 10 31. kg in the same size box:

E1

34 2

31

6 626 10

8 9 11 10 2 00 1=

× ⋅( )×( ) ×

−

−

.

. .

J s

kg 001 51 10 9 41

10 218

−−

( ) = × =m

J eV. .

(c) The electron has a much higher energy because it is much less massive.

FIG. P41.9


468 Chapter 41

*P41.11 Eh

mLnn =

⎛⎝⎜

⎞⎠⎟

2

22

8

E1

34 2

27

6 626 10

8 1 67 10 2 00 1=

× ⋅( )×( ) ×

−

−

.

. .

J s

kg 008 22 10

14 214

−−

( ) = ×m

J.

E1 0 513= . MeV E E2 14 2 05= = . MeV E E3 19 4 62= = . MeV

Yes; the energy differences are ~1 MeV, which is a typical energy for a g -ray photon as radiated by an atomic nucleus in an excited state.

P41.12 (a) The energies of the confi ned electron are Eh

m Lnn

e

=2

22

8. Its energy gain in the quantum

jump from state 1 to state 4 is h

m Le

2

22 2

84 1−( ) and this is the photon

energy: h

m Lhf

hc

e

2

2

15

8= =

λ. Then 8 152m cL he = λ and L

h

m ce

=⎛⎝⎜

⎞⎠⎟

15

8

1 2λ

.

(b) Let ′λ represent the wavelength of the photon emitted: hc h

m L

h

m L

h

m Le e e′= − =

λ

2

22

2

22

2

284

82

12

8.

Then hc

hc

h m L

m L he

eλλ′ =

( )=

2 2

2 2

15 8

8 12

5

4 and ′ =λ λ1 25. .

*P41.13 (a) From ∆x∆p ≥ h/2 with ∆x = L, the uncertainty in momentum must be at least ∆p ≈ h/2L .

(b) Its energy is all kinetic, E = p2/2m = (∆p)2/2m ≈ h2/8mL2 = h2/(4p)28mL2.

Compared to the actual h2/8mL2, this estimate is too low by 4p2 ≈ 40 times. The actual wave function does not have the particular (Gaussian) shape of a minimum-uncertainty wave function. The result correctly displays the pattern of dependence of the energy on the mass and on the length of the well.

P41.14 (a) x xL

x

Ldx

Lx

x

Ld

L

= ⎛⎝

⎞⎠ = −⎛⎝

⎞⎠∫

2 2 2 1

2

1

2

42

0

sin cosπ π

xxL

0∫

xL

x

L

L x

L

x

L

x

L

L

= − +⎡⎣⎢

⎤⎦⎥

1

2

1

16

4 4 42

0

2

2ππ π π

sin cos00 2

L L=

(b) Probability = ⎛⎝

⎞⎠ = −∫

2 2 1 1

42

0 490

0 510

L

x

Ldx

Lx

L

L

L

L

sin sin.

. ππ

44

0 490

0 510π x

L L

L⎡⎣⎢

⎤⎦⎥ .

.

Probability = − −( ) = × −0 0201

42 04 1 96 5 26 10 5. sin . sin . .

ππ π

(c) Probability x

L

x

L L

L

−⎡⎣⎢

⎤⎦⎥

= × −1

4

43 99 10

0 240

0 2602

ππ

sin ..

.

(d) In the n = 2 graph in the text’s Figure 41.4(b), it is more probable to fi nd the particle

either near xL=4

or xL= 3

4 than at the center, where the probability density is zero.

Nevertheless, the symmetry of the distribution means that the average position is

L

2.



P41.15 Normalization requires

ψ 21dx

all space∫ = or A

n x

Ldx

L2 2

0

1sinπ⎛

⎝⎞⎠ =∫

An x

Ldx A

LL2 2

0

2

21sin

π⎛⎝

⎞⎠ = ⎛

⎝⎞⎠ =∫ or A

L= 2

*P41.16 (a) The probability is ψ π π1

2

0

32

0

32 11

2dx

L

x

Ldx

L

x

L

L L/ /

sin cos∫ ∫= ⎛⎝⎞⎠ = − ⎛⎛

⎝⎞⎠

⎡⎣⎢

⎤⎦⎥∫ dx

L

0

3/

= − ⎛

⎝⎞⎠

⎡⎣⎢

⎤⎦⎥= − ⎛1

2

2 1

3

1

2

2

30

3

Lx

L x

L

L

ππ

ππ

sin sin/

⎝⎝⎞⎠ = − =1

3

0 866

20 196

..

π

(b) Classically, the particle moves back and forth steadily, spending equal time intervals in

each third of the line. Then the classical probability is 0.333, signifi cantly larger .

(c) The probability is ψ π99

2

0

32

0

32 99 11

1dx

L

x

Ldx

L

L L/ /

sin cos∫ ∫= ⎛⎝

⎞⎠ = − 998

0

3 π x

Ldx

L⎛⎝

⎞⎠

⎡⎣⎢

⎤⎦⎥∫

/

= − ⎛⎝

⎞⎠

⎡⎣⎢

⎤⎦⎥= −1

198

198 1

3

1

1980

3

Lx

L x

L

L

ππ

πsin s

/

iin .661

30 0 333π( ) = − =

in agreement with the classical model .

*P41.17 In 0 ≤ ≤x L, the argument 2π x

L of the sine function ranges from 0 to 2π . The probability

density 2 22

L

x

L⎛⎝⎞⎠

⎛⎝

⎞⎠sin

π reaches maxima at sinθ = 1 and sinθ = −1 . These points are at

2

2

π πx

L= and

2 3

2

π πx

L= .

Therefore the most probable positions of the particle are

at andxL

xL= =

4

3

4.


