Threshold Frequency From Work Function
A photoemissive metal requires a minimum energy of 3.3 eV to release an electron from its surface. Below which threshold frequency of incident radiation will photoemission completely stop for this metal?
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About This Question
- Subject
- physics
- Chapter
- dual nature of radiation and matter
- Topic
- threshold frequency from work function
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
8.0×1014 Hz
The threshold frequency ν0 is the smallest frequency whose photon energy just matches the work function, so that hν0=ϕ0 and any lower frequency cannot eject electrons. Rearranging, ν0=hϕ0. First convert the work function to joules: ϕ0=3.3×1.6×10−19=5.28×10−19 J. Then ν0=6.6×10−345.28×10−19=8.0×1014 Hz. The option 4.0×1014 Hz corresponds to half the work function. The option 1.6×1015 Hz doubles it. The option 2.0×1014 Hz underestimates the work function fourfold. Radiation of frequency below ν0, however intense, produces no photoelectrons, a frequency threshold that classical physics could not justify and which Einstein's photon model resolved, as covered in the NCERT dual-nature chapter. A check shows 8.0×1014 Hz lies just beyond the visible into the near-ultraviolet, consistent with a moderately high work function of 3.3 eV. A key conceptual point is that the existence of a sharp frequency threshold, completely independent of how bright the source is, was utterly inexplicable on the classical wave theory, which predicted that any frequency should eventually eject electrons if the light were intense enough. The experimental fact that dim blue light works while brilliant red light fails was decisive evidence for the photon picture, and the threshold frequency is simply the work function rendered in frequency units.
This medium difficulty physics question is from the chapter dual nature of radiation and matter, covering the topic of threshold frequency from work function. It appeared in the 2025 exam.
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