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

(b) : `phi_(0A)=(hv_(0))/(e)eV=((6.6xx10^(-34))xx(1.8xx10^(14)))/(1.6xx10^(-19))eV`
=0.74eV
`phi_(0B)=((6.6xx10^(-34))xx(2.2xx10^(14)))/(1.6xx10^(-19))eV=0.91eV`
Since the incident energy 0.825 eV is greater than 0.74 eV and less than 0.91 eV, so photoelectrone are emitted from metal A only.