Whenever a photon is emitted by hydrogen in balmer series it is followed by another in lyman series what wavelength does this latter photon correspond to?

Answer

A, B

Explanation

The second wave length is from Balmer to lyman i.e. from
`n=2` to `n=1`
`n_1=2`, `n_2 =1`
`1/lambda = R((1)/(n_1^2) - (1)/(n_2^2))
`1/lambda = 1.097 xx 10^7[1/4-1]`
`=-1.097 xx 3/4 xx 10^7`
`lambda = (4)/(1.097 xx 3) xx 10^-7`
`=1.215 xx 10^-7`
`=1215. xx 10^-9 = 122 nm`