Practice question

**De Broglie wavelength** λ = h/p = h/(m v) = h/√(2 m e V) for electron accelerated through V, for V=100 V λ= h/√(2 m e×100)=1.227/√V nm=0.1227 nm, for 50 V 0.173 nm, for particle mass 3.0×10⁻³⁰ kg v=10⁶ m/s λ=6.63×10⁻³⁴/(3×10⁻³⁰×10⁶)=2.21×10⁻¹⁰ m=0.221 nm, inversely proportional to momentum, property inversely proportional to de Broglie wavelength is momentum. p = (h/λ) = (6.63 × 10⁻³⁴/0.15 × 10⁻⁹) = 4.42 × 10⁻²⁴ kg m/s . v = (p/m) = (4.42 × 10⁻²⁴/9.11 × 10⁻³¹) ≈ 4.852 × 10⁶ m/s . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