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)]`

Options

A. `0.75`
B. `7.5xx10^(-2)`
C. `7.5xx10^(-13)`
D. `7.5xx10^(-21)`

Answer

D

Explanation

Given `d=10^(-3)m,I=100xx10^(-6)A=10^(-4) A`
`A=(10^(-2)m)^(2)=10^(-4)m^(2)`
Intensity,`I=100 Wm^(-1)`
`lambda=660nm=660xx10^(-9)m`
Volume of `6.02xx10^(26)` sodium atoms`=(23kg)/(0.97kg//m^(3))=23.7 m^(3)`
Volume of target=`(10^(-2))(10^(-2))(10^(-3))=10^(-7) m^(3)`
Number of sodium atoms in the target =`(6.02xx10^(26))/23.7xx10^(-7)=2.5410^(18)`
Let `n` be the number of photons falling per second on the target.
Each of each photon`=(hc)/lambda`
Total energy falling per second on target`=(nhc)/lambda=IA`
`thereforen=(IAlambda)/(hc)=(100xx10^(-4)xx660xx10^(-9))/(6.62xx10^(-34)xx(3xx10^(8)))=3.3xx10^(16)`
Number of electrons emitted per second by all the atoms in the target if one electron is emitted by each atom for one incident photon.
`=(2.54xx10^(18))(3.3xx10^(16))=8.4xx10^(34)`
Expected photocurrent =`(8.4xx10^(34))(1.6xx10^(-19))=1.34xx10^(16) A`
Observed photocurrrent=`100muA`
Probability of photo emission by single photon incident on a single atom `P=(100muA)/(1.34xx10^(16) A)=7.5xx10^(-21)`
Thus the probability of emission by single photon on a single atom is very much less than `1`, the probability of absorption of two photons by single atoms is negligible.