Photoelectric emission occurs only when the incident light has more than a certain minimum
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Frequency
In the photoelectric effect an electron absorbs the energy of a single photon, \( E = h\nu \). It can escape only if this energy is at least the work function \( \phi_0 \) of the metal.
\( \displaystyle h\nu \ge \phi_0 \;\Rightarrow\; \nu \ge \nu_0 = \frac{\phi_0}{h} \)
So emission needs a frequency above the threshold frequency \( \nu_0 \).
Option (1) Intensity — Incorrect. Intensity changes only the number of photons; below \( \nu_0 \) no electrons are emitted however intense the light is.
Option (2) Angle of Incidence — Incorrect. The angle of incidence does not change the energy of each photon.
Option (3) Speed — Incorrect. All light travels at the same speed \( c \) in vacuum, so speed cannot set a threshold.
Option (4) Frequency — Correct. The photon energy \( h\nu \) depends on the frequency, so emission needs \( \nu \gt \nu_0 \).
NCERT (2026-27):
Effect of Frequency of Incident Radiation on Stopping Potential – “… (iii) For a given photosensitive material, there exists a certain minimum cut-off frequency of the incident radiation, called the threshold frequency, below which no emission of photoelectrons takes place, no matter how intense the incident light is.”