ISM chapter41 tủ tài liệu training

41
Quantum Mechanics
CHAPTER OUTLINE
41.1

The Wave Function

41.2

Analysis Model: Quantum Particle Under Boundary Conditions

41.3

The Schrödinger Equation

41.4

A Particle in a Well of Finite Height

41.5

Tunneling Through a Potential Energy Barrier

41.6

Applications of Tunneling

41.7

The Simple Harmonic Oscillator

* An asterisk indicates a question or problem new to this edition.

ANSWERS TO OBJECTIVE QUESTIONS
OQ41.1

Answer (b). Fewer particles are reflected as the height of the
potential barrier decreases and approaches the energy of the
particles. By Equations 41.22 and 41.23, the transmission coefficient

T ≈ e −2CL , where C = 2m (U − E )  , increases as U − E decreases, so
the reflection coefficient R = 1 − T ≈ 1 − e −2CL decreases as U − E
decreases.
OQ41.2

The ranking is answer (b) > (a) > (c) > (e) > (d). From Equation 41.14,
consider the quantity
⎛ h2 ⎞ 2
E=⎜
n :
⎝ 8mL2 ⎟⎠

(a)

⎡
⎤ 2 1 ⎛ h2
⎞
h2
nm −1 ⎟
⎢
2 ⎥ ( 1) =

⎜
9 ⎝ 8m1
⎠
⎣ 8m1 ( 3 nm ) ⎦

948
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Chapter 41

OQ41.3

(b)

⎡
⎤ 2 4 ⎛ h2
⎞
h2
nm −1 ⎟
⎢
2 ⎥(2) =
⎜
9 ⎝ 8m1
⎠
⎣ 8m1 ( 3 nm ) ⎦

(c)

⎡

⎤ 2 1 ⎛ h2
⎞
h2
nm −1 ⎟
⎢
2 ⎥ ( 1) =
⎜
18 ⎝ 8m1
⎠
⎢⎣ 8 ( 2m1 ) ( 3 nm ) ⎥⎦

(d)

⎡
⎤ 2
( 0 )2
⎢
2 ⎥ ( 1) = 0
⎣ 8m1 ( 3 nm ) ⎦

(e)

⎡
⎤ 2 1 ⎛ h2
⎞
h2
nm −1 ⎟
⎢
2 ⎥ ( 1) =
⎜

36 ⎝ 8m1
⎠
⎣ 8m1 ( 6 nm ) ⎦

(a)

True. Examples: An electron has mass and charge, but it can
also display interference effects.

(b)

False. An electron has rest energy ER = mec2.

(c)

True. A moving electron possesses kinetic energy.

949

(d) True. p = meu.
OQ41.4

(e)

True.

(a)

True. Examples: A photon behaves as a particle in the
photoelectric effect and as a wave in double-slit interference.

(b)

True. A photon cannot have rest energy (mass) because it is
never at rest: it travels at the speed of light.

(c)

True. E = hf.

(d) True. p = E/c.
(e)

True.

OQ41.5

Answer (d). The probability of finding the particle is at the antinodes
(places of greatest amplitude) of the standing wave.

OQ41.6

Compare the ground state wave functions in Figures 41.4 and 41.7 in
the text. In the square well with infinitely 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/2L, and its
energy p2/2m = h2/8mL2. In the well with walls of only finite height,
the wave function has nonzero amplitude at the walls, and it extends
outside the walls.
(i)

Answer (a). The ground state wave function extends somewhat
outside the walls of the finite well, so the particle’s wavelength
is longer.

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950

Quantum Mechanics
(ii) Answer (b). The particle’s momentum in its ground state is
smaller because p = h/λ and the wave function has a larger
wavelength.
(iii) Answer (b). The particle has less energy because is has smaller
momentum.

OQ41.7

Answer (e). From the relation between the square of the wave
function and the probability P of finding the particle in the interval
Δx = (7 nm − 4 nm) = 3 nm, we have
2

ψ Δx = P
OQ41.8

→

ψ=

P
0.48
=
= 0.40 nm −1
Δx
3 nm

Answer (a). Because of the exponential tailing of the wave function
within the barrier, the tunneling current is more sensitive to the
width of the barrier than to its height. Notice that the exponent term
CL in the transmission coefficient T ≈ e −2CL , where

C = 2m (U − E )  , decreases more if L decreases than if U decreases
by the same percentage.
OQ41.9

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

OQ41.10

Answer (d). The probability of finding the particle is greatest at the
place of greatest amplitude of the wave function. The next most
likely place is point b, after that, points a and e appear to be equally
probable. The particle would never be found at point c.

ANSWERS TO CONCEPTUAL QUESTIONS
CQ41.1

Consider the Heisenberg uncertainty principle. It implies that
electrons initially moving at the same speed and accelerated by an
electric field 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.”

CQ41.2

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 infinite, but equal to the
width of the region. In such a case, the uncertainty product ΔxΔpx
would be zero, violating the uncertainty principle. This contradiction
proves that the minimum energy of the particle is not zero.

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Chapter 41

951

CQ41.3

The motion of the quantum particle does not consist of moving
through successive points. The particle has no definite 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 infinite speed
to cross the node.

CQ41.4

(a)

ψ (x) becomes infinite as x → ∞ .

(b)

ψ (x) is discontinuous and becomes infinite at x = π/2, 3π/2,…

CQ41.5

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

CQ41.6

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.

CQ41.7

Both (d) and (e) are not physically significant. Wave function (d) is
not acceptable because ψ is not single-valued. Wave function (e) is
not acceptable because ψ is discontinuous (as is its slope).

