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NEET Chemistry
QUESTION #7364
Question 1
The threshold frequency of a metallic surface is \(5 \times 10^{13}\) s\(^{-1}\). When light of frequency \(1 \times 10^{14}\) s\(^{-1}\) falls on it, the maximum kinetic energy of the emitted electrons is:
Correct Answer Explanation
Einstein's photoelectric equation: \(KE_{max} = h(\nu - \nu_0)\)
\(KE = 6.63 \times 10^{-34} \times (10^{14} - 5\times10^{13})\)
\(= 6.63 \times 10^{-34} \times 5 \times 10^{13}\)
\(= 33.15 \times 10^{-21}\)
\(\approx 3.3 \times 10^{-20}\) J
The excess energy above the threshold (work function) appears entirely as kinetic energy of the emitted photoelectron.
\(KE = 6.63 \times 10^{-34} \times (10^{14} - 5\times10^{13})\)
\(= 6.63 \times 10^{-34} \times 5 \times 10^{13}\)
\(= 33.15 \times 10^{-21}\)
\(\approx 3.3 \times 10^{-20}\) J
The excess energy above the threshold (work function) appears entirely as kinetic energy of the emitted photoelectron.
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