Ultra violet light of wavelength 66.26 nm and intensity $2 \mathrm{w}/{\mathrm{m}}^2$ falls on potassium surface by which photoelectrons are ejected out. If only 0.1% of the incident photons produce photoelectrons and surface area of metal surface is $4m^2$ , how many electrons are emitted per second ?

Number of photons falling on metal surface

$\mathrm{Energy} \mathrm{per} \mathrm{quanta}=\frac{\mathrm{hc}}{\mathrm{\lambda }}=\frac{12400}{66.26\times 10}\mathrm{eV}=1.87\times {10}^1\times 1.6\times {10}^{-19}\mathrm{Joule}$

$$\begin{gathered} =3\times {10}^{-18}\mathrm{Joule} \\ {\mathrm{n}}_{\mathrm{p}}=\frac{\mathrm{Intensity}\times \mathrm{Area}}{\mathrm{Energy}\mathrm{per}\mathrm{quanta}}=\frac{2\times 4}{3\times {10}^{-18}}=2.67\times {10}^{18}/\mathrm{s} \end{gathered}$$

$$\begin{gathered} \mathrm{From} \mathrm{the} \mathrm{question}, \\ \Rightarrow {\mathrm{n}}_{\mathrm{e}}=0.1\mathrm{\%}\mathrm{of}{\mathrm{n}}_{\mathrm{p}}=\frac{0.1}{100}\times 2.67\times {10}^{18}/\mathrm{s}=2.67\times {10}^{15}/\mathrm{s}. \end{gathered}$$