CQ41.8

Newton’s 1st and 2nd laws are used to determine the motion of a
particle of large mass. The Schrödinger equation is not used to
determine the motion of a particle of small mass; rather, it is used to
determine the state of the wave function of a particle of small mass.
In particular, the states of atomic electrons are confined-wave states
whose wave functions are solutions to the Schrödinger equation.
Anything that we can know about a particle comes from its wave
function.

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952

Quantum Mechanics

SOLUTIONS TO END-OF-CHAPTER PROBLEMS
Section 41.1
P41.1

(a)

The Wave Function
The wave function,

ψ ( x ) = Ae (

i 5×1010 x

) = A cos 5 × 1010 x + iA sin 5 × 1010 x
(
)
(
)

will go through one full cycle between x1 = 0 and (5.00 × 1010)x2 =
2 π . The wavelength is then

λ = x 2 − x1 =

2π
= 1.26 × 10 –10  m
10
−1
5.00 × 10  m

To say the same thing, we can inspect Ae
wave number is k = 5.00 × 1010 m–1 = 2π/λ .
(b)

) to see that the

Since λ = h/p, the momentum is

p=
(c)

(

i 5 × 1010 x

h 6.626 × 10 –34  J ⋅ s
=
= 5.27 × 10 –24  kg ⋅ m/s
λ
1.26 × 10 –10  m

The electron’s kinetic energy is

K=

1
p2
mu2 =
 
2
2m

( 5.27 × 10  kg ⋅ m/s )
=

2 ( 9.11 × 10  kg )
–24

–31

2

⎛
⎞
1 eV
⎜⎝ 1.602 × 10 –19  J ⎟⎠ = 95.3 eV

[We use u to represent the speed of a particle with mass in chapters
39, 40, and 41.]
P41.2

(a)

See ANS. FIG. P41.2 for a graph of
−3 <

x
< 3.
a

f (x)
= e − x /a for the range
A

ANS. FIG. P41.2

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Chapter 41
(b)

953

Normalization requires
2

∫ ψ dx = 1:

all space

∞

∞

−∞

0

2 −2 x /a
dx = 2 ∫ A 2 e −2 x /a dx = 1
∫Ae

− aA 2 e −2 x /a

∞

0

= aA 2 = 1 → A =

1
a

(c)

a −2 x /a
a
e −2 x /a
e
P= ∫
dx = 2 ∫
dx = −e −2 x/a 0 = −e −2 + 1 = 0.865
a
a
−a
0

(a)

Normalization requires

a

P41.3

2

∫

ψ dx = 1:

all space

1.00

2 2
∫ A x dx = 1
0

A2 x 3
3
0.400

(b)

1.00

=
0

A 23
=1
3

0.400

→

A= 3

P = ∫ 3x 2 dx = x 3 0.300 = ( 0.400 ) − ( 0.300 ) = 0.037 0
2

2

0.300

(c)

The expectation value is
1.00

x =

∫ ψ * xψ dx = ∫0

all space

P41.4

3x 4
3x dx =
4

1.00

= 0.750

3

0

The probability is given by

P=

a

∫ ψ ( x)

−a

P=

2

a

a
⎛ a ⎞ ⎛ 1⎞
⎛ x⎞
= ∫
dx = ⎜ ⎟ ⎜ ⎟ tan −1 ⎜ ⎟
2
2
⎝

⎠
⎝
⎠
⎝ a⎠
π a
−a π (x + a )

a

−a

1
1 ⎡π ⎛ π ⎞ ⎤ 1
⎡⎣ tan −1 1 − tan −1 ( −1) ⎤⎦ = ⎢ − ⎜ − ⎟ ⎥ =
π
π ⎣ 4 ⎝ 4⎠⎦ 2

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954

Quantum Mechanics

Section 41.2
P41.5

(a)

Analysis Model: Quantum Particle

Under Boundary Conditions
The energy of a quantum particle confined to a line segment is

En =

h 2 n2
8mL2

Here we have for the ground state

(6.626 × 10 J · s ) (1)
=
8 ( 1.67 × 10 kg ) ( 2.00 × 10
–34

E1

2

–27

2
–14

m)

2

= 8.22 × 10 –14 J = 0.513 MeV

and for the first and second excited states, which are states 2 and 3,

E2 = 4E1 = 2.05 MeV
(b)

and E3 = 9E1 = 4.62 MeV

They do; the MeV is the natural unit for energy radiated by an
atomic nucleus.
Stated differently: Scattering experiments show that an atomic
nucleus is a three-dimensional object always less than 15 fm in
diameter. This one-dimensional box 20 fm long is a good model in
energy terms.

P41.6

From Equation 41.14, the allowed energy levels of a particle in a box is
⎛ h2 ⎞ 2
En = ⎜
n ,
⎝ 8mL2 ⎟⎠

(a)

n = 1, 2, 3,...

For L = 1.00 nm,

⎛ h2 ⎞ 2
En = ⎜

n
⎝ 8mL2 ⎟⎠
2
⎤
6.626 × 10−34 J ⋅ s )
(
⎛
1 eV
⎞⎡
⎢
⎥ 2
=⎜
2 n
−19 ⎟
−31
−9
⎝ 1.60 × 10 J ⎠ ⎢ 8 ( 9.11 × 10 kg ) ( 1.00 × 10 m ) ⎥
⎣
⎦

= 0.377n2 = 6 eV
n≈4

P41.7

(b)

For n = 4, En = 0.377 ( 4 ) = 6.03 eV

(a)

From Equation 41.14, the allowed energy levels of an electron in a
box is

2

⎛ h2 ⎞ 2
En = ⎜
n
⎝ 8me L2 ⎟⎠

n = 1, 2, 3, . . .

