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JEE Main Chemistry Class 11 Structure of Atom Part 2 Questions
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© examsnet.com
Question : 32
Total: 91
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to
h
2
x
m
a
0
2
. The value of
10
x
is .......... (
a
0
is radius of Bohr's orbit) (Nearest integer)
[Given,
π
=
3.14
]
[27 Aug 2021 Shift 1]
Your Answer:
Validate
Solution:
∵ Angular moment is given as,
m
v
r
=
n
h
2
π
∴ Kinetic energy
=
n
2
h
2
8
π
2
m
r
2
=
4
h
2
8
π
2
m
(
4
a
0
)
2
=
(
4
8
π
2
×
16
)
h
2
m
a
0
2
=
h
2
x
m
a
0
2
x
=
4
8
π
2
×
16
⇒ x = 315.507
10x = 3155.07
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