A and B are two metals with threshold frequencies \( 1.8 \times 10^{14} \) Hz and \( 2.2 \times 10^{14} \) Hz. Two identical photons of energy \( 0.825 \) eV each are incident on them. Then photoelectrons are emitted in (Take \( h = 6.6 \times 10^{-34} \) Js):
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For the photoelectric effect:
- A photon must have energy greater than the work function to emit electrons.
- The threshold frequency determines the metal’s ability to emit electrons.
Step 1: The energy of a photon is given by:
\[
E = h \nu
\]
Step 2: Calculating the threshold energy:
\[
E_A = h \times 1.8 \times 10^{14}, \quad E_B = h \times 2.2 \times 10^{14}
\]
Step 3: Comparing with the incident photon energy, we find that only A emits photoelectrons.