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Chapter 41

955

Substituting numerical values,
2
⎡
⎤
6.626 × 10−34 J ⋅ s )
(
⎥ 2
En = ⎢
2 n
−31

−9
⎢ 8 ( 9.11 × 10 kg ) ( 0.100 × 10 m ) ⎥
⎣
⎦

= ( 6.02 × 10−18 J ) n2 = ( 37.7 eV ) n2

ANS. FIG. P41.7
(b)

When the electron falls from higher level ni to lower level nf , it
emits energy
⎛ h2 ⎞ 2
ΔEn = ⎜
ni − n2f = ( 37.7 eV ) ni2 − n2f
⎝ 8me L2 ⎟⎠

(

)

(

)

by emitting a photon of wavelength

hc
8me cL2
λ=

=
ΔEn h ni2 − n2f

(

=

)

8 ( 9.109 × 10−31 kg ) ( 2.998 × 108 m/s ) ( 0.100 × 10−9 m )

(6.626 × 10

−34

(

J ⋅ s ) ni2 − n2f

2

)

⎛ 1 nm ⎞
× ⎜ −9 ⎟
⎝ 10 m ⎠
=

33.0 nm

(n

2
i

− n2f

)

For example, for the transition 4 → 3, the wavelength is

λ=

33.0 nm
= 4.71 nm
( 4 )2 − ( 3 )2

The wavelengths produced by all possible transitions are:
Transition 4 → 3 4 → 2 4 → 1 3 → 2 3 → 1 2 → 1

λ (nm)

4.71

2.75

2.20

6.59

4.12

11.0

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956
P41.8

Quantum Mechanics
The energy of the photon is

E=

hc 1 240 eV ⋅ nm ⎛ 1 mm ⎞
−4
=
⎜⎝ 6
⎟⎠ = 2.05 × 10 eV
λ
6.06 mm
10 nm

The allowed energies of the proton in the box are

⎛ h2 ⎞ 2
En = ⎜
n
⎝ 8mL2 ⎟⎠

2
⎡
⎤⎛
6.626 × 10−34  J ⋅ s )
(
1 eV
⎞ 2
⎥
=⎢
2 ⎜
−19
⎟n
−27
−9
⎢ 8 ( 1.673 × 10  kg ) ( 1.00 × 10  m ) ⎥ ⎝ 1.602 × 10  J ⎠
⎣
⎦

= ( 2.05 × 10−4  eV ) n2

The smallest possible energy for a transition between states is from
n = 1 to n = 2, which has energy

ΔEn = ( 2.05 × 10−4 eV ) ( 2 2 − 12 ) = 6.14 × 10−4 eV
The photon does not have enough energy to cause this transition. The
photon energy would be sufficient to cause a transition from n = 0 to
n = 1, but the n = 0 state does not exist for the particle in a box.
P41.9

From Equation 41.14,

ΔE =

hc ⎛ h2 ⎞ 2
3h2
2
⎡
⎤
=⎜
2
−
1
=
⎦ 8m L2
λ ⎝ 8me L2 ⎟⎠ ⎣
e

Solving for the length of the box then gives

L=
=

3hλ
8me c
3 ( 6.626 × 10−34 J ⋅ s ) ( 694.3 × 10−9 m )
8 ( 9.11 × 10−31 kg ) ( 3.00 × 108 m/s )

= 7.95 × 10−10 m = 0.795 nm
P41.10

From Equation 41.14,

ΔE =

hc ⎛ h2 ⎞ 2
3h2
2
⎡
⎤
=⎜
2
−
1
=
⎦ 8m L2
λ ⎝ 8me L2 ⎟⎠ ⎣
e

Solving for the length of the box then gives
L=

3hλ
8me c

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Chapter 41
P41.11

957

From Equation 41.14, the allowed energy levels of a particle in a box is
⎛ h2 ⎞ 2
En = ⎜
n = n2E1
⎝ 8mL2 ⎟⎠

n = 1, 2, 3, …

For a proton (m = 1.673 × 10–27 kg) in a 10.0-fm wide box:

(6.626 × 10 J ⋅ s )
=
8 ( 1.673 × 10 kg ) ( 10.0 × 10
2

−34

E1

−27

−15

m)

2

⎛
1 eV
⎞

= 3.28 × 10−13 J ⎜
= 2.05 × 106 eV = 2.05 MeV
−19 ⎟
⎝ 1.602 × 10 J ⎠
(a)

The energy of the emitted photon is
E = ΔEn = E2 − E1 = ( 2 ) E1 − E1 = 3E1 = 6.14 MeV
2

(b)

The wavelength of the photon is
hc 1 240 eV ⋅ nm
=
E 6.14 × 106 eV
= 2.02 × 10−4 nm = 2.02 × 10−13 m = 202 × 10−15 m = 202 fm

λ=

(c)
P41.12

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

The ground state energy of a particle (mass m) in a 1-dimensional box
h2
of width L is E1 =
.

