A and B are two metals with threshold frequencies `1.8xx10^(14)Hz and 2.2xx10^(14)Hz`. Two identical photons of energy of 0.825 eV each are incident on them. Then photoelectrons are emitted in take `h=6.6xx10^(-34)j//s`

Options

A. B alone
B. A alone
C. neither A nor B
D. Both A and B

Answer

b

Explanation

`phi_(OA)=(hv_(0))/e eV`
`=((6.6xx10^(-34))xx(1.8xx10^(14)))/(1.6xx10^(-19))=0.74 eV`
`phi_(OB)=((6.6xx10^(-34))xx(2.2xx10^(14)))/(1.6xx10^(-19))=0.91 eV`
Since incident energy 0.825 eVis greater than 0.74 eV and less than 0.91 eV, so photoelectrons are emitted from a metal A only.