M40 YOUN7066 13 ISM c40 tủ tài liệu training

40

QUANTUM MECHANICS

40.1.

IDENTIFY: Using the momentum of the free electron, we can calculate k and ω and use these to express
its wave function.
SET UP: Ψ( x, t ) = Aeikx e − iω t , k = p/ , and ω = k 2 /2m.
EXECUTE: k =

ω=

p

=−

4.50 × 10−24 kg ⋅ m/s

= −4.27 × 1010 m −1.

k 2 (1.055 × 10−34 J ⋅ s)(4.27 × 1010 m −1 ) 2
=
= 1.05 × 1017 s −1.
2m
2(9.108 × 10−31 kg)
10

40.2.

1.055 × 10−34 J ⋅ s

−1

17

−1

Ψ( x, t ) = Ae− i[4.27 ×10 m ) x e−i[1.05 ×10 s ]t .
EVALUATE: The wave function depends on position and time.
IDENTIFY: Using the known wave function for the particle, we want to find where its probability function
is a maximum.
Ψ( x, t ) = A [eikxe− iωt − e2ikxe−4iωt ][e− ikx e+ iωt − e−2ikx e+4iω t ].
2

SET UP:

2

Ψ( x, t ) = A (2 − [e− i ( kx − 3ωt ) + e + i ( kx − 3ωt ) ]) = 2 A (1 − cos(kx − 3ωt )).
2

2

2

2

2

EXECUTE: (a) For t = 0, Ψ( x, t ) = 2 A (1 − cos(kx)). Ψ( x, t )

this happens when kx = (2n + 1)π , n = 0,1,… . Ψ( x, t )
(b) t =

2π

ω

2

2

2

is a maximum when cos( kx) = −1 and

is a maximum for x =

π 3π
k

,

k

, etc.

2

and 3ωt = 6π . Ψ( x, t ) = 2 A (1 − cos( kx − 6π )). Maximum for kx − 6π = π , 3π ,... , which

gives maxima when x =

7π 9π
,
.
k
k

(c) From the results for parts (a) and (b), vav =

ω − ω1
7π /k − π /k 3ω
with ω 2 = 4ω , ω1 = ω ,
=
. vav = 2
k
2π /ω
k2 − k1

3ω
.
k
EVALUATE: The expressions in part (c) agree.
IDENTIFY: Use the wave function from Example 40.1.

k2 = 2k and k1 = k gives vav =

40.3.

2

2

Ψ( x, t ) = 2 A {1 + cos[( k2 − k1 ) x − (ω 2 − ω1 )t ]}. k2 = 3k1 = 3k . ω =

SET UP:
2

k2
, so ω 2 = 9ω1 = 9ω .
2m

2

Ψ( x, t ) = 2 A {1 + cos(2kx − 8ωt )}.
2

2

EXECUTE: (a) At t = 2π /ω , Ψ( x, t ) = 2 A {1 + cos(2kx − 16π )}. Ψ( x, t )

2

is maximum for

cos(2kx − 16π ) = 1. This happens for 2kx − 16π = 0, 2π ,... . Smallest positive x where Ψ( x, t )

2

is a

8π
maximum is x = .
k
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

40-1


40-2

Chapter 40

8π /k 4ω
ω − ω1 8ω 4ω
=
. vav = 2
=
=
.
2π /ω
k
k2 − k1 2k
k
EVALUATE: The two expressions agree.
IDENTIFY: We have a free particle, described in Example 40.1.
ω − ω1
( k22 − k12 )

(k2 + k1)(k2 − k1)
p
SET UP and EXECUTE: vav = 2
=
=
=
( k2 + k1 ) = av .
k2 − k1 2m k2 − k1
2m
k2 − k1
2m
m
EVALUATE: This is the same as the classical physics result, v = p/m = mv/m = v.
(b) From the result of part (a), vav =

40.4.

40.5.

2

IDENTIFY and SET UP: ψ ( x) = A sin kx. The position probability density is given by ψ ( x) = A2 sin 2 kx.
EXECUTE: (a) The probability is highest where sin kx = 1 so kx = 2π x/λ = nπ /2, n = 1, 3, 5,…
x = nλ /4, n = 1, 3, 5,… so x = λ /4, 3λ /4, 5λ /4,…
2

(b) The probability of finding the particle is zero where ψ = 0, which occurs where sin kx = 0 and
kx = 2π x/λ = nπ , n = 0, 1, 2,…
x = nλ /2, n = 0,1, 2,… so x = 0, λ /2, λ , 3λ /2,…
EVALUATE: The situation is analogous to a standing wave, with the probability analogous to the square of

the amplitude of the standing wave.
40.6.

IDENTIFY and SET UP:

2

Ψ = Ψ ∗Ψ
2

2

EXECUTE: Ψ ∗ = ψ ∗ sin ωt , so Ψ = Ψ ∗Ψ = ψ ∗ψ sin 2 ωt = ψ sin 2 ωt. Ψ

2

is not time-independent, so

Ψ is not the wavefunction for a stationary state.
EVALUATE: Ψ = ψ eiωφ = ψ (cos ωt + i sin ωt ) is a wavefunction for a stationary state, since for it
2

2

Ψ = ψ , which is time independent.
40.7.

IDENTIFY: Determine whether or not −
SET UP: −

2

d 2ψ 1

2m dx 2
2

2

d 2ψ

2m dx 2

+ U ψ 1 = E1ψ 1 and −

2

+ U ψ is equal to Eψ , for some value of E.

d 2ψ 2

2m dx 2

+ U ψ 2 = E2ψ 2

d 2ψ

+ U ψ = BE1ψ 1 + CE2ψ 2 . If ψ were a solution with energy E, then
2m dx 2
BE1ψ 1 + CE2ψ 2 = BEψ 1 + CEψ 2 or B ( E1 − E )ψ 1 = C ( E − E2 )ψ 2 . This would mean that ψ 1 is a constant

EXECUTE: −

multiple of ψ 2 , and ψ 1 and ψ 2 would be wave functions with the same energy. However, E1 ≠ E2 , so this
is not possible, and ψ cannot be a solution to Eq. (40.23).
EVALUATE: ψ is a solution if E1 = E2 ; see Exercise 40.9.
40.8.

IDENTIFY: Apply the Heisenberg Uncertainty Principle in the form ΔxΔpx ≥ /2.
SET UP: The uncertainty in the particle position is proportional to the width of ψ ( x ).
EXECUTE: The width of ψ ( x) is inversely proportional to α . This can be seen by either plotting the
function for different values of α or by finding the full width at half-maximum. The particle’s uncertainty
in position decreases with increasing α .
(b) Since the uncertainty in position decreases, the uncertainty in momentum must increase.
EVALUATE: As α increases, the function A(k ) in Eq. (40.19) must become broader.

40.9.

IDENTIFY: Determine whether or not −

2

d 2ψ

2m dx 2

+ U ψ is equal to Eψ .

SET UP: ψ 1 and ψ 2 are solutions with energy E means that −

−

2

d 2ψ 2

2m dx 2

2

d 2ψ 1

2m dx 2

+ U ψ 1 = Eψ 1 and

+ U ψ 2 = Eψ 2 .

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


Quantum Mechanics

EXECUTE: Eq. (40.23):

40-3

− 2 d 2ψ
+ U ψ = Eψ . Let ψ = Aψ 1 + Bψ 2

2m dx 2

− 2 d2
( Aψ 1 + Bψ 2 ) + U ( Aψ 1 + Bψ 2 ) = E ( Aψ 1 + Bψ 2 )
2m dx 2
2 2
2 2
⎛
⎞
⎛
⎞
d ψ1
d ψ2
⇒ A⎜ −
+ Uψ 1 − Eψ 1 ⎟ + B ⎜ −
+ Uψ 2 − Eψ 2 ⎟ = 0. But each of ψ 1 and ψ 2 satisfy
2
2
⎜ 2m dx
⎟
⎜ 2m dx
⎟
⎝
⎠
⎝
⎠
Schrödinger’s equation separately so the equation still holds true, for any A or B.
EVALUATE: If ψ 1 and ψ 2 are solutions of the Schrodinger equation for different energies, then

⇒

ψ = Bψ 1 + Cψ 2 is not a solution (Exercise 40.7).
40.10.

IDENTIFY: To describe a real situation, a wave function must be normalizable.
2

ψ dV is the probability that the particle is found in volume dV. Since the particle must be

SET UP:

somewhere, ψ must have the property that

∫ψ

2

dV = 1 when the integral is taken over all space.

EXECUTE: (a) For normalization of the one-dimensional wave function, we have

1= ∫

∞

−∞

2

0

∞

0

−∞

0

−∞

ψ dx = ∫ ( Aebx )2dx + ∫ (Ae −bx )2 dx = ∫

∞

A2e2bx dx + ∫ A2e−2bx dx.
0

∞⎫
⎧ 2bx 0
e−2bx ⎪ A2
1= A ⎨
+
, which gives A = b = 2.00 m −1 = 1.41 m –1/2
⎬=
⎪⎩ 2b −∞ −2b 0 ⎪⎭ b
(b) The graph of the wavefunction versus x is given in Figure 40.10.
2 ⎪e

(c) (i) P = ∫

+5.00 m

2

−0.500 m

ψ dx = 2∫

+5.00 m

0

A2e−2bx dx, where we have used the fact that the wave function is an

even function of x. Evaluating the integral gives
− A2 −2b(0.500 m)
− (2.00 m −1 ) −2.00
(e
− 1) =
(e
− 1) = 0.865
b
2.00 m −1
There is a little more than an 86% probability that the particle will be found within 50 cm of the origin.
0
0
A2
2.00 m −1
1

(ii) P = ∫ (Aebx )2 dx = ∫ A2e2bx dx =
=
= = 0.500
−
1
−∞
−∞
2b 2(2.00 m ) 2
P=

There is a 50-50 chance that the particle will be found to the left of the origin, which agrees with the fact
that the wave function is symmetric about the y-axis.
(iii) P = ∫

1.00 m

0.500 m

A2e−2bx dx

A2 −2(2.00 m−1 )(1.00 m) −2(2.00 m−1 )(0.500 m)
1
−e
(e
) = − (e−4 − e−2 ) = 0.0585
−2b
2
EVALUATE: There is little chance of finding the particle in regions where the wave function is small.
=

Figure 40.10

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


40-4

Chapter 40

40.11.

IDENTIFY and SET UP: The energy levels for a particle in a box are given by En =
EXECUTE: (a) The lowest level is for n = 1, and E1 =
(b) E =

(1)(6.626 × 10−34 J ⋅ s)2
8(0.20 kg)(1.3 m) 2

n2h2

8mL2

.

= 1.6 × 10−67 J.

