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

Answer

A

Explanation

`(K_(1))/(K_(2)) = (hv_(1) - phi_(1))/(hv_(2) - phi_(2)) = (1 - 0.5)/(2.5 - 0.5) = 1/4`