470 Chapter 41

*P41.18 (a) Probability = = ⎛⎝⎞⎠ = − ⎛

⎝⎞∫ ∫ψ π π

1

2

0

2

0

2 11

2dx

L

x

Ldx

L

x

L

� �

sin cos ⎠⎠⎡⎣⎢

⎤⎦⎥∫ dx

0

�

= − ⎛⎝

⎞⎠

⎡⎣⎢

⎤⎦⎥= − ⎛

⎝1

2

2 1

2

2

0Lx

L x

L L Lππ

ππ

sin sin� � �⎞⎞

⎠

(b)

1.2

1

0.8

0.6

0.4

0.2

00 0.5 1 1.5

Probability Curve for an InfinitePotential Well

L

Prob

abili

ty

FIG. P41.18(b)

(c) The wave function is zero for x < 0 and for x > L. The probability at l = 0 must be zero because the particle is never found at x < 0 or exactly at x = 0. The probability at l = L must be 1 for normalization: the particle is always found somewhere at x < L.

(d) The probability of fi nding the particle between x = 0 and x = � is 2

3, and between x = �

and x L= is 1

3.

Thus, ψ1

2

0

2

3dx

�

∫ =

∴ − ⎛⎝

⎞⎠ =

� �L L

1

2

2 2

3ππ

sin , or u u− =1

22

2

3ππsin

This equation for �L

can be solved by homing in on the solution with a calculator, the

result being �L= 0 585. , or � = 0 585. L to three digits.



P41.19 (a) The probability is P dxL

x

Ldx

L

L L

= = ⎛⎝⎞⎠ = −∫ ∫ψ π π2

0

32

0

32 2 1

2

1

2

2sin cos

xx

Ldx

L⎛⎝

⎞⎠∫

0

3

Px

L

x

L

L

= −⎛⎝

⎞⎠ = −⎛⎝

⎞⎠ = −1

2

2 1

3

1

2

2

3

1

30

3

ππ

ππ

sin sin33

40 196

π⎛⎝⎜

⎞⎠⎟= .

(b) The probability density is symmetric about xL=2

.

Thus, the probability of fi nding the particle between

xL= 2

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

the probability of fi nding it in the range

L

xL

3

2

3≤ ≤ is P = − ( ) =1 00 2 0 196 0 609. . . .

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

space is Pclassical =1

3. The result of part (a) is signifi cantly smaller ,

because of the curvature of the graph of the probability density.

Section 41.3 The Schrödinger Equation

P41.20 ψ x A kx B kx( ) = +cos sin ∂∂= − +ψ

xkA kx kB kxsin cos

∂∂= − −

2

22 2ψ

xk A kx k B kxcos sin − −( ) = − +( )2 2

2

mE U

mEA kx B kx

� �ψ cos sin

Therefore the Schrödinger equation is satisfi ed if

∂∂= −⎛⎝

⎞⎠ −( )

2

2 2

2ψ ψx

mE U

� or − +( ) = −⎛⎝

⎞⎠ +(k A kx B kx

mEA kx B kx2

2

2cos sin cos sin

�))

This is true as an identity (functional equality) for all x if Ek

m= �2 2

2.

P41.21 We have ψ ω= −( )Aei kx t so ∂∂=ψ ψ

xik and

∂∂= −

2

22ψ ψ

xk .

We test by substituting into Schrödinger’s equation: ∂∂= − = − −( )

2

22

2

2ψ ψ ψx

km

E U�

.

Since kp

h

p22

2

2

2

2

2

2 2= ( )

=( ) =π

λπ

� and E U

p

m− =

2

2

Thus this equation balances.

FIG. P41.19(b)


472 Chapter 41

P41.22 (a) Setting the total energy E equal to zero and rearranging the Schrödinger equation to isolate the potential energy function gives

U xm

d

dx( ) = ⎛

⎝⎜⎞⎠⎟

�2 2

22

1

ψψ

If ψ x Axe x L( ) = − 2 2

Then d

dxAx AxL

e

L

x L2

23 2

44 62 2

ψ = −( )−

or d

dx

x L

Lx

2

2

2 2

4

4 6ψ ψ=−( ) ( )

and U xmL

x

L( ) = −

⎛⎝⎜

⎞⎠⎟

�2

2

2

22

46

(b) See the figure to the right.

P41.23 Problem 41 in Chapter 16 helps students to understand how to draw conclusions from an identity.

(a) ψ x Ax

L( ) = −

⎛⎝⎜

⎞⎠⎟

12

2 d

dx

Ax

L

ψ = − 22

d

dx

A

L

2

2 2

2ψ = −

Schrödinger’s equation d

dx

mE U

2

2 2

2ψ ψ= − −( )�

becomes − = − −⎛⎝⎜

⎞⎠⎟+

−( ) −2 21

2 12 2

2

2 2

2 2 2A

L

mEA

x

L

m x A x L

� �� 22

2 2 2

( )−( )mL L x

− = − + −12 2

2

2 2

2

4L

mE mEx

L

x

L� �

This will be true for all x if both 1

2 2L

mE=�

and mE

L L�2 2 4

10− =

both these conditions are satisfi ed for a particle of energy EL m= �2

2

(b) For normalization, 1 1 122

2

2

2

22

2

4

4= −⎛⎝⎜

⎞⎠⎟

= − +⎛⎝⎜

⎞⎠−

∫ Ax

Ldx A

x

L

x

LL

L

⎟⎟−∫ dxL

L

12

3 5

2

3 5

2

32

3

2

5

42= − +⎡

⎣⎢⎤⎦⎥= − + + −

−

A xx

L

x

LA L L

LL

L

L

LLL

AL

AL

+⎡⎣⎢

⎤⎦⎥= ⎛

⎝⎞⎠ =

5

16

15

15

162

(c) P dxL

x

L

x

Ldx

L

L

L

L

= = − +⎛⎝⎜

⎞⎠⎟− −

∫ ψ 2

3

3 2

2

4

43

315

161

2∫∫ = − +⎡

⎣⎢⎤⎦⎥

= −

−

15

16

2

3 5

30

16 3

2

3

2

5

53

3

Lx

x

L

x

L

L

L

L

L

LL L

81 1 215+⎡

⎣⎢

⎤⎦⎥

P = =47

810 580.

