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Two photons of energy 4 eV and 4.5 eV are incident on two metals A and B respectively.The maximum kinetic energy for an ejected electron is T_(A) for A and T_(B)=T_(A)-1.5 eV for the metal B.The relation between the de-Broglie wavelengths of the ejected electron of A and B are lambda_(B)=2lambda_(B) .The work function of the metal B is |
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Answer» 1.5 eV `lambda=(h)/(p)=(h)/(SQRT(2m_(e)K))[because p=sqrt(2m_(e)K)]` `therefore lambda prop (1)/(sqrt(K))` [`because` h,2 and `m_(e)` are same] `(lambda_(A))/(lambda_(B))=sqrt((K_(B))/(K_(A)))=sqrt((T_(B))/(T_(A)))` `therefore (1)/(2)=sqrt((T_(A)-1.5)/(T_(A)))` `therefore (1)/(4)=(T_(A)-1.5)/(T_(A))=1-(1.5)/(T_(A))` (By taking SQUARE) `therefore (1.5)/(T_(A))=1-(1)/(4)=(3)/(4)` `therefore T_(A)=(1.5xx4)/(3)` `therefore T_(A)=2EV` Now `T_(B)=(T_(A)-1.5)eV` `therefore phi_(B)=hf-T_(B)=(4.5-0.5)eV` `therefore phi_(B)=4.0 eV` |
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