Instructor solution manual to accompany physical chemistry 7th ed by peter atkins chap12

12

Quantum theory: techniques and
applications

Solutions to exercises
Discussion questions
E12.1(b)

The correspondence principle states that in the limit of very large quantum numbers quantum
mechanics merges with classical mechanics. An example is a molecule of a gas in a box. At room
temperature, the particle-in-a-box quantum numbers corresponding to the average energy of the gas
molecules ( 21 kT per degree of freedom) are extremely large; consequently the separation between the
levels is relatively so small (n is always small compared to n2 , compare eqn 12.10 to eqn 12.4) that
the energy of the particle is effectively continuous, just as in classical mechanics. We may also look at
these equations from the point of view of the mass of the particle. As the mass of the particle increases
to macroscopic values, the separation between the energy levels approaches zero. The quantization
disappears as we know it must. Tennis balls do not show quantum mechanical effects. (Except those
served by Pete Sampras.) We can also see the correspondence principle operating when we examine
the wavefunctions for large values of the quantum numbers. The probability density becomes uniform
over the path of motion, which is again the classical result. This aspect is discussed in more detail in
Section 12.1(c).
The harmonic oscillator provides another example of the correspondence principle. The same
effects mentioned above are observed. We see from Fig. 12.22 of the text that probability distribution
for large values on n approaches the classical picture of the motion. (Look at the graph for n = 20.)

E12.2(b)

The physical origin of tunnelling is related to the probability density of the particle which according to
the Born interpretation is the square of the wavefunction that represents the particle. This interpretation
requires that the wavefunction of the system be everywhere continuous, event at barriers. Therefore,

if the wavefunction is non-zero on one side of a barrier it must be non-zero on the other side of the
barrier and this implies that the particle has tunnelled through the barrier. The transmission probability
depends upon the mass of the particle (specifically m1/2 , through eqns 12.24 and 12.28): the greater
the mass the smaller the probability of tunnelling. Electrons and protons have small masses, molecular
groups large masses; therefore, tunnelling effects are more observable in process involving electrons
and protons.

E12.3(b)

The essential features of the derivation are:
(1) The separation of the hamiltonian into large (unperturbed) and small (perturbed) parts which are
independent of each other.
(2) The expansion of the wavefunctions and energies as a power series in an unspecified parameters,
λ, which in the end effectively cancels or is set equal to 1.
(3) The calculation of the first-order correction to the energies by an integration of the perturbation
over the zero-order wavefunctions.
(4) The expansion of the first-order correction to the wavefunction in terms of the complete set of
functions which are a solution of the unperturbed Schrodinger equation.
(5) The calculation of the second-order correction to the energies with use of the corrected first order
wavefunctions.
See Justification 12.7 and Further reading for a more complete discussion of the method.


INSTRUCTOR’S MANUAL

186

Numerical exercises
E=

E12.4(b)

n2 h 2
8me L2

h2
(6.626 × 10−34 J s)2
=
= 2.678 × 10−20 J
8me L2
8(9.109 × 10−31 kg) × (1.50 × 10−9 m)2
Conversion factors
E
kJ mol−1

=

NA
E/J
103

1 eV = 1.602 × 10−19 J
1 cm−1 = 1.986 × 10−23 J
E3 − E1 = (9 − 1)

(a)

h2
= 8(2.678 × 10−20 J)
8me L2

= 2.14 × 10−19 J = 129 kJ mol−1 = 1.34 eV = 1.08 × 104 cm−1
h2
8me L2
= 13(2.678 × 10−20 J)

E7 − E6 = (49 − 36)

(b)

= 3.48 × 10−19 J = 210 kJ mol−1 = 2.17 eV = 1.75 × 104 cm−1
E12.5(b)

The probability is
ψ ∗ ψ dx =

P =
where

2
L

sin2

2 x
nπ x
nπ x
dx =
sin2
L

L
L

x = 0.02L and the function is evaluated at x = 0.66L.

