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

Answer

C

Explanation

`E = hv = h (c)/(lamda) :. (E_(1))/(E_(2)) = (lamda_(2))/(lamda_(1))`
i.e., `(lamda_(1))/(lamda_(2)) = (E_(2))/(E_(1)) = (50eV)/(25 eV) = 2 or lamda_(1) = 2 lamda_(2)`