Homework 7 Answer Key
November 4, 2014
1
Q 8.20
How many vibrational degrees of freedom do each of the following molecules have: NH3 ,
HCN, C2 H6 , C60 ?
There are 3n − 6 vibrational modes for a nonlinear molecule and 3n − 5 modes for a linear molecule
NH3 : 6, HCN: 4, C2 H6 : 18, C60 : 174.
2
P 8.1
The 1 H3 5Cl molecule can be described by a Morse potential with De = 7.41 × 10−19 J. The
force constant k for this molecule is 516 N m −1 and ν = 8.97 × 1013 s− 1.
a.
Calculate the lowest four energy levels for a Morse potential.
2
1
(hν)2
1
1
En = hν n +
−
n+
= 6.626 × 10−34 J s−1 × 8.97 × 1013 s−1 × n +
2
4De
2
2
2
(6.626 × 10−34 J s−1 × 8.97 × 1013 s−1 )2
1
−
J n+
4 × 7.41 × 10−19
2
2
1
1
−20
−21
En = 5.944 × 10
× n+
J − 1.191 × 10
× n+
J
2
2
E0 = 2.94 × 10−20 J
E1 = 8.65 × 10−20 J
E2 = 1.41 × 10−19 J
E3 = 1.93 × 10−19 J
ν0→1 =
8.65 × 10−20 J − 2.94 × 10−20 J
E1 − E0
=
= 8.61 × 1013 s−1
h
6.626 × 10−34 J s−1
ν0→2 =
E2 − E0
1.41 × 10−19 J − 2.94 × 10−20 J
=
= 1.69 × 1014 s−1
h
6.626 × 10−34 J s−1
ν0→3 =
E3 − E0
1.93 × 10−19 J − 2.94 × 10−20 J
=
= 2.48 × 1014 s−1
h
6.626 × 10−34 J s−1
ν0→2 − 2ν0→1
Error(ν0→2 ) =
= −2.1%
ν0→2
ν0→3 − 3ν0→1
Error(ν0→3 ) =
= −4.4%
ν0→3
1
3
P 8.7
The rotational constant for 14 N2 determined from mucrowave spectrocopy is 1.99824 cm−1 .
Calculate the bond length in 14 N2 to the maximum number of significant figures consistent
with this information.
s
h
h
; r0 =
B=
8π 2 µr02
8π 2 µB
r0 =
q
14.003074007
8π 2 × (14.003074007
6.6260696×10−34 J s
kg amu−1 ×1.99824 cm−1 ×2.99792458×1010 cm s−1
amu×14.003074007 amu
×1.6605402×10−27
amu+14.003074007 amu)
r0 = 1.09769 × 10−10 m
4
P 8.35
Calculate the depth De of the Morse potential and bond energy of
12
C16 O.
2
1
1
cm−1 − 13.461 n +
cm−1
ν̃(n) = 2170.21 n +
2
2
ν̃0 = 2170.21 cm−1
Convert the expression to one in terms of energy
2
1
1
cm−1 − 13.461 × hν n +
cm−1
E(n) = 2170.21 × hν n +
2
2
and compare it with Equation (19.5) for the Morse potential:
2
1
(hν)2
1
En = hν n +
−
n+
2
4De
2
from which we see that
(hν)2
4De
hν 2
hcν̃ 2
De =
=
4 × 13.461 × c
4 × 13.461
6.626 × 10−34 J s−1 × 2.998 × 1010 cm s−1 × (2170.21) cm−1
= 1.738 × 10−18 J
=
4 × 13.461
1
D0 = De − hcν̃0 = 1.738 × 10−18 J × 6.626 × 10−34 J s−1 × 2.998 × 1010 cm s−1 × 2170.21 cm−1
2
= 1.717 × 10−18 J
13, 461 × hc =
5
Q 9.10
Explain the different degree to which the 1s, 2s, and 3s total energy eigenfunctions penetrate
into the classically forbidden region
Just as for the finite depth box, wave functions for which E − V0 is small (3s) penetrate further into
the barrier than wave functions for which E − V0 is large(1s). The 2s is intermediate between these two
extremes.
2
6
P 9.29
Show that the total energy eigenfunctions ψ210 (r, θ, φ) and ψ211 (r, θ, φ) are orthogonal. Do
you have to integrate over all three variables to show this?
Z ∞ 2
Z π
Z 2π
r
1
2
+iφ
cos θ sin θ dθ
e
dφ
e−r/a0 dr
dτ = √ √
3
ao
32 64πa0 0
0
0
h 3 iπ
Rπ
This integral is zero because 0 cos θ sin2 θ dθ = sin3 θ = 0 − 0 = 0.
0
It is sufficient to evaluate over θ.
Z Z Z
7
∗
ψ210
(τ )ψ211 (τ )
P 9.30
Calculate hri and the most probable value of r for the H atom in its ground state. Explain
why they differ with a drawing.
Z 2π Z π
Z ∞
1
− 2r
hri = 3
dφ
sin θ dθ
r3 e a0 dr
πa0 0
0
0
Z
4 ∞ 3 − a2r
r e 0 dr
hri = 3
a0 0
Using the standard integral
R∞
0
rn e−αr =
n!
αn+1 ,
hri =
4 6a40
3
= a0
a30 16
2
The most probable value of r is found by setting the derivative of the radial probability distribution
equal to zero.
r
2
2
R10
(r)
=4
1
a0
3
r2 e− 2r/a0
d 2 −
2r2 −
r
−2r/a0
−
(r e 2r/a0 ) = 2re
−
e 2r/a0 = 2e 2r/a0 r 1 −
=0
dr
a0
a0
Solutions are r = 0, r = a0 , and r = 0 corresponds to the minimum rather than most probable.
2
The mean and most probable values are not equal because r2 R10
(r) is not symmetric with respect
to r about its maximum value.
8
P 9.33
Using the result of Problem 9.12, calculate the probability of finding the electron in the 1s
state outside a sphere of radius 0.75a0 , 02.5a0 , and 4.5a0 .
2r
2r
The probability of finding the electron inside the sphere of radius r is 1 − e− a0 − a2r0 1 + ar0 e− a0 .
The probability of finding it outside the sphere of radius r is
2r
2r
2r
2r
2r
r
2r
r
1 − 1 − e− a0 −
1+
e− a0 = e− a0 +
1+
e− a0 .
a0
a0
a0
a0
Evaluating this function at 0.75a0 , 02.5a0 , and 4.5a0 gives 0.81, 0.12 and 6.2 × 10−3 , respectively
3