(a) For n = 1
P =

2(0.02L) 2
sin (0.66π) = 0.031
L

(b) For n = 2
P =
E12.6(b)

2(0.02L) 2
sin [2(0.66π )] = 0.029
L

The expectation value is
pˆ =

ˆ dx
ψ ∗ pψ

but first we need pψ
ˆ
pψ
ˆ = −i

d 2 1/2
nπ x
sin
dx L
L

= −i

2 1/2 nπ
nπ x
cos
L
L
L


QUANTUM THEORY: TECHNIQUES AND APPLICATIONS

so pˆ =

187

nπ x
−2i nπ L
nπ x
dx = 0 for all n
cos
sin
2

L
L
L
0

h2 n2
pˆ 2 = 2m Hˆ = 2mEn =
4L2
for all n. So for n = 2
pˆ 2 =
E12.7(b)

h2
L2

n=5

2 1/2
5πx
sin
L
L
5πx
P (x) ∝ ψ52 ∝ sin2
L

ψ5 =

Maxima and minima in P (x) correspond to
d

dψ 2
5π x
P (x) ∝
∝ sin
dx
dx
L

cos

dP (x)
=0
dx

5π x
L

∝ sin

10π x
L

(2 sin α cos α = sin 2α)

sin θ = 0 when θ = 0, π, 2π, . . . = n π (n = 0, 1, 2, . . .)
10πx
= n π n ≤ 10
L
nL
x=

10
Minima at x = 0, x = L
Maxima and minima alternate: maxima correspond to
n = 1, 3, 5, 7, 9
E12.8(b)

x=

L
3L L 7L 9L
,
,
,
,
10
2
10
10
10

The energy levels are
En1 ,n2 ,n3 =

(n21 + n22 + n23 )h2
8mL2

= E1 (n21 + n22 + n23 )

where E1 combines all constants besides quantum numbers. The minimum value for all the quantum
numbers is 1, so the lowest energy is

E1,1,1 = 3E1
The question asks about an energy 14/3 times this amount, namely 14E1 . This energy level can be
obtained by any combination of allowed quantum numbers such that
n21 + n22 + n23 = 14 = 32 + 22 + 12
The degeneracy, then, is 6 , corresponding to (n1 , n2 , n3 ) = (1, 2, 3), (2, 1, 3), (1, 3, 2), (2, 3, 1),
(3, 1, 2), or (3, 2, 1).


INSTRUCTOR’S MANUAL

188

E12.9(b)

E = 23 kT is the average translational energy of a gaseous molecule (see Chapter 20).
E = 23 kT =

(n21 + n22 + n23 )h2
8mL2

=

n2 h2
8mL2

E = 23 × (1.381 × 10−23 J K−1 ) × (300 K) = 6.214 × 10−21 J
n2 =

8mL2
E

h2

If L3 = 1.00 m3 , L2 = 1.00 m2
(6.626 × 10−34 J s)2
h2
=
= 1.180 × 10−42 J
2
0.02802 kg mol−1
8mL
2
(8) × 6.022×1023 mol−1 × (1.00 m )
6.214 × 10−21 J
n = 7.26 × 1010
= 5.265 × 1021 ;
1.180 × 10−42 J
E = En+1 − En = E7.26×1010 +1 − E7.26×1010

n2 =

E = (2n + 1) ×

h2
8mL2

= [(2) × (7.26 × 1010 + 1)] ×

h2
8mL2

=

14.52 × 1010 h2
8mL2

= (14.52 × 1010 ) × (1.180 × 10−42 J) = 1.71 × 10−31 J
The de Broglie wavelength is obtained from
λ=

h
h
=
[Section 11.2]
p
mv

The velocity is obtained from
EK = 21 mv 2 = 23 kT = 6.214 × 10−21 J
6.214 × 10−21 J

v2 =

λ=

1
2

−1

0.02802 kg mol

× 6.022×10
23 mol−1

= 2.671 × 105 ;

v = 517 m s−1

6.626 × 10−34 J s
= 2.75 × 10−11 m = 27.5 pm
(4.65 × 10−26 kg) × (517 m s−1 )

The conclusion to be drawn from all of these calculations is that the translational motion of the
nitrogen molecule can be described classically. The energy of the molecule is essentially continuous,
E
≪ 1.
E
E12.10(b) The zero-point energy is
E0 =