1 2
2E
2(1.2 × 10−67 J)

=
= 1.3 × 10−33 m/s. If the ball has this speed the time it
mv so v =
2
m
0.20 kg

would take it to travel from one side of the table to the other is
1.3 m
t=
= 1.0 × 1033 s.
1.3 × 10−33 m/s
(c) E1 =

40.12.

h2
2

, E2 = 4 E1, so ΔE = E2 − E1 = 3E1 = 3(1.6 × 10−67 J) = 4.9 × 10−67 J.

8mL
(d) EVALUATE: No, quantum mechanical effects are not important for the game of billiards. The discrete,
quantized nature of the energy levels is completely unobservable.
IDENTIFY: Solve Eq. (40.31) for L.
SET UP: The ground state has n = 1.

EXECUTE: L =

40.13.

EVALUATE: The value of L we calculated is on the order of the diameter of a nucleus.
IDENTIFY: An electron in the lowest energy state in this box must have the same energy as it would in the
ground state of hydrogen.
nh 2
SET UP: The energy of the n th level of an electron in a box is En =
.
8mL2
EXECUTE: An electron in the ground state of hydrogen has an energy of −13.6 eV, so find the width
corresponding to an energy of E1 = 13.6 eV. Solving for L gives

L=

40.14.

h
(6.626 × 10−34 J ⋅ s)
=
= 6.4 × 10−15 m
−27
6
−19
8mE1
8(1.673 × 10
kg)(5.0 × 10 eV)(1.602 × 10
J/eV)

h
(6.626 × 10−34 J ⋅ s)
=

= 1.66 × 10−10 m.
8mE1
8(9.11 × 10−31 kg)(13.6 eV)(1.602 × 10−19 J/eV)

EVALUATE: This width is of the same order of magnitude as the diameter of a Bohr atom with the
electron in the K shell.
c
IDENTIFY and SET UP: The energy of a photon is E = hf = h . The energy levels of a particle in a box

λ

are given by Eq. (40.31).
EXECUTE: (a) E = (6.63 × 10−34 J ⋅ s)

L=

(3.00 × 108 m/s)
(122 × 10

−9

m)

= 1.63 × 10−18 J. Δ E =

h2
8mL2

(n12 − n22 ).

h 2 (n12 − n22 )
(6.63 × 10−34 J ⋅ s) 2 (22 − 12 )
=
= 3.33 × 10−10 m.
8mΔ E
8(9.11 × 10−31 kg)(1.63 × 10−18 J)

(b) The ground state energy for an electron in a box of the calculated dimensions is
h2
(6.63 × 10−34 J ⋅ s) 2
E=
=
= 5.43 × 10−19 J = 3.40 eV (one-third of the original
2
−31
−10
2
8mL 8(9.11 × 10 kg)(3.33 × 10
m)
photon energy), which does not correspond to the −13.6 eV ground state energy of the hydrogen atom.

40.15.

EVALUATE: (c) Note that the energy levels for a particle in a box are proportional to n 2 , whereas the
energy levels for the hydrogen atom are proportional to − 12 . A one-dimensional box is not a good model
n
for a hydrogen atom.
IDENTIFY and SET UP: Eq. (40.31) gives the energy levels. Use this to obtain an expression for E2 − E1

and use the value given for this energy difference to solve for L.

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


Quantum Mechanics

EXECUTE: Ground state energy is E1 =

h2

4h 2
E
=
. The energy
;
first
excited
state
energy
is
2
8mL2
8mL2

separation between these two levels is Δ E = E2 − E1 =
L = 6.626 × 10−34 J ⋅ s
40.16.

3h 2

2

8mL

. This gives L = h

3
8(9.109 × 10

−31

40-5

kg)(3.0 eV)(1.602 × 10−19 J/1 eV)

3
=
8mΔ E

= 6.1 × 10−10 m = 0.61 nm.

EVALUATE: This energy difference is typical for an atom and L is comparable to the size of an atom.
IDENTIFY: The energy of the absorbed photon must be equal to the energy difference between the two states.
9π 2 2
. The ground state energy is
SET UP and EXECUTE: The second excited state energy is E3 =
2mL2

E1 =

π2

2

4π 2 2 hc
E
.
.
E
=
1.00
eV,
so
E
=
9.00
eV.
For
the
transition
Δ
=
= Δ E.
1
3
λ
mL2
2mL2

hc (4.136 × 10−15 eV ⋅ s)(2.998 × 108 m/s)

=
= 1.55 × 10−7 m = 155 nm.
ΔE
8.00 eV
EVALUATE: This wavelength is much shorter than those of visible light.
IDENTIFY: If the given wave function is a solution to the Schrödinger equation, we will get an identity
when we substitute that wave function into the Schrödinger equation.
2
⎛ nπ x ⎞ −iEnt /
sin ⎜
SET UP: We must substitute the equation Ψ ( x, t ) =
into the one-dimensional
⎟e
L
⎝ L ⎠

λ=

40.17.

Schrödinger equation −

2

d 2ψ ( x)
dx 2

2m

+ U ( x)ψ ( x) = Eψ ( x).

EXECUTE: Taking the second derivative of Ψ( x, t ) with respect to x gives

Substituting this result into −

2m

d 2ψ ( x)
dx 2

+ U ( x)ψ ( x) = Eψ ( x), we get

2

dx 2

2

⎛ nπ ⎞
= −⎜
⎟ Ψ ( x, t ).
⎝ L ⎠

2

⎛ nπ ⎞
⎜
⎟ Ψ ( x, t ) = E Ψ ( x, t )
2m ⎝ L ⎠

2

⎛ nπ ⎞
⎜
⎟ , the energies of a particle in a box.
2m ⎝ L ⎠
EVALUATE: Since this process gives us the energies of a particle in a box, the given wave function is a
solution to the Schrödinger equation
IDENTIFY: Find x where ψ 1 is zero and where it is a maximum.

which gives En =

40.18.

2

2

d 2 Ψ ( x, t )

2
⎛πx⎞
sin ⎜
⎟.
L
⎝ L ⎠
EXECUTE: (a) The wave function for n = 1 vanishes only at x = 0 and x = L in the range 0 ≤ x ≤ L.
(b) In the range for x, the sine term is a maximum only at the middle of the box, x = L/2.
EVALUATE: (c) The answers to parts (a) and (b) are consistent with the figure.
IDENTIFY and SET UP: For the n = 2 first excited state the normalized wave function is given by

SET UP: ψ 1 =

40.19.

2
2 2 ⎛ 2π x ⎞
2
⎛ 2π x ⎞
2
sin ⎜
⎟ . ψ 2 ( x ) dx = sin ⎜
⎟ dx. Examine ψ 2 ( x ) dx and find where
L
L
L
L
⎝
⎠
⎝
⎠
it is zero and where it is maximum.
2
⎛ 2π x ⎞
EXECUTE: (a) ψ 2 dx = 0 implies sin ⎜
⎟=0
⎝ L ⎠
Eq. (40.35). ψ 2 ( x) =

2π x
= mπ , m = 0, 1, 2, … ; x = m( L/2)

L
For m = 0, x = 0; for m = 1, x = L/2; for m = 2, x = L
The probability of finding the particle is zero at x = 0, L/2, and L.

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


40-6

Chapter 40
2
⎛ 2π x ⎞
(b) ψ 2 dx is maximum when sin ⎜
⎟ = ±1
⎝ L ⎠
2π x
= m(π /2), m = 1, 3, 5, … ; x = m( L/4)
L
For m = 1, x = L/4; for m = 3, x = 3L/4
The probability of finding the particle is largest at x = L/4 and 3L/4.

(c) EVALUATE: The answers to part (a) correspond to the zeros of ψ

2

shown in Figure 40.12 in the

textbook and the answers to part (b) correspond to the two values of x where ψ
2

40.20.

IDENTIFY: Evaluate
SET UP:

d ψ

2

in the figure is maximum.

and see if Eq. (40.25) is satisfied. ψ ( x) must be zero at the walls, where U → ∞.

dx 2

d
d
sin kx = k cos kx.
cos kx = − k sin kx.
dx
dx
d 2ψ

2m
= − k 2ψ , and for ψ to be a solution of Eq. (40.25), k 2 = E 2 .
dx 2
(b) The wave function must vanish at the rigid walls; the given function will vanish at x = 0 for any k ,
but to vanish at x = L, kL = nπ for integer n.
EXECUTE: (a)

40.21.

n 2π 2

2

nπ
, so kn =
and ψ = A sin kx is the same as ψ n in
L
2mL2
Eq. (40.32), except for a different symbol for the normalization constant
(a) IDENTIFY and SET UP: ψ = A cos kx. Calculate dψ 2 /dx 2 and substitute into Eq. (40.25) to see if this
EVALUATE: From Eq. (40.31), En =

equation is satisfied.
EXECUTE: Eq. (40.25): −

h 2 d 2ψ
8π 2m dx 2

= Eψ

dψ
= A(− k sin kx) = − Ak sin kx
dx
d 2ψ
dx 2

= − Ak ( k cos kx ) = − Ak 2 cos kx

Thus Eq. (40.25) requires −
This says

h2k 2
2

8π m

= E; k =

h2
8π 2m

(− Ak 2 cos kx) = E ( A cos kx).

2mE
2mE
=
(h/2π )

ψ = A cos kx is a solution to Eq. (40.25) if k =

2mE

.

(b) EVALUATE: The wave function for a particle in a box with rigid walls at x = 0 and x = L must
satisfy the boundary conditions ψ = 0 at x = 0 and ψ = 0 at x = L. ψ (0) = A cos0 = A, since cos0 = 1.

Thus ψ is not 0 at x = 0 and this wave function isn’t acceptable because it doesn’t satisfy the required
40.22.

boundary condition, even though it is a solution to the Schrödinger equation.
IDENTIFY: The energy levels are given by Eq. (40.31). The wavelength λ of the photon absorbed in an
hc
.
atomic transition is related to the transition energy ΔE by λ =
ΔE
SET UP: For the ground state n = 1 and for the third excited state n = 4.
EXECUTE: (a) The third excited state is n = 4, so

ΔE = (42 − 1)

h2
8mL2

=

15(6.626 × 10−34 J ⋅ s) 2
8(9.11 × 10−31 kg)(0.125 × 10−9 m) 2

= 5.78 × 10−17 J = 361 eV.

© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


Quantum Mechanics

40-7

hc (6.63 × 10−34 J ⋅ s)(3.0 × 108 m/s)
=
= 3.44 nm
ΔE
5.78 × 10−17 J
EVALUATE: This photon is an x ray. As the width of the box increases the transition energy for this
transition decreases and the wavelength of the photon increases.
h
h
IDENTIFY and SET UP: λ = =
. The energy of the electron in level n is given by Eq. (40.31).
p
2mE
(b) λ =

40.23.

