Radiation of two photons energies, twice and five times the work function of metal are incident successively on the metal surface. The ratio of the maximum velocity of photoelectrons emitted in the two cases will be
Answer & explanation
Correct answer: option 2
$ \text{Energy of Photon 1} , E = 2\phi_0 = \phi_0 + \frac{1}{2}mv_1^2 \Rightarrow \phi_0 = \frac{1}{2}mv_1^2$
$ \text{Energy of Photon 2} , E' = 5\phi_0 = \phi_0 + \frac{1}{2}mv_2^2 \Rightarrow 4\phi_0 = \frac{1}{2}mv_2^2$
$ \Rightarrow \frac{v_1^2}{v_2^2} = \frac{\phi_0}{4\phi_0}$
$\Rightarrow \frac{v_1}{v_2} = 1:2$