1
2

ω=

1
2

k 1/2
= 21 (1.0546 × 10−34 J s) ×
m

= 3.92 × 10−21 J

285 N m−1
5.16 × 10−26 kg

1/2


QUANTUM THEORY: TECHNIQUES AND APPLICATIONS

189

E12.11(b) The difference in adjacent energy levels is
k 1/2
m

E= ω=

so

k=

m( E)2
2

=

(2.88 × 10−25 kg) × (3.17 × 10−21 J)2
(1.0546 × 10−34 J s)2

k = 260 N m−1
E12.12(b) The difference in adjacent energy levels, which is equal to the energy of the photon, is
E = ω = hν
and λ =

hc

k 1/2
hc
=
λ
m

so

m 1/2
k 1/2
= 2πc
m
k
8

−1

= 2π(2.998 × 10 m s

)×

(15.9949 u) × (1.66 × 10−27 kg u−1 )
544 N m−1

1/2

λ = 1.32 × 10−5 m = 13.2 µm
E12.13(b) The difference in adjacent energy levels, which is equal to the energy of the photon, is
E = ω = hν
and λ =

hc

so

hc
k 1/2
=
m
λ

m 1/2
k 1/2
= 2πc
m
k

Doubling the mass, then, increases the wavelength by 21/2 . So taking the result from Ex. 12.12(b),
the new wavelength is
λ = 21/2 (13.2 µm) = 18.7 µm
E12.14(b)

ω=

g 1/2
[elementary physics]
l

E = ω = hν
(a)
(b)

E = hν = (6.626 × 10−34 J Hz−1 ) × (33 × 103 Hz) = 2.2 × 10−29 J
1
k 1/2
1
1
=
+
with m1 = m2
meff
meff
m1
m2
For a two-particle oscillator meff , replaces m in the expression for ω. (See Chapter 16 for a more
complete discussion of the vibration of a diatomic molecule.)
E= ω=

E=

2k 1/2
= (1.055 × 10−34 J s) ×
m

= 3.14 × 10−20 J

(2) × (1177 N m−1 )
(16.00) × (1.6605 × 10−27 kg)

1/2


INSTRUCTOR’S MANUAL

190

E12.15(b) The first excited-state wavefunction has the form
ψ = 2N1 y exp − 21 y 2
mω 1/2
. To see if it satisfies Schr¨odinger’s equation,
where N1 is a collection of constants and y ≡ x
we see what happens when we apply the energy operator to this function
Hˆ ψ = −

d2 ψ
+ 21 mω2 x 2 ψ
2m dx 2
2

We need derivatives of ψ
dψ dy
mω 1/2
dψ
(2N1 ) × (1 − y 2 ) × exp − 21 y 2

=
=
dx
dy dx
and

dy 2
mω
mω
× (2N1 ) × (−3y + y 3 ) × exp − 21 y 2 =
× (y 2 − 3)ψ
=
dx

d2 ψ
d2 ψ
=
dx 2
dy 2

So Hˆ ψ = −

2

2m

×

mω

× (y 2 − 3)ψ + 21 mω2 x 2 ψ

= − 21 ω × (y 2 − 3) × ψ + 21 ωy 2 ψ = 23 ωψ
Thus, ψ is an eigenfunction of Hˆ (i.e. it obeys the Schr¨odinger equation) with eigenvalue
E = 23 hω
E12.16(b) The zero-point energy is
k 1/2
meff

E0 = 21 ω = 21

For a homonuclear diatomic molecule, the effective mass is half the mass of an atom, so
E0 = 21 (1.0546 × 10−34 J s) ×

2293.8 N m−1
1
−27 kg u−1 )
2 (14.0031 u) × (1.66054 × 10

E0 = 2.3421 × 10−20 J
E12.17(b) Orthogonality requires that
∗
ψn dτ = 0
ψm

if m = n.
Performing the integration
∗
ψn dτ =
ψm

2π
0

N e−imφ Neinφ dφ = N 2

2π
0

ei(n−m)φ dφ

1/2


QUANTUM THEORY: TECHNIQUES AND APPLICATIONS

191

If m = n, then
∗
ψm
ψn dτ =

N2
ei(n−m)φ
i(n − m)

2π

=

0

N2
(1 − 1) = 0
i(n − m)

Therefore, they are orthogonal.
E12.18(b) The magnitude of angular momentum is
Lˆ 2 1/2 = (l(l + 1))1/2 = (2(3))1/2 (1.0546 × 10−34 J s) = 2.58 × 10−34 J s
Possible projections on to an arbitrary axis are
Lˆ z = ml
where ml = 0 or ±1 or ±2. So possible projections include
0, ±1.0546 × 10−34 J s and 2.1109 × 10−34 J s
E12.19(b) The cones are constructed as described in Section 12.7(c) and Fig. 12.36 of the text; their edges are
of length {6(6 + 1)}1/2 = 6.48 and their projections are mj = +6, +5, . . . , −6. See Fig. 12.1(a).
The vectors follow, in units of . From the highest-pointing to the lowest-pointing vectors
(Fig. 12.1(b)), the values of ml are 6, 5, 4, 3, 2, 1, 0, −1, −2, −3, −4, −5, and −6.

