Taking the wavelength of first Balmer line in hydrogen spectrum (`n = 3` to `n = 2`) as 660 nm, the wavelength of the `2^(nd)` Balmer line (`n = 4` to `n = 2`) will be:

Explanation

`1/(lamda_(1))=R(1/(2^(2))-1/(3^(2)))`=(5R)/36`
`1/(lamda_(2))=R(1/(2^(2))-1/(4^(2)))=(3R)/16`
`:.(lamda_(2))/(lamda_(1))=80/108`
`lamda_(2)=80/108lamda_(1)=80/108xx660=488.9nm`