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