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. Photon energy E = h v = 6.63 × 10⁻³⁴ × 7.5 × 10¹⁴ = 4.9725 × 10⁻¹⁹ J . E = (4.9725 × 10⁻¹⁹/1.6 × 10⁻¹⁹) ≈ 3.11 eV . Kₘₐₓ = E -