Two photons of energies twice and thrice the work function of a metal are incident on the metal surface. Then the ratio of maximum velocities of the photoelectrons emitted in the two cases respectively, is:

Let v₁ and v₂ be the maximum velocities of photoelectrons in the two cases. The energies of photon are given by

            $E_1=W_0+\frac{1}{2}m{v}_1^2$                       (i)

And     $E_2=W_0+\frac{1}{2}m{v}_2^2$                        (ii)

Given $E_1=2W_0\text{ and }{E}_2=3W_0$. substituting these in (i) and (ii), we have

         $2W_0=W_0+\frac{1}{2}m{v}_1^2 \Rightarrow \frac{1}{2}m{v}_1^2=W_0$   (iii)

and   $3W_0=W_0+\frac{1}{2}m{v}_2^2 \Rightarrow \frac{1}{2}m{v}_2^2=2W_0$  (iv)

From (iii) and . (iv), we get  $\frac{v_1}{v_2}=\frac{1}{\sqrt{2}}.$