Photoelectrons are emitted when photons of energy $4.2 \mathrm{eV}$ are incident on a photosensitive…
Photoelectrons are emitted when photons of energy $4.2 \mathrm{eV}$ are incident on a photosensitive metallic sphere of radius $10 \mathrm{~cm}$ and work function $2.4 \mathrm{eV}$. The number of photoelectrons emitted before the emission is stopped is
$\left[\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \text { SI unit; } \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\right]$
$1.25 \times 10^6$
$1.25 \times 10^8$
$1.25 \times 10^2$
$1.25 \times 10^4$
Solution
$\mathrm{h} v=4.2 \mathrm{eV}, \phi=2.4 \mathrm{eV},(\mathrm{KE})_{\max }=4.2-2.4=1.8 \mathrm{eV}$
The electrons will not be able to escape from the surface when its potential becomes $1.8 \mathrm{~V}$.
$\begin{aligned}
& \mathrm{V}=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}} \\
& 1.8=9 \times 10^9 \times \frac{\mathrm{q}}{0.1} \\
& \mathrm{q}=2 \times 10^{-11} \mathrm{C}
\end{aligned}$
The number of photoelectrons emitted before the emission is stopped is,
$\mathrm{n}=\frac{\mathrm{q}}{\mathrm{e}}=\frac{2 \times 10^{-11}}{1.6 \times 10^{-19}}=1.25 \times 10^8$
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