FIG. P41.22(b)



P41.24 (a) ψ π1

2x

L

x

L( ) = ⎛

⎝⎞⎠cos ; P x x

L

x

L1 1

2 22( ) = ( ) = ⎛⎝⎞⎠ψ π

cos

ψ π2

2 2x

L

x

L( ) = ⎛

⎝⎞⎠sin ; P x x

L

x

L2 2

2 22 2( ) = ( ) = ⎛⎝

⎞⎠ψ π

sin

ψ π3

2 3x

L

x

L( ) = ⎛

⎝⎞⎠cos ; P x x

L

x

L3 3

2 22 3( ) = ( ) = ⎛⎝

⎞⎠ψ π

cos

P41.25 (a) With ψ x A kx( ) = ( )sin

d

dxA kx Ak kxsin cos= and

d

dxAk kx

2

22ψ = − sin

Then − = + =( )( )(

� �2 2

2

2 2 2 2

2 22 2

4

4 2m

d

dx

k

mA kx

h

m

ψ ππ λ

sin))= = = =ψ ψ ψ ψ ψp

m

m u

mmu K

2 2 22

2 2

1

2

(b) With ψ πλ

x Ax

A kx( ) = ⎛⎝

⎞⎠ =sin sin

2, the proof given in part (a) applies again.

Section 41.4 A Particle in a Well of Finite Height

P41.26

xx

00n = 1

n = 2

n = 3

ψ 2ψ

− L2

− L2

L2

L2

∞∞ ∞∞ (b)

FIG. P41.24(b)

FIG. P41.26


474 Chapter 41

P41.27 (a) See figure to the right.

(b) The wavelength of the transmitted wave traveling to the left is the same as the original wavelength, which equals

2L .

Section 41.5 Tunneling Through a Potential Energy Barrier

P41.28 C =×( ) −( ) ×( ) ⋅− −2 9 11 10 5 00 4 50 1 60 1031 19. . . . kg m ss

J s

m

1 055 10

3 62 10

34

9 1

.

.

× ⋅= ×

−

−

T e CL= = − ×( ) ×( )⎡⎣ ⎤− − −2 9 1 122 3 62 10 950 10exp . m m ⎦⎦ = −( )

= × −

exp .

.

6 88

1 03 10 3T

P41.29 From problem 28, C = × −3 62 109 1. m

We require 10 2 3 62 106 9 1− −= − ×( )⎡⎣ ⎤⎦exp . m L .

Taking logarithms, − = − ×( )−13 816 2 3 62 109 1. . m L

New L = 1 91. nm

∆L = − =1 91 0 950 0 959. . .nm nm nm

*P41.30 T e CL= −2 where Cm U E

=−( )2

�

(a) 22 2 9 11 10 0 01 1 6 10

1 055 10

31 19

CL =×( ) × ×( )

×

− −

−

. . .

. 3341010 0 102−( ) = . T e= =−0 102 0 903. .

(b) 22 2 9 11 10 1 6 10

1 055 1010

31 19

34CL =×( ) ×( )×

− −

−−

. .

.110 1 02( ) = . T e= =−1 02 0 359. .

(c) 22 2 6 65 10 10 1 6 10

1 055 10

27 6 19

3CL =×( ) × ×( )

×

− −

−

. .

. 441510 0 875−( ) = . T e= =−0 875 0 417. .

(d) 22 2 8 1

1 055 100 02 1 52 1034

33CL =( )( )×

( ) = ×−.. . T e e= = =− × − × −1 52 10 10 1 52 10 10 633 33

10. (ln )( . /ln ) ..59 1032×

FIG. P41.27(a)

FIG. P41.28



P41.31 T e CL= −2 where Cm U E

=−( )2

�

22 2 9 11 10 8 00 10

1 055 102

31 19

34CL =×( ) ×( )

×

− −

−

. .

..000 10 4 5810×( ) =− .

(a) T e= =−4 58 0 010 3. . , a 1% chance of transmission.

(b) R T= − =1 0 990. , a 99% chance of refl ection.

P41.32 The original tunneling probability is T e CL= −2 where

Cm U E

=−( )( ) = × × −( ) ×−2 2 2 9 11 10 12 1 61 2 31

�π . .kg 20 110

6 626 101 448 1 10

19 1 2

3410 1

−

−−( )

× ⋅= ×

J

J sm

..

The photon energy is hfhc= = ⋅ =λ

1 2402 27

eV nm

546 nmeV. , to make the electron’s new kinetic

energy 12 2 27 14 27+ =. . eV and its decay coeffi cient inside the barrier

′ =× × −( ) ×( )− −

C2 2 9 11 10 20 14 27 1 6 1031 19π . . .kg J

11 2

3410 1

6 626 101 225 5 10

..

× ⋅= ×−

−

J sm

Now the factor of increase in transmission probability is

e

ee e

C L

CLL C C

− ′

−− ′( ) × × ×= =

−2

22 2 10 0 223 109 10m . mm− = =

1 4 45 85 9e . .

Section 41.6 Applications of Tunneling

P41.33 With the wave function proportional to e CL− , the transmission coeffi cient and the tunneling

current are proportional to ψ 2, to e CL−2 .

Then, I

I

e0 500

0 515

2 10 0 0 500.

.

