Answer
(a) so stopping potential $V_{0}=0.64375V$
(b) cut off frequency $f_{0}=8.446455\times10^{14}Hz$
Work Step by Step
Given that work function of the material $\phi_{0}=3.5eV$
wavelength of the incident $\lambda= 300nm=300\times10^{-9}m=3\times10^{-7}m$
(a) calculation of stopping potential
$E=hf=\frac{hc}{\lambda}$
putting $h=6.63\times10^{-34} J.s$, $c=3\times10^{8}m/s$,$\lambda=3\times10^{-7}m$
$E=hf=\frac{6.63\times10^{-34} J.s\times3\times10^{8}m/s}{3\times10^{-7}m}=6.63\times10^{-19} J$
$1.6\times10^{-19} J$ is equal to $1eV$
$1J$ is equal to $\frac{1}{1.6\times10^{-19}}eV$
so $6.63\times10^{-19} J$ is equal to $\frac{6.63\times10^{-19}}{1.6\times10^{-19}}eV$= $4.14375eV$
Energy of incident light $E=hf=6.63\times10^{-19} J=4.14375eV$
From equation 27.7
$hf= K_{max} + \phi_{0}$
$K_{max}=hf-\phi_{0}$
putting the values of $\phi_{0}$ and we will get $hf$ we will get
$K_{max}=4.14375eV-3.5eV=0.64375eV$
from equation 27.5 $K_{max}=eV_{0}$
so stopping potential $V_{0} =\frac{K_{max}}{e}=\frac{0.64375eV}{e}=0.64375V$
(b) Calculation of cutoff frequency
From Equation 27.8 cut off frequency is given by
$f_{0}=\frac{\phi_{0}}{h}$
$h=6.63\times10^{-34} J.s=\frac{6.63\times10^{-34} }{1.6\times10^{-19} }eV.s=4.14375\times10^{-15}eV.s$
putting $h=4.14375\times10^{-15}eV.s$ and $\phi_{0}=3.5eV$ in above equation we will get
$f_{0}=\frac{3.5eV}{4.14375\times10^{-15}eV.s}=0.8446455\times10^{15}/s=8.446455\times10^{14}Hz$
