A metal surface is illuminated first with monochromatic light of wavelength λ and then with light of wavelength 2λ. Both radiations produce photoelectric emission. The corresponding maximum photoelectron kinetic energies are 5.0 eV and 1.5 eV respectively. Take hc = 1240 eV·nm. If the threshold wavelength of the metal is N nm, find N.
Physics · Photoelectric effect and dual nature of matter · question based on photoelectric effect
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Question

A metal surface is illuminated first with monochromatic light of wavelength λ and then with light of wavelength 2λ. Both radiations produce photoelectric emission. The corresponding maximum photoelectron kinetic energies are 5.0 eV and 1.5 eV respectively. Take hc = 1240 eV·nm. If the threshold wavelength of the metal is N nm, find N.

Physics · Photoelectric effect and dual nature of matter · question based on photoelectric effect

G1 Physics · Accepted Answer

For wavelength λ: K₁ = hc ÷ λ − φ. For wavelength 2λ: K₂ = hc ÷ (2λ) − φ. Eliminating hc ÷ λ gives the useful relation K₁ = 2K₂+φ. Substitute the measured kinetic energies: 5.0 = 2(1.5)+φ. Therefore, φ = 2.0 eV. At the threshold wavelength λ₀, the emitted electron has zero maximum kinetic energy, so φ = hc ÷ λ₀. Hence, λ₀ = 1240 ÷ 2.0 = 620 nm. Therefore N = 620.