Consider a thin target (m thickness) of sodium, which produces a photocurrent of 100 $\mathrm{\mu }$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.

Given,

The volume of sodium target =A$×d$

We know that $6×{10}^{26}$ atoms of Na weigh = 23 kg

So, volume $6×{10}^{26}$ Na atoms =

The volume occupied by one Na-atom

Number of Na-atoms in the target $\left({n}_{Na}\right)$$=\frac{{10}^{-7}}{3.95×{10}^{-26}}=2.53×{10}^{18}$

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/$\lambda$

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

$N=P×n×\left({n}_{Na}\right)\phantom{\rule{0ex}{0ex}}=P×\left(3.3×{10}^{16}\right)×2.53×{10}^{18}\right)$

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.