Question

In the photoelectric experiment, if we use a monochromatic light, the photoelectric current vs voltage curve is as shown. If work function of the metal is $2eV$, the power of light used in Watt is . (Assume number of photoelectrons emitted are ${10}^{-3}\%$ of number of photons incident on metal)

Answer 1

The energy of incident photons is given by

$hv=e{V}_s+{\phi }_0=(5+2)eV=7eV$

($V_s$ is stopping potiential and ${\phi }_0$ is work function)

Saturation Current $i={10}^{-5}=\frac{n_{e}e}{t}$
where $n_e$ is number of electrons
Given
$\frac{n_e}{n_p}\times 100={10}^{-3}$

$\Rightarrow \frac{n_p}{t}=\frac{n_e}{t}\times {10}^5$
Substituting $\Rightarrow \frac{n_p}{t}=\frac{1}{e}$

Power $=\frac{n_p}{t}\times Energyofonephoton$
$\Rightarrow P=\frac{1}{e}\times 7eW=7W$

$\therefore$ $P=7W$