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 photoelectrons from them is `lambda_(A):lambda_(B)=1.2`,then find the work function of `A` and `B`

Answer

2

Explanation

Debroglie wavelenght
`lambda=(h)/sqrt(2km)rArrlambdaprop(1)/(sqrt(k))(K=k,E=T),(lambda_(B))/(lambda_(A))=sqrt((T_(A))/(T_(B))`
`2=sqrt((T_(A))/(T_(A))-1.5)rArrT_(A)=2eV`
`rArrW_(A)=4.25-T_(A)=2.25eV`
`rArrT_(B)=T_(A)-1.5=2-1.5=0.5eV`
`rArrW_(B)=4.7-T_(B)=4.7-0.5=4.2eV`