Consider a thin target (10-2m square, 10-3m thickness) of sodium, which produces a photocurrent of 100 μA when the light of intensity 100 W/m2 (λ=660 nm) falls on it. Find the probability that a photoelectron is produced when a photon strikes a sodium atom.

Hint: The intensity of the radiation gives the energy falling on the metal per second.

Step 1: Find the no. of atoms in the Na-target.

Given,

A=102m2=102×102m2=104m2d=103mi=100×106A=104AIntensity , I=100W/m2λ=660nm=660×109mρNa=0.97 kg/m3

Avogadro's number=6×1026 kg atom

The volume of sodium target =A×d

                   =10-4×10-3=10-7 m3

We know that 6×1026 atoms of Na weigh = 23 kg 

So, volume 6×1026 Na atoms =230.97 m3

The volume occupied by one Na-atom=230.97×(6×1026)=3.95×10-26 m3

Number of Na-atoms in the target (nNa)=10-73.95×10-26=2.53×1018

Step 2: Find the no. of photons falling on the target per sec.

Let n be the number of photons falling per second on the target. 

The energy of each photon=hc/λ

Total energy falling per second on target =nhcλ=lAn=IAλhcn=100×10-4×(660×10-9)(662×10-34×(3×108)=3.3×1016

Step 3: Find the probability of the emission of electrons.

Let P be the probability of emission per atom-photon. 

The number of photoelectrons emitted per second 

                N=P×n×nNa=P×(3.3×1016)×2.53×1018)

Now, according to the question, 

             i=100 μA =100×10-6=10-4  A

Current, i=Ne104=P×(3.3×1016)×(2.53×1018)×(1.6×1019)P=104(3.3×1016)×(2.53×1018)×(1.6×1019)=7.48×1021

Thus, the probability of emission by a single photon on a single atom is very much less than 1. It is due to this reason, the absorption of two photons by an atom is negligible.