A silver ball of radius 4.8 cm is suspended by a thread within a...

The energy of a single incident photon is given by:

$E = \dfrac{hc}{\lambda} = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{200 \times 10^{-9}} = 9.945 \times 10^{-19} \text{ J}$

The total number of photons striking the surface is:

$n = \dfrac{E_{\text{total}}}{E} = \dfrac{1.0 \times 10^{-7}}{9.945 \times 10^{-19}} \approx 1.005 \times 10^{11}$

Given that one out of every $10^4$ photons ejects a photoelectron, the total number of ejected electrons is:

$n_e = n \times 10^{-4} = 1.005 \times 10^{11} \times 10^{-4} = 1.005 \times 10^7$

The total positive charge developed on the silver ball due to the emission of electrons is:

$q = n_e \times e = 1.005 \times 10^7 \times 1.6 \times 10^{-19} \approx 1.608 \times 10^{-12} \text{ C}$

The electric potential at the surface of the spherical ball is:

$V_{\text{surface}} = \dfrac{1}{4\pi\epsilon_0} \dfrac{q}{R} = \dfrac{9 \times 10^9 \times 1.608 \times 10^{-12}}{0.048} \approx 0.3 \text{ V}$

Since silver is a conducting material, the electric field inside the ball is zero, which means the electric potential is constant throughout its volume. Therefore, the potential at the centre of the ball is equal to the potential at its surface:

$V_{\text{centre}} = V_{\text{surface}} = 0.3 \text{ V}$

Answer: $0.3 \text{ V}$ at the surface, $0.3 \text{ V}$ at the centre