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VITEEE 2018 Solved Paper
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© examsnet.com
Question : 6
Total: 125
Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelength
λ
1
:
λ
2
emitted in the two cases is
7/5
27/20
27/5
20/7
Validate
Solution:
Here, for wavelength
λ
1
n
1
=
4
and
n
2
=
3
and for
λ
2
,
n
1
=
3
and
n
2
=
2
We have
h
c
λ
=
−
13.6
[
1
n
2
2
−
1
n
1
2
]
so, for
λ
1
⇒
h
c
λ
1
=
−
13.6
[
1
(
4
)
2
−
1
(
3
)
2
]
h
c
λ
1
=
−
13.6
[
7
144
]
.
.
.
(i)
simiarly, for
λ
2
⇒
h
c
λ
1
=
−
13.6
[
1
(
3
)
2
−
1
(
2
)
2
]
h
c
λ
1
=
−
13.6
[
5
36
]
.
.
.
(ii)
Hence, from Eqs. (i) and (ii), we get
∴
λ
1
λ
2
=
−
13.6
[
7
144
]
−
13.6
[
5
36
]
=
20
7
© examsnet.com
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