Question

# The first line of the Balmer series in the hydrogen spectrum has a wavelength of 6564˚A. Calculate the wavelength of the first line of Lyman series in the same spectrum.

A
1215˚A
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B
1222˚A
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C
1346˚A
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D
122.5˚A
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Solution

## The correct option is A 1215˚AThe wavelength of the spectral lines of hydrogen spectrum are given by the formula 1λ=R(1n21−1n22) Where R = Rydberg ConstantFor the first line of the Balmer series, nf=2 and ni=31λ1=R(122−132)1λ1=5R36 ................... (1)For the first line of the Lyman series, nf=1 and ni=21λ′1=R(112−122)1λ′1=3R4 .......................(2)From equation (1) and (2), we getλ1λ′1=5R36 X 43Rλ1λ′1=527That is, λ1=527λ′1Given that λ′1=6564˚ASo, λ1=527 X 6564λ1=1215˚A

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