The intensity of the incident radiation is extremely dim, but its frequency is 1.5 times the threshold frequency of the metal. What can we predict about the photoelectric emission in this case?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → The emission will occur anyway.
Whether electrons are emitted depends only on the frequency of the radiation. Emission takes place when \( \nu \gt \nu_0 \), however dim the light. The intensity only fixes how many photons, and so how many photoelectrons, arrive per second.
Option (1) The emission will occur only if the intensity is increased. — Incorrect. Each photon already has energy \( 1.5\,h\nu_0 \), more than the work function. A brighter beam only increases the number of photoelectrons per second, that is, the photocurrent.
Option (2) The emission will not take place. — Incorrect. Emission fails only below the threshold frequency. Here the frequency is above it.
Option (3) The emission will occur anyway. — Correct. With \( \nu=1.5\,\nu_0 \), each photon has energy \( h\nu=1.5\,\phi_0 \), which is more than the work function \( \phi_0 \). An electron that absorbs one photon escapes at once. So emission is instantaneous even for very dim light. The dim light gives only a small photocurrent.
Option (4) Cannot predict. — Incorrect. The frequency is known relative to the threshold frequency, and that alone decides whether emission takes place.
NCERT (2026-27):
Effect of Frequency of Incident Radiation on Stopping Potential – “Note that in all the above experiments, it is found that, if frequency of the incident radiation exceeds the threshold frequency, the photoelectric emission starts instantaneously without any apparent time lag, even if the incident radiation is very dim. It is now known that emission starts in a time of the order of 10–9 s or less.”