Question

The work function of potassium metal is $2.30\, eV$. When light of frequency $8 \times 10^{14} \,Hz$ is incident on the metal surface, photoemission of electrons occurs. The stopping potential (in volt) of the electrons will be equal to

Answer 1

Energy of the incident photon, $E=h v$
$E =\left(6.6 \times 10^{-34} \, J \,s \right)\left(8 \times 10^{14} \,Hz \right) $
$=52.8 \times 10^{-20} \,J =\frac{52.8 \times 10^{-20}}{1.6 \times 10^{-19}} eV =3.3 \,eV$
According to Einstein's photoelectric equation,
$e V_{s}=h v-\phi_{0} $
$e V_{s}=3.3 eV -2.3 eV =1.0 \,eV$
Stopping potential, $V_{s}=1,0 \,V$