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MHT CET 2019 2nd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

The maximum velocity of the photoelectron emitted by the metal surface is ' $v$ '. Charge and mass of the photoelectron is denoted by ' $e$ ' and ' $m$ ' respectively. The stopping potential in volt is

A
$\frac{v^2}{2\left(\frac{m}{e}\right)}$
B
$\frac{v^2}{2\left(\frac{e}{m}\right)}$
C
$\frac{v^2}{\left(\frac{e}{m}\right)}$
D
$\frac{v^2}{\left(\frac{m}{e}\right)}$
MHT CET Subjects
EXAM MAP