Two photons of equal energy 8 MeV each, strikes two different metal sheets having work functions 4 MeV and 6 MeV, respectively. The ratio of maximum velocity of electrons emitted from two metal sheets respectively is:

$\sqrt{2}:1$

The correct answer is Option (3) → $\sqrt{2}:1$

The maximum kinetic energy of the emitted electron is -

$K_{max}=E_{photon}-\phi$

$∴K_{max_1}=8-4=4MeV$

$K_{max_2}=8-6=2MeV$

Also,

$K_{max}=\frac{1}{2}mv^2$

Hence,

$\frac{V_{max_1}}{V_{max_2}}=\sqrt{\frac{K_{max_1}}{K_{max_2}}}$

$=\sqrt{\frac{4}{2}}=\sqrt{2}$