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:
Answer & explanation
Correct answer: option 3
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,
$\frac{V_{max_1}}{V_{max_2}}=\sqrt{\frac{K_{max_1}}{K_{max_2}}}$
$=\sqrt{\frac{4}{2}}=\sqrt{2}$