EXECUTE: (a) E1 =

h2
2

8mL

is twice the width of the box. p1 =
4h 2

(b) E2 =

p2 =

h

λ2

(c) E3 =

h

⇒ λ1 =

2mh 2 /8mL2
h

λ1

=

= 2 L = 2(3.0 × 10−10 m) = 6.0 × 10−10 m. The wavelength

(6.63 × 10−34 J ⋅ s)
6.0 × 10−10 m

= 1.1 × 10−24 kg ⋅ m/s.

⇒ λ 2 = L = 3.0 × 10−10 m. The wavelength is the same as the width of the box.

8mL2

= 2 p1 = 2.2 × 10−24 kg ⋅ m/s.
9h 2
2

8mL

2
L = 2.0 × 10−10 m. The wavelength is two-thirds the width of the box.
3

⇒ λ3 =

p3 = 3 p1 = 3.3 × 10−24 kg ⋅ m/s.
EVALUATE: In each case the wavelength is an integer multiple of λ /2. In the n th state, pn = np1.
40.24.

IDENTIFY: To describe a real situation, a wave function must be normalizable.

ψ

SET UP:

2

dV is the probability that the particle is found in volume dV. Since the particle must be

somewhere, ψ must have the property that

∫ψ

2

dV = 1 when the integral is taken over all space.

EXECUTE: (a) In one dimension, as we have here, the integral discussed above is of the form
∞

∫−∞ ψ ( x)

2

dx = 1.

(b) Using the result from part (a), we have

∞

∫−∞

(eax ) 2 dx = ∫

∞

−∞

e2ax dx =

e2 ax
2a

∞

= ∞. Hence this wave
−∞

function cannot be normalized and therefore cannot be a valid wave function.
(c) We only need to integrate this wave function of 0 to ∞ because it is zero for x < 0. For normalization we
have 1 = ∫

∞

−∞

40.25.

2

∞

∞

0

0

ψ dx = ∫ (Ae-bx )2 dx = ∫ A2e−2bx dx =

A2e−2bx
−2b

∞

=
0

A2
A2
, which gives
= 1, so A = 2b .
2b
2b

EVALUATE: If b were negative, the given wave function could not be normalized, so it would not be allowable.
2 2
d ψ
IDENTIFY: Compare −
+ U ψ to Eψ and see if there is a value of k for which they are equal.
2m dx 2
SET UP:
EXECUTE:

d2
dx 2

sin kx = − k 2 sin kx.

(a) Eq. (40.23):

Left-hand side:

− 2 d 2ψ
+ U ψ = Eψ .
2m dx 2

2 2
⎛ 2k 2
⎞
− 2 d2
k
+
=
+
=
+ U 0 ⎟ψ . But
(
A
sin
kx
)
U
A
sin
kx
A
sin
kx
U
A
sin

kx
⎜⎜
0
0
⎟
2m dx 2
2m
2
m
⎝
⎠

2 2

2 2
k
k
+ U 0 > U 0 > E if k is real. But
+ U 0 should equal E. This is not the case, and there is no k
2m
2m

for which this ψ

2

is a solution.

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40-8

Chapter 40

40.26.

2 2

k
+ U 0 = E is consistent and so ψ = A sin kx is a solution of Eq. (40.23) for this case.
2m
EVALUATE: For a square-well potential and E < U 0 , Eq. (40.23) with U = U 0 applies outside the well
and the wave function has the form of Eq. (40.40).
h
p2
IDENTIFY: λ = . p is related to E by E =
+U.
p
2m
(b) If E > U 0 , then

SET UP: For x > L, U = U 0 . For 0 < x < L, U = 0.
EXECUTE: For 0 < x < L, p = 2mE = 2m(3U 0 ) and λin =

h
h
=
. Thus, the ratio of the

2 m( E − U 0 )
2m(2U 0 )

p = 2m( E − U 0 ) = 2m(2U 0 ) and λout =
wavelengths is

40.27.

h
. For x > L,
2m(3U 0 )

2m(3U 0 )
3
λout
.
=
=
2
λin
2m(2U 0 )

EVALUATE: For x > L some of the energy is potential and the kinetic energy is less than it is for
0 < x < L, where U = 0. Therefore, outside the box p is less and λ is greater than inside the box.
IDENTIFY: Figure 40.15b in the textbook gives values for the bound state energy of a square well for
which U 0 = 6 E1-1DW .
SET UP: E1-1DW =

π2

2

2mL2

.

EXECUTE: E1 = 0.625E1-1DW = 0.625

π2

2

2mL2

; E1 = 2.00 eV = 3.20 × 10−19 J.

1/2

40.28.

⎛
⎞
0.625
−10
L = π ⎜⎜
m.
⎟⎟ = 3.43 × 10
−31
−19
2(9.109

10
kg)(3.20
10
J)
×
×
⎝
⎠
EVALUATE: As L increases the ground state energy decreases.
IDENTIFY: The energy of the photon is the energy given to the electron.
SET UP: Since U 0 = 6 E1-1DW we can use the result E1 = 0.625 E1-1DW from Section 40.4. When the
electron is outside the well it has potential energy U 0 , so the minimum energy that must be given to the
electron is U 0 − E1 = 5.375 E1-1DW .
EXECUTE: The maximum wavelength of the photon would be
hc
hc
8mL2c
8(9.11 × 10−31 kg)(1.50 × 10−9 m) 2 (3.00 × 108 m/s)
=
=
=
λ=
U 0 − E1 (5.375)(h 2 /8mL2 ) (5.375)h
(5.375)(6.63 × 10−34 J ⋅ s)

40.29.

= 1.38 × 10−6 m.
EVALUATE: This photon is in the infrared. The wavelength of the photon decreases when the width of the
well decreases.

d 2ψ
2mE
IDENTIFY: Calculate
and compare to − 2 ψ .
dx 2
d
d
SET UP:
sin kx = k cos kx.
cos kx = − k sin kx.
dx
dx
EXECUTE: Eq. (40.37): ψ = Asin

2mE

x + B cos

2mE

x.

d 2ψ

2mE
2mE
−2mE
⎛ 2mE ⎞
⎛ 2mE ⎞
= − A ⎜ 2 ⎟ sin

x − B ⎜ 2 ⎟ cos
x=
(ψ ). This is Eq. (40.38), so this ψ is a
2
dx
⎝
⎠
⎝
⎠
solution.
EVALUATE: ψ in Eq. (40.38) is a solution to Eq. (40.37) for any values of the constants A and B.
2

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Quantum Mechanics
40.30.

40-9

IDENTIFY: The longest wavelength corresponds to the smallest energy change.
h2
SET UP: The ground level energy level of the infinite well is E1-1DW =
, and the energy of the
8mL2
photon must be equal to the energy difference between the two shells.
EXECUTE: The 400.0 nm photon must correspond to the n = 1 to n = 2 transition. Since U 0 = 6 E1-1DW ,

we have E2 = 2.43E1-1DW and E1 = 0.625 E1-1DW . The energy of the photon is equal to the energy
difference between the two levels, and E1-1DW =
Eγ = E2 − E1 ⇒
L=
40.31.

hc

λ

= (2.43 − 0.625) E1-1DW =

h2
8mL2

1.805 h 2
8mL2

, which gives

. Solving for L gives

(1.805) hλ
(1.805)(6.626 × 10−34 J ⋅ s)(4.00 × 10−7 m)
=
= 4.68 × 10−10 m = 0.468 nm.
8mc
8(9.11 × 10−31 kg)(3.00 × 108 m/s)

EVALUATE: This width is approximately half that of a Bohr hydrogen atom.

IDENTIFY: Find the transition energy ΔE and set it equal to the energy of the absorbed photon. Use
E = hc/λ , to find the wavelength of the photon.
SET UP: U 0 = 6 E1-1DW , as in Figure 40.15 in the textbook, so E1 = 0.625 E1-1DW and E3 = 5.09 E1-1DW

with E1-1DW =

π2

2

2mL2

. In this problem the particle bound in the well is a proton, so m = 1.673 × 10−27 kg.

EXECUTE: E1-1DW =

π2

2
2

2mL

=

π 2 (1.055 × 10−34 J ⋅ s) 2
2(1.673 × 10−27 kg)(4.0 × 10−15 m) 2

= 2.052 × 10−12 J. The transition energy

is Δ E = E3 − E1 = (5.09 − 0.625) E1-1DW = 4.465 E1-1DW . Δ E = 4.465(2.052 × 10−12 J) = 9.162 × 10−12 J
The wavelength of the photon that is absorbed is related to the transition energy by ΔE = hc/λ , so
hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
=
= 2.2 × 10−14 m = 22 fm.
ΔE
9.162 × 10−12 J
EVALUATE: The wavelength of the photon is comparable to the size of the box.

λ=

40.32.

IDENTIFY: The tunneling probability is T = Ge −2κ L , with G = 16

E ⎛
E ⎞
T = 16 ⎜1 −
⎟e
U0 ⎝ U0 ⎠

−2 2 m (U 0 − E )

L

2m(U 0 − E )
E ⎛
E ⎞
. so
⎜1 −

⎟ and κ =
U0 ⎝ U0 ⎠

.

SET UP: U 0 = 30.0 × 106 eV, L = 2.0 × 10−15 m, m = 6.64 × 10 −27 kg.
EXECUTE: (a) U 0 − E = 1.0 × 106 eV (E = 29.0 × 106 eV), T = 0.090.
(b) If U 0 − E = 10.0 × 106 eV (E = 20.0 × 106 eV), T = 0.014.
EVALUATE: T is less when U 0 − E s 10.0 MeV than when U 0 − E is 1.0 MeV.
40.33.

IDENTIFY: The tunneling probability is T = 16
SET UP:

E ⎛
E ⎞ −2 L
⎜1 −
⎟e
U0 ⎝ U0 ⎠

2 m (U 0 − E ) /

.

E
6.0 eV
=
and E − U 0 = 5 eV = 8.0 × 10−19 J.
U 0 11.0 eV

EXECUTE: (a) L = 0.80 × 10−9 m:

⎛ 6.0 eV ⎞⎛
6.0 ev ⎞ −2(0.80 × 10−9 m)
T = 16 ⎜
⎟⎜1 −
⎟e
⎝ 11.0 eV ⎠⎝ 11.0 eV ⎠

2(9.11 × 10−31 kg)(8.0 × 10−19 J) /1.055 × 10−34 J ⋅ s

= 4.4 × 10−8.

(b) L = 0.40 × 10−9 m: T = 4.2 × 10−4.
EVALUATE: The tunneling probability is less when the barrier is wider.

