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?
Given:
Intensity of light, I = 10-5 Wm-2
Surface area of photo-cell, A = 2 cm2 = 0.0002 m2
Work function of sodium, Φ0 = 2eV = 2 × 1.6 × 10-19 = 3.2 × 10-19J
By knowing the effective area of each sodium atom we can find the absorption of incident energy,
Effective area of each sodium atom, ANa = 10-20 m2
Number of layers of sodium, n = 5
Number of atoms absorbing radiation, NNa = n × (ANa/ A)
→ NNa = 5 × (10-20 m2/ 0.0002 m2)
→ NNa = 1017
So we conclude that 1017 atoms are effectively absorbing radiation.
Energy absorbed per atom, E = I/NNa
E = 10-5Js-1/1017
→ E = 2 × 10-26 Js-1
Times required for photo electric emission, t is give by,
t = Φ0/E
t = 3.2 × 10-19 J / 2 × 10-26 Js-1
t = 1.6 × 107 s
t = 0.5 years
Hence the time requires to initiate photoelectric emission is 0.5 years which is impractical hence, the wave model stands in disagreement with the experimental results.