When photon oFIGURE energy 4.0eV strikes the surFIGUREace oFIGURE a metal, the ejected photoelectrons have maximum kinetic energy `T_(A)` eV and de-Broglie wavelength `lambda_(A)`. The maximum kinetic energy oFIGURE photoelectrons liberated FIGURErom another metal B by photon oFIGURE energy 4.50eV is `T_(B)=(T_(1)-1.5)eV`. IFIGURE the de-Broglie wavelength oFIGURE these photoelectrons `lambda_(B)=2lambda_(A)`, then the work, FIGUREunction oFIGURE metal B is:

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`