. .nm

nm

nm nm( )( ) =

− ( )( ))

− ( )( )( )= =

ee2 10 0 0 515

20 0 0 015 1 35. .. . .nm nm

P41.34 With transmission coeffi cient e CL−2 , the fractional change in transmission is

e e

e

L L− ( ) − ( ) +( )

−

−2 10 0 2 10 0 0 002 00

2 10

. . .nm nm nm

... . . . %0

20 0 0 002 001 0 0392 3 92nm( )− ( )= − = =

Le

FIG. P41.31


476 Chapter 41

Section 41.7 The Simple Harmonic Oscillator

P41.35 ψ ω= −( )Be m x2 2� so d

dx

mx

ψ ω ψ= − ⎛⎝⎞⎠�

and d

dx

mx

m2

2

22ψ ω ψ ω ψ= ⎛⎝

⎞⎠ + −⎛⎝

⎞⎠� �

Substituting into the Schrödinger equation gives

m

xm mE m

xω ψ ω ψ ψ ω� � � �

⎛⎝

⎞⎠ + −⎛⎝

⎞⎠ = − ⎛⎝

⎞⎠ + ⎛⎝

⎞⎠

22

2

22 22ψ

which is satisfi ed provided that E = �ω2

.

P41.36 Problem 41 in Chapter 16 helps students to understand how to draw conclusions from an identity.

ψ = −Axe bx2

so d

dxAe bx Aebx bxψ = −− −2 2

2 2

and d

dxbxAe bxAe b x e bbx bx bx

2

22 32 4 4 6

2 2 2ψ ψ= − − + = −− − − ++ 4 2 2b x ψ

Substituting into the Schrödinger equation, − + = − ⎛⎝⎞⎠ + ⎛⎝

⎞⎠6 4

22 22

2b b xmE m

xψ ψ ψ ω ψ� �

For this to be true as an identity, it must be true for all values of x.

So we must have both − = −62

2bmE

�and

4 2

2

bm= ⎛⎝⎞⎠ω�

(a) Therefore bm= ω2�

(b) and Eb

m= =3 3

2

2��ω

(c) The wave function is that of the first excited state .

P41.37 The longest wavelength corresponds to minimum photon energy, which must be equal to the spacing between energy levels of the oscillator:

hc k

mλω= =� � so λ π π= = ×( ) × −

2 2 3 00 109 11 108

31

cm

k.

.m s

kg

8.99 N m

⎛⎛⎝⎜

⎞⎠⎟=

1 2

600 nm



P41.38 (a) With ψ ω= −( )Be m x2 2� , the normalization condition ψ 21dx

xall∫ =

becomes 1 2 22 2 2 2

0

22 2

= = =− ( )

−∞

∞−( )

∞

∫ ∫B e dx B e dx Bm x m xω ω� � 11

2

πωm �

where Table B.6 in Appendix B was used to evaluate the integral.

Thus, 1 2= Bm

πω�

and Bm=⎛⎝⎜

⎞⎠⎟ωπ �

1 4

.

(b) For small d, the probability of fi nding the particle in the range − < <δ δ2 2

x is

ψ δ ψ δ δ ωπδ

δ2

2

22 2 0

1 2

0dx B em

−

−∫ = ( ) = =⎛⎝⎜

⎞⎠⎟�

P41.39 (a) For the center of mass to be fi xed, m u m u1 1 2 2 0+ = . Then

u u u um

mu

m m

mu= + = + = +

1 2 11

21

2 1

21 and u

m u

m m12

1 2

=+

Similarly, um

mu u= +2

12 2

and um u

m m21

1 2

=+

. Then

1

2

1

2

1

2

1

2

1

21 12

2 22 2 1 2

2 2

1 2

2m u m u kxm m u

m m

m+ + =+( )

+ 22 12 2

1 2

22

1 2 1 2

1 2

2

1

2

1

2

m u

m mkx

m m m m

m m

+( )+

=+( )

+( )uu kx u kx2 2 2 21

2

1

2

1

2+ = +µ

(b) d

dxu kx

1

2

1

202 2µ +⎛

⎝⎞⎠ = because energy is constant

01

22

1

22= + = + = +µ µ µu

du

dxk x

dx

dt

du

dxkx

du

dtkx

Then µ a kx= − , akx= −µ

. This is the condition for simple harmonic motion, that the

acceleration of the equivalent particle be a negative constant times the excursion from

equilibrium. By identifi cation with a x= −ω 2 , ωµ

π= =kf2 and f

k= 1

2π µ.


478 Chapter 41

P41.40 (a) With x = 0 and px = 0, the average value of x2 is ∆x( )2 and the average value of px2

is ∆px( )2. Then ∆∆

xpx

≥ �2

requires

Ep

m

k

p

p

m

k

px

x

x

x

≥ + = +2 2

2

2 2

22 2 4 2 8

� �

(b) To minimize this as a function of px2 , we require

dE

dp m

k

px x2

2

401

2 81

1= = + −( )�

Then k

p mx

�2

48

1

2= so p

mk mkx2

2 1 22

8 2=⎛⎝⎜

⎞⎠⎟=� �

and Emk

m

k

mk

k

m

k

m≥( ) + = +� �

��

2 2

2

8 4 4

2

Ek

mmin = =� �2 2

ω

Additional Problems

P41.41 Suppose the marble has mass 20 g. Suppose the wall of the box is 12 cm high and 2 mm thick. While it is inside the wall,

U mgy= = ( )( )( ) =0 02 9 8 0 12 0 023 5. . . .kg m s m J2

and E K mu= = = ( )( ) =1

2

1

20 02 0 8 0 006 42 2. . .kg m s J

Then Cm U E

=−( )

=( )( )

× −

2 2 0 02 0 017 1

1 055 10 34�. .

.

kg J

JJ sm

⋅= × −2 5 1032 1.

and the transmission coeffi cient is

e e e eCL− − ×( ) ×( ) − × −= = =−

2 2 2 5 10 2 10 10 10 2 332 3 29. . 00 4 3 10 4 3 10 1029 29 30

10 10. . ~×( ) − × −= =

P41.42 (a) λ = = × −2 2 00 10 10L . m

(b) ph= = × ⋅

×= ×

−

−−

λ6 626 10

3 31 1034

1024.