8mL2
(a)

For a proton (m = 1.67 × 10–27 kg) in a 0.200-nm wide box:

(6.626 × 10 J ⋅ s )
=
8 ( 1.67 × 10 kg ) ( 2.00 × 10
−34

E1

2

−27

−10

m)

2

= 8.22 × 10−22 J = 5.13 × 10−3 eV
(b)

For an electron (m = 9.11 × 10–31 kg) in the same size box:

(6.626 × 10 J ⋅ s )
=
8 ( 9.11 × 10 kg ) ( 2.00 × 10

−34

E1

−31

2

−10

m)

2

= 1.51 × 10−18 J = 9.41 eV
(c)

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

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958

Quantum Mechanics

P41.13

E1 = 2.00 eV = 3.20 × 10–19 J. For the ground state,

E1 =
(a)

h2
8me L2

The length of the region is
L=

h
6.626 × 10−34 J ⋅ s
=
8meE1
8 ( 9.11 × 10−31 kg ) ( 3.20 × 10−19 J )

= 4.34 × 10−10 m = 0.434 nm

(b)

⎛ h2 ⎞ 2
For the excited states, En = ⎜
n = n2E1 . For the first excited
2⎟
⎝ 8m L ⎠
e

state, ΔE = E2 − E1 = 4E1 − E1 = 3E1 = 6.00 eV
P41.14

(a)

The classical kinetic energy of the particle is
K=

2
1 2 1
mv = ( 4.00 × 10−3 kg ) ( 1.00 × 10−3 m/s )
2
2

= 2.00 × 10−9 J

(b)

The length L can be found from

⎛ h2 ⎞ 2
E=⎜
n
⎝ 8mL2 ⎟⎠
Solving,
6.626 × 10−34 J ⋅ s )
(
h2
L=n
=2
8mE
8 ( 4.00 × 10−3 kg ) ( 2.00 × 10−9 J )
2

= 1.66 × 10−28 m

(c)

No. The length of the box would have to be much smaller than
the size of a nucleus ( ~ 10 –14 m) to confine the particle.

*P41.15

(a)

h2
n2 . Its
The energies of the confined electron are En =
2
8me L
energy gain in the quantum jump from state 1 to state 4 is
h2
h2 15
hc
2
2
4
−
1
,
(
)
and this is the photon energy

= hf = .
2
2
8me L
8me L
λ

⎛ 15hλ ⎞
Then 8me cL = 15hλ and L = ⎜
⎝ 8me c ⎟⎠
2

12

.

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Chapter 41
(b)

959

Let λ ′ represent the wavelength of the photon emitted:

hc
h2
h2
12h2

2
2
=
4
−
2
=
λ ′ 8me L2
8me L2
8me L2
2
2
hc λ ′ h 15 ( 8me L ) 5
=
= and λ ′ = 1.25λ .
λ hc
8me L2 12h2
4

Then
P41.16

(a)

(b)




=

From ΔxΔp ≥ , with Δx = L, Δp ≥
, so the uncertainty
2
2Δx 2L

in momentum must be at least Δp ≈
.
2L

Its energy is all kinetic, so
p 2 (Δp)2
2
h2
E=
=
≈
=
2m
2m
8mL2 (4π )2 8mL2

(c)

Compare the result of part (b) to the result h2/8mL2 for the wave
function as a standing wave. This estimate is too low by 4π 2 ≈ 40
times, but it correctly displays the pattern of dependence of the
energy on the mass and on the length of the well.
∞

P41.17

(a)

2

∫ ψ dx = 1 becomes

−∞

L4
1 + cos ⎡⎣ 2 ( 2π x L ) ⎤⎦
⎛ 2π x ⎞
2
A ∫ cos ⎜
dx = A ∫
dx = 1
⎟
⎝
⎠
L
2
−L 4
−L 4
L4

2

A2
2

2

L
⎡
⎛ 4π x L ⎞ ⎤
x
+
cos
⎜⎝
⎟
⎢
4π
L ⎠ ⎥⎦
⎣

L4

=1
−L 4

A2 ⎛ L ⎞
2
⎜⎝ ⎟⎠ = 1 → A =
2 2
L

(b)

The probability of finding the particle between 0 and
L8

2

∫ ψ dx = A
0

L8
2

∫
0

L
is
8

A2 ⎡
L
⎛ 2π x ⎞
⎛ 4π x ⎞ ⎤
cos ⎜
dx
=
x+
cos ⎜
⎟
⎢
⎝ L ⎠
⎝ L ⎟⎠ ⎥⎦
2 ⎣

4π

L8

2

0

1 ⎛ 4⎞ ⎡L L
⎛π⎞⎤ 1 1
sin ⎜ ⎟ ⎥ = +
= 0.409
⎜⎝ ⎟⎠ ⎢ +
⎝ 2 ⎠ ⎦ 4 2π
2 L ⎣ 8 4π

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960
P41.18

Quantum Mechanics
Normalization requires

∫

2

ψ dx = 1 :

all space

L
1 − cos ⎡⎣ 2 (π x L ) ⎤⎦
⎛ nπ x ⎞
2
A
sin
dx
=
A
dx = 1
⎜⎝
⎟
∫0
∫0
L ⎠
2
L

2

2

=

A2
2

A2
=
2
A=

L

L
⎡
⎛ 2π x ⎞ ⎤
⎢ x − 2π sin ⎜⎝ L ⎟⎠ ⎥ = 1
⎣
⎦0
L

L
A2 L
⎡
⎤
⎢⎣ L − 2π sin 2π ⎥⎦ = 2 = 1
0

2
L
L

P41.19

(a)

