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422
Chapter 42
(b)
91
2
11
1
(0.0644 10 m)(1.03 10 rad/s)
66.3 m/s (carbon)
vr
ω
−
==
×
×
=
2
22
2
(0.0484 10 m)(1.03 10 rad/s)
49.8 m/s (oxygen)
−
×
×
=
(c)
12
2 /
6.10 10
s
T
πω
−
×
EVALUATE:
From the information in Example 42.3 we can calculate the vibrational period to be
14
r
2/
2
/
1
.
51
0
s
.
Tm
k
π
−
′
=
×
The rotational motion is over an order of magnitude slower than the
vibrational motion.
42.8.
r
hc
Ek
m
,
λ
′
Δ=
=
=
and solving for
,
k
′
2
r
2
205 N m.
π
c
km
⎛⎞
′
⎜⎟
⎝⎠
42.9.
IDENTIFY
and
SET UP:
The energy of a rotational level with quantum number
l
is
2
(1
)/
2
l
El
l
I
=+
=
(Eq.(42.3)).
2
r
,
Im
r
=
with the reduced mass
r
m
given by Eq.(42.4). Calculate
I
and
E
Δ
and then use
/
Eh
c
to find
.
EXECUTE: (a)
26
27
27
12
L
iH
r
26
27
L
(1.17 10
kg)(1.67 10
kg)
1.461 10
kg
1.17 10
kg 1.67 10
kg
mm
m m
m
mmmm
−−
−
××
=
=
×
++
×
+
×
7
9
24
7
2
r
(1.461 10
kg)(0.159 10 m)
3.694 10
kg m
r
−
×
×
=
×
⋅
3:
3(4)
6
2
lE
II
⎛⎞ ⎛⎞
=
⎜⎟ ⎜⎟
⎝⎠ ⎝⎠
4:
4(5)
10
2
=
23
4
2
21
3
43
47
2
(1.055 10
J s)
4
4
1.20 10
J
7.49 10 eV
3.694 10
kg m
EE E
I
−
−
⎛⎞ ⎛
⎞
×⋅
Δ= − =
=
=
×
=
×
⎜⎟ ⎜
⎟
⎝⎠ ⎝
⎠
=
(b)
/
c
so
15
8
3
(4.136 10
eV)(2.998 10 m/s)
166 m
hc
E
μ
−
−
=
Δ×
EVALUATE:
LiH has a smaller reduced mass than CO and
is somewhat smaller here than the
calculated for
CO in Example 42.2
42.10.
IDENTIFY:
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.
SET UP:
The vibrational energy is
r
n
k
En
n
m
′
⎛⎞⎛⎞
⎜⎟⎜⎟
⎝⎠⎝⎠
and the rotational energy is
2
)
2
l
l
I
=
.
EXECUTE:
For a vibrational transition, we have
v
r
k
E
m
′
=
, so we first need to find
m
r
. The energy for a
rotational transition is
[]
R
2
2(2 1) 1(1 1)
2
E
+−+=
. Solving for
I
and using the fact that
I = m
r
r
0
2
, we have
2
2
r0
R
2
mr
E
=
Δ
=
, which gives
()
(
)
(
)
34
16
2
r
2
2
94
0R
2 1.055 10
J s 6.583 10
eV s
2
0.8860 10 m
8.841 10 eV
m
rE
×
⋅
Δ
=
= 2.0014
×
10
–28
kg
Now look at the vibrational transition to find the force constant.
2
28
2
rv
v
2
2
16
r
2.0014 10
kg (0.2560
eV)
6.583 10
eV s
mE
k
m
−
−
×
Δ
′
′
⇒
=
=
= 30.27 N/m
EVALUATE:
This would be a rather weak spring in the laboratory.
42.11.
(a)
2
2
1
)
)
,(
)
2
ll
l
EE
E
l
l
l
l
I
I
−
+−
⇒
+− + =
=
=
(b)
ΔΔ
l
f.
h
ππ
I
=
=
=
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423
42.12.
IDENTIFY:
Find
E
Δ
for the transition and compute
λ
from
/
.
Eh
c
Δ=
SET UP:
From Example 42.2,
2
(1
) ,
2
l
El
l
I
=+
=
with
2
3
0.2395 10 eV.
2
I
−
=×
=
From Example 42.3,
0.2690 eV
E
is the spacing between vibrational levels. Thus
1
2
()
,
n
En
ω
=
with
0.2690 eV.
=
=
By Eq.(42.9),
2
1
2
(
1
)
.
2
nl
EE E n
l
l
I
=+=+
++
=
=
EXECUTE: (a)
01
a
n
d
12
nn
ll
=→=
For
2
1
2
0,
1,
2
.
2
i
nlE
I
⎛⎞
==
=
+
⎜⎟
⎝⎠
=
=
For
2
3
2
1,
2,
6
.
2
f
E
I
=
+
=
=
2
3
4
0.2690 eV
4(0.2395 10 eV)
0.2700 eV
2
fi
EE E
I
−
− =
+
=
+
×
=
=
=
hc
E
=Δ
so
15
8
6
(4.136 10
eV s)(2.998 10 m/s)
4.592 10 m
4.592 m
0.2700 eV
hc
E
μ
−
−
×⋅×
=
×
=
Δ
(b)
a
n
d
21
For
2
1
2
0,
2,
6
.
2
i
I
=
+
=
=
For
2
3
2
1,
2
.
2
f
I
=
+
=
=
2
3
4
0.2690 eV
0.2680 eV
2
I
−
−
=
−
×
=
=
=
15
8
6
(4.136 10
eV s)(2.998 10 m/s)
4.627 10 m
4.627 m
0.2680 eV
hc
E
−
−
=
×
=
Δ
(c)
a
n
d
32
For
2
1
2
0,
3,
12
.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics

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