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

Answer

A

Explanation

(a) `E_(1)=25 eV, E_(2)=50 eV`
`E_(1)=(hc)/(lambda_(1))` and `E_(2)=(hc)/(lambda_(2))`
or `(E_(1))/(E_(2))=(lambda_(2))/(lambda_(1))` or `(25)/(50)=(lambda_(2))/(lambda_(1))` or `lambda_(1)=2 lambda_(2)`