InterviewSolution
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InterviewSolution
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The work function of a metal is 3.0 V. It is illuminated by a light of wave length 3xx10^(-7)m.Calculate i) threshold frequency, ii) the maximum energy of photoelectrons, iii) the stopping potential. (h=6.63xx10^(-34)Js and c=3xx10^(8)ms^(-)). |
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Answer» Solution :`W=hv_(0)=3.0eV=3xx1.6xx10^(-19)J` THRESHOLD frequency `v_(0)=(W)/(H)=(3xx1.6xx10^(-19))/(6.63xx10^(-34))=0.72xx10^(15)Hz` (ii) Maximum kinetic energy `(E_(max))=h(v-v_(0))` `lamda-3XX10^(-7)m,v=(c)/(lamda)=(3xx10^(8))/(3xx10^(-7))=1xx10^(15)Hz` `K_(max)=h(v-v_(0))` `=6.63xx10^(-34)(1-0.72)xx10^(15)J=1.86xx10^(-19)J`. (III) `K_(max)=eV_(0)` where `V_(0)` is stopping potential in volt and e is the charge of electron `V_(0)=(K_(max))/(e)`. Here `K_(max)=1.86xx10^(-19)J` and `e=1.6xx10^(-19)C" "V_(0)=(1.86xx10^(-19)J)/(1.6xx10^(-19)C)=1.16V` |
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