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40-10

40.34.

Chapter 40

IDENTIFY: The transmission coefficient is T = 16

E ⎛
E ⎞ −2

⎜1 −
⎟e
U0 ⎝ U0 ⎠

2 m (U 0 − E ) L /

.

SET UP: E = 5.0 eV, L = 0.60 × 10−9 m, and m = 9.11 × 10−31 kg
EXECUTE: (a) U 0 = 7.0 eV ⇒ T = 5.5 × 10−4 .
(b) U 0 = 9.0 eV ⇒ T = 1.8 × 10−5 .
(c) U 0 = 13.0 eV ⇒ T = 1.1 × 10−7.
40.35.

EVALUATE: T decreases when the height of the barrier increases.
IDENTIFY and SET UP: Use Eq. (39.1), where K = p 2 /2m and E = K + U .
EXECUTE: λ = h/p = h/ 2mK , so λ K is constant. λ1 K1 = λ 2 K 2 ; λ1 and K1 are for x > L where

K1 = 2U 0 and λ 2 and K 2 are for 0 < x < L where K 2 = E − U 0 = U 0 .
K2
U0
λ1
1
=
=
=
K1
λ2
2U 0
2

40.36.

EVALUATE: When the particle is passing over the barrier its kinetic energy is less and its wavelength is
larger.
IDENTIFY: The probability of tunneling depends on the energy of the particle and the width of the barrier.
E ⎛
E ⎞
SET UP: The probability of tunneling is approximately T = Ge−2κ L , where G = 16 ⎜ 1 −
⎟ and
U0 ⎝ U0 ⎠

κ=

2m(U 0 − E )

.

EXECUTE: G = 16

κ=

2m(U 0 − E )

E ⎛
E ⎞
50.0 eV ⎛ 50.0 eV ⎞
⎜1 −
⎟ = 16
⎜1 −

⎟ = 3.27.
U0 ⎝ U0 ⎠
70.0 eV ⎝ 70.0 eV ⎠

=

2(1.67 × 10−27 kg)(70.0 eV − 50.0 eV)(1.60 × 10−19 J/eV)
(6.63 × 10−34 J ⋅ s)/2π

= 9.8 × 1011 m −1

Solving T = Ge−2κ L for L gives
1
1
⎛ 3.27 ⎞
−12
L=
ln(G /T ) =
ln ⎜
m = 3.6 pm.
⎟ = 3.6 × 10
11
−
1
2κ
2(9.8 × 10 m ) ⎝ 0.0030 ⎠
If the proton were replaced with an electron, the electron’s mass is much smaller so L would be larger.
EVALUATE: An electron can tunnel through a much wider barrier than a proton of the same energy.
40.37.

IDENTIFY and SET UP: The probability is T = Ae−2κ L , with A = 16

E ⎛
E ⎞
⎜1 −
⎟ and κ =
U0 ⎝ U0 ⎠

2m(U 0 − E )

.

E = 32 eV, U 0 = 41 eV, L = 0.25 × 10−9 m. Calculate T.

EXECUTE: (a) A = 16

κ=

κ=

E ⎛
E ⎞
32 ⎛ 32 ⎞
⎜1 −
⎟ = 16 ⎜1 − ⎟ = 2.741.
U0 ⎝ U0 ⎠
41 ⎝ 41 ⎠

2m(U 0 − E )

2(9.109 × 10−31 kg)(41 eV − 32 eV)(1.602 × 10−19 J/eV)
1.055 × 10−34 J ⋅ s
−1

= 1.536 × 1010 m −1

−9

T = Ae−2κ L = (2.741)e−2(1.536 ×10 m )(0.25 ×10 m) = 2.741e−7.68 = 0.0013
(b) The only change in the mass m, which appears in κ .
10

κ=

2m(U 0 − E )

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Quantum Mechanics

κ=

2(1.673 × 10−27 kg)(41 eV − 32 eV)(1.602 × 10−19 J/eV)
1.055 × 10−34 J ⋅ s

Then T = Ae−2κ L = (2.741)e−2(6.584 × 10

11

40.38.

= 6.584 × 1011 m −1

= 2.741e−392.2 = 10−143

EVALUATE: The more massive proton has a much smaller probability of tunneling than the electron does.
d 2ψ
IDENTIFY: Calculate
and insert the result into Eq. (40.44).
dx 2
SET UP:

2
2
d −δ x 2
d 2 −δ x 2
e
e
= −2δ xe−δ x and
= (4δ 2 x 2 − 2δ )e−δ x
2
dx
dx

EXECUTE: Let

Eq. (40.44) if E =
EVALUATE: E =

40.39.

m −1 )(0.25 × 10−9 m)

40-11

mk ′ /2 = δ , and so
2

m
1
2

δ=

1
2

k ′/m =

d 2ψ
dψ
= (4 x 2δ 2 − 2δ )ψ , and ψ is a solution of
= −2 xδψ and
dx
dx 2

1
ω.
2

ω agrees with Eq. (40.46), for n = 0.

IDENTIFY and SET UP: The energy levels are given by Eq. (40.46), where ω =
EXECUTE: ω =

k′
.
m

k′
110 N/m
=
= 21.0 rad/s
m
0.250 kg

The ground state energy is given by Eq. (40.46):
1
1
E0 = ω = (1.055 × 10−34 J ⋅ s)(21.0 rad/s) = 1.11 × 10−33 J(1 eV/1.602 × 10−19 J) = 6.93 × 10−15 eV
2
2
1
1⎞
⎛
⎞
⎛
En = ⎜ n + ⎟ ω , E( n + 1) = ⎜ n + 1 + ⎟ ω
2⎠

2⎠
⎝
⎝
The energy separation between these adjacent levels is
Δ E = En +1 − En = ω = 2 E0 = 2(1.11 × 10−33 J) = 2.22 × 10−33 J = 1.39 × 10−14 eV.
40.40.

40.41.

EVALUATE: These energies are extremely small; quantum effects are not important for this oscillator.
IDENTIFY: The energy of the absorbed photon must be equal to the energy difference between the two states.
hc (4.136 × 10−15 eV ⋅ s)(2.998 × 108 m/s)
=
= 0.1433 eV. ΔE = ω .
SET UP and EXECUTE: ΔE =
λ
8.65 × 10−6 m
ω 0.1433 eV
=
= 0.0717 eV.
E0 =
2
2
EVALUATE: The energy of the photon is not equal to the energy of the ground state, but rather it is the
energy difference between the two states.
IDENTIFY: We can model the molecule as a harmonic oscillator. The energy of the photon is equal to the
energy difference between the two levels of the oscillator.
SET UP: The energy of a photon is Eγ = hf = hc/λ , and the energy levels of a harmonic oscillator are

1⎞

⎛
given by En = ⎜ n + ⎟
2⎠
⎝

k′ ⎛
1⎞
= ⎜ n + ⎟ ω.
m ⎝
2⎠

EXECUTE: (a) The photon’s energy is Eγ =

hc

λ

=

(6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)

(b) The transition energy is ΔE = En + 1 − En = ω =

we get k ′ =

4π 2c 2m

k′
2π c
=

, which gives
m
λ

= 0.21 eV.
k′
. Solving for k ′,
m

4π 2 (3.00 × 108 m/s) 2 (5.6 × 10−26 kg)

= 5,900 N/m.
(5.8 × 10−6 m) 2
EVALUATE: This would be a rather strong spring in the physics lab.

λ2

=

5.8 × 10−6 m

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40-12
40.42.

Chapter 40
IDENTIFY: The photon energy equals the transition energy for the atom.

SET UP: According to Eq. (40.46), the energy released during the transition between two adjacent levels
is twice the ground state energy E3 − E2 = ω = 2 E0 = 11.2 eV.
EXECUTE: For a photon of energy E ,

c hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
=
=
= 111 nm.
f
E
(11.2 eV)(1.60 × 10−19 J/eV)
EVALUATE: This photon is in the ultraviolet.
IDENTIFY and SET UP: Use the energies given in Eq. (40.46) to solve for the amplitude A and maximum
speed vmax of the oscillator. Use these to estimate Δ x and Δpx and compute the uncertainty product
E = hf ⇒ λ =

40.43.

Δ xΔpx .
EXECUTE: The total energy of a Newtonian oscillator is given by E = 12 k ′A2 where k′ is the force

(

constant and A is the amplitude of the oscillator. Set this equal to the energy E = n + 12
level that has quantum number n, where ω =
A=

k′
, and solve for A:
m

1
k ′A2
2

(

= n + 12

)

)

ω of an excited

ω.

(2n + 1) ω
2
. The total energy of the Newtonian oscillator can also be written as E = 12 mvmax
. Set
k′

(

this equal to E = n + 12

)

ω and solve for vmax :

2
1
mvmax
2

(

= n + 12

)

ω. vmax =

(2n + 1) ω
. Thus the
m

maximum linear momentum of the oscillator is pmax = mvmax = (2n + 1) mω . Now A/ 2 represents the
uncertainty Δ x in position and that pmax / 2 is the corresponding uncertainty Δ px in momentum. Then
the uncertainty product is
⎛ 1 (2n + 1) ω ⎞ ⎛ 1
⎞ (2n + 1) ω m (2n + 1) ω ⎛ 1 ⎞
Δ xΔ px = ⎜⎜
=
⎜ ⎟ = (2n + 1) .
⎟⎟ ⎜⎝ 2 (2n + 1) mω ⎟⎠ =
′
k
2

k′
2
2
2
⎝ω ⎠
⎝
⎠
EVALUATE: For n = 0 this gives Δ xΔ px = /2, in agreement with the result derived in Section 40.5. The
uncertainty product Δ xΔ px increases with n.
40.44.

IDENTIFY: Compute the ratio specified in the problem.
ω
k′
SET UP: For n = 0, A =
. ω=
.
k′
m
2

⎛
mk ′ 2 ⎞
ω⎞
⎛
A ⎟⎟ = exp ⎜ − mk ′ ⎟ = e −1 = 0.368. This is consistent with what is
= exp ⎜⎜ −
k′ ⎠
⎝
ψ (0)

⎝
⎠
shown in Figure 40.27 in the textbook.

EXECUTE: (a)

2

ψ (2 A)2

⎛
⎞
mk ′
ω⎞
⎛
(2 A) 2 ⎟⎟ = exp ⎜ − mk ′ 4 ⎟ = e−4 = 1.83 × 10−2. This figure cannot be read this
= exp ⎜⎜ −
k′ ⎠
⎝
ψ (0)
⎝
⎠
precisely, but the qualitative decrease in amplitude with distance is clear.
EVALUATE: The wave function decays exponentially as x increases beyond x = A.
IDENTIFY: We model the atomic vibration in the crystal as a harmonic oscillator.
1 ⎞ k′ ⎛
1⎞
⎛
SET UP: The energy levels of a harmonic oscillator are given by En = ⎜ n + ⎟
= ⎜ n + ⎟ ω.

