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Question
The wavelength of the second line of the Balmer series in the hydrogen spectrum is 4861 Å. Calculate the wavelength of the first line of the same series.
Solution
For the first line in Balmer series:
`1/lambda = R(1/2^2 - 1/3^2) = (5R)/36` ..........(i)
For second Balmer line:
`1/4861 = R(1/2^2 - 1/4^2) = (3R)/16` ............(ii)
Dividing equation (ii) by (i),
`lambda/4861 = (3R)/16 xx 36/(5R)`
`lambda = 4861 xx 27/20 = 6562` Å
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