Q. 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 $4 \times 10^{6}m/s.$ When the photon energy is increases to $5hv_{0}$ then maximum velocity of photo electron will be
A) $2 \times 10^{6}m/s$
B) $2 \times 10^{7}m/s$
C) $8 \times 10^{5}m/s$
D) $8 \times 10^{6}m/s$
Explanation
(d) From $E = W_{0} + \frac{1}{2}mv_{\max}^{2}$
$\Rightarrow 2hv_{0} = hv_{0} + \frac{1}{2}mv_{1}^{2} \Rightarrow hv_{0} = \frac{1}{2}mv_{1}^{2}\ \ \ \ \ldots(i)
$and $5hv_{0} = hv_{0} + \frac{1}{2}mv_{2}^{2} \Rightarrow 4hv_{0} = \frac{1}{2}mv_{2}^{2}\ \ $…(ii)
Dividing equation (ii) by (i) $\left( \frac{v_{2}}{v_{1}} \right)^{2} = \frac{4}{1}
$$\Rightarrow v_{2} = 2v_{1} = 2 \times 4 \times 10^{6} = 8 \times 10^{6}m/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.