Q. The threshold frequency for a photosensitive metal is 3.3 × 10¹⁴ Hz. If light of frequency 8.2 × 10¹⁴ Hz is incident on this metal, the cut-off voltage for the photoelectric emission is nearly
A) 5 V
B) 1 V
C) 2 V
D) 3 V
Explanation
[Solution from source] eV₀ = h(ν − ν₀) = 6.63 × 10⁻³⁴ × (8.2 − 3.3) × 10¹⁴ = 3.25 × 10⁻¹⁹ J, so V₀ = 3.25 × 10⁻¹⁹/1.6 × 10⁻¹⁹ ≈ 2 V. Using the incident frequency alone (ignoring the threshold) would give about 3.4 V, which is not the cut-off voltage. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.