The energies `E_(1)` and `E_(2)`of two radiations are 25 eV and 50 eV respectively. The relation beween their wavelength i.e.`lamda_(1)` and `lamda_(2)` will be

Options

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

Answer

B

Explanation

`E_(1)=(hc)/(lamda_(1)),E_(2)=(hc)/(lamda_(2))`
`:.(E_(1))/(E_(2))=(lamda_(2))/(lamda_(1))` or `(lamda_(2))/(lamda_(1))=25/50=1/2`