PHY 2048: Answer Key #21 - Edubirdie

University:
Santa Fe College
Course:
PHY 2048 | General Physics 1 with Calculus
Academic year:

2024
Views:

215

Pages:

5
Author:

Jabin F.

prob. potential V(n) = (Vo|x|)/a •, Whose vod are constant, Find the value of minimum energ of the particle. Given: V(x) = V₀|x|/a Hamiltonian: H = p²/2m + V(x) H = p²/2m + V₀|x|/a Uncertainty Principle: ΔxΔp ≥ ħ/2 Energy (E): E = = (1/2m) * + (V₀/a) * <|x|> Minimization: dE/d = 0 Calculations: dE/d = (1/2m) * d/d + (V₀/a) = 0 d/d = -8mα (1/2m) * (-8mα) + (V₀/a) = 0 -4α + V₀/a = 0 α = V₀ / (4a) Given: V(x) = (V₀/a) * |x| Calculations: Δx = (ħ²a / m²v₀²)^(1/3) d²E/d² > 0 Minimum Energy (E_min): E_min = (1/2m) * (ħ²a / m²v₀²)^(2/3) + (V₀/a) * (ħ²a / m²v₀²)^(1/3) E_min = (1/2) * (ħ²v₀²)^(1/3) / (ma²)^(1/3) + (V₀/a) * (ħ²a / m²v₀²)^(1/3) E_min = (3/2) * (ħ²v₀²)^(1/3) / (ma²)^(1/3) Comparison: E_min ∝ (V₀/a)^(2/3) Problem: Calculate the ground state energy of a particle of mass m moving under a 1-D potential V(x) = V₀ * cosh(x/L), where V₀ and L are constants. Solution: V(x) = V₀ * (e^(x/L) + e^(-x/L)) / 2 Approximation for x << L: V(x) ≈ V₀ * (1 + (1/2) * (x/L)²) Hamiltonian: H = p²/2m + V(x) Energy: E = (1/2m) * + V₀ + (1/2) * (V₀/L²) * Ground State Energy (E_min): E_min = (1/2) * ħω + V₀ Given: E_min = (1/2) * ħ * √(V₀ / mL²) + V₀ Problem: Calculate the ground state energy and first Bohr radius of an electron in a hydrogen atom using the Heisenberg Uncertainty Principle (HUP). Solution: E = p²/2m - e² / (4πε₀r) ΔxΔp ≈ ħ Δp ≈ ħ / Δx E ≈ (ħ²/2m) / (Δx)² - e² / (4πε₀(Δx)) dE/d(Δx) = 0 Calculations: dE/d(Δx) = (-ħ²/2m) * (2/Δx³) + (e²/4πε₀) * (1/Δx²) = 0 (ħ²/2m) * (2/Δx³) = (e²/4πε₀) * (1/Δx²) Δx = (ħ²/2m) * (4πε₀/e²) Δx = (ħ²/me²) * (4πε₀) Δx = a₀ (Bohr radius) E_min = (1/2) * ħ² / (ma₀²) - e² / (4πε₀a₀) E_min = (-e⁴m) / (2ħ² * (4πε₀)²)

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