Practice question

**De Broglie wavelength calculations** for electron moving at 2×10⁶ m/s λ= h/(m v)=6.63×10⁻³⁴/(9.11×10⁻³¹×2×10⁶)=3.64×10⁻¹⁰ m=0.364 nm, for mass 5×10⁻³⁰ kg v=2×10⁵ m/s λ=6.63×10⁻³⁴/(1×10⁻²⁴)=6.63×10⁻¹⁰ m, showing λ decreases with increasing m v. E = h v = 6.63 × 10⁻³⁴ × 5.5 × 10¹⁴ = 3.6465 × 10⁻¹⁹ J . E = (3.6465 × 10⁻¹⁹/1.6 × 10⁻¹⁹) ≈ 2.28 eV . Kₘₐₓ = (1/2) m vₘₐₓ² = (1/2) × 9.11 × 10⁻³¹ × (4.0 × 10⁵)² = 7.288 × 10⁻²⁰ J . Kₘₐₓ = (7.288 × 10⁻²⁰/1.6 × 10⁻¹⁹) ≈ 0.455 eV . Φ₀ = E - Kₘₐₓ = 2.28 - 0.455 ≈ 1.825 eV . Applying