Intensity of incident light, I = 10^-5 W m^-2
Surface area ofa sodium photocell, A = 2 cm² = 2 x 10^-4 m² and incident power of the light, P = I x A = 10^-5 x 2 x 10^-4 = 2 x 10^-9 W
Work function of the metal, \(\phi_{0}\) = 2 eV = 2 x 1.6 x 10^-19 = 3.2 x 10^-19 J

Step 2: Find the number of conduction electrons available.
Number of layers of sodium that absorbs the incident energy, n = 5
We know that the effective atomic area of a sodium atom, A_e is 10^-20 m²
Hence, the number of conduction electrons in n layers is given as: 

$\begin{array}{rl}{\mathrm{n}}^{\mathrm{\prime }}& =\mathrm{n}\times \frac{\mathrm{A}}{{\mathrm{A}}_{\mathrm{c}}}\\ & =5\times \frac{2\times {10}^{-4}}{{10}^{-20}}={10}^{17}\end{array}$

Step 3: Find the amount of energy absorbed per second per electron.
The incident power is uniformly absorbed by all the electrons continuously. Hence, the amount of energy absorbed per second per electron is:

$\begin{array}{rl}\mathrm{E}& =\frac{\mathrm{P}}{{\mathrm{n}}^{\mathrm{\prime }}}\\ & =\frac{2\times {10}^{-9}}{{10}^{17}}=2\times {10}^{-26}\mathrm{J}/\mathrm{s}\end{array}$

Step 4: Find the time required for photoelectric emission.
The time required for photoelectric emission:

$\begin{array}{rl}\mathrm{t}& =\frac{{\mathrm{\varphi }}_0}{\mathrm{E}}\\ & =\frac{3.2\times {10}^{-19}}{2\times {10}^{-26}}=1.6\times {10}^7\mathrm{s}\approx 0.507\mathrm{years}\end{array}$

The time required for the photoelectric emission is nearly half a year, which is not practical. Hence, the wave picture is in disagreement with the given experiment.