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$\mathrm{~A}$ and $\mathrm{B}$ are two metals with threshold frequencies $1.8 \times 10^{14} \mathrm{~Hz}$ and $2.2 \times 10^{4} \mathrm{~Hz}$. Two identical photons of energy $0.825 \mathrm{eV}$ each are incident on them. Then photoelectrons are emitted by $\left(\right.$ Take $\left.\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)$
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A alone
$\begin{aligned} \text { Threshold energy of A E }_{\mathrm{A}}=\mathrm{hv}_{\mathrm{A}} \\=& 6.6 \times 10^{-34} \times 1.8 \times 10^{14} \\=& 11.88 \times 10^{-20} \mathrm{~J} \\=& \frac{11.88 \times 10^{-20}}{1.6 \times 10^{-19}} \mathrm{eV}=0.74 \mathrm{eV} \end{aligned}$
Similarly, $\mathrm{E}_{\mathrm{B}}=0.91 \mathrm{eV}$
As the incident photons have energy greater than $\mathrm{E}_{\mathrm{A}}$ but less than $\mathrm{E}_{\mathrm{B}}$ So, photoelectrons will be emitted from metal A only.
Similarly, $\mathrm{E}_{\mathrm{B}}=0.91 \mathrm{eV}$
As the incident photons have energy greater than $\mathrm{E}_{\mathrm{A}}$ but less than $\mathrm{E}_{\mathrm{B}}$ So, photoelectrons will be emitted from metal A only.
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