2
m
2⎠
⎝
⎠
⎝
EXECUTE: (a) The ground state energy of a simple harmonic oscillator is
(b)

40.45.

ψ ( A)

2

E0 =

1
1
ω=
2
2

12.2 N/m
k ′ (1.055 × 10−34 J ⋅ s)
=
= 9.43 × 10−22 J = 5.89 × 10−3 eV
2
m
3.82 × 10−26 kg

(b) E4 − E3 = ω = 2 E0 = 0.0118 eV, so λ =

hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
=
= 106 μm
E
1.88 × 10−21 J

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Quantum Mechanics

40-13

(c) En +1 − En = ω = 2 E0 = 0.0118 eV
EVALUATE: These energy differences are much smaller than those due to electron transitions in the
hydrogen atom.
40.46.

2

IDENTIFY: For a stationary state, Ψ is time independent.
SET UP: To calculate Ψ ∗ from Ψ , replace i by −i.
EXECUTE: For this wave function, Ψ ∗ = ψ 1∗eiω1t + ψ 2∗eiω2t , so

Ψ = Ψ ∗Ψ = (ψ 1∗eiω1t + ψ 2∗eiω2t )(ψ 1e−iω1t + ψ 2e−iω2t ) = ψ 1∗ψ 1 + ψ 2∗ψ 2 + ψ 1∗ψ 2ei (ω1 −ω2 )t + ψ 2∗ψ 1ei (ω2 −ω1 )t .
2

The frequencies ω1 and ω 2 are given as not being the same, so Ψ

2

is not time-independent, and Ψ is

not the wave function for a stationary state.
EVALUATE: If ω1 = ω 2 , then Ψ is the wave function for a stationary state.
40.47.

IDENTIFY: We know the wave function of a particle in a box.
1
1
SET UP and EXECUTE: (a) Ψ( x, t ) =
ψ 1( x)e− iE1t/ +
ψ 3 ( x)e−iE3t/ .
2
2
1
1
Ψ∗ ( x, t ) =
ψ 1( x)e+ iE1t/ +
ψ 3 ( x)e+ iE3t/ .
2
2
1
1⎡
2
⎛ [ E − E1 ]t ⎞ ⎤

Ψ ( x, t ) = [ψ 12 + ψ 32 + ψ 1ψ 3 (ei ( E3 − E1 )t / + e−i ( E3 − E1 )t / )] = ⎢ψ 12 + ψ 32 + 2ψ 1ψ 3 cos ⎜ 3
⎟⎥ .
2
2⎣
⎝
⎠⎦

ψ1 =

9π 2 2
π2 2
4π 2 2
2
⎛ 3π x ⎞
E
E
E
.
=
and
E
=
,
so
−
=
sin ⎜
.
1
3

1
⎟ 3
L
2mL2
mL2
2mL2
⎝ L ⎠

2
⎛πx⎞
sin ⎜
⎟. ψ 3 =
L
⎝ L ⎠
2

Ψ ( x, t ) =

2
1 ⎡ 2⎛πx⎞
⎛ π x ⎞ ⎛ 3π x ⎞ ⎛ 4π t ⎞ ⎤
2 ⎛ 3π x ⎞
⎢sin ⎜
⎟ ⎥ . At x = L/2,
⎟ + sin ⎜
⎟ + 2sin ⎜
⎟ sin ⎜
⎟ cos ⎜⎜
L ⎢⎣
⎝ L ⎠

⎝ L ⎠
⎝ L ⎠ ⎝ L ⎠ ⎝ mL2 ⎟⎠ ⎥⎦

⎛πx⎞
⎛π ⎞
⎛ 3π x ⎞
⎛ 3π
sin ⎜
⎟ = sin ⎜ ⎟ = 1. sin ⎜
⎟ = sin ⎜
⎝ L ⎠
⎝2⎠
⎝ L ⎠
⎝ 2
(b) ωosc =

E3 − E1

=

⎛ 4π 2 t ⎞ ⎤
2⎡
2
⎞
=
−
Ψ
=
−
1.

(
x
,
t
)
1
cos
.
⎢
⎜⎜
⎟
2 ⎟
⎟⎥
L ⎢⎣
⎠
⎝ mL ⎠ ⎥⎦

4π 2

.
mL2
EVALUATE: Note that Δ E = ω .

40.48.

IDENTIFY: Carry out the calculations specified in the problem.
∞ −α 2 k 2

∫0 e

SET UP: A standard integral is
EXECUTE: (a) B ( k ) = e−α

⇒ kh =

1

α

2 2

k

cos( kx) dk =

π − x 2 /4α 2
e
.
2α

. B (0) = Bmax = 1. B (kh ) =

2 2
1
= e −α kh ⇒ ln(1/2) = −α 2 kh2
2

ln(2) = wk .
∞

(b) ψ ( x) = ∫ e−α

2 2

k

0

cos kxdk =

π
2α

(e − x

2

/4α 2

). ψ ( x) is a maximum when x = 0.

1
− x2
⇒ h2 = ln(1/2) ⇒ xh = 2α ln 2 = wx
4α
2
4α
h ⎛1
h
h ln 2

⎛ hwk ⎞
⎞
(d) w p wx = ⎜
= (2ln 2) .
ln2 ⎟ (2α ln2) =
(2ln 2) =
⎟ wx =
⎜
π
2π ⎝ α
2π
⎝ 2π ⎠
⎠
EVALUATE: The Heisenberg Uncertainty Principle says that ΔxΔpx ≥ /2. If Δx = wx and Δpx = w p ,
(c) ψ ( xh ) =

π

when e− xh /4α =
2

2

then the uncertainty principle says wx w p ≥ /2. So our result is consistent with the uncertainty principle
since (2ln 2) > /2.

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40-14
40.49.

Chapter 40
∞

IDENTIFY: Evaluate ψ ( x) = ∫ B( k )cos kx dk for the function B (k ) specified in the problem.
0

SET UP:

1

∫ cos kx dk = x sin kx.
∞

k0 ⎛

0

0

EXECUTE: (a) ψ ( x) = ∫ B (k )cos kxdk = ∫

k

0
1 ⎞
sin kx
sin k0 x

=
⎜ ⎟ cos kxdk =
k
k
x
k0 x
0
⎝ 0⎠
0

(b) ψ ( x ) has a maximum value at the origin x = 0. ψ ( x0 ) = 0 when k0 x0 = π so x0 =

π
k0

. Thus the width of

2π
2π
. If k0 =
, wx = L. B (k ) versus k is graphed in Figure 40.49a. The graph of
L
k0
ψ ( x) versus x is in Figure 40.49b.

this function wx = 2 x0 =

(c) If k0 =

π

L

, wx = 2 L.

⎛ hw ⎞ ⎛ 2π ⎞ hwk hk0
EVALUATE: (d) w p wx = ⎜ k ⎟ ⎜
=
= h. If Δx = wx and Δpx = w p , then the uncertainty
⎟=
k0
⎝ 2π ⎠ ⎝ k0 ⎠ k0

principle states that w p wx ≥ . For us, no matter what k0 is, w p wx = h, which is greater than /2.
2

Figure 40.49
40.50.

IDENTIFY: If the given wave function is a solution to the Schrödinger equation, we will get an identity
when we substitute that wave function into the Schrödinger equation.
SET UP: The given function is ψ ( x) = Aeikx , and the one-dimensional Schrödinger equation is

−

d 2ψ ( x)
2m

dx 2

+ U ( x)ψ ( x) = Eψ ( x).

EXECUTE: Start with the given function and take the indicated derivatives: ψ ( x ) = Aeikx .
2
2
dψ ( x )
d 2ψ ( x)
d 2ψ ( x)
2 2 ikx
2 ikx d ψ ( x )
2
Ai
k
e
Ak
e
.
k
(
x
).
k 2ψ ( x).
ψ
= Aikeikx .
=
=
−
=
−
−
=

dx
2m dx 2
2m
dx 2
dx 2
Substituting these results into the one-dimensional Schrödinger equation gives
2 2
k
ψ ( x) + U 0ψ ( x) = E ψ ( x).
2m

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Quantum Mechanics

EVALUATE: ψ ( x) = A eikx is a solution to the one-dimensional Schrödinger equation if E − U 0 =

or k =

2m ( E − U 0 )
2

40-15
2 2

k
2m

. (Since U 0 < E was given, k is the square root of a positive quantity.) In terms of the

particle’s momentum p: k = p/ , and in terms of the particle’s de Broglie wavelength λ: k = 2π /λ .
40.51.

IDENTIFY: Let I refer to the region x < 0 and let II refer to the region x > 0, so ψ I ( x) = Aeik1 x + Be− ik1 x

dψ I dψ II
=
at x = 0.
dx
dx

and ψ II ( x) = Ceik2 x . Set ψ I (0) = ψ II (0) and
d ikx
(e ) = ikeikx .
dx

SET UP:

dψ I dψ II
=
at x = 0 gives ik1 A − ik1B = ik2C. Solving
dx
dx
⎛k −k ⎞
⎛ 2k2 ⎞
this pair of equations for B and C gives B = ⎜ 1 2 ⎟ A and C = ⎜
⎟ A.
+

k
k
⎝ 1 2⎠
⎝ k1 + k2 ⎠
EXECUTE: ψ I (0) = ψ II (0) gives A + B = C.

EVALUATE: The probability of reflection is R =

T=
40.52.

C2
A2

=

4k12

(k1 + k2 ) 2

B2
A

2

=

(k1 − k2 ) 2

(k1 + k2 ) 2

. The probability of transmission is

. Note that R + T = 1.

IDENTIFY: For a particle in a box, En =

n2h2
8mL2

.

SET UP: Δ En = En +1 − En
( n + 1) 2 − n 2

2n + 1

2 1
+ 2 . This is never larger than it is for n = 1, and R1 = 3.
n
n
n
n
EVALUATE: (b) Rn approaches zero as n becomes very large. In the classical limit there is no

EXECUTE: (a) Rn =

2

=

2

=

quantization and the spacing of successive levels is vanishingly small compared to the energy levels.
Therefore, Rn for a particle in a box approaches the classical value as n becomes very large.
40.53.

IDENTIFY and SET UP: The energy levels are given by Eq. (40.31): En =

transition and set Δ E = hc/λ , the energy of the photon.
EXECUTE: (a) Ground level, n = 1, E1 =

energy is Δ E = E2 − E1 =
This gives

hc

λ

λ = 1.92 × 10

=

−5

3h 2
2

8mL

3h 2
8mL2

. λ=

n2h2
8mL2

. Calculate Δ E for the

h2

4h 2
.
=
=
2,
. The transition
n
E
First
excited
level,
2
8mL2
8mL2

. Set the transition energy equal to the energy hc/λ of the emitted photon.

