Question

The stopping potential in an experiment on photoelectric effect is $1.5V$. What is the maximum kinetic energy of the photoelectrons emitted? Calculate in Joules.

Answer 1

Photoelectric cut-off voltage , $V_o=1.5V$

$K.E=eV$

where e = charge of an electron =$1.6\times {10}^{-19}C$

so,

$K.E=1.6\times {10}^{-19}\times 1.5$

$K.E=2.4\times {10}^{-19}J$