1.5 mW of 400 nm light is directed at a photoelectric cell. If 0.1% of the incident photons produce photoelectrons, the current in the cell is:
Correct answer: A. 0.48 μA
- A. 0.48 μA
- B. 0.42 mA
- C. 0.48 mA
- D. 0.42 μA
Explanation
Photon energy E = hc/λ = 1240/400 ≈ 3.1 eV = 4.96 × 10⁻¹⁹ J. Number of photons n = P/E = (1.5 × 10⁻³)/(4.96 × 10⁻¹⁹) ≈ 3.02 × 10¹⁵. Photoelectrons = 0.001n. Current = nq = (0.001 × 3.02 × 10¹⁵) × (1.6 × 10⁻¹⁹) ≈ 0.48 μA.
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