Answer
work function of the metal is
$\phi_{0}=4.5175eV$
Work Step by Step
wavelength of the incident $\lambda= 250nm=250\times10^{-9}m=2.5\times10^{-7}m$
$E=hf=\frac{hc}{\lambda}$
putting $h=6.63\times10^{-34} J.s$, $c=3\times10^{8}m/s$,$\lambda=2.5\times10^{-7}m$
$E=hf=\frac{6.63\times10^{-34} J.s\times3\times10^{8}m/s}{2.5\times10^{-7}m}=7.956\times10^{-19} J$ .........equation(1)
Maximum kinetic energy of photoelectron is given by $K_{max}=\frac{1}{2}mv^{2}$
given that $v=4.0\times10^{5}m/s$, mass of electron $m=9.1\times10^{-31}kg$
$K_{max}=\frac{1}{2}\times9.1\times10^{-31}kg \times(4.0\times10^{5}m/s)^{2}$
$K_{max}=0.728\times10^{-19}J$ .......equation(2)
From equation 27.7
$hf= K_{max} + \phi_{0}$
$\phi_{0}=hf- K_{max}$
putting the values from equation (1) and (2)
$\phi_{0}=7.956\times10^{-19} J-0.728\times10^{-19}J$
$\phi_{0}=7.228\times10^{-19} J$
$1.6\times10^{-19} J$ is equal to $1eV$
$1J$ is equal to $\frac{1}{1.6\times10^{-19}}eV$
so $7.228\times10^{-19} J$ is equal to $\frac{7.228\times10^{-19}}{1.6\times10^{-19}}eV$= $4.5175eV$
$\phi_{0}=4.5175eV$