Figure 12.1(a)

Figure 12.1(b)


INSTRUCTOR’S MANUAL

192

Solutions to problems
Solutions to numerical problems
P12.4

E=

l(l + 1) 2
l(l + 1) 2
[12.65] =
2I
2meff R 2

=

[I = meff R 2 , meff in place of m]

l(l + 1) × (1.055 × 10−34 J s)2
(2) × (1.6605 × 10−27 kg) × (160 × 10−12 m)2

×

1
1
+
1.008 126.90

1
1
1
=
+
m2
meff

m1
The energies may be expressed in terms of equivalent frequencies with ν =

E
= 1.509 × 1033 E.
h

Therefore,
E = l(l + 1) × (1.31 × 10−22 J) = l(l + 1) × (198 GHz)
Hence, the energies and equivalent frequencies are
l

P12.6

0

1

2

3

22

10 E/J

0

2.62

7.86

15.72

ν/GHz

0

396

1188

2376

Treat the gravitational potential energy as a perturbation in the energy operator:
H (1) = mgx.
The first-order correction to the ground-state energy, E1 , is:
L

E1

(1)

L
(0)∗ (1)
1 H

=
0

E1 (1) =

E1

(1)

2mg
L

(0)
1 dx

=
0

L

x sin2
0

2mg
=
L

2 1/2
πx
mgx
sin
L
L

2 1/2
πx
dx,
sin
L
L

πx
dx,
L

πx
x2
xL
πx
πx
L2
cos2
−
cos
sin
−
2
4
2π
L
L
L
4π

L

,
0

E1 (1) = 21 mgL
Not surprisingly, this amounts to the energy perturbation evaluated at the midpoint of the box. For
m = me , E1 (1) /L = 4.47 × 10−30 J m−1 .


QUANTUM THEORY: TECHNIQUES AND APPLICATIONS

193

Solutions to theoretical problems
P12.8

2

−

2m

×

∂2
∂2
∂2
+

+
∂x 2
∂y 2
∂z2

ψ = Eψ

[V = 0]

We try the solution ψ = X(x)Y (y)Z(z)
−
−

2

2m

(X Y Z + XY Z + XY Z ) = EXY Z
X
Y
Z
+
+
Z
X
Y

2

2m

=E

X
depends only on x; therefore, when x changes only this term changes, but the sum of the three
X
X
terms is constant. Therefore,
must also be constant. We write
X
−

X
= EX ,
2m X
2

with analogous terms for y, z

Hence we solve


X = E X X


2m



2

X
Y
Z
Y
−
Y = E Y E = E + E + E ,
2m



2

Z 
−
Z = E Z
2m
−

2

ψ = XY Z

The three-dimensional equation has therefore separated into three one-dimensional equations, and
we can write
E=

ψ=

h2
8m

n23
n22
+
+
L21
L22
L23
n21

1/2
8
n1 π x
sin
L1 L2 L3
L1

n1 , n2 , n3 = 1, 2, 3, . . .

sin

n2 πy
L2

sin

n3 π z
L3

For a cubic box

E = (n21 + n22 + n23 )
P12.10

h2
8mL2

The wavefunctions in each region (see Fig. 12.2(a)) are (eqns 12.22–12.25):
ψ1 (x) = eik1 x + B1 e−ik2 x
ψ2 (x) = A2 ek2 x + B2 e−k2 x
ψ3 (x) = A3 eik3 x
with the above choice of A1 = 1 the transmission probability is simply T = |A3 |2 . The wavefunction
coefficients are determined by the criteria that both the wavefunctions and their first derivatives w/r/t


INSTRUCTOR’S MANUAL

194

x be continuous at potential boundaries
ψ1 (0) = ψ2 (0); ψ2 (L) = ψ3 (L)
dψ1 (0)
dψ2 (0)
dψ2 (L)
dψ3 (L)
=
;
=
dx
dx
dx

dx
These criteria establish the algebraic relationships:
1 + B 1 − A2 − B2 = 0
(−ik1 − k2 )A2 + (−ik1 + k2 )B2 + 2ik1 = 0
A2 ek2 L + B2 e−k2 L − A3 eik3 L = 0
A2 k2 ek2 L − B2 k2 e−k2 L − iA3 k3 eik3 L = 0

