The number of photoelectrons emitted from a metal surface for incident light of frequency v (higher than the threshold frequency \( v_0 \)) is directly proportional to:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → intensity of light
In the photon picture each photon ejects at most one electron. The intensity of light, for a given frequency, decides the number of photons striking the surface per second, so above the threshold the number of photoelectrons (the photocurrent) is proportional to the intensity. The frequency decides only the energy of each photon, and hence the maximum kinetic energy of the electrons.
Option (1) Threshold frequency — Incorrect. The threshold frequency is a property of the metal; it fixes whether emission happens at all, not how many electrons come out.
Option (2) intensity of light — Correct. More intense light means more photons per second, and so more photoelectrons per second.
Option (3) frequency of light — Incorrect. Raising the frequency raises the energy of each electron, not their number.
Option (4) \( \left(v - v_0\right) \) — Incorrect. \( h(v - v_0) \) is the maximum kinetic energy of a photoelectron; it has nothing to do with how many are emitted.
NCERT (2026-27):
Effect of Frequency of Incident Radiation on Stopping Potential – “… (i) For a given photosensitive material and frequency of incident radiation (above the threshold frequency), the photoelectric current is directly proportional to the intensity of incident light (Fig. 11.2).”