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

Given, `" "A=10^(-2)m^(2)xx10^(-2)m^(2)`
`implies" "=10^(-4)m^(2)`
`" "d=10^(-3)m`
`" "i=100xx10^(-6)A=10^(-4)A`
Intensity, `" "I=100W//m^(2)`

`implies" "lamda=660mn=660xx10^(-9)m`
`implies" "p_(Na)=0.97kg//m^(3)`
Avogadro's number `=6xx10^(26)` kg atom
Volume of sodium target `=Axxd`
`" "=10^(-4)xx10^(-3)`
`implies" "=10^(-7)m^(3)`
We know that `6xx10^(26)` atoms of Na weights = 23 kg
So, volume of `6xx10^(6)` Na atoms `=23/(0.97)m^(3)`
Volume occupied by one Na atom `=(23)/(0.97xx(6xx10^(26)))=3.95xx10^(-26)m^(3)`
Number of Na atoms in target `("^(n)Na)`
`" "=(10^(-7))/(3.95xx10^(-26))=2.53xx10^(18)`
Let n be the number of photons faling per second on the target. Energy of each photon `= hc//lamda`
Total energy falling per second on targer `=(nhc)/(lamda)=IA`
`therefore" "n=(IAlamda)/(hc)`
`implies" "=(100xx10^(-4)xx(600xx10^(-9)))/((6.62xx10^(-34))xx(3xx10^(8)))=3.3xx10^(16)`
Let P be the probability of emissio per atom per photon. The number of photoelectrons emitted per second
`" "N=Pxxnxx("^(n)Na)`
`" "=Pxx(3.3xx10^(16))xx(2.53xx10^(18))`
Now, according to question,
`" "i=100muA=100xx10^(-6)=10^(-4)A.`
Current, `" "i=Ne`
`therefore" "10^(-4)=Pxx(3.3xx10^(16))xx(2.53xx10^(18))xx(1.6xx10^(-19))`
`implies" "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 his reason, the absorption of two photons by an atom is negligible.