Radiation of wavelength 300   nm and intensity 100   W   m - 2 falls on the surface of a…

Power of radiation is expressed as, $\text{Power}=\frac{N\mathrm{hc}}{\lambda }$

Here, $N$is the number of photons per unit time per unit area.

$$\begin{gathered} \Rightarrow P=\frac{N\mathrm{hc}}{\lambda } \\ \Rightarrow N=\frac{P\lambda }{\mathrm{hc}}=\frac{100 \mathrm{W} {\mathrm{m}}^{-2}\times 300\times {10}^{-9} \mathrm{m}}{6.6\times {10}^{-34} \mathrm{J} \mathrm{s}\times 3\times {10}^8 \mathrm{m} {\mathrm{s}}^{-1}}=1.51\times {10}^{20} {\mathrm{m}}^{-2} {\mathrm{s}}^{-1} \end{gathered}$$

Given that $2\%$of the incident, photons produce photoelectron therefore, number of photoelectrons produced by photons is $=\frac{2}{100}\times 1.51\times {10}^{20} {\mathrm{m}}^{-2} {\mathrm{s}}^{-1}$

Now, the number of photoelectrons emitted from an area of $2{ \mathrm{cm}}^2$ of the surface in $1 \mathrm{s}$is nearly 

$$\begin{gathered} n=\left(1 \mathrm{s}\right)\left(2\times {10}^{-4} {\mathrm{m}}^2\right)\times \left(3.02\times {10}^{18} {\mathrm{m}}^{-2} {\mathrm{s}}^{-1}\right) \\ n=6.04\times {10}^{14} \end{gathered}$$