A and B are two metals with threshold frequencies $1.8×10^{14}Hz$ and $2.2×10^{14}Hz$. Two identical photons of energy 0.825 eV each are incident on them. Then photoelectrons are emitted in: (Take $h = 6.6×10^{-34}Js$)

A alone

$\phi_{0A}=\frac{hv_0}{e}eV=\frac{(6.6×10^{-34})×(1.8×10^{14})}{1.6×10^{-19}}eV=0.74eV$

$\phi_{0B}=\frac{(6.6×10^{-34})×(2.2×10^{14})}{1.6×10^{-19}}eV=0.91eV$

Since the incident energy 0.825 eV is greater than 0.74 eV and less than 0.91 eV, so photoelectrons are emitted
from metal only.