NEET Physics Photoelectric Effect MCQ Question
If the work function of a metal is 2.14 eV and light of frequency 6 × 10^14 Hz is incident on it, what is the maximum kinetic energy of the emitted photoelectrons?
0.37 eV
86 eV
14 eV
14 eV
Correct Answer
Detailed Explanation
The maximum kinetic energy of the emitted photoelectrons is given by KE_max = hν - φ, where φ is the work function. Using h = 4.14 × 10^-15 eV·s, KE_max = (6 × 10^14 Hz) × (4.14 × 10^-15 eV·s) - 2.14 eV = 0.37 eV.
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