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. Kₘₐₓ = e V₀ = 0.8 eV = 0.8 × 1.6 × 10⁻¹⁹ = 1.28 × 10⁻¹⁹ J . Kₘₐₓ = (1/2) m vₘₐₓ² ⇒ vₘₐₓ = √((2 Kₘₐₓ/m)) = √((2 × 1.28 × 10⁻¹⁹/9.11 × 10⁻³¹)) ≈ 5.3 × 10⁵ m/s . Applying E = h f = h c/λ, p = h/λ, K_max