Question

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 light energy of 1.0 × 10⁻⁷ J falls on the surface. Assuming that on average, one photon out of every ten thousand is able to eject a photoelectron, find the electric potential at the surface of the ball, assuming zero potential at infinity. What is the potential at the centre of the ball?

Answer 1

Given:
Radius of the silver ball, r = 4.8 cm
Wavelength of the ultra violet light, λ = 200 nm = 2 × 10⁻⁷ m
Total energy of light, E = 1.0 × 10⁻⁷ J
We are given that one photon out of every ten thousand is able to eject a photoelectron.
Energy of one photon,
$E\text{'}=\frac{hc}{\lambda }$ ,
where h = Planck's constant
c = speed of light
$\lambda$ = wavelength of light
On substituting the respective values in the above formula, we get:
$$\begin{gathered} E\text{'}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{2\times {10}^{-7}} \\ =9.945\times {10}^{-19} \end{gathered}$$
Number of photons,
$n=\frac{E}{E\text{'}}=\frac{1\times {10}^{-7}}{9.945\times {10}^{-19}}=1\times {10}^{11}$

Number of photoelectrons
$=\frac{1\times {10}^{11}}{{10}^4}=1\times {10}^7$
The amount of positive charge developed due to the outgoing electrons,
$$\begin{gathered} q=1\times {10}^7\times 1.6\times {10}^{-19} \\ =1.6\times {10}^{-12}\mathrm{C} \end{gathered}$$

Potential developed at the centre as well as on surface,
$V=\frac{Kq}{r}$,
where K = $\frac{1}{4{\mathrm{\pi \epsilon }}_0}$
$\therefore V=\frac{9\times {10}^9\times 1.6\times {10}^{-12}}{4.8\times {10}^{-2}}=0.3\mathrm{V}$

Potential inside the silver ball remains constant. Therefore, potential at the centre of the sphere is 0.3 V.