.J s

2.00 10 mkg m s⋅

(c) Ep

m= =

2

20 172. eV



P41.43 (a) See the figure.

(b) See the figure.

(c) y is continuous and ψ → 0 as x→ ±∞. The function can be normalized. It describes a particle bound near x = 0.

(d) Since y is symmetric,

ψ ψ2 2

0

2 1dx dx−∞

∞ ∞

∫ ∫= =

or 22

212 2

0

20A e dx

Ae ex−

∞−∞∫ =

−⎛⎝⎜

⎞⎠⎟

−( ) =α

α

This gives A = α .

(e) P a e dxx

x−( )→( )

−

=

= ( ) =−⎛⎝

⎞⎠∫1 2

2 2

0

1 2

22

2α αα

α αα1 2 ee e− −−( ) = −( ) =2 2 11 1 0 632α α .

*P41.44 If we had n = 0 for a quantum particle in a box, its momentum would be zero. The uncertainty in its momentum would be zero. The uncertainty in its position would not be infi nite, but just equal to the width of the box. Then the uncertainty product would be zero, to violate the uncertainty principle. The contradiction shows that the quantum number cannot be zero. In its ground state the particle has some nonzero zero-point energy.

*P41.45 (a) With ground state energy 0.3 eV, the energy in the n = 2 state is 22 × 0.3 eV = 1.2 eV. The energy in state 3 is 9 × 0.3 eV = 2.7 eV. The energy in state 4 is 16 × 0.3 eV = 4.8 eV. For the transition from the n = 3 level to the n = 1 level, the electron loses energy (2.7 – 0.3) eV = 2.4 eV. The photon carries off this energy and has wavelength hc/E = 1240 eV⋅nm/2.4 eV = 517 nm .

(b) For the transition from level 2 to level 1, the photon energy is 0.9 eV and the photon wavelength is l = hc/E = 1240 eV ⋅ nm/0.9 eV = 1.38 mm . This photon, with wavelength greater than 700 nm, is infrared .

For level 4 to 1, E = 4.5 eV and l = 276 nm ultraviolet .

For 3 to 2, E = 1.5 eV and l = 827 nm infrared .

For 4 to 2, E = 3.6 eV and l = 344 nm near ultraviolet .

For 4 to 3, E = 2.1 eV and l = 590 nm yellow-orange visible .

FIG. P41.43(a) FIG. P41.43(b)



480 Chapter 41

P41.46 (a) Use Schrödinger’s equation

∂∂= − −( )

2

2 2

2ψ ψx

mE U

� with solutions

ψ11 1= + −Ae Beik x ik x [region I ]

ψ 22= Ceik x [region II ]

Where kmE

1

2=�

and km E U

2

2=

−( )�

Then, matching functions and derivatives at x = 0

ψ ψ1 0 2 0( ) = ( ) gives A B C+ =

and d

dx

d

dx

ψ ψ1

0

2

0

⎛⎝

⎞⎠ =

⎛⎝

⎞⎠

gives k A B k C1 2−( ) =

Then Bk k

k kA= −

+1

12 1

2 1

and Ck k

A=+

2

1 2 1

Incident wave Aeikx refl ects Be ikx− , with probability RB

A

k k

k k

k k

k k

= =−( )+( )

=−( )+( )

2

22 1

2

2 1

2

1 2

2

1 2

2

1

1

(b) With E = 7 00. eV

and U = 5 00. eV

k

k

E U

E2

1

2 00

7 000 535= − = =.

..

The refl ection probability is R = −( )+( ) =

1 0 535

1 0 5350 092 0

2

2

.

..

The probability of transmission is T R= − =1 0 908.

FIG. P41.46(a)



P41.47 Rk k

k k

k k

k k=

−( )+( )

=−( )+( )

1 2

2

1 2

22 1

2

2 1

2

1

1

�2 2

2

k

mE U= −

for constant U

�2

12

2

k

mE=

since U = 0 (1)

�2

22

2

k

mE U= − (2)

Dividing (2) by (1), k

k

U

E22

12 1 1

1

2

1

2= − = − =

so

k

k2

1

1

2=

and therefore, R =−( )+( )

=−( )+( )

=1 1 2

1 1 2

2 1

2 10 029 4

2

2

2

2 .

P41.48 (a) The wave functions and probability densities are the same as those shown in the two lower curves in Figure 41.4 of the textbook.

(b) P dx1 1

2

0 150

22

1 00= = ⎛⎝

⎞⎠∫ ψ

π

. .sin

nm

0.350 nm

nm

xxdx

x

1 00

2 002

1 00

0 150

0 350

.

..

.

.

nm

nm

⎛⎝

⎞⎠

= ( ) −

∫nnm

4 nm nm

0.350 n

ππ

sin. .

2

1 00 0 150

x⎛⎝

⎞⎠

⎡⎣⎢

⎤⎦⎥

mm

In the above result we used sin sin2

2

1

42axdx

x

aax∫ = ⎛⎝

⎞⎠ −⎛⎝⎞⎠ ( ).

Therefore, P xx

1 1 001 00 2

1 00= ( ) − ⎛

⎝⎞⎠

⎡⎣⎢

⎤..

sin.

nmnm

2 nmππ

⎦⎦⎥0 150. nm

0.350 nm

P1 1 00 0 350 0 1501 00

0= ( ) − −. . ..

sin .nm nm nmnm

2π7700 0 300 0 200π π( ) − ( )[ ]{ } =sin . .

(c) Px

dxx

22

0 150

0 3502

1 00

2

1 002 00

2= ⎛

⎝⎞⎠ =∫.

sin.

..

. π −− ⎛⎝

⎞⎠

⎡⎣⎢

⎤⎦⎥

1 00

8

4

1 00 0 150

0 350.

sin. .

.