8mcL2 8(9.109 × 10−31 kg)(2.998 × 108 m/s)(4.18 × 10−9 m)2
=
.
3h
3(6.626 × 10−34 J ⋅ s)

m = 19.2 μm.

(b) Second excited level has n = 3 and E3 =
−

4h 2

8mL2

The transition energy is

hc 5h 2
8mcL2 3
=
.
so
λ
=
= (19.2 μ m) = 11.5 μm.
5h
5
8mL2 8mL2 8mL2 λ 8mL2
EVALUATE: The energy spacing between adjacent levels increases with n, and this corresponds to a

shorter wavelength and more energetic photon in part (b) than in part (a).

Δ E = E3 − E2 =

9h 2

9h 2

=

5h 2

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40-16
40.54.

Chapter 40
IDENTIFY: The probability of finding the particle between x1 and x2 is
SET UP: For the ground state ψ 1 =

2
πx
sin . sin 2 θ = 12 (1 − cos 2θ ).
L
L

x2

∫x

2

ψ dx.

1

1

∫ cos α x dx = α sin α x.
L/ 4

EXECUTE: (a)

2 L /4
πx
2 L /4 1 ⎛
2π x ⎞
1⎛
L
2π x ⎞
sin 2
dx = ∫
sin
⎜1 − cos
⎟ dx = ⎜ x −
⎟
L ∫0

L
L 0 2⎝
L ⎠
L⎝
2π
L ⎠0

=

1 1
−
, which is
4 2π

about 0.0908.
L/ 2

(b) Repeating with limits of L/4 and L/2 gives

1⎛
L
2π x ⎞
1 1
sin
, about 0.409.
⎜x−
⎟ = +
L⎝
2π
L ⎠ L / 4 4 2π

(c) The particle is much likely to be nearer the middle of the box than the edge.
EVALUATE: (d) The results sum to exactly

1
.
2

Since the probability of the particle being anywhere in the

box is unity, the probability of the particle being found between x = L/2 and x = L is also

1
.
2

This means

that the particle is as likely to be between x = 0 and L/2 as it is to be between x = L/2 and x = L.
(e) These results are consistent with Figure 40.12b in the textbook. This figure shows a greater probability

near the center of the box. It also shows symmetry of ψ
40.55.

2

about the center of the box.

IDENTIFY: The probability of the particle being between x1 and x2 is

x2

∫x

| ψ |2 dx, where ψ is the

1

normalized wave function for the particle.
2
⎛πx⎞
sin ⎜
⎟.
L
⎝ L ⎠
EXECUTE: The probability P of the particle being between x = L/4 and x = 3L/4 is
3 L /4
2 3L / 4 2 ⎛ π x ⎞
2
P=∫
ψ 1 dx = ∫
sin ⎜
⎟ dx. Let y = π x/L; dx = ( L/π ) dy and the integration limits become
L/ 4
L L /4
⎝ L ⎠
π /4 and 3π /4.
(a) SET UP: The normalized wave function for the ground state is ψ 1 =

3π /4

P=

2 ⎛ L ⎞ 3π /4 2
2 ⎡1
1
⎤
sin y dy = ⎢ y − sin 2 y ⎥
⎜ ⎟
π ⎣2
L ⎝ π ⎠ ∫π /4
4
⎦π / 4

P=

2 ⎡ 3π π 1 ⎛ 3π ⎞ 1 ⎛ π ⎞ ⎤
− − sin
+ sin
π ⎢⎣ 8 8 4 ⎜⎝ 2 ⎟⎠ 4 ⎜⎝ 2 ⎟⎠ ⎥⎦

P=

1
1
2⎛π 1
1 ⎞ 1 1
− ( −1) + (1) ⎟ = + = 0.818. (Note: The integral formula ∫ sin 2 y dy = y − sin 2 y was used.)
2
4

π ⎜⎝ 4 4
4 ⎠ 2 π

(b) SET UP: The normalized wave function for the first excited state is ψ 2 =
EXECUTE: P = ∫

3 L /4

L/ 4

2

ψ 2 dx =

2
⎛ 2π x ⎞
sin ⎜
⎟.
L
⎝ L ⎠

2 3 L /4 2 ⎛ 2π x ⎞
sin ⎜
⎟ dx. Let y = 2π x/L; dx = ( L/2π ) dy and the integration
L ∫L /4
⎝ L ⎠

limits become π /2 and 3π /2.
P=

3π /2

2 ⎛ L ⎞ 3π /2 2
1 ⎡1
1
1 ⎛ 3π π ⎞
⎤
= ⎜
− ⎟ = 0.500
⎜
⎟ ∫π / 2 sin y dy = ⎢ y − sin 2 y ⎥
π ⎣2
L ⎝ 2π ⎠
4
⎦π /2 π ⎝ 4 4 ⎠

(c) EVALUATE: These results are consistent with Figure 40.11b in the textbook. That figure shows that ψ

2

is more concentrated near the center of the box for the ground state than for the first excited state; this is
consistent with the answer to part (a) being larger than the answer to part (b). Also, this figure shows that for
the first excited state half the area under ψ

2

curve lies between L/4 and 3L/4, consistent with our answer

to part (b).

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Quantum Mechanics
40.56.

40-17

2

IDENTIFY: The probability is ψ dx, with ψ evaluated at the specified value of x.
SET UP: For the ground state, the normalized wave function is ψ 1 = 2/L sin(π x/L) .
EXECUTE: (a) (2/L) sin 2 (π /4)dx = dx/L.
(b) (2/L) sin 2 (π /2)dx = 2dx/L
(c) (2 L )sin 2 (3π /4) = dx/L
EVALUATE: Our results agree with Figure 40.12b in the textbook. ψ

2

is largest at the center of the box,

2

40.57.

at x = L/2. ψ is symmetric about the center of the box, so is the same at x = L/4 as at x = 3L/4.
IDENTIFY and SET UP: The normalized wave function for the n = 2 first excited level is
2
⎛ 2π x ⎞

2
sin ⎜
⎟ . P = ψ ( x) dx is the probability that the particle will be found in the interval x to x + dx.
L
⎝ L ⎠
EXECUTE: (a) x = L/4

ψ2 =

2
⎛ ⎛ 2π ⎞⎛ L ⎞ ⎞
sin ⎜ ⎜
⎟⎜ ⎟ ⎟ =
L
⎝ ⎝ L ⎠⎝ 4 ⎠ ⎠
P = (2/L) dx
(b) x = L/2

ψ ( x) =

ψ ( x) =

2
⎛ ⎛ 2π ⎞⎛ L ⎞ ⎞
sin ⎜ ⎜
⎟⎜ ⎟ ⎟ =
L
⎝ ⎝ L ⎠⎝ 2 ⎠ ⎠

2

⎛π ⎞
sin ⎜ ⎟ =
L
⎝2⎠

2
.
L

2
sin(π ) = 0.
L

P=0
(c) x = 3L/4
2
⎛ ⎛ 2π ⎞⎛ 3L ⎞ ⎞
sin ⎜ ⎜
⎟⎜ ⎟ ⎟ =
L
⎝ ⎝ L ⎠⎝ 4 ⎠ ⎠
P = (2/L)dx

ψ ( x) =

2
2
⎛ 3π ⎞
sin ⎜ ⎟ = −
.

L
L
2
⎝ ⎠

EVALUATE: Our results are consistent with the n = 2 part of Figure 40.12 in the textbook. ψ
40.58.

2

is zero at

the center of the box and is symmetric about this point.
IDENTIFY: The impulse applied to a particle equals its change in momentum.
SET UP: For a particle in a box, the magnitude of its momentum is p = k =

nh
(Eq. 40.29).
2L

nπ hn
=
. At x = 0 the initial momentum at the wall is
L
2L
hn
hn ˆ
pinitial = − iˆ and the final momentum, after turning around, is pfinal = +
i . So,
2L

2L
hn ˆ
hn ˆ ⎛ hn ˆ ⎞
hn
i and the final
Δp=+
i − ⎜ − i ⎟ = + iˆ. At x = L the initial momentum is pinitial = +
2L
2L
L
⎝ 2L ⎠

EXECUTE: Δ p = pfinal − pinitial . p = k =

momentum, after turning around, is pfinal = −
40.59.

hn ˆ
hn ˆ hn ˆ
hn
i . So, Δ p = −
i−
i = − iˆ.
2L
2L
2L
L

EVALUATE: The impulse increases with n.
IDENTIFY: Carry out the calculations that are specified in the problem.

SET UP: For a free particle, U ( x ) = 0 so Schrödinger’s equation becomes

d 2ψ ( x )
dx 2

=−

2m
h2

Eψ ( x ).

EXECUTE: (a) The graph is given in Figure 40.59.
2 2
dψ ( x )
d 2ψ ( x)
2m
κ
(b) For x < 0: ψ ( x ) = e +κ x .
= κ e+κ x .
= κ 2e+κ x . So κ 2 = − 2 E ⇒ E = −
.
dx
dx
2m

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40-18

Chapter 40
(c) For x > 0: ψ ( x ) = e−κ x .

d 2ψ ( x)
dψ ( x )
2m
− 2κ 2
.
= −κ e−κ x .
= κ 2e−κ x . So again κ 2 = − 2 E ⇒ E =
dx
dx
2m

Parts (b) and (c) show ψ ( x ) satisfies the Schrödinger’s equation, provided E =

− 2κ 2
.
2m

dψ ( x )
is discontinuous at x = 0. (That is, it is negative for x > 0 and positive for x < 0.)
dx
Therefore, this ψ is not an acceptable wave function; dψ /dx must be continuous everywhere, except
where U → ∞.
EVALUATE: (d)

Figure 40.59

40.60.

IDENTIFY: We start with the penetration distance formula given in the problem.
SET UP: The given formula is η =

2m(U 0 − E )

.

EXECUTE: (a) Substitute the given numbers into the formula:
1.055 × 10−34 J ⋅ s
=
= 7.4 × 10−11 m
η=
−31
−19
2m(U 0 − E )
2(9.11 × 10 kg)(20 eV − 13 eV)(1.602 × 10
J/eV)
(b) η =

40.61.

1.055 × 10−34 J ⋅ s
2(1.67 × 10−27 kg)(30 MeV − 20 MeV)(1.602 × 10−13 J/MeV)

= 1.44 × 10−15 m

EVALUATE: The penetration depth varies widely depending on the mass and energy of the particle.
IDENTIFY: Eq. (40.38) applies for 0 ≤ x ≤ L. Eq. (40.40) applies for x < 0 and x > L. D = 0 for x < 0

and C = 0 for x > L.

