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 c/λ) = (1240/500) = 2.48 eV . Kₘₐₓ = e V₀ = 0.6 eV . Φ₀ = E - Kₘₐₓ = 2.48 - 0.6 = 1.88 eV . λ₀ = (h c/Φ₀) = (1240/1.88) ≈ 659.57 nm . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h c/λ₀, λ = h/√(2 m e V) and f_max = e