Q. The threshold frequency for certain metal is $3.3 \times 10^{14}$ Hz. If light of frequency $8.2 \times 10^{14}$ Hz is incident on the metal, the cut-off voltage of the photoelectric current will be
A) 4.9 V
B) 3.0 V
C) 2.0 V
D) 1 V
Explanation
(c) From relation
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ eV_{s} = h\left( v - v_{0} \right)$
or $V_{s} =$ threshold or cut off voltage
$= \frac{h}{e}\left( v - v_{0} \right)$
$= \ \frac{6.6 \times 10^{- 34}}{1.6 \times 10^{- 19}}\ (8.2 - 3.3) \times 10^{14}$
$\ = \ \frac{6.6 \times 4.9 \times 10^{- 1}}{1.6}\ = \ 2V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.