The wavelength of the radiation emitted by a hydrogen atom in the electronic transition from n=3 to n=2 is `lambda`. For the same transition in the singly ionized helium, the wavelength of the emitted radiation is

Options

A. `lambda//4`
B. `lambda//2`
C. `2 lambda`
D. `4 lambda`

Answer

D

Explanation

`(1)/(lambda) prop Z^(2)`
`(lambda_(2))/(lambda_(1))=((Z_(1))/(Z_(2)))^(2) implies (lambda_(2))/(lambda)=((4)/(1))^(2) implies lambda_(2)=4 lambda`