A and B are two metals with threshold frequencies `1.8 xx 10^(14)Hz` and `2.2 xx 10^(14)Hz`. Two identical photons of energy 0.825 eV each are incident on them. Then, photoelectrons are emitted by (Taking, `h = 6.6 xx 10^(-34)J-s`)

Options

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

Answer

B

Explanation

Energy of electron needed if emitted from
`A = (hepsilon)/(e)eV`
`therefore E_(A) = (6.6 xx 10^(-34)xx(1.8 xx10^(14)))/(1.6 xx 10^(-19))=0.74 eV`

Again `E_B = ((6.6 xx 10^(-34))xx(2.2 xx10^(14)))/(1.6 xx 10^(-19))= 0.91 eV`
Incident energy 0.825 eV is greather than `E_A` (0.74 eV) but less than `E_B` (0.91 eV).
Hence, photoelectrons are emitted in A alone.