Question

A light whose frequency is equal to $6\times {10}^{14}Hz$ is incident on a metal whose work function is $2eV$ $(h=6.63\times {10}^{-34}Js,1eV=1.6\times {10}^{-19}J)$. The maximum energy of electrons emitted will be:

• A. $2.49eV$
• B. $4.49eV$
• C. $0.49eV$
• D. $5.49eV$

Answer 1

Answer: (C)

$0.49eV$

Absorbed energy $=$ Threshold energy $+$ Kinetic energy of photoelectrons
Absorbed energy $=hv$
$=6.626\times {10}^{-34}\times 6\times {10}^{14}$
$=3.9756\times {10}^{-19}J$
$=\frac{3.9756\times {10}^{-19}}{1.6\times {10}^{-19}}=2.49eV$
$2.49=2eV+$ Kinetic energy of photoelectron
Kinetic energy of photoelectron $=0.49eV$.