The energies `E_1` and `E_2` of two radiations are `25 eV` and `50 e V` respectively. The relation between their wavelengths, i.e., `lamda_1` and `lamda_2` will be.

Options

A. `lamda_(1)=1/2lamda_(2)`
B. `lamda_(1)=lamda_(2)`
C. `lamda_(1)=2lamda_(2)`
D. `lamda_(1)=4lamda_(2)`

Answer

C

Explanation

`E=hc//lamda`
`Eprop1//lamda` `E_(1)/E_(2)=lamda_(2)/lamda_(1)`
`25/50=lamda_(2)/lamda_(1)`
`lamda_(1)=2lamda_(2)`.