Q. A photon of energy $8\ eV$ is incident on a metal surface of threshold frequency $1.6 \times 10^{15}Hz,$ then the maximum kinetic energy of photoelectrons emitted is $(h = 6.6 \times 10^{- 34}Js)$
A) $4.8\ eV$
B) $2.4\ eV$
C) $1.4\ eV$
D) $0.8\ eV$
Explanation
(c) Work function $W_{0} = hv_{0} = 6.6 \times 10^{- 34} \times 1.6 \times 10^{15}$
$= 1.056 \times 10^{- 18}J = 6.6\ eV$
From $E = W_{0} + K_{\max} \Rightarrow K_{\max} = E - W_{0} = 1.4\ eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.