Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively, illuminate a metal of work function 0.5 eV. The ratio of the maximum KE of the emitted electrons will be:

$1: 4$

The correct answer is Option (2) → $1: 4$

According to Photo-electric equation -

$K.E_{max}=E_{photon}-\phi$

$K.E_1=E_1-\phi=1-0.5=0.5eV$

$K.E_2=E_2-\phi=2.5-0.5=2.0eV$

$\frac{K.E_1}{K.E_2}=\frac{0.5}{2.0}=\frac{1}{4}$