Photoelectric Effect
The work function of a metal surface is 2.3 eV. What is the kinetic energy of the photoelectrons emitted when light of frequency 8.0 × 10^14 Hz strikes the surface? (h = 6.626 × 10^-34 J·s, 1 eV = 1.6 × 10^-19 J)
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About This Question
- Subject
- chemistry
- Chapter
- atomic structure
- Topic
- photoelectric effect
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
1.01 eV
The photoelectric effect is explained by Einstein's equation: KE = hν - φ, where hν is the energy of the incident photon and φ is the work function of the metal. This equation tells us that photoelectrons are only emitted when the photon energy exceeds the binding energy of surface electrons, and any excess appears as kinetic energy. Calculating photon energy: hν = 6.626 × 10^-34 × 8.0 × 10^14 = 5.301 × 10^-19 J. Converting to eV: 5.301 × 10^-19 / 1.6 × 10^-19 = 3.313 eV. Applying Einstein's equation: KE = 3.313 - 2.3 = 1.013 eV ≈ 1.01 eV. Option 0.75 eV would arise if the work function were incorrectly taken as 2.56 eV. Option 3.31 eV is the photon energy itself, forgetting to subtract the work function. Option 0.43 eV would result from an arithmetic error in the frequency-energy conversion. This is a direct application of Einstein's photoelectric equation, a cornerstone JEE topic linking quantum theory to experimental observation. Plausibility check: photon energy (3.31 eV) exceeds work function (2.3 eV), confirming emission does occur and the positive KE is physically valid.
This medium difficulty chemistry question is from the chapter atomic structure, covering the topic of photoelectric effect. It appeared in the 2025 exam.
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