422
Chapter 42
(b)
9
12
1
1
1
(0.0644
10
m)(1.03
10
rad/s)
66.3 m/s (carbon)
v
r
ω
−
=
=
×
×
=
9
12
2
2
2
(0.0484
10
m)(1.03
10
rad/s)
49.8 m/s (oxygen)
v
r
ω
−
=
=
×
×
=
(c)
12
2
/
6.10
10
s
T
π ω
−
=
=
×
E
VALUATE
:
From the information in Example 42.3 we can calculate the vibrational period to be
14
r
2
/
2
/
1.5
10
s.
T
m
k
π ω
π
−
′
=
=
=
×
The rotational motion is over an order of magnitude slower than the
vibrational motion.
42.8.
r
hc
E
k
m ,
λ
′
Δ
=
=
=
and solving for
,
k
′
2
r
2
205 N m.
π
c
k
m
λ
⎛
⎞
′ =
=
⎜
⎟
⎝
⎠
42.9.
I
DENTIFY
and
S
ET
U
P
:
The energy of a rotational level with quantum number
l
is
2
(
1)
/ 2
l
E
l l
I
=
+
=
(Eq.(42.3)).
2
r
,
I
m r
=
with the reduced mass
r
m
given by Eq.(42.4). Calculate
I
and
E
Δ
and then use
/
E
hc
λ
Δ
=
to find
.
λ
E
XECUTE
:
(a)
26
27
27
1
2
Li
H
r
26
27
1
2
Li
H
(1.17
10
kg)(1.67
10
kg)
1.461
10
kg
1.17
10
kg
1.67
10
kg
m m
m m
m
m
m
m
m
−
−
−
−
−
×
×
=
=
=
=
×
+
+
×
+
×
2
27
9
2
47
2
r
(1.461
10
kg)(0.159
10
m)
3.694
10
kg m
I
m r
−
−
−
=
=
×
×
=
×
⋅
2
2
3:
3(4)
6
2
l
E
I
I
⎛
⎞
⎛
⎞
=
=
=
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
=
=
2
2
4:
4(5)
10
2
l
E
I
I
⎛
⎞
⎛
⎞
=
=
=
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
=
=
2
34
2
21
3
4
3
47
2
(1.055
10
J s)
4
4
1.20
10
J
7.49
10
eV
3.694
10
kg m
E
E
E
I
−
−
−
−
⎛
⎞
⎛
⎞
×
⋅
Δ
=
−
=
=
=
×
=
×
⎜
⎟
⎜
⎟
×
⋅
⎝
⎠
⎝
⎠
=
(b)
/
E
hc
λ
Δ
=
so
15
8
3
(4.136
10
eV)(2.998
10
m/s)
166
m
7.49
10
eV
hc
E
λ
μ
−
−
×
×
=
=
=
Δ
×
E
VALUATE
:
LiH has a smaller reduced mass than CO and
λ
is somewhat smaller here than the
λ
calculated for
CO in Example 42.2
42.10.
I
DENTIFY
:
The vibrational energy of the molecule is related to its force constant and reduced mass, while the
rotational energy depends on its moment of inertia, which in turn depends on the reduced mass.
S
ET
U
P
:
The vibrational energy is
r
1
1
2
2
n
k
E
n
n
m
ω
′
⎛
⎞
⎛
⎞
=
+
=
+
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
=
=
and the rotational energy is
2
(
1)
2
l
E
l l
I
=
+
=
.
E
XECUTE
:
For a vibrational transition, we have
v
r
k
E
m
′
Δ
=
=
, so we first need to find
m
r
. The energy for a
rotational transition is
[
]
2
2
R
2
2(2
1)
1(1
1)
2
E
I
I
Δ
=
+
−
+
=
=
=
. Solving for
I
and using the fact that
I = m
r
r
0
2
, we have
2
2
r 0
R
2
m r
E
=
Δ
=
, which gives
(
)(
)
(
) (
)
34
16
2
r
2
2
9
4
0
R
2
1.055
10
J
s
6.583
10
eV s
2
0.8860
10
m
8.841
10
eV
m
r
E
−
−
−
−
×
⋅
×
⋅
=
=
Δ
×
×
=
= 2.0014
×
10
–28
kg
Now look at the vibrational transition to find the force constant.
(
)
(
)
(
)
2
28
2
r
v
v
2
2
16
r
2.0014
10
kg (0.2560
eV)
6.583
10
eV s
m
E
k
E
k
m
−
−
×
Δ
′
′
Δ
=
⇒
=
=
×
⋅
=
=
= 30.27 N/m
E
VALUATE
:
This would be a rather weak spring in the laboratory.
42.11.
(a)
2
2
2
2
2
2
1
(
1)
(
1)
,
(
)
2
2
2
l
l
l l
l l
l
E
E
E
l
l
l
l
I
I
I
I
−
+
−
=
=
⇒
Δ
=
+
−
+
=
=
=
=
=
(b)
2
2
E
E
l
f
.
h
π
π
I
=
=
=
=
=
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Molecules and Condensed Matter
423
42.12.
I
DENTIFY
:
Find
E
Δ
for the transition and compute
λ
from
/
.
E
hc
λ
Δ
=
S
ET
U
P
:
From Example 42.2,
2
(
1)
,
2
l
E
l l
I
=
+
=
with
2
3
0.2395
10
eV.
2
I
−
=
×
=
From Example 42.3,
0.2690 eV
E
Δ
=
is the spacing between vibrational levels. Thus
1
2
(
)
,
n
E
n
ω
=
+
=
with
0.2690 eV.
ω
=
=
By Eq.(42.9),
2
1
2
(
)
(
1)
.
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 Spring '06
 Buchler
 Physics, Photon, Light, Trigraph, Condensed matter physics, ev, Field electron emission

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