Answer
$1.06$ $eV$
Work Step by Step
For light of frequency $f$ and wavelength $\lambda$, the photon energy is
$E =hf$
or, $E=\frac{ch}{\lambda}$,
in which $c$ is the speed of light in vacuum, and $h$ is the Planck's constant.
In our case, the wavelength of the incident ultraviolet light is $\lambda=254$ $nm$ $=254\times 10^{-9}$ $m$
Therefore, the energy of the incident a photon
$E=\frac{3\times 10^{8}\times 6.63\times 10^{-34}}{254\times 10^{-9}}$ $J$
or, $E\approx 7.83\times 10^{-19}$ $J$
or, $E=\frac{7.83\times 10^{-19}}{1.6\times 10^{-19}}$ $eV$
or, $E\approx 4.89$ $eV$
If $f_{0}$ be cutoff frequency, then the work function ($\phi$) of the silver metal is
$\phi=hf_{0}$
or, $\phi=\frac{ch}{\lambda_{0}}$, where $\lambda_{0}$ the corresponding cutoff wavelength.
$\lambda_{0}=325$ $nm$ $=325\times 10^{-9}$ $m$
$\therefore \phi=\frac{3\times 10^{8}\times 6.63\times 10^{-34}}{325\times 10^{-9}}$ $J$
or, $\phi= 6.12\times 10^{-19}$ $J$
or, $\phi=\frac{6.12\times 10^{-19}}{1.6\times 10^{-19}}$ $eV$
or, $\phi\approx 3.83$ $eV$
Therefore, the maximum kinetic energy of electrons ejected from the silver surface is
$K=E-\phi$
or, $K=4.89-3.83$ $eV$
or, $K=1.06$ $eV$
