Einstein's Photoelectric Equation
Ultraviolet light of wavelength 400 nm illuminates a metal whose work function is 2.0 eV, ejecting electrons from its surface. What is the maximum kinetic energy with which the photoelectrons leave the metal?
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About This Question
- Subject
- physics
- Chapter
- dual nature of radiation and matter
- Topic
- einstein's photoelectric equation
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
1.1 eV
Einstein's photoelectric equation expresses energy conservation for a single photon-electron interaction: the photon energy hν splits into the work function ϕ0 needed to free the electron and the leftover kinetic energy, giving Kmax=hν−ϕ0=λhc−ϕ0. The photon energy for 400 nm light is 4001240=3.1 eV using hc=1240 eV nm. Subtracting the work function, Kmax=3.1−2.0=1.1 eV. The option 3.1 eV is just the photon energy and forgets to subtract ϕ0. The option 5.1 eV wrongly adds the work function instead of subtracting it. The option 0.9 eV uses an incorrect photon energy. Because the calculated value is positive, emission does indeed occur, confirming the incident photons exceed the threshold. This single-photon energy balance, which classical wave theory could not reproduce, earned Einstein the Nobel Prize and is the heart of the dual-nature chapter. A consistency check shows Kmax<hν, exactly as the conservation statement demands. It is important to recognise that increasing the intensity of this 400 nm beam would liberate more electrons per second but would not change the 1.1 eV maximum energy of any single one, because that energy is fixed by frequency alone. The complete failure of the classical wave picture to predict this frequency dependence, together with the instantaneous nature of emission, is precisely what forced the adoption of the photon model.
This medium difficulty physics question is from the chapter dual nature of radiation and matter, covering the topic of einstein's photoelectric equation. It appeared in the 2025 exam.
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