Light of intensity ${10}^{-5}$Wm^–2 falls on sodium photo cell of surface area 2cm² and work function 2eV. Assuming that, only top 5 layers of sodium absorbs the incident energy and effective atomic area of the incident energy and effective atomic area of sodium is ${10}^{-20}$ m², the time required for photo emission in wave picture of light is nearly

$$\begin{gathered} Number of the electrons in 5 layers is given as: \\ \frac{5\times \text{ area of one layer }}{\text{ Total atomic area }} \\ =\frac{5\times 2\times {10}^{-4}{\mathrm{m}}^2}{{10}^{-20}{\mathrm{m}}^2}={10}^{17} \\ Incident power is given by: P=I\times A \\ {10}^{-5}\times 2\times {10}^{-4}{\mathrm{m}}^2=2\times {10}^{-9} \\ In the wave picture, incident power is uniformly absorbed by all the electrons continuously. Consequently, energy absorbed per second per electron: \\ \begin{array}{r}\mathrm{E}=\frac{\text{ Incident Power }}{\text{ No. of electrons }}\\ =\frac{2\times {10}^{-9}}{{10}^{17}}=2\times {10}^{-26}\mathrm{W}\end{array} \\ Time required for photoelectric emission: \frac{2\times 1.6\times {10}^{-19}}{2\times {10}^{-26}}=1.6\times {10}^7\mathrm{s} \end{gathered}$$