Respostas Problemas Capitulo 3 Física Quântica - Eisberg Resnick - 2 Edição

15) Δp.Δx ≥ ℏ/2 Δp = -h/λ². Δλ p = h/λ h/λ². Δx . Δλ ≥ ℏ/4π ΔλΔx ≥ λ²/4π 16) Δλ/λ = 10⁻⁴ a) Δλ = 10⁻⁵. λ = 10³.5.10⁴ Δλ = 5.10⁻¹¹ Δx. Δλ = λ²/4π Δx.5.10⁻¹¹ = (5.10⁻³)²/4π Δx = 3.98 .10⁸ Å b) Δλ = 10⁻⁷.5 = 5.10⁻⁷ Δx = λ²/4π. Δλ = 5²/4π.5.10⁻⁹ = 3,98.10⁹ Å c) Δλ = 10⁻⁵.5000 = 5000.10⁻⁷ Δx = λ²/4π. Δλ = 5000²/4π.5000.10⁻⁷ = 3,98.10¹⁰ Å 19) Δx ~ λ Δx ~ mυ Δx ~ h/ p Δx ~υm.υm p.Δx > h/4π p.Δx ≥ h/4π Δx ~ λ/4π 20) β = υ/c Δxmin = h/4πmac .(1-β²)¹/² = λc/4π .√(1-β²) a) β = υ/c Δxmin = h/4πmac. √1-β² Como Δp ≈ p Δxmin = h/4πmac. (√1-β²) b) Quando β = 0 Δxmin = h/4πmac Quando β = 1 Δxmin = 0 23) Δt.ΔE ≥ h/4π ΔE ≥ h/π.Δt = 4,135.10⁻¹⁵/4π.(10⁻¹²) = 3,3.10⁴ eV 25) Δp.Δx = h/4π E = (pₓ)²/2m + cX²/2 E = h²/16π²x² - 1/2m + cX²/2 a) X.p = h/4π Ee = h/4πx²/2m + cX²/2 Ee = h²/32π²x² + cX²/2 28 b) dE/ d(Δx) = -h²/16π₂. 1/Δx² + c(Δx) = 0 (Δx)² = h/4π.1/εmin Eₘᵢₙ = 1/2. [1/2π. (c/m)/e] = 1/2 h ν 27) Vₖ = ΔVₖ = ℏ/2mΔx x = Vₖ(2h/g)¹/² = h/2. 1/2m.Δx (2h/g)¹/² a) X = - Δx + γ = Δx + h/2m (2h/g)¹/². 1/Δx d(Δx) = 0 Δx = (h/2m (2h/g) ¹/²) b) H=10m H/g = 0,001kg xₘᵢₙ = (10⁻³/10³)¹/² xₘᵢₙ = 3.10⁻¹⁶ m