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

Explanation

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`