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Test Index
Atoms and Nuclei
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Section:
Physics
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© examsnet.com
Question : 40
Total: 51
The key feature of Bohr’s theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr’s quantization condition.
In a CO molecule, the distance between C (mass = 12 amu) and O (mass = 16 amu), where 1 amu =
5
3
×
10
−
27
k
g
, is close to
[JEE Adv 2010 P2]
2.4 ×
10
−
10
m
1.9 ×
10
−
10
m
1.3 ×
10
−
10
m
4.4 ×
10
−
10
m
Validate
Solution:
The moment of inertia is
I =
m
1
r
1
2
+
m
2
r
2
2
If d is the distance between the two atoms C and O, then
r
2
=
m
2
d
m
1
+
m
2
and
r
2
=
m
1
d
m
1
+
m
2
Therefore, the moment of inertia becomes as follows:
I =
m
1
(
m
2
d
m
1
+
m
2
)
2
+
m
2
(
m
1
d
m
1
+
m
2
)
2
=
m
1
m
2
d
2
m
1
+
m
2
Therefore,
d =
[
l
(
m
1
+
m
2
)
m
1
m
2
]
1
∕
2
=
[
1.87
×
10
−
46
(
12
+
16
)
12
×
16
×
(
5
∕
3
)
×
10
−
27
]
1
2
= 1.28 ×
10
−
10
m
~ 1.3 ×
10
−
10
m
© examsnet.com
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