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. Momentum p = m v = 9.11 × 10⁻³¹ × 2.0 × 10⁶ = 1.822 × 10⁻²⁴ kg m/s . λ = (h/p) = (6.63 × 10⁻³⁴/1.822 × 10⁻²⁴) ≈ 3.64 × 10⁻¹⁰ m = 0.364 nm . Applying E = h f = h c/λ, p = h/λ,