Answer
Work function of the surface is
$\phi_{0}=3.945\times10^{-19}J$
Work Step by Step
given maximum kinetic energy of photoelectrons is $K_{max}=6.0\times10^{-19}J$
wavelength of incident light is $\lambda=200nm=200\times10^{-9}m=2\times10^{-7}m$
$E=hf$, since $f=\frac{c}{\lambda}$
Energy of a Quanta of light is $E=hf=\frac{hc}{\lambda}$
$h=6.63\times10^{-34} J.s$, $c=3\times10^{8}m/s$, $\lambda=2\times10^{-7}m$
putting these values we will get
$E=hf=\frac{6.63\times10^{-34} J.s\times3\times10^{8}m/s}{2\times10^{-7}m}$
$E=hf=9.945\times10^{-19}J$
From equation 27.7
$hf= K_{max} + \phi_{0}$
where $hf$ is energy of incident photon, $K_{max}$ is maximum kinetic energy of photoelectron, $\phi_{0}$ is work function of the surface.
from above equation
$\phi_{0} =hf-K_{max}$
putting values of $K_{max}$ and $ hf$ in this equation we will get
$\phi_{0}=9.945\times10^{-19}J-6.0\times10^{-19}J$
$\phi_{0}=3.945\times10^{-19}J$
