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421
M
OLECULES AND
C
ONDENSED
M
ATTER
42.1.
(a)
41
9
23
3
2
2(7.9 10 eV)(1.60 10
J eV)
6.1K
2
3
3(1.38 10
J K)
K
Kk
T
T
k
−−
−
××
=
⇒
==
=
×
(b)
19
23
2(4.48 eV) (1.60 10
J eV)
34,600 K.
3(1.38 10
J K)
T
−
−
×
×
(c)
The thermal energy associated with room temperature (300 K) is much greater than the bond energy of
2
He
(calculated in part (a)), so the typical collision at room temperature will be more than enough to break up
2
He .
However, the thermal energy at 300 K is much less than the bond energy of
2
H , so we would expect it to remain
intact at room temperature.
42.2.
(a)
2
0
1
5.0 eV.
4
e
U
πε
r
=−
(b)
5.0 eV
(4.3 eV
3.5 eV)
4.2 eV.
−
+−=
−
42.3.
IDENTIFY:
The energy given to the photon comes from a transition between rotational states.
SET UP:
The rotational energy of a molecule is
2
(1
)
2
El
l
I
=+
=
and the energy of the photon is
E = hc/
λ
.
EXECUTE:
Use the energy formula, the energy difference between the
l
= 3 and
l
= 1 rotational levels of the
molecule is
[]
22
5
3(3 1) 1(1 1)
2
E
II
Δ=
+ − +
=
.
Since
Δ
E = hc/
,
we get
hc/
=
5
2
=
/
I
.
Solving for
I
gives
()
34
52
2
8
5 1.055 10
J s (1.780
nm)
5
4.981 10
kg m
2
23
.
0
01
0
m
/
s
I
c
π
−
−
×⋅
=
×
⋅
×
=
.
Using
I = m
r
r
0
2
, we can solve for
r
0
:
(
)
52
2
26
27
NH
0
26
27
4.981 10
kg m
2.33 10
kg 1.67 10
kg
2.33 10
kg 1.67 10
kg
Im
m
r
mm
−
×+
×
+
0
r
= 5.65
×
10
–13
m
EVALUATE:
This separation is much smaller than the diameter of a typical atom and is not very realistic. But we
are treating a
hypothetical
NH molecule.
42.4.
The energy of the emitted photon is
5
1.01 10 eV,
−
×
and so its frequency and wavelength are
51
9
34
(1.01 10 eV)(1.60 10
J eV)
2.44 GHz
(6.63 10
J s)
E
f
h
−
=
and
8
9
(3.00 10 m s)
0.123 m.
(2.44 10 Hz)
c
f
×
=
×
This frequency
corresponds to that given for a microwave oven.
42.5.
Let 1 refer to C and 2 to O.
26
26
120
1.993 10
kg,
2.656 10
kg,
0.1128 nm
r
=×
= ×
=
.
2
10
12
0.0644 nm (carbon)
m
rr
⎛⎞
⎜⎟
+
⎝⎠
;
1
20
0.0484 nm (oxygen)
m
+
(b)
4
6
2
11
2 2
1.45 10
kg m ;
r m
r
−
=+ =×
⋅
yes, this agrees with Example 42.2.
42.6.
Each atom has a mass
m
and is at a distance
2
L
from the center, so the moment of inertia is
4
4
2
2( )(
2)
2
2.21 10
kg m .
mL
m
L
−
×
⋅
42.7.
IDENTIFY
and
SET UP:
Set
1
KE
=
from Example 42.2. Use
2
1
2
KI
ω
=
to solve for
and
vr
=
to solve for
v
.
EXECUTE: (a)
From Example 42.2,
23
1
0.479 meV
7.674 10
J
E
−
×
and
46
2
1.449 10
kg m
I
−
⋅
2
1
2
=
and
=
gives
12
1
2
/
1.03 10 rad/s
EI
×
42
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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
.
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 Spring '11
 Shaefer

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