For a certain metal, monochromatic light of wavelength λ produces a stopping potential 3V₀. When the same metal is illuminated by light of wavelength 2λ, the stopping potential is V₀. If the threshold wavelength is αλ, what is α?
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Question

For a certain metal, monochromatic light of wavelength λ produces a stopping potential 3V₀. When the same metal is illuminated by light of wavelength 2λ, the stopping potential is V₀. If the threshold wavelength is αλ, what is α?

Options

G1 Physics · Accepted Answer

Einstein's photoelectric equation is hc ÷ λ - φ = eV(stop). For wavelength λ: hc ÷ λ - φ = 3eV₀. (1) For wavelength 2λ: hc ÷ (2λ) - φ = eV₀. (2) Multiply equation (2) by 3: 3hc ÷ (2λ) - 3φ = 3eV₀. Using equation (1), hc ÷ λ - φ = 3hc ÷ (2λ) - 3φ. Rearrange: 2φ = hc ÷ (2λ). Therefore, φ = hc ÷ (4λ). At threshold wavelength λ₀, φ = hc ÷ λ₀. Thus, hc ÷ λ₀ = hc ÷ (4λ). Hence, λ₀ = 4λ. Therefore, α = 4. Correct option: B.