The wavelength of radiation emitted is `lamda_0` when an electron in a hydrogen atom jumps from 3rd to 2nd orbit . If the same hydrogen atom , the electron jumps from 4th orbit to 2nd orbit , then the wavelength of the emitted radiation will be

Options

A. `10/25 lamda_0`
B. `25/16 lamda_0`
C. `27/20 lamda_0`
D. `20/27 lamda_0`

Answer

D

Explanation

`1/lamda_0 = R [ 1/2^2 - 1/3^2 ] = R [ 1/4 - 1/9] = (5R)/(36)`
`1/lamda = R [ 1/2^2 - 1/4^2]=R [ 1/4 - 1/16] = (3R)/(16)`
`therefore (1/lamda_0)/(1/lamda) = lamda/lamda_0 = (5R)/(36 )xx(16)/(3R) = 80/108`
`therefore lamda = 20/27 lamda_0`