The only `e^-` in the H-atom resides under ordinary conditions on the first orbit when energy is supplied, the `e^-` moves to higher energy shells depending upon the amount of energy absorbed. When an e emits energy i.e., the `e^-` returns to the lowest energy state, from this Lyman, Balmer, Paschen, Bracket, Pfund series are there, so different spectral lines in the spectra of atoms correspond to different transitions of `e^-` s from higher to lower energy levels:
If the shortest wavelength of H atom in Lyinan series x, then longest wavelength in Balmer series of `He^+` is

Options

A. `(36x)/(5)`
B. `(x)/(4)`
C. `(9x)/(5)`
D. `(5x)/(9)`

Answer

C

Explanation

Shortest wavelength `implies 1/(R_H) =x`
For Balmer series of `He^(+2)`
longest `implies 1/lambda = R_H [1/(2^2) - 1/(3^2)] xx 4 implies lambda implies (9x)/(5)`