Two radiations of photon energies 1 eV and 2.5 eV, successively illuminate a photosensitive metallic surface of work function 0.5 eV. The ratio of the maximum speeds of the emitted electrons is:

1 : 2

According to Einstein’s photoelectric equation $\frac{1}{2}mv^2_{max}=hv-\phi_0$

Where $\frac{1}{2}mv^2_{max}$ is the maximum kinetic energy of the emitted electrons, hv is the incident energy and $\phi_0$ is the work function of the metal.

$∴ \frac{1}{2}mv^2_{max_1} = 1eV - 0.5eV = 0.5eV$  (i)

and $\frac{1}{2}mv^2_{max_2} =2.5eV- 0.5eV= 2eV$  (ii)

Divide (i) and (ii), we get

$\frac{V^2_{max_1}}{V^2_{max_2}}=\frac{0.5}{2};\frac{V_{max_1}}{V_{max_2}}=\sqrt{\frac{0.5}{2}}=\frac{1}{2}$