Consider a thin target (`10^(-2)` m square ,`10^(-3)` m thickness) of sodium ,which produces a photocurrent of 100 `mu`A when a light of intensity `100W//m^(2)` (`lambda`=660nm) falls on it.Find the probability that a photoelectron is produced when a photon strikes a sodium atom. [Take density of Na=0.97 `kg//m^(3)`].

Explanation

If total no. of sodium atoms in N. in a given layer of sodium then no.of moles of this number is
`mu=(N.)/(N_(A))=(M)/(M_(0))`
Here for sodium molar mass of mass is `m_(0)`=23g/mol
`therefore N.(MN_(0))/(M_(0))` (Where V=volume ,`rho`=density) `=(10^(-4)xx10^(-3)xx0.97xx6.02xx10^(23))/(23xx10^(10^(-3)))`
`=0.254xx10^(-7+23+3)`
`N=2.54xx10^(18)` atoms ........(1)
Now ,intensity of incident radiation,
`I=(E_(n))/(At)=(nhf)/(At)=(nhc)/(Atlambda)`
`implies` No.of photons made incident on given layer per unit time.
`((n)/(t))=(IAlambda)/(hc)`
`=((100)(10^(-4))(660xx10^(-9)))/((6.625xx10^(-34))xx(3xx10^(8))`
`=3.321xx10^(16)` photon /second ........(2)
If no. of photoelectrons emitted from given layer in unit time is N then probability (P) of emission of photoelectron from this layer can be found out by its definition given below.
`P=(N)/(nxxN.)=("total no. of photoelectrons emitted")/(("total no. of incident photons")("Total no.of sodium atoms"))`
`therefore P=(N//t)/((n//t)xxN.)`
`therefore (N)/(t)=6.25xx10^(14)` Photoelectron/second.......(4)
From equation (1),(2),(3),(4),
`P=(6.25xx10^(14))/(3.321xx10^(16)xx2.54xx10^(18))`
`therefore P=7.41xx10^(-21)`
Thus ,above probabnility is nearly zero .It means that if one photon interacts with only one atom then we would get photoelectric current almost zero.But here we get it 100 `muA`. This means that each incident photon must be interacting with each electron inside the atom.Afterwards,whichever electron absorbs energy from incident photon ,more than its binding energy will get emitted .In this later case,we obtain quite high probability of emission of electrons which is quite consistent with experiment observation.