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 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?

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 absorbs the incident energy, $n=5$

We know that the effective atomic area of a sodium atom,$A_e$ is $10^{-20} m ^{2}$. Hence, the number of conduction electrons in n layers is given as:

$n^{\prime}= n \times \frac{A}{A_{e}}$

$=5 \times \frac{2 \times 10^{-4}}{10^{-20}}=10^{17}$

The incident power is uniformly absorbed by all the electrons continuously. Hence, the amount

of energy absorbed per second per electron is:

$E=\frac{P}{n^{\prime}}$

$=\frac{2 \times 10^{-9}}{10^{17}}=2 \times 10^{-26} J / s$

Time required for photoelectric emission:

$t=\frac{\phi_{0}}{E}$

$=\frac{3.2 \times 10^{-19}}{2 \times 10^{-26}}=1.6 \times 10^{7} s=0.507 \text { years }$

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.