2mE

d
d
d κx
d −κ x
sin kx = k cos kx.
cos kx = − k sin kx.
= −κ e −κ x .
e = κ eκ x .
e
dx
dx
dx
dx
EXECUTE: (a) We set the solutions for inside and outside the well equal to each other at the well
boundaries, x = 0 and L.
x = 0: B sin(0) + A = C ⇒ A = C , since we must have D = 0 for x < 0.

SET UP: Let k =

x = L: B sin

2mE L

.

+ A cos

2mE L

= + De−κ L since C = 0 for x > L.

This gives B sin kL + A cos kL = De−κ L , where k =

2mE

.

(b) Requiring continuous derivatives at the boundaries yields
dψ
x = 0:
= kB cos( k ⋅ 0) − kA sin(k ⋅ 0) = kB = κ Cek ⋅ 0 ⇒ kB = κ C.
dx

x = L: kB cos kL − kA sin kL = −κ De−κ L
EVALUATE: These boundary conditions allow for B, C, and D to be expressed in terms of an overall
normalization constant A.
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Quantum Mechanics

40.62.

IDENTIFY: T = Ge−2κ L with G = 16

40-19

2m(U 0 − E )
1 ⎛T ⎞
E ⎛
E ⎞
, so L = − ln ⎜ ⎟ .
⎜1 −
⎟ and κ =
2κ ⎝ G ⎠
U0 ⎝ U0 ⎠

SET UP: E = 5.5 eV, U 0 = 10.0 eV, m = 9.11 × 10−31 kg, and T = 0.0010.

2(9.11 × 10−31 kg)(4.5 eV)(1.60 × 10−19 J/eV)

EXECUTE: κ =

and G = 16

= 1.09 × 1010 m −1

5.5 eV ⎛
5.5 eV ⎞
⎜1 −
⎟ = 3.96,
10.0 eV ⎝ 10.0 eV ⎠

⎛ 0.0010 ⎞
−10

ln ⎜
⎟ = 3.8 × 10 m = 0.38 nm.
2(1.09 × 10 m ) ⎝ 3.96 ⎠
EVALUATE: The energies here are comparable to those of electrons in atoms, and the barrier width we
calculated is on the order of the diameter of an atom.
IDENTIFY and SET UP: When κ L is large, then eκ L is large and e−κ L is small. When κ L is small,
sinh κ L → κ L. Consider both κ L large and κ L small limits.

1

so L = −

40.63.

(1.054 × 10−34 J ⋅ s)

−1

10

⎡ (U sinh κ L) 2 ⎤
EXECUTE: (a) T = ⎢1 + 0
⎥
4 E (U 0 − E ) ⎥⎦
⎢⎣

−1

sinh κ L =

eκ L − e−κ L
2

For κ L

1, sinh κ L →

For κ L

1, 16 E (U 0 − E ) + U 02e2κ L → U 02e2κ L

T→

16 E (U 0 − E )
U 02e2κ L

(b) κ L =

⎡
eκ L
U 02e2κ L ⎤
and T → ⎢1 +
⎥
2
⎢⎣ 16 E (U 0 − E ) ⎥⎦

−1

=

16 E (U 0 − E )

16 E (U 0 − E ) + U 02e2κ L

⎛ E ⎞⎛
E ⎞ −2κ L
= 16 ⎜
, which is Eq. (40.42).
⎟⎜1 −
⎟e
U
U
0⎠
⎝ 0 ⎠⎝

L 2m(U 0 − E )

. So κ L

1 when L is large (barrier is wide) or U 0 − E is large. (E is small

compared to U 0 .)
(c) κ =

2m(U 0 − E )

; κ becomes small as E approaches U 0 . For κ small, sinh κ L → κ L and

⎡
U 02κ 2 L2 ⎤

T → ⎢1 +
⎥
⎣⎢ 4 E (U 0 − E ) ⎦⎥

−1

⎡ 2U 02 L2m ⎤
Thus T → ⎢1 +
⎥
4 E 2 ⎦⎥
⎣⎢

U 0 → E so

⎡ U 2 2m(U 0 − E ) L2 ⎤
= ⎢1 + 0 2
⎥
4 E (U 0 − E ) ⎦⎥
⎣⎢

−1

(using the definition of κ ).

−1

⎡ 2 EL2 m ⎤
U 02
→ E and T → ⎢1 +
⎥

E
4 2 ⎦⎥
⎣⎢

−1

−1

⎡ ⎛ kL ⎞2 ⎤
But k = 2 , so T → ⎢1 + ⎜ ⎟ ⎥ , as was to be shown.
⎢⎣ ⎝ 2 ⎠ ⎥⎦
EVALUATE: When κ L is large Eq. (40.41) applies and T is small. When E → U 0 , T does not approach unity.
2

40.64.

2mE

IDENTIFY: Compare the energy E of the oscillator to Eq. (40.46) in order to determine n.
SET UP: At the equilibrium position the potential energy is zero and the kinetic energy equals the total
energy.

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


40-20

Chapter 40
EXECUTE: (a) E =

1 2
mv = [n + (1/2)] ω = [n + (1/2)]hf , and solving for n,
2

1 2
mv
1 (1/2)(0.020 kg)(0.360 m/s) 2 1
− =
− = 1.3 × 1030.
n= 2
hf
2 (6.63 × 10−34 J ⋅ s)(1.50 Hz) 2
(b) The difference between energies is ω = hf = (6.63 × 10−34 J ⋅ s)(1.50 Hz) = 9.95 × 10−34 J. This energy

40.65.

is too small to be detected with current technology.
EVALUATE: This oscillator can be described classically; quantum effects play no measurable role.
IDENTIFY and SET UP: Calculate the angular frequency ω of the pendulum and apply Eq. (40.46) for the
energy levels.
2π
2π
EXECUTE: ω =
=
= 4π s −1
T
0.500 s
1
1

The ground-state energy is E0 = ω = (1.055 × 10−34 J ⋅ s)(4π s −1) = 6.63 × 10−34 J.
2
2
E0 = 6.63 × 10−34 J(1 eV/1.602 × 10−19 J) = 4.14 × 10−15 eV
1⎞
⎛
En = ⎜ n + ⎟ ω
2⎠
⎝

1⎞
⎛
En +1 = ⎜ n + 1 + ⎟ ω
2⎠
⎝
The energy difference between the adjacent energy levels is
Δ E = En +1 − En = ω = 2 E0 = 1.33 × 10−33 J = 8.30 × 10−15 eV.

40.66.

EVALUATE: These energies are much too small to detect. Quantum effects are not important for ordinary
size objects.
IDENTIFY: We model the electrons in the lattice as a particle in a box. The energy of the photon is equal
to the energy difference between the two energy states in the box.
n2h2
SET UP: The energy of an electron in the n th level is En =
. We do not know the initial or final
8mL2
levels, but we do know they differ by 1. The energy of the photon, hc/λ , is equal to the energy difference
between the two states.

hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
=
=
EXECUTE: The energy difference between the levels is Δ E =
λ
1.649 × 10−7 m

1.206 × 10−18 J. Using the formula for the energy levels in a box, this energy difference is equal to

40.67.

h2
h2
Δ E = ⎡⎣ n 2 − (n − 1) 2 ⎤⎦
=
(2
n
−
1)
.
8mL2
8mL2
⎞ 1 ⎛ (1.206 × 10−18 J)8(9.11 × 10−31 kg)(0.500 × 10−9 m)2
⎞
1 ⎛ Δ E8mL2
+ 1⎟ = ⎜
+ 1⎟ = 3.
Solving for n gives n = ⎜
2
−

34
2
⎜
⎟
⎜
⎟
2⎝ h
(6.626 × 10
J ⋅ s)
⎠ 2⎝
⎠
The transition is from n = 3 to n = 2.
EVALUATE: We know the transition is not from the n = 4 to the n = 3 state because we let n be the
higher state and n − 1 the lower state.
IDENTIFY: At a maximum, the derivative of the probability function is zero.
2
2
mk ′
2
2
SET UP and EXECUTE: ψ ( x) = Ce−α x , where α =
. ψ ( x) = C e−2α x . At values of x where
2

ψ ( x)

2

2

is a maximum,

2

2

2
d 2 ψ ( x)
d ψ ( x)
d ψ ( x)
2
< 0.
= 0 and
= C ( −2α x)e−2α x = 0. Only
2
dx
dx
dx

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Quantum Mechanics

solution is x = 0.

2

dx 2

2

2
2
d 2 ψ ( x)
2
2
= C ⎡⎢ −2α e−2α x + 4α 2 x e−2α x ⎤⎥ . At x = 0,
= C ( −2α ) < 0, so
⎣
⎦
dx 2

2

ψ ( x)

40.68.

d 2 ψ ( x)

40-21

is a maximum at x = 0.
EVALUATE: There is only one maximum, at x = 0, so the probability function peaks only there.
IDENTIFY: If the given wave function is a solution to the Schrödinger equation, we will get an identity
when we substitute that wave function into the Schrödinger equation.
SET UP: The given wave function is ψ 1( x) = A1xe−α

−

2

d ψ ( x)
2m

dx

2

+

2 2

x /2

and the Schrödinger equation is

2

k′x
ψ ( x) = E ψ ( x).
2

EXECUTE: (a) Start by taking the indicated derivatives: ψ 1( x) = A1xe−α

dψ 1 ( x )
= −α 2 x 2 A1e−α
dx

d 2ψ 1 ( x )
dx

2

d 2ψ 1( x)
dx 2

−

2 2

+ A1e−α

x /2

2 2

x /2

=−

.

− A1α 2 x 2 (−α 2 x )e−α

2 2

x /2

+ A1 (−α 2 x)e−α

2 2

x /2

.

⎡ −3α 2 + (α 2 ) 2 x 2 ⎤ ψ 1( x).
⎦
2m ⎣
d 2ψ ( x)
2m

dx 2

+

k ′ x2
ψ ( x) = E ψ ( x). Substituting the above result into that equation
2

2

2
⎡ −3α 2 + (α 2 ) 2 x 2 ⎤ ψ 1 ( x) + k ′ x ψ 1 ( x) = E ψ 1 ( x). Since α 2 = mω and ω = k ′ , the
⎣
⎦
2m
2

m

coefficient of x 2 is −

⎛ mω ⎞
(b) A1 = ⎜
⎟
⎝
⎠

3/4

2

2m

(α 2 )2 +

2

2
k′
mω 2
⎛ mω ⎞
=−
= 0.
⎜
⎟ +
2
2m ⎝

2
⎠

1/ 4

⎛4⎞
⎜ ⎟
⎝π ⎠

(c) The probability density function ψ

is ψ 1 ( x ) = A12 x 2e −α
2

2

d 2 ψ 1( x)

2

2

2

d 2 ψ 1 ( x)

2

d ψ 1 ( x)
1 d ψ 1 ( x)

= 0. At x = ± ,
= 0.
α
dx
dx

dx 2
d 2 ψ 1( x)

x

2 2
2 2
2 2
2 2
d ψ 1( x)
= A12 2 xe−α x + A12 x 2 (−α 2 2 x)e−α x = A12 2 xe−α x − A12 2 x3α 2e−α x .
dx

2

At x = 0,

2 2

2

2

At x = 0, ψ 1 = 0.

dx 2

.