ππ x

P xx

20 150

0 3

1 001 00

4

4

1 00= − ⎛

⎝⎞⎠

⎡⎣⎢

⎤⎦⎥

..

sin. .

.

ππ 550

1 00 0 350 0 1501 00

41 40 0= −( ) − ( ) −. . .

.sin . sin

ππ ..

.

600

0 351

π( )[ ]{ }=

(d) Using En h

mLn =2 2

28, we fi nd that E1 0 377= . eV and E2 1 51= . eV .

FIG. P41.47



482 Chapter 41

P41.49 (a) fE

h= = ( )

× ⋅( )×

−

−1 80

6 626 10

1 60 1034

19.

.

.eV

J s

J

11.00 eVHz

⎛⎝⎜

⎞⎠⎟= ×4 34 1014.

(b) λ = = ××

= × =−c

f

3 00 10

4 34 106 91 10 69

8

147.

..

m s

Hzm 11 nm

(c) ∆ ∆E t ≥ �2

so ∆∆ ∆

Et

h

t≥ =

( )= × ⋅

×

−

−

�2 4

6 626 10

4 2 00 10

34

6π π.

.

J s

ssJ eV( ) = × = ×− −2 64 10 1 65 1029 10. .

P41.50 (a) Taking L L Lx y= = , we see that the expression for E becomes

Eh

m Ln n

ex y= +( )

2

22 2

8

For a normalizable wave function describing a particle, neither nx nor ny can be zero. The ground state, corresponding to n nx y= = 1, has an energy of

Eh

m L

h

m Le e1 1

2

22 2

2

281 1

4, = +( ) =

The fi rst excited state, corresponding to either nx = 2, ny = 1 or nx = 1, ny = 2, has an energy

E Eh

m L

h

m Le e2 1 1 2

2

22 2

2

282 1

5

8, ,= = +( ) =

The second excited state, corresponding to nx = 2, ny = 2, has an energy of

Eh

m L

h

m Le e2 2

2

22 2

2

282 2, = +( ) =

Finally, the third excited state, corresponding to either nx = 1, ny = 3 or nx = 3, nx = 1, has an energy

E Eh

m L

h

m Le e1 3 3 1

2

22 2

2

281 3

5

4, ,= = +( ) =

(b) The energy difference between the second excited state and the ground state is given by

∆E E Eh

m L

h

m L

h

m L

e e

e

= − = −

=

2 2 1 1

2

2

2

2

2

2

4

3

4

, ,

E1, 3, E3, 1

E2, 2

E1, 2, E2, 1

E1, 1

energy

h2

meL2

0Energy level diagram

FIG. P41.50(b)



P41.51 x x dx2 2 2=−∞

∞

∫ ψ

For a one-dimensional box of width L, ψ πn L

n x

L= ⎛

⎝⎞⎠

2sin .

Thus, xL

xn x

Ldx

L L

n

L2 2 2

0

2 2

2 2

2

3 2= ⎛

⎝⎞⎠ = −∫ sin

ππ

(from integral tables).

P41.52 (a) ψ 21dx

−∞

∞

∫ =

becomes

Ax

Ldx A

L x

LL

L2 2

4

422

2

1

4

4cos sin

πππ π⎛

⎝⎞⎠ = ⎛

⎝⎞⎠ +

−∫

xx

LA

L

L

L⎛⎝

⎞⎠

⎡⎣⎢

⎤⎦⎥

= ⎛⎝⎞⎠⎡⎣⎢⎤⎦⎥=

− 4

42

2 21

ππ

or AL

2 4= and AL

= 2.

(b) The probability of fi nding the particle between 0 and L

8 is

ψ ππ

2

0

82 2

0

82 1

4

1

20 409dx A

x

Ldx

L L

∫ ∫= ⎛⎝

⎞⎠ = + =cos .

P41.53 For a particle with wave function

ψ xa

e x a( ) = −2 for x > 0

and 0 for x < 0

(a) ψ x( ) =20, x < 0 and ψ 2 22

xa

e x a( ) = − , x > 0 as shown

(b) Prob x x dx dx<( ) = ( ) = ( ) =−∞ −∞∫ ∫0 0 0

20 0

ψ

(c) Normalization ψ ψ ψx dx dx dx( ) = + =−∞

∞

−∞

∞

∫ ∫ ∫2 20

2

0

1

02

0 10

2

0

2

0dx

ae dx e ex a x a

−∞

−∞

− ∞ −∞∫ ∫+ ⎛⎝⎞⎠ = − = − −( ) ==

< <( ) = = ⎛⎝⎞⎠

= −

∫ ∫ −

−

1

022

0

2

0

Prob x a dxa

e dx

e

ax a

a

ψ

22

0

21 0 865x a ae= − =− .

FIG. P41.53


484 Chapter 41

P41.54 (a) The requirement that n

Lλ2= so p

h nh

L= =λ 2

is still valid.

E pc mc Enhc

Lmc

K E mc

n

n n

= ( ) + ( ) ⇒ = ⎛⎝⎞⎠ + ( )

= −

2 2 22

2 2

2

222

2 2 2

2= ⎛

⎝⎞⎠ + ( ) −nhc

Lmc mc

(b) Taking L = × −1 00 10 12. m, m = × −9 11 10 31. kg, and n = 1, we fi nd K1144 69 10= × −. J .

Nonrelativistic, Eh

mL1

2

2

34 2

318

6 626 10

8 9 11 10= =

× ⋅( )×(

−

−

.

.

J s

kg)) ×( ) = ×−

−

1 00 106 02 10

12 214

..

mJ.

Comparing this to K1, we see that this value is too large by 28 6. % .

P41.55 (a) Ue

d=

∈− + − + − +⎛

⎝⎞⎠ + −( )

⎡⎣⎢

⎤⎦⎥=−(2

041

1

2

1

31

1

21

7 3

π))∈

= −e

d

k e

de

2

0

2

4

7

3π

(b) From Equation 41.14, K Eh

m d

h

m de e

= = ( ) =22

8 9 361

2

2

2

2 .