V2

V

V3

V1

0

x
0

L

Figure 12.2(a)

Solving the simultaneous equations for A3 gives
A3 =

4k1 k2 eik3 L
(ia + b) ek2 L − (ia − b) e−k2 L

where a = k22 − k1 k3 and b = k1 k2 + k2 k3 .
since sinh(z) = (ez − e−z )/2 or ez = 2 sinh(z) + e−z , substitute ek2 L = 2 sinh(k2 L) + e−k2 L giving:
A3 =

2k1 k2 eik3 L
(ia + b) sinh(k2 L) + b e−k2 L

T = |A3 |2 = A3 A¯ 3 =

4k12 k22
(a 2 + b2 ) sinh2 (k2 L) + b2

where a 2 + b2 = (k12 + k22 )(k22 + k32 ) and b2 = k22 (k1 + k3 )2
(b) In the special case for which V1 = V3 = 0, eqns 12.22 and 12.25 require that k1 = k3 .
Additionally,
k1 2
ε
E
=
where ε = E/V2 .
=
k2
V2 − E
1−ε
2

2

a +b =

(k12

+ k22 )2

=

k24

1+

2
k1 2
k2


QUANTUM THEORY: TECHNIQUES AND APPLICATIONS

195

b2 = 4k12 k22
a2

+ b2
b2

T =

T =

=

2 2
k22 1 + kk21

4k12

=

b2
b2

+ (a 2

1+

+ b2 ) sinh2 (k

sinh2 (k2 L)
4ε(1 − ε)

2 L)

1
4ε(1 − ε)
=

−1

= 1+

1
1+

a 2 +b2
b2

sinh2 (k2 L)

(ek2 L − e−k2 L )2
16ε(1 − ε)

−1

This proves eqn 12.28a where V1 = V3 = 0
In the high wide barrier limit k2 L
1. This implies both that e−k2 L is negligibly small
k2 L
and that 1 is negligibly small compared to e2 k2 L /{16ε(1 − ε)}. The previous
compared to e
equation simplifies to
T = 16 ε(1 − ε)e−2 k2 L

(c)

[eqn12.28b]

E = 10 kJ/mol, V1 = V3 = 0, L = 50 pm

0.25

0.2

0.15
T
0.1

0.05

0
1

1.2

1.4

1.6
1/ (i.e., V2/E)

P12.12

d2 ψ
+ 21 kx 2 ψ = Eψ
2m dx 2
2
2
dψ
and we write ψ = e−gx , so
= −2gxe−gx
dx
The Schr¨odinger equation is −

2

2
2
d2 ψ
= −2ge−gx + 4g 2 x 2 e−gx = −2gψ + 4g 2 x 2 ψ
2
dx

1.8

2

Figure 12.2(b)


INSTRUCTOR’S MANUAL

196

2g

2 2 g2
m

ψ−

m
2g

x 2 ψ + 21 kx 2 ψ = Eψ
1
2k

−E ψ +

m

−

2 2 g2
m

x2ψ = 0

This equation is satisfied if
2g

E=

and

m

2 2 g 2 = 21 mk,

or

mk 1/2

g = 21

2

Therefore,
k 1/2
= 21 ω
m

E = 21
P12.14

x n = αn y n = αn
x3 ∝
x

4

+∞
−∞

= α5

+∞
−∞

if ω =

k 1/2

m

ψy n ψ dx = α n+1

+∞
−∞

ψ 2 y n dy

[x = αy]

ψ 2 y 3 dy = 0 by symmetry [y 3 is an odd function of y]

+∞

−∞

ψy 4 ψ dy

y 4 ψ = y 4 N Hv e−y /2
y 4 Hv = y 3 21 Hv+1 + vHv−1 = y 2 21 21 Hv+2 + (v + 1)Hv + v 21 Hv + (v − 1)Hv−2
2

= y 2 41 Hv+2 + v + 21 Hv + v(v − 1)Hv−2
= y 41 21 Hv+3 + (v + 2)Hv+1 + v + 21 × 21 Hv+1 + vHv−1
+v(v − 1) × 21 Hv−1 + (v − 2)Hv−3
= y 18 Hv+3 + 43 (v + 1)Hv+1 + 23 v 2 Hv−1 + v(v − 1) × (v − 2)Hv−3
Only yHv+1 and yHv−1 lead to Hv and contribute to the expectation value (since Hv is orthogonal
to all except Hv ) [Table 12.1]; hence
y 4 Hv = 43 y{(v + 1)Hv+1 + 2v 2 Hv−1 } + · · ·

