Q. A photon of energy 8 eV is incident on metal surface of threshold frequency $1.6 \times 10^{15}$ Hz. The kinetic energy of the photoelectrons emitted (in eV) $(Take\ h = 6 \times 10^{- 34}\ Js)$
A) 1.6
B) 6
C) 2
D) 1.2
Explanation
(c) $KE = hv - hv_{0}$
$= \ 8\ eV - \left( \frac{6 \times 10^{- 34} \times 1.6 \times 10^{15}}{1.6 \times 10^{- 19}}\ eV \right)$
= 8-6= 2 eV
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.