NEET Physics Photoelectric Effect MCQ Question
A photosensitive material has a work function of 2.5 eV. If light of frequency 8 × 10^14 Hz is incident on it, what is the maximum kinetic energy of the emitted photoelectrons? (Planck's constant h = 6.626 × 10^-34 J s and 1 eV = 1.6 × 10^-19 J)
0.8 eV
1.5 eV
2.0 eV
1.0 eV
Correct Answer
Detailed Explanation
The maximum kinetic energy is given by K_max = hν - φ₀. Converting 2.5 eV to Joules (2.5 × 1.6 × 10^-19 J), and plugging values into K_max = (6.626 × 10^-34 J s × 8 × 10^14 Hz) - (2.5 eV × 1.6 × 10^-19 J/eV), we find K_max ≈ 0.8 eV.
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