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Photons of energies 4.25eV and 4.7eV are incident on two metal surfaces A and B respectively. The maximum KE of emitted electrons are respectively T_(A)eV and T_(B)=(T_(A)-1.5)eV. The ratio de-Broglie wavelengths of photo electrons from them is lamda_(A):lamda_(B)=1:2, then find the work function of A and B |
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Answer» Solution :DE BROGLIE wavelength `LAMDA=(h)/(sqrt(2km))implieslamdaprop(1)/(sqrt(k))""(k=k.E=T)` `(lamda_(B))/(lamda_(A))=sqrt((T_(A))/(T_(B)))` `2=sqrt(T_(A)/(T_(A)-1.5))impliesT_(A)=2eV` `impliesW_(A)=4.25-T_(A)=2.25eV` `impliesT_(B)=T_(A)-1.5=2-1.5=0.5eV` `impliesW_(B)=4.7-T_(B)=4.7-0.5=4.2eV` |
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