Work Function Comparison
Two clean metal plates, caesium with work function 2.1 eV and platinum with work function 5.6 eV, are illuminated by the same violet light, and a student predicts the outcome.
Select the correct option:
Solution
Only caesium emits electrons because the photon energy exceeds its lower work function
NCERT establishes that a metal emits photoelectrons only when the incident photon energy exceeds its work function. Violet light of wavelength around 400 nm carries a photon energy of roughly 4001242≈3.1 eV. This comfortably exceeds caesium's 2.1 eV work function, so caesium emits electrons, but it falls well short of platinum's 5.6 eV barrier, so platinum does not respond. This is exactly why low-work-function metals like caesium are used in practical photocells. The option favouring only platinum is wrong because a higher work function makes emission harder, not easier. The option claiming both emit equally is wrong because emission depends critically on whether photon energy surpasses each metal's specific work function. The option claiming neither emits is wrong because caesium's low barrier is clearly overcome by the 3.1 eV violet photons. As stated in NCERT Class 12, Chapter 11, alkali metals with small work functions are photosensitive even to visible light. This selectivity is the operating principle behind practical photocells and light meters, where a low-work-function coating is chosen so that ordinary visible light is enough to trigger emission. It also underlies why photographic and light-sensing devices historically favoured alkali-metal photocathodes over refractory metals. A consistency check confirms that with 3.1 eV photons, only the metal whose work function lies below this value, namely caesium, will emit.
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About This Question
- Subject
- physics
- Chapter
- dual nature of matter and radiation
- Topic
- work function comparison
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
Only caesium emits electrons because the photon energy exceeds its lower work function
NCERT establishes that a metal emits photoelectrons only when the incident photon energy exceeds its work function. Violet light of wavelength around 400 nm carries a photon energy of roughly 4001242≈3.1 eV. This comfortably exceeds caesium's 2.1 eV work function, so caesium emits electrons, but it falls well short of platinum's 5.6 eV barrier, so platinum does not respond. This is exactly why low-work-function metals like caesium are used in practical photocells. The option favouring only platinum is wrong because a higher work function makes emission harder, not easier. The option claiming both emit equally is wrong because emission depends critically on whether photon energy surpasses each metal's specific work function. The option claiming neither emits is wrong because caesium's low barrier is clearly overcome by the 3.1 eV violet photons. As stated in NCERT Class 12, Chapter 11, alkali metals with small work functions are photosensitive even to visible light. This selectivity is the operating principle behind practical photocells and light meters, where a low-work-function coating is chosen so that ordinary visible light is enough to trigger emission. It also underlies why photographic and light-sensing devices historically favoured alkali-metal photocathodes over refractory metals. A consistency check confirms that with 3.1 eV photons, only the metal whose work function lies below this value, namely caesium, will emit.
This medium difficulty physics question is from the chapter dual nature of matter and radiation, covering the topic of work function comparison. It appeared in the 2025 exam.
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