Find the quantum number n corresponding to nth excited state of `He^(++)` ion if on transition to the ground state the ion emits two photons in succession with wavelength 108.5 nm and 30.4 nm. The ionization energy of the hydrogen atom is 13.6 eV.

Explanation

`lambda_(1) = 30.4 xx 10^(-7) cm`
`lambda_(2) = 108.5 xx 10^(-7) cm`
Let the excited state of `He^(Theta)` be `n_(2)` it comes from `n_(2)` to `n_(1)` and then from `n_(1)` to `1` to emit two successive photons
`(1)/(lambda_(2)) = R_(H)Z^(2)[(1)/((1)^(2)) - (1)/(n_(1))^(2)]`
`:. n_(1) = 2`
Now for `lambda_(2)n_(1) = 2,n_(2) = `?
`(1)/(108.5 xx 10^(-7)) = 109678 xx (2)^(2)[(1)/((2)^(2))- (1)/((n_(2))^(2))]`
`:. n_(2) = 3`
Hence the excited state of `He^(Theta)` is fifth orbit