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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.
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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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