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Asked in GUJCET 2013 · Work function and the photoelectric equation
Given: kinetic energy k₁ with wavelength λ₁ and k₂ with λ₂, on the same metal.
Idea: write Einstein's equation for each light and eliminate hc.
(hc)/(λ₁) = φ₀ + k₁, so hc = λ₁(φ₀ + k₁)
(hc)/(λ₂) = φ₀ + k₂, so hc = λ₂(φ₀ + k₂)
Equate: λ₁φ₀ + k₁λ₁ = λ₂φ₀ + k₂λ₂
φ₀(λ₂ - λ₁) = k₁λ₁ - k₂λ₂
φ₀ = (k₁λ₁ - k₂λ₂)/(λ₂ - λ₁)
Check: φ₀ = 2 eV, λ₁ = 248 nm (k₁ = 3 eV), λ₂ = 310 nm (k₂ = 2 eV) give (744 - 620)/(62) = 2 eV.
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