Consider a thin target (m thickness) of sodium, which produces a photocurrent of 100 A when the light of intensity 100 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.
The volume of sodium target =A
We know that atoms of Na weigh = 23 kg
So, volume Na atoms =
The volume occupied by one Na-atom
Number of Na-atoms in the target
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/
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
Now, according to the question,
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.