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BITSAT 2017 Slot 2 Solved Paper
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© examsnet.com
Question : 16
Total: 150
The total energy of an electron in the first excited state of hydrogen is about - 3.4 eV. Its kinetic energy in this state is
- 3.4 eV
- 6.8 eV
6.8 eV
3.4 eV
Validate
Solution:
We have, Kinetic energy of electron,
K
=
Z
e
2
8
π
ε
0
.
r
also, Potential energy of electron,
U
=
−
1
4
π
ε
0
.
Z
e
2
r
∴ Total energy
E
=
K
+
U
=
Z
e
2
8
π
ε
0
.
r
−
Z
e
2
4
π
ε
0
.
r
E
=
−
Z
e
2
8
π
ε
0
r
or
K
=
−
E
=
−
(
−
3.4
)
=
3.4
e
V
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