Light of intensity 10^–5  Wm^–2 falls on the sodium photocell of surface area 2 cm² and works function 2 eV. Assuming that, only the top 5 layers of sodium absorb the incident energy and the effective atomic area of sodium is 10^-20  m², the time required for photoemission in wave picture of light is nearly

Number of the electrons in 5 layers is given as:

$\mathrm{n}=\frac{5\times \text{ area of one layer }}{\text{ Total atomic area }}$

$\mathrm{n}=\frac{5\times 2\times {10}^{-4}{\mathrm{m}}^2}{{10}^{-20}{\mathrm{m}}^2}={10}^{17}$

Incident power is given by:

$\mathrm{P}=\mathrm{I}\times \mathrm{A}={10}^{-5}\times 2\times {10}^{-4}{\mathrm{m}}^2=2\times {10}^{-9}$

In the wave picture, incident power is uniformly absorbed by all the electrons continuously. Consequently, energy absorbed per second per electron:

$\mathrm{E}=\frac{\text{ Incident Power }}{\text{ No. of electrons }}$

$\mathrm{E}=\frac{2\times {10}^{-9}}{{10}^{17}}=2\times {10}^{-26}\mathrm{W}$

Time required for photoelectric emission:

$\mathrm{t}=\frac{2\times 1.6\times {10}^{-19}}{2\times {10}^{-26}}=1.6\times {10}^7\mathrm{s}\approx 0.5 \mathrm{yrs}$