2

Equation (40.44) is −

dx

x /2

= ⎡ −2α 2 + (α 2 ) 2 x 2 − α 2 ⎤ ψ 1( x) = ⎡ −3α 2 + (α 2 ) 2 x 2 ⎤ ψ 1( x).
⎣
⎦
⎣
⎦

dx 2

gives −

x /2

= − A1α 2 2 xe −α

d 2ψ 1 ( x)
2m

2 2

2 2

= A12 2e−α

2 2

+ A12 2 x(−α 2 2 x)e−α

= A12 2e−α

2 2

− A12 4 x 2α 2e−α

x

x

2 2

x

2 2

x

− A12 2(3 x 2 )α 2e−α

− A12 6 x 2α 2e−α

2 2

x

2 2

x

− A12 2 x3α 2 (−α 2 2 x)e−α

+ A12 8 x 4 (α 2 ) 2 e−α

2 2

x

2 2

x

.

. At x = 0,

2

> 0. So at x = 0, the first derivative is zero and the second derivative is positive. Therefore,
1

d 2 ψ 1( x)

2

1

< 0. So at x = ± , the
α
α
dx 2
first derivative is zero and the second derivative is negative. Therefore, the probability density function has
1
maxima at x = ± , corresponding to the classical turning points for n = 0 as found in the previous question.
α
the probability density function has a minimum at x = 0. At x = ±

,

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40-22

Chapter 40
EVALUATE: ψ 1 ( x) = A1xe−α

E=
40.69.

2 2/ 2

x

is a solution to Eq. (40.44) if −

2

2m

(−3α 2 )ψ 1( x) = E ψ 1 ( x) or

3 2α 2 3 ω
3 ω
. E1 =
=
corresponds to n = 1 in Equation (40.46).
2m
2
2

⎛λ⎞
IDENTIFY: For a standing wave in the box, there must be a node at each wall and n ⎜ ⎟ = L.
⎝2⎠
h
h
SET UP: p = so mv = .

λ

λ

EXECUTE: (a) For a standing wave, nλ = 2 L, and En =

p 2 (h/λ ) 2 n 2 h 2
.
=
=
2m
2m
8mL2

(b) With L = a0 = 0.5292 × 10−10 m, E1 = 2.15 × 10−17 J = 134 eV.
EVALUATE: For a hydrogen atom, En is proportional to 1/n 2 so this is a very poor model for a hydrogen
40.70.

atom. In particular, it gives very inaccurate values for the separations between energy levels.
IDENTIFY and SET UP: Follow the steps specified in the problem.
EXECUTE: (a) As with the particle in a box, ψ ( x ) = A sin kx, where A is a constant and k 2 = 2mE/ 2 .
Unlike the particle in a box, however, k and hence E do not have simple forms.
(b) For x > L, the wave function must have the form of Eq. (40.40). For the wave function to remain finite
as x → ∞, C = 0. The constant κ 2 = 2m(U 0 − E )/ , as in Eq. (40.40).
(c) At x = L, A sin kL = De −κ L and kA cos kL = −κ De −κ L . Dividing the second of these by the first gives
k cot kL = −κ , a transcendental equation that must be solved numerically for different values of the length

L and the ratio E/U 0 .
EVALUATE: When U 0 → ∞, κ → ∞ and
40.71.

cos(kL)
nπ
→ ∞. The solutions become k =
, n = 1, 2, 3,…, the
sin(kL)
L

same as for a particle in a box.
IDENTIFY: Require ψ (− L/2) = ψ ( L/2) = 0.
p2
.
λ
λ
2m
EXECUTE: (a) ψ ( x) = A sin kx and ψ (− L /2) = 0 = ψ (+ L /2)
SET UP: k =

2π

, p=

h

and E =

2nπ 2π
⎛ + kL ⎞ + kL
⇒ 0 = A sin ⎜
= nπ ⇒ k =
=

⎟⇒
λ
2
L
⎝ 2 ⎠

L
h nh
p 2 n 2 h 2 (2n) 2 h 2
⇒ pn = =
⇒ En =
=
=
, where n = 1, 2…
λ L
n
2m 2mL2
8mL2
(b) ψ ( x) = A cos kx and ψ ( − L /2) = 0 = ψ (+ L /2)

⇒λ =

π
(2n + 1)π 2π
⎛ kL ⎞ kL
⇒ 0 = A cos ⎜ ⎟ ⇒
= (2n + 1) ⇒ k =
=
2
2

2
L
λ
⎝ ⎠
2L
(2n + 1)h
⇒λ =
⇒ pn =
(2n + 1)
2L
⇒ En =

(2n + 1) 2 h 2

n = 0, 1, 2…
8mL2
(c) The combination of all the energies in parts (a) and (b) is the same energy levels as given in
n2h2
.
Eq. (40.31), where En =
8mL2
EVALUATE: (d) Part (a)’s wave functions are odd, and part (b)’s are even.

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Quantum Mechanics
40.72.

IDENTIFY and SET UP: Follow the steps specified in the problem.
p2
h
EXECUTE: (a) E = K + U ( x) =
+ U ( x) ⇒ p = 2m( E − U ( x)). λ = ⇒ λ ( x) =
2m
p

40-23

h
.
2m( E − U ( x))

(b) As U ( x) gets larger (i.e., U ( x) approaches E from below—recall k ≥ 0), E − U ( x )
gets smaller, so λ ( x ) gets larger.
(c) When E = U ( x ), E − U ( x ) = 0, so λ ( x ) → ∞.
b
dx
1 b
n
hn
⇒ ∫ 2m( E − U ( x)) dx = .
= ∫ 2m( E − U ( x)) dx =
a
2
2
2m( E − U ( x )) h a
(e) U ( x) = 0 for 0 < x < L with classical turning points at x = 0 and x = L. So,

(d)

b

∫a

b

dx

b

∫a λ ( x) = ∫a h/

2m(E − U (x)) dx = ∫

L

0

L

2mEdx = 2mE ∫ dx = 2mE L. So, from part (d),
0

2

hn
1 ⎛ hn ⎞
h2n2

.
⇒E=
⎜
⎟ =
2
2m ⎝ 2 L ⎠
8mL2
EVALUATE: (f) Since U ( x ) = 0 in the region between the turning points at x = 0 and x = L, the result is
2mE L =

the same as part (e). The height U 0 never enters the calculation. WKB is best used with smoothly varying
potentials U ( x).
40.73.

Perform the calculations specified in the problem.
SET UP: U ( x) = 12 k ′x 2 .
DENTIFY:

EXECUTE: (a) At the turning points E =
(b)

1
nh
⎛
⎞
2m ⎜ E − k ′x 2 ⎟dx = . To evaluate the integral, we want to get it into a form that matches
2
2
⎝
⎠

+ 2 E /k ′

∫−

2 E /k ′

the standard integral given.
Letting A2 =

⇒ mk ′ ∫

b

a

1 2
2E
.
k ′xTP ⇒ xTP = ±
2
k′

1
2mE
2E
⎛
⎞
2m ⎜ E − k ′x 2 ⎟ = 2mE − mk ′x 2 = mk ′
− x 2 = mk ′

− x2 .
′
′
2
mk
k
⎝
⎠

2E
2E
2E
,a=−
, and b = +
k′
k′
k′

mk ′
A − x dx = 2
2
2

⎡ 2E
= mk ′ ⎢
⎢⎣ k ′

2

b

⎡
⎛ x ⎞⎤
2
2
2
⎢ x A − x + A arcsin ⎜⎜ ⎟⎟ ⎥
⎢⎣
⎝ A ⎠ ⎥⎦ 0

⎛ 2E k ′ ⎞⎤
2E 2E 2E
2E
m⎛1⎞
arcsin ⎜
arcsin (1) = 2 E
−
+
⎟⎟ ⎥ = mk ′
⎜ ⎟.
⎜
′
′
′
′
k
k
k
k
k′ ⎝ 2 ⎠

⎝ 2 E k ′ ⎠ ⎥⎦

Using WKB, this is equal to

hn
m
hn
k′
h
, so E
π = . Recall ω =
, so E =
ω n = hω n.
2
k′
2
m
2π

ω⎛
1 ⎞⎞
⎛
⎜ recall E = ω ⎜ n + ⎟ ⎟ . It
2 ⎝
2 ⎠⎠
⎝
underestimates the energy. However, our approximation isn’t bad at all!
IDENTIFY and SET UP: Perform the calculations specified in the problem.
E
EXECUTE: (a) At the turning points E = A xTP ⇒ xTP = ± .

A
EVALUATE: (c) We are missing the zero-point-energy offset of

40.74.

(b)

+ E /A

∫− E /A

2m( E − A x )dx = 2∫

dy = −2mA dx when x =

E /A

0

2m( E − Ax) dx. Let y = 2m( E − Ax) ⇒

E
, y = 0, and when x = 0, y = 2mE. So
A

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


40-24

Chapter 40
E

2∫ A 2m( E − Ax)dx = −
0

0

1 0
2 3/2
2
hn
.
(2mE )3/2 . Using WKB, this is equal to
y1/2dy = −
y
=
∫
mE
2
3mA
3mA
mA
2
2 mE
2/3

2
hn

1 ⎛ 3mAh ⎞
2/3
⇒E=
(2mE )3/ 2 =
⎜
⎟ n .
3mA
2
2m ⎝ 4 ⎠
EVALUATE: (c) The difference in energy decreases between successive levels. For example:
So,

12/3 − 02/3 = 1, 22/3 − 12/3 = 0.59, 33/2 − 23 2 = 0.49,…
•

A sharp ∞ step gave ever-increasing level differences (~ n 2 ).

•

A parabola (~ x 2 ) gave evenly spaced levels (~ n).

•

Now, a linear potential (~ x) gives ever-decreasing level differences (~ n 2/3 ).

Roughly speaking, if the curvature of the potential (~ second derivative) is bigger than that of a parabola,
then the level differences will increase. If the curvature is less than a parabola, the differences will
decrease.

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.