The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying a stopping potential of $0.35 \ V$ when radiation of $256.7 \ nm$ is used. Calculate the work function for silver metal.

(N/A) According to the photoelectric effect equation:
$E = W_{0} + K.E_{max}$
where $E$ is the energy of the incident photon,$W_{0}$ is the work function,and $K.E_{max}$ is the maximum kinetic energy of the emitted photoelectron.
The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Using $h = 6.626 \times 10^{-34} \ J \ s$,$c = 3.0 \times 10^{8} \ m \ s^{-1}$,and $\lambda = 256.7 \times 10^{-9} \ m$:
$E = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{256.7 \times 10^{-9}} \ J = 7.744 \times 10^{-19} \ J$.
Converting to electron-volts $(eV)$:
$E = \frac{7.744 \times 10^{-19} \ J}{1.602 \times 10^{-19} \ J/eV} \approx 4.83 \ eV$.
The stopping potential is $0.35 \ V$,so the maximum kinetic energy is $K.E_{max} = 0.35 \ eV$.
Substituting these values into the equation:
$W_{0} = E - K.E_{max} = 4.83 \ eV - 0.35 \ eV = 4.48 \ eV$.
The work function for silver metal is $4.48 \ eV$.