KCET · Physics · Dual Nature of Matter
\(A\) 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 \(h=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}\) )
- A \(B\) alone
- B \(A\) alone
- C neither \(A\) nor \(B \quad\)
- D both \(A\) and \(B\)
Answer & Solution
Correct Answer
(B) \(A\) alone
Step-by-step Solution
Detailed explanation
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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