Consider a thn target (`10^(-2)` m square, `10^(-3)` m thinkness) of sodium, which produces a photocurrent of 100 `muA` when a light of intensity 100 `W//m^(2)(lamda=660 nm)` falls on it. Find the probability that a photoenectron is produced when a photon strikes a sodium atom.
`" "["Taken density of Na"=0.97kg//m^(3)]`

Explanation

Here, `A=10^(-2)m sq=10^(-2)xx10^(-2)sq m=10^(-4)m^(2) , d=10^(-3)m, i=100xx10^(-6)A=10^(-4)A`
Intensity, `I=100W//m^(2), lambda=600nm=600xx10^(-9)m`
Probability=? , `rho_(Na)=0.97xxkg//m^(3)`, Avogadro's number `=6xx10^(26)kg` atom
Volume of sodium target `=Axxd=10^(-4)xx10^(-3)=10^(-7)m^(3)`
We know that `6xx10^(26)` atoms of Na weighs=23g `:.` Volume of `6xx10^(26)`Na atoms `=23/0.97m^(3)`
Volume occupied by one Na atom `=23/(0.97xx(6xx10^(26)))=3.95xx10^(-26)m^(3)`
No. of Na atoms in the target `(n_(Na))=(10^(-7))/(3.95xx10^(-26))=2.53xx10^(18)`
Let n be the number of photons falling per second on the target.
Energy of each photon `=hc//lambda`
Total energy falling per sec on target =`(nhc)/lambda=IA :. n=(IAlambda)/(hc)=(100xx10^(-4)xx(600xx10^(-9)))/((6.62xx10^(-34))xx(3xx10^(8)))=3.3xx10^(16)`
Let P be the probability of emission per atom per photon.
The number of photoelectrons emitted per second `N=Pxxnxx(n_(Na))=Pxx(3.3xx10^(16))xx(2.53xx10^(18))`
As per equation, `i=100 muA=100xx10^(-6)=10^(-4)A`,
Current, `i=Ne :. 10^(-4)=Pxx(3.3xx10^(16))xx(2.53xx10^(18))xx(1.6xx10^(-19))`
or `P=(10^(-4))/((3.3xx10^(16))xx(2.53xx10^(18))xx(1.6xx10^(-19)))=7.48xx10^(-21)`
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.