A surface has work function 2.20 eV. Define its threshold frequency and threshold wavelength, calculate λ₀ using h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹ and 1 eV = 1.60 × 10⁻¹⁹ J, then decide whether 600 nm radiation can eject electrons. Explain whether greater intensity changes the conclusion.

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f₀ is the minimum emission frequency; λ₀ the maximum emission wavelength; λ₀ = 565 nm. 600 nm is too long, so increasing intensity does not overcome the threshold in the single-photon model. At threshold hf₀ = Φ and λ₀ = c/f₀ = hc/Φ. Φ = 2.20 × 1.60 × 10⁻¹⁹ = 3.52 × 10⁻¹⁹ J. λ₀ = (6.63 × 10⁻³⁴)(3.00 × 10⁸)/(3.52 × 10⁻¹⁹) = 5.65 × 10⁻⁷ m. 600 nm > 565 nm means photon energy is below Φ. Greater intensity adds photons but does not increase energy per photon at fixed wavelength; no ordinary single-photon emission is predicted.

Explain the reversed wavelength inequality

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