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?
Answers
Let us first calculate the total number of recipient electrons in 5 layers of sodium.
consider each sodium atom has one free electron .
from experiment, effective atomic area of sodium, Ae = 10^-20 m²
so, number of conduction electrons = 5 × area of each layer /effective atomic area of sodium atom
= 5 × (2 × 10^-4 m²)/(10^-20m²)
= 10^17
incident power = incident intensity × area
= 10^-5 W/m² × 2 × 10^-4 m²
= 2 × 10^-9 W
as incident energy is equally distributed among all conduction electrons.
so, energy to each conduction electron per second = 2 × 10^-9/10^17
= 2 × 10^-26 W
time required for emission by each electron, t = work function/energy to each conduction electron per second
= 2 × 1.6 × 10^-19J/(2 × 10^-26 J/s)
= 1.6 × 10^7 sec
= 0.5 years
where experimental observations shows that emission of photoelectrons is instantaneous ≈ 10^-9 sec
Thus, wave theory fails to explain photoelectric effect.