Question

Find the number of photons emitted per second by a 25W source of monochromatic light of wavelength 6600AA. What is the photoelectric current assuming 3% efficiency for photoelectric effect?

• A. \(\dfrac{25}{3}\times10^{19}\,\text{s}^{-1},\,0.4\,\text{A}\)
• B. \(\dfrac{25}{4}\times10^{19}\,\text{s}^{-1},\,6.2\,\text{A}\)
• C. \(\dfrac{25}{2}\times10^{19}\,\text{s}^{-1},\,0.8\,\text{A}\)
• D. None of these

Hint:

Photon energy depends only on wavelength, not on intensity.

Answer 1

The Correct Option is A

Step 1: Energy of one photon: E = (hc)/(λ) = frac6.6×10⁻34×3×10⁸6.6×10⁻7 = 3×10⁻19J
Step 2: Number of photons per second: N = \frac253×10⁻19 = (25)/(3)×10¹9s⁻1
Step 3: Photoelectric current (3% efficiency): I = 0.03Ne = 0.03×(25)/(3)×10¹9×1.6×10⁻19 = 0.4A