Practice question

**Work function** Φ minimum energy to escape, for Φ=2.2 eV λ₀=1240/2.2=563.6 nm, Φ=3.0 eV threshold frequency f₀=Φ/h=3×1.6×10⁻¹⁹/6.63×10⁻³⁴=7.24×10¹⁴ Hz, stopping potential V₀ = (h f - Φ)/e, for 0.7 V K_max=0.7 eV, for 1.5 V K_max=1.5 eV, intensity increase increases photocurrent not K_max, frequency below threshold no emission regardless of intensity. Φ₀ = h v₀ = 6.63 × 10⁻³⁴ × 5.0 × 10¹⁴ = 3.315 × 10⁻¹⁹ J . E = h v = 6.63 × 10⁻³⁴ × 7.0 × 10¹⁴ = 4.641 × 10⁻¹⁹ J . Kₘₐₓ = E - Φ₀ = 4.641 × 10⁻¹⁹ - 3.315 × 10⁻¹⁹ = 1.326 × 10⁻¹⁹ J .