A photosensitive metallic surface has work function, $hv_0$. If photons of energy $2hv_0$ fall on this surface, the electrons come out with a maximum velocity of 4 × 106 m/s. When the photon energy is increased to $5hv_0$, then maximum velocity of photoelectrons will be –
Answer & explanation
Correct answer: option 2
$\frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2}=\frac{K_1}{K_2}=\frac{hv_1-\phi}{hv_2-\phi}=\frac{2hv_0-hv_0}{5hv_0-hv_0}=\frac{1}{4}$
$⇒\frac{v_1^2}{v_2^2}=\frac{1}{4}⇒v_2=2v_1=8×10^6m/s$