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

Answer

C

Explanation

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