Question

Find the number of photons emitted per second by a 25 watt 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:

Photoelectric current depends on: I = e × (efficiency) × (photon rate) Higher intensity means more current, not higher energy electrons.

Answer 1

The Correct Option is A

Step 1: Energy of one photon: E=(hc)/(λ) =frac6.63×10⁻34×3×10⁸6.6×10⁻7 ≈3.0×10⁻19J
Step 2: Number of photons emitted per second: N=(P)/(E)=\frac253.0×10⁻19 =(25)/(3)×10¹9s⁻1
Step 3: Effective electrons emitted (3% efficiency): Nₑ=0.03N
Step 4: Photoelectric current: I=eNₑ=1.6×10⁻19×0.03×(25)/(3)×10¹9 ≈0.4A