InterviewSolution
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InterviewSolution
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If the wavelength of light incident on a photoelectric cell be reduced from `4000Å` to `3600Å`, then what will be the change in the cut off potential? `(h=6.6xx10^(-34)Js , e=1.6xx10^(-19)C)` |
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Answer» Let `phi_0` be the work function of the surface of the photoelectric cell to be exposed by light. Maximum KE of emitted photoelectron will be `K_(max)=(hc)/lambda-phi_0` If the `V_0` is the cut off potential, then `eV_0=K_(max)=(hc)/lambda-phi_0` or `V_0=(hc)/(e lambda)-(phi_0)/e` When the incident light of wavelength `lambda_1` be changed to wavelength `lambda_(2)`, then the change (increase) in cut off potential will be `DeltaV_0=((hc)/(elambda_2)-(phi_0)/e)-((hc)/(elambda_1)-(phi_0)/e)` `=(hc)/e (1/(lambda_1)-1/(lambda_1))` `=((6.6xx10^(-34))xx(3xx10^8))/((1.6xx10^(-19)))[(10^(10))/3600-(10^(10))/4000]` `=0.34V` |
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