A silver ball of radius 4.8 cm is suspended by a thread in a vacuum chamber. Ultraviolet light of wavelength 200 nm is incident on the ball for some time during which a total light energy of 1.0 x 10^-7 J falls on the surface. Assuming that on the average, one photon out of ten thousand photons is able to eject a photoelectron, find the electric potential (in x 10^-1 V) at the surface of the ball assuming zero potential at infinity.

Given $\mathrm{\lambda }=200\mathrm{nm}=2\times {10}^{-7}\mathrm{m}$

Energy of one photon is

         $\frac{\mathrm{hc}}{\mathrm{\lambda }}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{2\times {10}^{-7}}=9.945\times {10}^{-19}$

Number of photons incident on surface,  $n_p=\frac{1\times {10}^{-7}}{9.945\times {10}^{-19}}=1\times {10}^{11}$

Hence, number of photoelectrons emitted is   $n_e=\frac{1\times {10}^{11}}{{10}^4}=1\times {10}^7$

Net amount of +ve charge developed due to the outgoing electrons,  $q=ne=1\times {10}^7\times 1.6\times {10}^{-19}=1.6\times {10}^{-12} \mathrm{C}.$

Now potential developed on the silver ball due to these charges is 

$V_{\text{photoelectric}}=\frac{\mathrm{Kq}}{\mathrm{r}}=\frac{9\times {10}^9\times 1.6\times {10}^{-12}}{4.8\times {10}^{-2}}=3\times {10}^{-1}\mathrm{V}$