The wavelength of the energy emitted when electron come from fourth orbit to second orbit in hydrogen is `20.397 cm`. The wavelength of energy for the same transition in `He^(+)` is

Options

A. `5.099 cm^(-1)`
B. `20.497 cm^(-1)`
C. `40.994 cm^(-1)`
D. `81.988 cm^(-1)`

Answer

A

Explanation

`E ( = (hc)/(lambda)) prop (Z^(2))/(n^(2)) rArr lambda prop (1)/(Z^(2))`
Hence `lambda_(He^(+)) = (20.397)/(4) = 5.099 cm`