Question 14 Marks
A silver ball of radius $4.8\ cm$ is suspended by a thread in a vacuum chamber. Ultraviolet light of wavelength $200$ run is incident on the ball for some time during which a total light energy of $1.0 \times 10^{-7}J$ falls on the surface. Assuming that on the average one photon out of every ten thousand is able to eject a photoelectron, find the electric potential et the surface of the bell assuming zero potential at infinity. What is the potential at the centre of the bell?
Answer
View full question & answer→$\lambda=200\text{nm}=2\times10^{-7}\text{m}E$ of one photon $=\frac{\text{hc}}{\lambda}$
$=\frac{6.63\times10^{-34}\times3\times10\times^{8}}{2\times10^{-7}}=9.945\times10^{-19}$
No. of photons
$=\frac{1\times10^{-7}}{9.945\times10^{-19}}=1\times10^{11}\text{no.s}$
Hence, No.of photo electrons
$=\frac{1\times10^{11}}{10^{4}}=1\times10^{7}$
Net amount of positive charge $'q\ ’$ developed due to the outgoing electrons
$= 1 \times 10^7 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-12} C.$
Now potential developed at the centre as well as at the surface due to these charger
$=\frac{\text{kq}}{\text{r}}$
$=\frac{9\times10\times1.6\times10^{-12}}{4.8\times10^{-2}}$
$=3\times10^{-1}\text{v}$
$=0.3\text{v.}$
$=\frac{6.63\times10^{-34}\times3\times10\times^{8}}{2\times10^{-7}}=9.945\times10^{-19}$
No. of photons
$=\frac{1\times10^{-7}}{9.945\times10^{-19}}=1\times10^{11}\text{no.s}$
Hence, No.of photo electrons
$=\frac{1\times10^{11}}{10^{4}}=1\times10^{7}$
Net amount of positive charge $'q\ ’$ developed due to the outgoing electrons
$= 1 \times 10^7 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-12} C.$
Now potential developed at the centre as well as at the surface due to these charger
$=\frac{\text{kq}}{\text{r}}$
$=\frac{9\times10\times1.6\times10^{-12}}{4.8\times10^{-2}}$
$=3\times10^{-1}\text{v}$
$=0.3\text{v.}$