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?

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

Surface area of a 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, φ_O = 2 eV = 2 x 1.6 x 10^-19 = 3.2 x 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 m².

Hence, the number of conduction electrons in n layers is given as:

n’ = n x A/Ae = 5 x (2 x 10^-4) /(10^-20) = 10¹⁷

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

E = P/n’

= 5 x (2 x 10^-9)/(10¹⁷) = 2 x 10⁻²⁶ J/s

Time required for photoelectric emission:

t = φ_O/E = (3.2 x 10^–19) /( 2 x 10⁻²⁶)

=1.6 x 10⁷ s ≈ 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.