(c) E U K= + and dE

dd= 0 for a minimum:

7

3 180

2

2

2

3

k e

d

h

m de

e

− =

dh

k e m

h

m k ee e e e

=( )( ) = =

×( )−3

7 18 42

6 626 102

2

2

2

34.22

31 9 19 242 9 11 10 8 99 10 1 60 10( ) ×( ) ×( ) ×( )− −. . . C

== 0 049 9. nm

(d) Since the lithium spacing is a, where Na V3 = , and the density is Nm

V, where m is the

mass of one atom, we get:

aVm

Nm

m= ⎛⎝⎞⎠ =

⎛⎝⎜

⎞⎠⎟= × −1 3 1 3 271 66 10

density

kg. ××⎛⎝⎜

⎞⎠⎟

= × =−7

5302 80 10 0 280

1 3

10

kgm m nm. .

The lithium interatomic spacing of 280 pm is 5.62 times larger than the answer to (c). Thus it is of the same order of magnitude as the interatomic spacing 2d here.



P41.56 (a) ψ ω= −( )Bxe m x2 2�

d

dxBe Bx

mxe Bem x m xψ ωω ω= + −⎛⎝

⎞⎠ =−( ) −( )2 22 2

22� �

�−−( ) −( )− ⎛

⎝⎞⎠

= −

m x m xBm

x e

d

dxBx

m

ω ωω

ψ ω

2 2 2

2

2

2 2� �

�

�� ⎛

⎝⎞⎠ − ⎛

⎝⎞⎠ −−( ) −( )xe B

mxe B

mm x m xω ωω ω2 22 2

2 ⎛⎛⎝

⎞⎠ −⎛⎝

⎞⎠

= − ⎛⎝

⎞⎠

−( )xm

xe

d

dxB

m

m x2 2

2

2

2

3

ω

ψ ω

ω

�

�

�

xxe Bm

x em x m x−( ) −( )+ ⎛⎝

⎞⎠

ω ωω22

3 22 2� �

�

Substituting into the Schrödinger equation, we have

− ⎛⎝

⎞⎠ + ⎛

⎝⎞⎠

−( ) −( )3 22

3 22 2

Bm

xe Bm

x em x m xω ωω ω

� �� == − + ⎛⎝

⎞⎠

−( ) −( )22

22

2 22 2mEBxe

mx Bxem x m x

� �� ω ωω

This is true if − = −32ω E

�; it is true if E = 3

2

�ω.

(b) We never fi nd the particle at x = 0 because ψ = 0 there.

(c) y is maximized if d

dxx

mψ ω= = − ⎛⎝

⎞⎠0 1 2

�, which is true at x

m= ± �

ω.

(d) We require ψ 21dx

−∞

∞

∫ = :

1 2 22 2 2 2 22 2

= = =−( )

−∞

∞−( )∫ ∫B x e dx B x e dx Bm x m xω ω� � 11

4 23

2 1 2 3 2

3 2

πω

πωm

B

m��

( )=

( )

Then Bm m= ⎛⎝

⎞⎠ =

⎛⎝⎜

⎞⎠⎟

2 41 2

1 4

3 4 3 3

3

1 4

πω ω

π� �.

(e) At xm

= 2�ω

, the potential energy is

1

2

1

2

422 2 2m x m

mω ω

ωω= ⎛

⎝⎞⎠ =

�� . This is larger than

the total energy 3

2

�ω, so there is zero classical probability of fi nding the particle here.

(f) Probability = = ( ) =−( ) −( )ψ δ δω ω2 22

2 22 2

dx Bxe B x em x m x� �

Probability = ⎛⎝

⎞⎠⎛⎝

⎞⎠ =−( ) ( )δ

πω

ωδ ωπ

ω ω2 481 2

3 24m

me

mm m

��

�� ⎛⎛

⎝⎞⎠

−1 2

4e


486 Chapter 41

P41.57 (a) ψ 2

0

1dxL

∫ = : Ax

L

x

L

x

L2 2 216

28sin sin sin sin

π π π⎛⎝⎞⎠ +

⎛⎝

⎞⎠ +

⎛⎝⎞⎠

221

0

π x

Ldx

L⎛⎝

⎞⎠

⎡⎣⎢

⎤⎦⎥=∫

A

L L x

L

x

Ldx2

0216

28

2⎛⎝⎞⎠ +

⎛⎝⎞⎠ +

⎛⎝⎞⎠⎛⎝

⎞⎠sin sin

π πLL

AL x

L

x

Ldx

∫⎡

⎣⎢

⎤

⎦⎥ =

+ ⎛⎝⎞⎠

⎛⎝⎞⎠

1

17

2162 2sin cos

π π

00

2 3

0

17

2

16

3

L

x

x L

AL L x

L∫⎡

⎣⎢

⎤

⎦⎥ = + ⎛

⎝⎞⎠

⎡

⎣⎢

=

=

ππ

sin⎤⎤

⎦⎥ = 1

AL

2 2

17= , so the normalization constant is A

L= 2

17.

(b) ψ 21dx

a

a

−∫ = : A

x

aB

x

aA B

x

a2 2 2 2

22

2cos sin cos

π π π⎛⎝⎞⎠ +

⎛⎝⎞⎠ +

⎛⎝⎞⎠ ssin

π x

adx

a

a⎛⎝⎞⎠

⎡⎣⎢

⎤⎦⎥=

−∫ 1

The fi rst two terms are A a2 and B a2 . The third term is:

2

22

2 2A B

x

a

x

a

x

acos sin cos

π π π⎛⎝⎞⎠

⎛⎝⎞⎠

⎛⎝⎞⎠

⎡⎣⎢

⎤⎦⎥

ddx A Bx

a

x

adx

a

a

a

a

a

− −∫ ∫= ⎛

⎝⎞⎠⎛⎝⎞⎠

=

42 2

8

2cos sinπ π

AA B x

a a

a

3 203

ππ

cos ⎛⎝⎞⎠ =−

so that a A B2 2 1+( ) = , giving A Ba

2 2 1+ = .

