Practice question

**Photoelectric effect** demonstrates particle nature of light, photon energy E = h f, work function Φ = h f₀ minimum energy to eject electron, threshold frequency f₀ = Φ/h, threshold wavelength λ₀ = hc/Φ, hc=1240 eV·nm, maximum kinetic energy K_max = h f - Φ = e V₀, V₀ stopping potential, photocurrent proportional to intensity (number of photons) when f>f₀, saturation current depends on intensity, frequency determines K_max not current. E = (h c/λ) = (1240/450) ≈ 2.76 eV . Kₘₐₓ = e V₀ = 0.9 eV . Φ₀ = E - Kₘₐₓ = 2.76 - 0.9 = 1.86 eV . v₀ = (Φ₀/h) =