Wave picture of radiation state that incident energy is uniformly distributed among all the electrons continuously. Let us first calculate the total number of recipient electrons in 5 layers of sodium.
Consider each sodium atom has one electron free as conduction electron.
Effective atomic area $=10^{-20} \mathrm{~m}^2$
Number of conduction electrons in 5 layers
$$
n=\frac{5 \times \text { area of each layer }}{\text { effective area of each atom }}
$$
Wave picture of radiation state that incident energy is uniformly distributed among all the electrons continuously. Let us first calculate the total number of recipient electrons in 5 layers of sodium.
Consider each sodium atom has one electron free as conduction electron.
Effective atomic area $=10^{-20} \mathrm{~m}^2$
Number of conduction electrons in 5 layers
$$
n=\frac{5 \times \text { area of each layer }}{\text { effective area of each atom }}
$$
or $n=\frac{5 \times 2 \times 10^{-4}}{10^{-20}}=10^{17}$
Total energy incident for unit time.
Incident power $=$ incident intensity $\times$ area
$$
=10^{-5} \times 2 \times 10^{-4}=2 \times 10^{-9} \mathrm{~W}
$$
As incident energy is equally distributed among all conduction electrons.
Energy to each conduction electron per second
$$
=\frac{2 \times 10^{-9}}{10^{17}}=2 \times 10^{-26} \mathrm{~W}
$$
Time required for emission by each electron $t=\frac{\text { total work function energy }}{\text { energy received per second }}$
$$
\begin{aligned}
t &=\frac{2 \mathrm{eV}}{2 \times 10^{-26}}=\frac{2 \times 1.6 \times 10^{-19}}{2 \times 10^{-26}} \\
&=1.6 \times 10^7=0.5 \text { year }
\end{aligned}
$$
where experimental observation shows that emission of photoelectrons is instantaneous $=10^{-9} \mathrm{sec}$
Thus wave picture fails to explain photoelectric effect.