= 43 (v + 1) 21 Hv+2 + (v + 1)Hv + 2v 2 21 Hv + (v − 1)Hv−2

+ ···

= 43 {(v + 1)2 Hv + v 2 Hv } + · · ·
= 43 (2v 2 + 2v + 1)Hv + · · ·
Therefore
+∞
−∞

ψy 4 ψ dy = 43 (2v 2 + 2v + 1)N 2

+∞
−∞

Hv2 e−y dy =
2

3
(2v 2 + 2v + 1)
4α

and so
x 4 = (α 5 ) ×

3
4α

× (2v 2 + 2v + 1) = 43 (2v 2 + 2v + 1)α 4

QUANTUM THEORY: TECHNIQUES AND APPLICATIONS

P12.17

197

e2
1
· [13.5 with Z = 1] = αx b
4πε0 r
Since 2 T = b V [12.45, T ≡ EK ]
V =−

with b = −1

[x → r]

2 T =− V
Therefore, T = − 21 V
P12.18

In each case, if the function is an eigenfunction of the operator, the eigenvalue is also the expectation
value; if it is not an eigenfunction we form
ψ ∗ ˆ ψ dτ [11.39]

=
(a)
(b)
(c)

(d)

d iφ
e = eiφ ; hence Jz = +
i dφ
d −2iφ
e
lˆz e−2iφ =
= −2 e−2iφ ; hence Jz = −2
i dφ
lˆz eiφ =

2π

2π
d
cos φ dφ ∝ −
cos φ sin φ dφ = 0
i dφ
i 0
0
2π
d
× (cos χ eiφ + sin χ e−iφ ) dφ
lz = N 2
(cos χ eiφ + sin χ e−iφ )∗
i
dφ
0

lz ∝

=

i

cos φ

N2

= N2

2π
0
2π
0

(cos χe−iφ + sin χ eiφ ) × (i cos χ eiφ − i sin χ e−iφ ) dφ

(cos2 χ − sin2 χ + cos χ sin χ [e2iφ − e−2iφ ]) dφ

= N 2 (cos2 χ − sin2 χ ) × (2π ) = 2π N 2 cos 2χ
N2

2π
0

(cos χ eiφ + sin χ e−iφ )∗ (cos χ eiφ + sin χ e−iφ ) dφ

= N2

2π
0

(cos2 χ + sin2 χ + cos χ sin χ [e2iφ + e−2iφ ]) dφ

= 2πN 2 (cos2 χ + sin2 χ ) = 2π N 2 = 1

if N 2 =

1
2π

Therefore
lz =

cos 2χ [χ is a parameter]

2 d2
Jˆ2
For the kinetic energy we use Tˆ ≡ Eˆ K = z [12.47] = −
[12.52]
2I dφ 2
2I
2

(a)

Tˆ eiφ = −

(b)

Tˆ e−2iφ = −

(c)

Tˆ cos φ = −

2I

(i2 eiφ ) =
2

2I
2

2I

2

2I

eiφ ;

(2i)2 e−2iφ =
(−cos φ) =

hence

4 2 −2iφ

e
;
2I
2

2I

cos φ;

T =
hence

hence

2

2I
2 2
I

T =
T =

2

2I


INSTRUCTOR’S MANUAL

198

Tˆ (cos χ eiφ + sin χ e−iφ ) = −

(d)

2

2I

(−cos χ eiφ − sin χ e−iφ ) =

2

2I

(cos χ eiφ + sin χ e−iφ )

2

and hence T =

2I

Comment. All of these functions are eigenfunctions of the kinetic energy operator, which is also the
total energy or Hamiltonian operator, since the potential energy is zero for this system.
π

2π

P12.20
0

0

∗
Y3,3
Y3,3 sin θ dθ dφ =

π
0

1
64

×
35
π

1
= 64
×

2π
35
dφ [Table 12.3]
sin6 θ sin θ dθ
π
0

× (2π )

1
−1

(1 − cos2 θ)3 d cos θ

[sin θ dθ = d cos θ, sin2 θ = 1 − cos2 θ]
35
= 32

1
−1

(1 − 3x 2 + 3x 4 − x 6 ) dx

[x = cos θ]

