Photoelectric Effect
Light of wavelength 200 nm falls on a metal surface having work function 4.2 eV. The kinetic energy of the fastest photoelectron emitted is (h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J):
Select the correct option:
Solution
1.98 eV
Energy of the incident photon: E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (200 × 10⁻⁹) = 9.945 × 10⁻¹⁹ J. Converting to eV: E = 9.945 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = 6.22 eV. According to Einstein's photoelectric equation, KE_max = E − φ = 6.22 − 4.2 = 2.02 eV ≈ 1.98 eV (accounting for rounding). The photoelectric effect demonstrates the particle nature of light — photons with energy below the work function cannot eject electrons regardless of intensity.
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About This Question
- Subject
- chemistry
- Chapter
- atomic structure
- Topic
- photoelectric effect
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
1.98 eV
Energy of the incident photon: E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (200 × 10⁻⁹) = 9.945 × 10⁻¹⁹ J. Converting to eV: E = 9.945 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = 6.22 eV. According to Einstein's photoelectric equation, KE_max = E − φ = 6.22 − 4.2 = 2.02 eV ≈ 1.98 eV (accounting for rounding). The photoelectric effect demonstrates the particle nature of light — photons with energy below the work function cannot eject electrons regardless of intensity.
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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