Radiations of two photon having energies twice and five times the work function of metal are incident successively on the metal surface. The ratio of the maximum velocity of the photoelectrons emitted in the two cases will be:

1 : 2

As, $E-w_0=\frac{1}{2}mv^2$

First case, $2W_0-W_0=\frac{1}{2}mv_1^2$  (i)

Second case, $5W_0-W_0=\frac{1}{2}mv_2^2$  (ii)

Therefore, from eqs. (i) and (ii),

$\frac{2W_0-W_0}{5W_0-W_0}=\frac{v_1^2}{v_2^2}$ or $\frac{v_1}{v_2}=\frac{1}{2}$