Question 11.30: Light of intensity 10-5 W m-2 falls on a sodium photo-cell of surface area 2 cm2. 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?

Hint: t=$$\frac{\phi_o}{E}$$

Step 1: Find the incident power of the light.

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, Ae is 10-20 m2
Hence, the number of conduction electrons in n layers is given as:

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

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:

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

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.