The expectation value is x = ∫ ψ * xψ dx:
0

L
L
⎧⎪ 1 − cos ⎡⎣ 2 (π x L ) ⎤⎦ ⎫⎪
2
2
⎛ 2π x ⎞
x = ∫ x sin 2 ⎜
dx
=
x⎨
⎬ dx
⎟
∫
⎝
⎠
L
L
L
2
0
0
⎪⎭
⎩⎪
L

4π x ⎞
1 ⎛

= ∫ x ⎜ 1 − cos
⎟ dx
L0 ⎝
L ⎠
From integral tables, we find that

1 x2
x =
L 2
(b)

L

0

L

4π x
4π x ⎤
1 L2 ⎡ 4π x
L
−
sin
+ cos
=
2 ⎢
⎥
L 16π ⎣ L
L
L ⎦0 2

The probability of finding the particle in the range
0.490L ≤ x ≤ 0.510L is
0.510L
0.510L
1 − cos ⎡⎣ 2 ( 2π x L ) ⎤⎦
2
2
2 ⎛ 2π x ⎞
P=
sin ⎜
dx =
dx
⎟
∫
∫
⎝ L ⎠
L 0.490L
L 0.490L
2
0.510L

=

1⎡
L
⎛ 4π x ⎞ ⎤
x−
sin ⎜
⎢

⎝ L ⎟⎠ ⎥⎦ 0.490L
L⎣
2π

= 0.020 −
(c)

1
sin 2.04π − sin 1.96π ) = 5.26 × 10−5
(
4π

The probability of finding the particle in the range
0.240L ≤ x ≤ 0.260L is
0.260L

1⎡
L
⎛ 4π x ⎞ ⎤
P = ⎢x −
sin ⎜
= 3.99 × 10−2
⎟
⎥
⎝ L ⎠ ⎦ 0.240L
L⎣
2π

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 41
(d)

961

In the n = 2 graph in the text’s Figure 41.4(b), it is more probable
to find the particle either near x = L 4 or x = 3L 4 than at the
center, where the probability density is zero. Nevertheless, the
symmetry of the distribution means that the average position is
x = L 2.

P41.20

P41.21

(a)

The most probable positions of the particle are x = L/4, L/2, and
3L/4.

(b)

We look for sin (3πx/L) taking on its extreme values 1 and –1 so
that the squared wave function is as large as it can be. The result
can also be found by studying Figure 41.4b. The most probable
locations are at the antinodes of the standing wave pattern n = 3,
which has three antinodes that are equally spaced, one at the
center, and two a distance L/4 from either end.

(a)

The probability of finding the electron between x = 0 and
x = 0.100 nm = L/3 is
L/3

∫ψ

2
1

0

2
dx =
L

L/3

∫
0

2
⎛ π x⎞
sin ⎜
dx =
⎟
⎝ L ⎠
L

L/3

∫

2

0

1 − cos ⎡⎣ 2 (π x L ) ⎤⎦
dx
2

L/3

=

1⎡
L
⎛ 2π x ⎞ ⎤
x−
sin ⎜
⎢
⎝ L ⎟⎠ ⎥⎦ 0
L⎣
2π

=

1 1
⎛ 2π ⎞ 1 0.866

−
sin ⎜
= −
= 0.196
⎝ 3 ⎟⎠ 3
3 2π
2π

(b)

Classically, the particle moves back and forth steadily, spending
equal time intervals in each third of the line. The classical
probability is 0.333, which is significantly larger.

(c)

The probability is
L/3

∫
0

ψ 99

2

2
dx =
L

L/3

∫
0

1
⎛ 99π x ⎞
sin ⎜
dx =
⎟
⎝ L ⎠
L
2

L/3

⎡

⎛ 198π x ⎞ ⎤
⎟ dx
L ⎠ ⎥⎦

∫ ⎢⎣1 − cos ⎜⎝
0

L/3

1⎡
L
⎛ 198π x ⎞ ⎤

= ⎢x −
sin ⎜
⎝ L ⎟⎠ ⎥⎦ 0
L⎣
198π
=

1
1
1
−
sin ( 66π ) = − 0 = 0.333
3 198π
3

The probability is 0.333 for both classical and quantum models.

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


962

P41.22

Quantum Mechanics

(a)

2
⎛ π x⎞

. The probability of
sin ⎜
⎝ L ⎟⎠
L
finding the particle between x = 0 and x =  is
From Equation 41.13, ψ 1 ( x ) =



∫ ψ 1 dx =
2

0



2
2 1 − cos ⎡⎣ 2 (π x L ) ⎤⎦
2 ⎛ π x⎞
sin
dx
=
dx
⎜⎝
⎟
L ∫0
L ⎠
L ∫0
2


1⎡
L

1
⎛ 2π x ⎞ ⎤
⎛ 2π  ⎞
= ⎢x −
sin ⎜
= −
sin ⎜
⎟
⎥
⎝ L ⎠⎦
⎝ L ⎟⎠
L⎣
2π
L 2π
0
(b)

The probability function is sketched in ANS. FIG. P41.22(b).

ANS. FIG. P41.22(b)
(c)

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

found somewhere in the range 0 < x < L.

