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AP EAMCET Engineering 2015 Code C Paper
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© examsnet.com
Question : 96
Total: 160
The ratio of longest wavelength lines in the Balmer and Paschen series of hydrogen spectrum is
5
36
7
20
7
144
5
27
Validate
Solution:
For largest wavelength of Balmer series electron jumps from 3rd to 2nd orbit
1
λ
B
=
R
[
1
2
2
−
1
3
2
]
1
λ
B
=
R
×
5
36
→(1)
For largest wavelength of Paschan series electron jumps from 4th to 3rd orbit
1
λ
P
=
R
[
1
3
2
−
1
4
2
]
1
λ
p
=
R
[
7
9
×
16
]
→(2)
(
2
)
(
1
)
(
1
λ
P
)
(
1
λ
B
)
=
R
(
7
9
×
16
)
R
(
5
36
)
λ
B
λ
P
=
7
9
×
10
×
36
5
=
λ
B
λ
P
=
7
20
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