Given that
Light intensity I = 10^{-5} W m^{-2}
Surface area of sodium photo-cell A = 2 cm^2 = 2 * 10^{-4} m^2
Number of layers n = 5
Work function of the metal \phi = 2 eV
Step 1: Finding the number of conduction electrons
Effective atomic area of a sodium atom A_e = 10^{-20} m^2
Number of conduction electrons n_e = n * \frac{A}{A_e}
n_e = 5 * \frac{2 * 10^{-4}}{10^{-20}}
n_e = 10^{17}
Step 2: The amount of energy consumed per second of the electron
Power of the incident light P = I * A
P = 10^{-5} * 2 * 10^{-4}
P = 2* 10^{-9} W
Relation between power and energy of photon
P = nE
Number of photons per second E=\frac{P}{n_e}
E = \frac{2* 10^{-9}}{10^{17}}
E = 2* 10^{-26} J/s
Step 3: Time needed for photoelectric emissions
Time t = \frac{\phi}{E}
t = \frac{2 ev}{2* 10^{-26}} = \frac{2 * 1.6 * 10^{-19}}{2* 10^{-26}}
t =1.67* 10^{7} s
t = 0.507 years
The time needed for photoelectric emission is almost half a year, which is not realistic. Thus, the description of the wave is in conflict with the experiment.