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Standard XII
Physics
Question
The wavelength of the second line of Balmer series in the hydrogen spectrum is
4861
˚
A
. The wavelength of the first line is
27
20
×
4861
˚
A
20
27
×
4861
˚
A
20
×
4861
˚
A
4861
˚
A
A
20
27
×
4861
˚
A
B
4861
˚
A
C
27
20
×
4861
˚
A
D
20
×
4861
˚
A
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Solution
Verified by Toppr
For the first line in balmer series:
1
λ
=
R
(
1
2
2
−
1
3
2
)
=
5
R
36
For second balmer line:
1
4861
=
R
(
1
2
2
−
1
4
2
)
=
3
R
16
Divide both equations:
λ
4861
=
3
R
16
×
36
5
R
λ
=
4861
×
27
20
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˚
A
. The wavelength of the first line is
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