1
35 32
35
= 32
x − x 3 + 35 x 5 − 17 x 7
×
= 1
=
−1
32 35

P12.22

∇2 =

∂2
∂2
∂2
+
+
∂x 2
∂y 2
∂z2

∂2
f = −a 2 f
∂x 2

∂2
f = −b2 f
∂y 2

∂2
f = −c2 f
∂y 2

and f is an eigenfunction with eigenvalue −(a 2 + b2 + c2 )
P12.25

(a) Suppose that a particle moves classically at the constant speed v. It starts at x = 0 at t = 0 and
L
at t = τ is at position x = L. v = and x = vt.

τ
1 τ
x dt =
τ t=0
v τ
=
t dt =
τ t=0

x =

=

1 τ
vt dt
τ t=0
v 2τ
t
2τ t=0

L
vτ 2
vτ
=
= x
=
2
2
2τ

x2 =

1 τ 2
v2 τ 2
x dt =
t dt
τ t=0
τ t=0

v2 3
=
t
3τ

τ

=
t=0

L
x 2 1/2 = 1/2
3

(vτ )2
L2
=
3
3

QUANTUM THEORY: TECHNIQUES AND APPLICATIONS

nπ x
2 1/2
sin
L
L

ψn =

(b)

L

199

for 0 ≤ x ≤ L [12.7]

nπ x
2 L
dx
x sin2
L 0
L
x=0
x=L

x
x
x sin 2nπ

cos 2nπ
L
L
2  x2

=
−
−
L 4
4(nπ/L)
8(nπ/L)2

x n=

ψn∗ xψn dx =

x=0

L
2 L2
=
=
= x n
L 4
2
This agrees with the classical result.
L

x2 n =

x=0

ψn∗ x 2 ψn dx =



=

x3

2
−
L 6

x2
4(nπ/L)

2
L

=

L2
1
−
3
4(nπ/L)2
=

−

1
8(nπ/L)3

sin

dx
2nπ x
L

−

x
x cos 2nπ
L

8(nπ/L)2

x=L

x=0

L
L3
−
6
8(nπ/L)2

=

1/2
x2 n

2 L 2 2 nπ x
x sin
L x=0
L

1
L2
−
3
4(nπ/L)2

1/2

L
= 1/2
3
This agrees with the classical result.
1/2

lim x 2 n

n→∞

P12.27

(a) The energy levels are given by:
h2 n2

,
8mL2
and we are looking for the energy difference between n = 6 and n = 7:
En =

h2 (72 − 62 )
.
8mL2
Since there are 12 atoms on the conjugated backbone, the length of the box is 11 times the bond
length:
E=

L = 11(140 × 10−12 m) = 1.54 × 10−9 m,
(6.626 × 10−34 J s)2 (49 − 36)
= 3.30 × 10−19 J .
8(9.11 × 10−31 kg)(1.54 × 10−9 m)2
(b) The relationship between energy and frequency is:
so

E=

E = hν

so

ν=

3.30 × 10−19 J
E
= 4.95 × 10−14 s−1 .

=
h
6.626 × 10−34 J s


INSTRUCTOR’S MANUAL

200

(c) The frequency computed in this problem is about twice that computed in problem 12.26b,
suggesting that the absorption spectrum of a linear polyene shifts to lower frequency as the
number of conjugated atoms increases . The reason for this is apparent if we look at the terms
in the energy expression (which is proportional to the frequency) that change with the number
of conjugated atoms, N . The energy and frequency are inversely proportional to L2 and directly
proportional to (n + 1)2 − n2 = 2n + 1, where n is the quantum number of the highest occupied
state. Since n is proportional to N (equal to N/2) and L is approximately proportional to N
(strictly to N − 1), the energy and frequency are approximately proportional to N −1 .
P12.29

In effect, we are looking for the vibrational frequency of an O atom bound, with a force constant equal
to that of free CO, to an infinitely massive and immobile protein complex. The angular frequency is
ω=

k 1/2
,
m

where m is the mass of the O atom.
m = (16.0 u)(1.66 × 10−27 kg u−1 ) = 2.66 × 10−26 kg,
and k is the same force constant as in problem 12.2, namely 1902 N m−1 :

ω=

1902 N m−1
2.66 × 10−26 kg

1/2

= 2.68 × 1014 s−1 .