InterviewSolution
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InterviewSolution
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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)]`A. `0.75`B. `7.5xx10^(-2)`C. `7.5xx10^(-13)`D. `7.5xx10^(-21)` |
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Answer» Correct Answer - D 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. |
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