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37.1
MOLECULES AND SOLIDS
EXERCISES
Section 37.2 Molecular Energy Levels
16.
The energies of rotational states (above the
j
=
0 state) are given by Equation 37.2, where for the HCl molecule,
h
2
263
/.
I
=
meV (from Example 37.1). Thus,
El
l
Il
l
rot
meV.
=+
= +
()
/
.
12
1
2
6
3
2
1
2
h
For
l
=
012
,, , and 3,
E
rot
meV
meV and
=
0 263
789
1578
,.
,
.
.
17.
INTERPRET
The oxygen molecule must absorb a photon in order to make a transition to the excited rotational
energy state. We are interested in the wavelength of the photon.
DEVELOP
Using Equation 37.2, the difference in energy between the
l
=
1and
l
=
0 states is
∆
E
II
rot
−
=
hh
22
2
11 1
0
The photon wavelength corresponding to this transition is
λ
=
hc
E
∆
rot
EVALUATE
Substituting the value of
I
given in the problem, we get
ππ
==
=
=
××
−
hc
E
hc
I
cI
∆
2
84
3
1
0
/
(/
m s)(1.95 10
6
2
34
1 055 10
348
kg m
Js
mm
⋅
×⋅
=
−
)
.
.
ASSESS
Transition between adjacent rotational energy levels requires absorbing a photon in the microwave
region (frequency
f
,
10
11
Hz).
18.
The energy difference between adjacent rotational levels is proportional to the upper
j
value (see Example 37.1 and
the solution to Problem 30), so
=
×
×
−
hc
E
cI l
//
(
/
.
∆
2
2
3 10
1 75 10
7
h
ms)(
kg
⋅
×
⋅
=
−
mJ
s
2
)/ ( .
)
5 1 055 10
34
62 5
.m
.
µ
19.
INTERPRET
The gas molecules must absorb a photon in order to make a transition to the excited rotational state.
We are given the wavelength of the photon, and asked to find the rotational inertia of the molecule.
DEVELOP
Using Equation 37.2, the difference in energy between the
l
=
1and
l
=
0 states is
∆
E
10
2
0
→
−
=
The energy of the absorbed photon equals the difference in energy:
∆
Eh
f
h
c
rot
Thus,
h
2
5
1240
168
738 10
I
E
hc
=
⋅
=×
→
−
∆
eV nm
cm
e
.
.VJ
−
118 10
23
.
We can find
I
readily from the above equation.
EVALUATE
The rotational inertia is
I
E
×
=
→
−
−
h
2
34
2
23
1 055 10
9
∆
(.
.
.
)
J
41 10
46
−
kg m
2
ASSESS
The value of
I
is reasonable for a molecule. We can estimate the bond length of the molecule using
Im
R
=
2
. With
m
,
10
26
−
kg, we get
R
,
03
.nm
, which is also a reasonable value.
20.
The spacing between adjacent vibrational energy levels is
h
ω
×
⋅
×
=
−
hf
)
4 136
10
1 32
10
15
14
eV s
Hz)
0 546
.e
V
.
37
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View Full Document37.2
Chapter 37
21.
INTERPRET
This problem is about the vibrational motion of the N molecule.
2
We’re given the energy difference between the adjacent levels, and asked about its corresponding classical
vibrational frequency.
DEVELOP
The quantized vibrational energy levels are given by Equation 37.3:
En
vib
=+
(/
)
12
h
ω
Therefore, the energy difference between the adjacent levels is
∆
Eh
f
vib
==
h
.
EVALUATE
The classical vibrational frequency is
f
E
h
×⋅
=×
−
∆
vib
eV
eV s
0 293
4 136
10
708 10
15
13
.
.
.H
z
ASSESS
The frequency associated with vibrational motion is
f
vib
Hz,
,
10
13
which is higher than that associated
with rotation,
f
rot
Hz.
,
10
11
This means that a more energetic photon must be absorbed by the molecule in order to
excite the vibrational modes.
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 Fall '10
 Pheong

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