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In an experiment of photoelectric emission for incident light of 4000Å, the stopping potential is 2V. If the wavelength of incident light is made 3000Å, then stopping potential will be : |
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Answer» less than 2 VOLT `(hc)/(lambda_(2))-(phi_(0))/(e)=V_(0_(2)) ""...(2)` From eqs (2)-(1) `rArr V_(0_(2))-V_(0_(1))=(hc)/(elambda_(2))-(hc)/(elambda_(1))""...(3)` `lambda_(1)=400xx10^(-10)m` `lambda_(2)=3000xx10^(-10)m""...(4)` `V_(0_(1))=2` volt `V_(0_(2))=?` From eqs (3) and (4), `V_(0_(2))2+(6.62xx10^(-34)xx3xx10^(8))/(1*6xx10^(-19)) xx[(1)/(3xx10^(-7))-(1)/(4XX10^(-7))]` `V_(0_(2))=2+1=3` volts. |
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