P41.58 (a) x xa

e dxax0

1 22

0= ⎛⎝⎞⎠ =−

−∞

∞

∫ π, since the integrand is an odd function of x.

(b) x xa

x e dxax1

3 1 2

240

2

=⎛⎝⎜

⎞⎠⎟

=−

−∞

∞

∫ π, since the integrand is an odd function of x.

(c) x x dx x x x x x01 0 1

2

0 1 0 1

1

2

1

2

1

2= +( ) = + + ( ) ( )−∞

∞

∫ ψ ψ ψ ψ ddx−∞

∞

∫

The fi rst two terms are zero, from (a) and (b). Thus:

x xa

ea

xe dxax ax01

1 42

3 1 4

22 24= ⎛⎝⎞⎠

⎛⎝⎜

⎞⎠⎟

− −

−∞

∞

π π∫∫ ∫=⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

−∞

22

22 1

2 1 2

2

0

2 1 2

2ax e dx

a

ax

π

π 44

1

2

3

1 2πa

a

⎛⎝⎞⎠

=

, from Table B.6



P41.59 With one slit open P1 1

2= ψ or P2 2

2= ψ

With both slits open, P = +ψ ψ1 2

2

At a maximum, the wave functions are in phase Pmax = +( )ψ ψ1 2

2

At a minimum, the wave functions are out of phase Pmin = −( )ψ ψ1 2

2

Now P

P1

2

1

2

2

2 25 0= =ψψ

. , so ψψ

1

2

5 00= .

and

P

Pmax

min

.

.=

+( )−( )

=+( )ψ ψ

ψ ψ

ψ ψ

ψ1 2

2

1 2

22 2

25 00

5 00 22 2

2

2

2

6 00

4 00

36 0

16 02 25

−( )= ( )( ) = =

ψ.

.

.

..

P41.2 1

2

P41.4 (a) 4 (b) 6 03. eV

P41.6 9.56 × 1012

P41.8 3

8

1 2h

m ce

λ⎛⎝⎜

⎞⎠⎟

P41.10 (a) 5 13. meV (b) 9 41. eV (c) The much smaller mass of the electron requires it to have much more energy to have the same momentum.

P41.12 (a) 15

8

1 2h

m ce

λ⎛⎝⎜

⎞⎠⎟

(b) 1 25. λ

P41.14 (a) L

2 (b) 5 26 10 5. × − (c) 3 99 10 2. × − (d) See the solution.

P41.16 (a) 0.196 (b) The classical probability is 0.333, signifi cantly larger. (c) 0.333 for both classical and quantum models.

P41.18 (a) � �L L− ⎛

⎝⎞⎠

1

2

2

ππ

sin (b) See the solution. (c) The wave function is zero for x < 0 and for x > L.

The probability at l = 0 must be zero because the particle is never found at x < 0 or exactly at x = 0. The probability at l = L must be 1 for normalization. This statement means that the particle is always found somewhere at x < L. (d) l = 0.585L

P41.20 See the solution; �2 2

2

k

m

P41.22 (a) �2

2

2

22

46

mL

x

L−

⎛⎝⎜

⎞⎠⎟

(b) See the solution.

ANSWERS TO EVEN PROBLEMS


488 Chapter 41

P41.24 (a) ψ π1

2x

L

x

L( ) = ⎛

⎝⎞⎠cos P x

L

x

L122( ) = ⎛⎝⎞⎠cos

π ψ π

2

2 2x

L

x

L( ) = ⎛

⎝⎞⎠sin

P xL

x

L222 2( ) = ⎛⎝

⎞⎠sin

π ψ π3

2 3x

L

x

L( ) = ⎛

⎝⎞⎠cos P x

L

x

L322 3( ) = ⎛⎝

⎞⎠cos

π

(b) See the solution.

P41.26 See the solution.

P41.28 1 03 10 3. × −

P41.30 (a) 0.903 (b) 0.359 (c) 0.417 (d) 10 6 59 1032− ×.

P41.32 85.9

P41.34 3 92. %

P41.36 (a) See the solution. bm= ω2�

(b) E = 3

2�ω (c) first excited state

P41.38 (a) Bm=⎛⎝⎜

⎞⎠⎟ωπ �

1 4

(b) δ ωπm

�⎛⎝⎜

⎞⎠⎟

1 2


P41.42 (a) 2 00 10 10. × − m (b) 3 31 10 24. × ⋅− kg m s (c) 0 172. eV


P41.46 (a) See the solution. (b) 0 092 0. , 0 908.

P41.48 (a) See the solution. (b) 0 200. (c) 0 351. (d) 0 377. eV, 1 51. eV

P41.50 (a) h

m Le

2

24,

5

8

2

2

h

m Le

, h

m Le

2

2 , 5

4

2

2

h

m Le

(b) See the solution, 3

4

2

2

h

m Le

P41.52 (a) 2

L (b) 0 409.

P41.54 (a) nhc

Lm c mc

2

22 4 2⎛

⎝⎞⎠ + − (b) 46 9. fJ; 28 6. %

P41.56 (a) 3

2

�ω (b) x = 0 (c) ± �

mω (d)

4 3 3

3

1 4m ωπ �

⎛⎝⎜

⎞⎠⎟

(e) 0 (f ) 81 2

4δ ωπ

me

�⎛⎝

⎞⎠

−

P41.58 (a) 0 (b) 0 (c) 2 1 2a( )−


Solucionario serway cap 41

Engineering

Transcript of Solucionario serway cap 41