(d) The probability of finding the particle between x = 0 and x =  is
2
1
, and between x =  and x = L is .
3
3


Thus,

2

2

∫ ψ 1 dx = 3
0
∴


1
⎛ 2π  ⎞ 2
−
sin ⎜
=
⎝ L ⎟⎠ 3
L 2π

or, defining u =


,
L

u−

1
2
sin 2π u =
2π
3

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 41

963

This equation for u can be solved by homing in on the solution

with a calculator, the result being u = = 0.585 , or  = 0.585L
L
to three digits.
P41.23

(a)

The probability is

L/3

P=

∫

2

ψ 1 dx =

0

=

2
L

L/3

∫
0

2
L

L/3

∫ sin
0

2

⎛ π x⎞
⎜⎝
⎟ dx
L ⎠

1 − cos ⎡⎣ 2 (π x L ) ⎤⎦
dx
2

1⎡
L
⎛ 2π x ⎞ ⎤
= ⎢x −
sin ⎜
⎝ L ⎟⎠ ⎥⎦
L⎣
2π
=

L3

=
0

1 ⎡L L
⎛ 2π
−
sin ⎜

⎢
⎝ 3
L ⎣ 3 2π

⎞⎤
⎟⎠ ⎥
⎦

1 1
⎛ 2π ⎞
−
sin ⎜
⎝ 3 ⎟⎠
3 2π

⎛1
3⎞
=⎜ −
= 0.196
⎝ 3 4π ⎟⎠
(b)

L
. Thus, the
2
2L
probability of finding the particle between x =
and x = L is the
3
same, 0.196. Therefore, the probability of finding it in the range

L
2L
is P = 1.00 − 2 ( 0.196 ) = 0.609 .
≤x≤
3
3

The probability density is symmetric about x =

ANS. FIG. P41.23(b)

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


964

Quantum Mechanics

Section 41.3
P41.24

The Schrödinger Equation

From

i kx − ω t )

ψ = Ae (

[1]

we evaluate

dψ
i kx − ω t )
= ikAe (
dx
and

d 2ψ
i kx − ω t )
= − k 2 Ae (
2
dx

[2]

We substitute equations [1] and [2] into the Schrödinger equation, so
that Equation 41.15,

−

 2 d 2ψ
+ Uψ = Eψ
2m dx 2

becomes the test equation

)

(

⎛ 2 ⎞
i ( kx − ω t )
i kx − ω t )
2
+ 0 = EAe (
⎜⎝ − 2m ⎟⎠ −k Ae

[3]

)
The wave function ψ = Ae (
is a solution to the Schrödinger
equation if equation [3] is true. Both sides depend on A, x, and t in the
same way, so we can cancel several factors, and determine that we
have a solution if
i kx − ω t

2 k 2
=E
2m
But this is true for a nonrelativistic particle with mass in a region
where the potential energy is zero, since
2

2

2 k 2
1 ⎛ h ⎞ ⎛ 2π ⎞

=
⎟ =
⎜
⎟ ⎜
2m
2m ⎝ 2π ⎠ ⎝ λ ⎠

(h/λ )2
p2
=
2m
2m




using de Broglie's equation

=

m 2 u2 1
= 2 mu2 = K = K + U = E 

2m
recall U=0

where K is the kinetic energy. Therefore, the given wave function
does satisfy Equation 41.15.
P41.25

(a)

Given the function

ψ ( x ) = A cos kx + Bsin kx
Its derivative with respect to x is
∂ψ
= −kA sin kx + kBcos kx
∂x
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 41

965

And its second derivative is

∂ 2ψ
= −k 2 A cos kx − k 2 Bsin kx
2
∂x
= −k 2 ( A cos kx + Bsin kx ) = −k 2ψ
The Schrödinger equation is satisfied if

 2 d 2ψ
−
+ Uψ = Eψ ,
2m dx 2
−

2
( −k 2ψ ) = Eψ
2m

where U = 0:

2 k 2
ψ = Eψ
2m

→

This is true as an identity (functional equality) for all x if
2 k 2
, which is true because E = K + U = K + 0 = K, and
E=
2m
2

2

2

p2
1 ⎛ h⎞
2 k 2
1 ⎛ h ⎞ ⎛ 2π ⎞
=
=

=K
=
⎜ ⎟ ⎜ ⎟
⎜ ⎟
2m ⎝ λ ⎠
2m
2m 2m ⎝ 2π ⎠ ⎝ λ ⎠

P41.26

2 k 2
.
2m

(b)

From part (a), E =

(a)

These are standing wave patterns with nodes at the ends and n
antinodes.
For n = 1, the wave function is

ψ 1 ( x) =

2
⎛ π x⎞
cos ⎜
⎝ L ⎟⎠

L

and the probability density is

P1 ( x ) = ψ 1 ( x ) =
2

2
⎛ π x⎞
cos 2 ⎜
⎝ L ⎟⎠
L

For n = 2, the wave function is

ψ 2 ( x) =

2
⎛ 2π x ⎞
sin ⎜
⎝ L ⎟⎠
L

and the probability density is

P2 ( x ) = ψ 2 ( x ) =
2

2
⎛ 2π x ⎞

sin 2 ⎜
⎝ L ⎟⎠
L

For n = 3, the wave function is

ψ 3 ( x) =

2
⎛ 3π x ⎞
cos ⎜
⎝ L ⎟⎠
L

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


966

Quantum Mechanics
and the probability density is

P3 ( x ) = ψ 3 ( x ) =
2

(b)

2
⎛ 3π x ⎞
cos 2 ⎜

⎝ L ⎟⎠
L

The wave functions and probability densities are shown in ANS.
FIG. P41.26(b).

