A and B are two metals with threshold frequencies $1.8 \times {10}^{14} \mathrm{Hz}$ and $2.2 \times {10}^{11} \mathrm{Hz}.$ Two identical photons of energy $0.825 \mathrm{eV}$ each are incident on them. Then photoelectrons are emitted in _____. (Take $\mathrm{h}= 6.6 \times {10}^{-34} \mathrm{Js}$)

$$\begin{gathered} {\varphi }_{0A} (\mathrm{in} \mathrm{eV}) = \frac{h{\nu }_0}{e} = \frac{\left(6.6 \times {10}^{-34}\right) \times \left(1.8 \times {10}^{14}\right)}{1.6 \times {10}^{-19}} \\ = 0.74 \mathrm{eV} \end{gathered}$$

$$\begin{gathered} {\varphi }_{0B}(\mathrm{in} \mathrm{eV}) = \frac{\left(6.6 \times {10}^{-34}\right) \times \left(2.2 \times {10}^{14}\right)}{1.6 \times {10}^{-19}}eV \\ = 0.91 eV \end{gathered}$$

Since the incident energy $0.825 \mathrm{eV}$ is greater than $0.74 \mathrm{eV}$ and less than $0.91 \mathrm{eV},$ so photoelectrons are emitted from metal $\mathrm{A}$ only.