Light of intensity $10^{-5} \; W m^{-2}$ falls on a sodium photo-cell of surface area $2 \; cm^2$. Assuming that the top $5$ layers of sodium absorb the incident energy,estimate the time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about $2 \; eV$. What is the implication of your answer?

(N/A) Intensity of incident light,$I = 10^{-5} \; W m^{-2}$.
Surface area of a sodium photocell,$A = 2 \; cm^2 = 2 \times 10^{-4} \; m^2$.
Incident power of the light,$P = I \times A = 10^{-5} \times 2 \times 10^{-4} = 2 \times 10^{-9} \; W$.
Work function of the metal,$\phi_0 = 2 \; eV = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} \; J$.
Number of layers of sodium that absorb the incident energy,$n = 5$.
The effective atomic area of a sodium atom,$A_e$,is approximately $10^{-20} \; m^2$.
The number of conduction electrons in $n$ layers is $n' = n \times \frac{A}{A_e} = 5 \times \frac{2 \times 10^{-4}}{10^{-20}} = 10^{17}$.
The incident power is uniformly absorbed by all the electrons continuously. The energy absorbed per second per electron is $E = \frac{P}{n'} = \frac{2 \times 10^{-9}}{10^{17}} = 2 \times 10^{-26} \; J/s$.
Time required for photoelectric emission,$t = \frac{\phi_0}{E} = \frac{3.2 \times 10^{-19}}{2 \times 10^{-26}} = 1.6 \times 10^7 \; s \approx 0.507 \; \text{years}$.
The time required for photoelectric emission is nearly half a year,which is not practical. Hence,the wave picture is in disagreement with the experimental observation of instantaneous emission.