ANS. FIG. P41.26(b)
P41.27

(a)

Setting the total energy E equal to zero and rearranging the
Schrödinger equation to isolate the potential energy function
gives

⎛  2 ⎞ d 2ψ
⎜⎝ 2m ⎟⎠ dx 2 + U ( x )ψ = 0
⎛  2 ⎞ 1 d 2ψ
U ( x) = ⎜
⎝ 2m ⎟⎠ ψ dx 2
If

ψ ( x ) = Axe − x

2

L2

Then,
2

2

−x L
d 2ψ
3
2 e
=
4Ax
−
6AxL
(
) L4
dx 2

or

2
2
d 2ψ ( 4x − 6L )
=
ψ ( x)
dx 2
L4

and U ( x ) =

⎞
 2 ⎛ 4x 2
− 6⎟

2 ⎜
2
2mL ⎝ L
⎠

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 41
(b)

967

U(x) is sketched in ANS. FIG. P41.27(b).

ANS. FIG. P41.27(b)
P41.28

(a)

⎛
x2 ⎞
ψ ( x) = A ⎜ 1 − 2 ⎟ →
L ⎠
⎝

dψ
2Ax
=− 2
→

dx
L

d 2ψ
2A
=− 2
2
dx
L

Schrödinger’s equation:

−

 2 d 2ψ
+ Uψ = Eψ
2m dx 2

becomes
− 2 x 2 )
⎛
⎛
(
 2 ⎛ 2A ⎞
x2 ⎞
x2 ⎞
−
−
+
A

1
−
=
EA
1
−
⎜
⎟
⎜⎝
2m ⎝ L2 ⎠ mL2 ( L2 − x 2 ) ⎜⎝
L2 ⎟⎠
L2 ⎟⎠
2 2
2
2
⎛
 2 ⎛ 2 ⎞ ( − x ) ( L − x )
x2 ⎞
−
−
+
=
E
1
−
⎜⎝
2 m ⎜⎝ L2 ⎟⎠
L2 ⎟⎠
mL4 ( L2 − x 2 )

− 2 x 2 )
⎛
(
2
x2 ⎞
+
=
E
1
−
⎜⎝
mL2
mL4
L2 ⎟⎠
⎛
2 ⎛
x2 ⎞
x2 ⎞
1
−
=
E
1
−
⎜⎝
mL2 ⎜⎝
L4 ⎟⎠
L2 ⎟⎠

2

E= 2 .
Lm

This will be true for all x if
(b)

Note that the wave function ψ ( x ) is an even function; therefore,
we may write the normalization condition as
2

2

L

2

L
⎛
x2 ⎞
x2 ⎞
2⎛
∫ ψ dx = 1 = ∫ A ⎜⎝ 1 − L2 ⎟⎠ dx = 2 ∫ A ⎜⎝ 1 − L2 ⎟⎠ dx
−L
−L
0
L

2

L

⎛
2x 2 x 4 ⎞
= 2A 2 ∫ ⎜ 1 − 2 + 4 ⎟ dx
L
L ⎠
⎝
0

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


968

Quantum Mechanics
Solving,
L

⎡
2x 3 x 5 ⎤
2
L⎤
⎡
1 = 2A ⎢ x − 2 + 4 ⎥ = 2A 2 ⎢ L − L + ⎥
3L 5L ⎦ 0
3
5⎦
⎣
⎣
2

⎛ 16L ⎞
= A2 ⎜
⎝ 15 ⎟⎠
(c)

→

A=

15
16L

As in part (b), because the wave function is an even function, the
probability is
L3

P=

∫

−L 3

ψ 2 dx =

L3

2
∫ ψ dx = 2
0

15
16L

L3

∫
0

⎛
2x 2 x 4 ⎞
1
−
+ 4 ⎟ dx
⎜⎝
L2
L ⎠

L3

Section 41.4
P41.29

=

15 ⎡
2x 3 x 5 ⎤
x
−
+
8L ⎢⎣

3L2 5L5 ⎥⎦ 0

=

15 ⎡ L 2L
L ⎤ 47
−
+
=
= 0.580
8L ⎢⎣ 3 81 1215 ⎥⎦ 81

A Particle in a Well of Finite Height

(a)

For n = 4, the wave function has two maxima and two minima
(four extrema), as shown in the left-hand panel of ANS. FIG.
P41.29.

(b)

For n = 4, the probability function has four maxima. as shown in
the right-hand panel of ANS. FIG. P41.29.

ANS. FIG. P41.29
P41.30

(a)

See ANS. FIG. P41.30(a).

ANS. FIG. P41.30(a)

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 41
(b)

Section 41.5
P41.31

969

The wavelength inside the box is 2L. The wave function
penetrates the wall, but the wavelength of the transmitted wave
traveling to the left is the same, 2L , because U = 0 on both sides
of the wall, so the energy and momentum and, therefore, the
wavelength, are the same.

Tunneling Through a Potential Energy Barrier

The decay constant for the wave function inside the barrier is:

C=
=

2m (U − E )


2 ( 9.11 × 10 –31  kg ) ( 10.0 eV − 5.00 eV ) ( 1.60 × 10 –19  J/eV )
6.626 × 10 –34  J ⋅ s/2π

= 1.14 × 1010  m –1
(a)

The approximate probability of transmission is
−2 1.14 × 1010 m –1 )( 2.00 × 10 –10 m )
T ≈ e −2CL = e (
= 0.010 3

or a 1% chance of transmission.

P41.32

(b)

R = 1 − T = 0.990 , a 99% chance of reflection.

