Question

The wavelength of the first line in Balmer series in the hydrogen spectrum is 'λ'. What is the wavelength of the second line in the same series?

Options

• `20/27 lambda`

• `3/16 lambda`

• `5/36 lambda`

• `3/4 lambda`

Answer 1

`20/27 lambda`

Explanation:

Wavelength in Balmer series of hydrogen spectrum is given by relation,

`1/lambda = "R"[1/2^2 - 1/"n"^2]`  where, n = 3, 4, 5

So, wavelength of first line,

`1/lambda_1 = "R"[1/4 - 1/3^2]`

`= "5R"/36`

`=> "R" = 36/(5lambda)` ..(∴ λ₁ = λ, given)  ...(i)

Similarly wavelength of second line,

`1/lambda_2 = "R"[1/4 - 1/4^2] = "3R"/16`

`=> lambda_2 = 16/"3R"`   ...(ii)

From Eqs. (i) and (ii), we get

`lambda_2 = 20/27 lambda`

Hence, the wavelength of the second line in the same series is `20/27 lambda`.