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_0) _(A ) =( hupsilon _0)/(e) eV`
`=(6.6xx106(-34)xx(1.8xx10^(14)))/(1.6 xx10^(-19))=0.74 eV`
`(phi_0)_E = ((6.6 xx10^(-34)) xx(2.2xx10^(14)) )/(1.6 xx10^(-19)) = 0.91 eV`
Since the incident energy 0.825 eV is greater than 0.74 eV and less than 0.91 eV, therefore photoelectrons are emitted from metal A only.