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Consider a thin target (10^(-2)m square,10^(-3)m thickness) of sodium, which produces a photocurrent of 100 muA when a light of intensity 100W//m^(2)(lambda=660nm) falls on it. Find the probability that a photoelectron is produce when a photon strikes a sodium atom. [Take density of Na=0.97kg//m^(3), Avogadro's number =6xx10^(26)"atom"]. |
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Answer» Solution :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. |
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