Chemistry 350
Assignment 2 Solutions
October 13, 2004
1. The energies and degeneracies of the two lowest electronic states of atomic iodine are:
e
1
= 0
cm

1
,
g
e
1
= 4;
e
2
= 7603.2 cm

1
,
g
e
2
= 2. What temperature is required so that 10% of the
atoms are in the excited state? Disregard any contribution from higher electronic states.
Answer:
As in any question like this, this requires a simple calculation of the Boltzmann factor,
p
j
, for the desired state. We can assume we can state the percentage to as many significant
digits as we desire, so we’ll choose six here.
p
j
=
e

e
j
/kT
∑
i
e

e
i
/kT
0
.
100000
=
2
e

7603
.
2
cm

1
/
(0
.
69509
cm

1
K

1
)
T
4 + 2
e

7603
.
2
cm

1
/
(0
.
69509
cm

1
K

1
)
T
ln
0
.
400000
1
.
80000
=

7603
.
2
cm

1
(0
.
69509
cm

1
K

1
)
T
T
=
7272
.
5
K
2. Show that the fraction of harmonic oscillators in the ground vibrational state is given by
f
0
=
1

e

hν
e
/kT
Calculate
f
0
for
15
N
2
(g) (
k
f
= 2243 N m

1
) at 300, 500 and 1500 K.
Answer:
Using the harmonic oscillator partition function, we find that for the
v
’th vibrational energy
level,
p
v
=
e

βhν
e
v
e

βhν
e
/
2
(1

e

βhν
e
)
e

βhν
e
/
2
=
e

βhν
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 Spring '10
 Leroy
 Thermodynamics, Atom, Electron, Excited state, qtrans

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