Light of two different frequencies whose photons have energies 1eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum kinetic energy of emitted electrons will be:

Options

A. `1:5`
B. `1:4`
C. `1:2`
D. `1:1`

Answer

B

Explanation

`(k_("max"_1))/(k_("max"_2))=(E_1-phi_(0_1))/(E_2-phi_(0_2))= (1eV -0.5eV)/(2.5 eV -0.5eV)`
`=(0.5)/2=1/4`