Answer
Maximum kinetic energy of the photoelectrons ejected from the surface is
$K_{max}=1.63\times10^{-19}J$
Work Step by Step
given that work function of the surface $\phi_{0}$ =$5.0\times10^{-19}J$
wavelength of incident light is $\lambda=300nm=300\times10^{-9}m=3\times10^{-7}m$
$E=hf$, since $f=\frac{c}{\lambda}$
Energy of a Quanta of light is $E=hf=\frac{hc}{\lambda}$
$h=6.63\times10^{-34} J.s$, $c=3\times10^{8}m/s$, $\lambda=3\times10^{-7}m$
putting these values we will get
$E=hf=\frac{6.63\times10^{-34} J.s\times3\times10^{8}m/s}{3\times10^{-7}m}$
$E=hf=6.63\times10^{-19}J$
From equation 27.7
$hf= K_{max} + \phi_{0}$
where $hf$ is energy of incident photon, $K_{max}$ is maximium kinetic energy of photelectron, $\phi_{0}$ is work function of the surface.
$K_{max} =hf-\phi_{0}$
putting values of $hf$ and $\phi_{0}$ in this equation we will get
$K_{max}=6.63\times10^{-19}J-5.0\times10^{-19}J$
$K_{max}=1.63\times10^{-19}J$