Light of intensity
Question:

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 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?

Solution:

Intensity of incident light, I = 10−5 W m−2

Surface area of a sodium photocell, A = 2 cm2 = 2 × 10−4 m2

Incident power of the light, P = I × A

= 10−5 × 2 × 10−4

= 2 × 10−9 W

Work function of the metal, $\phi_{0}=2 \mathrm{eV}$

= 2 × 1.6 × 10−19

= 3.2 × 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, Ae is 10−20 m2.

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:

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

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

Time required for photoelectric emission:

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

$=\frac{3.2 \times 10^{-19}}{2 \times 10^{-26}}=1.6 \times 10^{7} \mathrm{~s} \approx 0.507$ 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.