Class 12 Physics
CBSE and UP Board
Dual Nature of Radiation and Matter
Chapter-11 Exercise 11.30
NCERT Solutions for Class 12 Physics Chapter 11 Question-30
Additional Exercise
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?
Share
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, φO = 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, Ae 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 = φO/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.