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

Options

A. 1.5 eV
B. 3 eV
C. 4eV
D. 4.5 eV

Answer

C

Explanation

de-Broglie wavelength
`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` Now `T_(B)=hf-phi_(B)`
`therefore phi_(B)=hf-T_(B)=(4.5-0.5)eV`
`therefore phi_(B)=4.0 eV`