(a)

T = e −2CL , where
C=
=

2m(U − E )


2 ( 9.11 × 10−31 kg )( 5.00 − 4.50 )( 1.60 × 10−19 J )

1.055 × 10−34 J ⋅ s

= 3.62 × 109 m −1
and T = e −2CL = exp ⎡⎣ −2 ( 3.62 × 109 m −1 ) ( 950 × 10−12 m ) ⎤⎦

= exp ( −6.88 ) = 1.03 × 10−3
(b)

We require e −2CL = 10−6. Taking logarithms,

−2CL = ln 10−6 = −6 ln 10
L=

3 ln 10
3 ln 10
=
= 1.91 × 10−9 m = 1.91 nm
9
−1
C
3.62 × 10 m

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


970
P41.33

Quantum Mechanics
The original tunneling probability is T = e −2CL , where

C=
=

2m(U − E )


2 ( 9.11 × 10−31 kg )( 20.0 − 12.0 )( 1.60 × 10−19 J )
6.626 × 10−34 J ⋅ s 2π

= 1.448 1 × 1010 m −1
hc 1 240 eV ⋅ nm
=
= 2.27 eV, to make the
λ
546 nm
electron’s new kinetic energy 12.0 + 2.27 = 14.27 eV and its decay
coefficient inside the barrier

The photon energy is hf =

C′ =
=

2m(U − E )


2 ( 9.11 × 10−31 kg )( 20.0 − 14.27 )( 1.60 × 10−19 J )
6.626 × 10−34 J ⋅ s 2π

= 1.225 5 × 1010 m −1
Now the factor of increase in transmission probability is

e −2 C ′L
2 ( 1.00×10−9 m )( 0.223×1010 m −1 )
2L ( C − C ′ )
=
e
=
e
= e 4.45 = 85.9
e −2CL

Section 41.6
P41.34

Applications of Tunneling

With the wave function proportional to e–CL, the transmission
2
coefficient and the tunneling current are proportional to ψ , to e–2CL.
Then,
I ( 0.500 nm ) e −2 (10.0 nm)( 0.500 nm)
= −2 (10.0 nm)( 0.515 nm) = e 20.0( 0.015) = 1.35
I ( 0.515 nm ) e

P41.35

With transmission coefficient e–2CL, the fractional change in
transmission is

−2 10.0 nm )L
−2 10.0 nm )( L+0.002 00 nm )
e (
−e (
−20.0 0.002 00 )
= 1− e (
−2( 10.0 nm )L
e
= 0.039 2 = 3.92%

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 41

Section 41.7
P41.36

(a)

971

The Simple Harmonic Oscillator
2

The wave function is given by ψ = Axe −bx , so
2
2
dψ
= Ae −bx − 2bx 2 Ae −bx

dx

and
2
2
2
d 2ψ
= ⎡⎣ −2bxAe −bx − 4bxAe −bx ⎤⎦ + 4b 2 x 3 e −bx = −6bψ + 4b 2 x 2ψ
2
dx

Substitute into Equation 41.24:

−

 2 d 2ψ 1
+ mω 2 x 2ψ = Eψ
2m dx 2 2

−

2
1
⎡⎣ −6bψ + 4b 2 x 2ψ ⎤⎦ + mω 2 x 2ψ = Eψ
2m
2

3b 2
2b 2  2 2
1

ψ−
x ψ = − mω 2 x 2ψ + Eψ
m
m
2
For this to be true as an identity, the coefficients of like terms
must be the same for all values of x. So we must have both

2b 2  2 1
= mω 2
m
2
mω
b=
2

→ b2 =
and

m2ω 2
4 2

and

3b 2
=E
m

3b 2
3

E=
= ω
m
2

(b)

Therefore,

(c)

1⎞
3
⎛
The energy levels are En = ⎜ n + ⎟ ω = ω , so n = 1, which
⎝
2⎠
2
corresponds to 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. From E = ω , we have

hc
k
h
=

=
λ
m 2π

k
m

or
⎛ 9.11 × 10−31 kg ⎞
m
8
λ = 2π c
= 2π ( 3.00 × 10 m/s ) ⎜
⎝ 8.99 N/m ⎟⎠
k

12

= 600 nm

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


972
P41.38

Quantum Mechanics
The longest wavelength corresponds to minimum photon energy,
which must be equal to the spacing between energy levels of the
oscillator, which is (from Equation 41.28)

E = ω

hc
k
h
=
=
λ
m 2π

λ = 2π c
P41.39

(a)

k
m

m
k

2
− mω 2 )x 2
, the normalization condition ∫ ψ dx = 1
With ψ = Be (
all x

becomes
∞

∞

−∞

0

−2 mω 2 )x 2
− mω  x 2
1 = ∫ B2 e (
dx = 2B2 ∫ e ( ) dx

= 2B2

1
π
π
= B2
2 mω 
mω

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

⎛ mω ⎞
Thus, B = ⎜
⎝ π  ⎟⎠
(b)

14

.

For small δ, the probability of finding the particle in the range
δ
δ
− < x < is
2
2
δ 2

∫ ψ dx ≈ δ ψ ( 0 )
2

−δ 2

P41.40

(a)

2

⎛ mω ⎞
= δB e = δ ⎜
⎝ π  ⎟⎠

12

2 −0

For the center of mass to be fixed, m1u1 + m2 u2 = 0 . Then

u = u1 + u2 = u1 +

m1
m + m1
u1 = 2
u1
m2
m2

and

u1 =

m2 u
m1 + m2

Also,
u=

⎛ m + m1 ⎞
m2
u2 + u2 = ⎜ 2
u2
m1
⎝ m1 ⎟⎠

→

u2 =

m1u
m1 + m2

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.