Practice question
Question
An alpha-particle with 4.0 MeV kinetic energy approaches a gold nucleus (Z = 79). What is the distance
of closest approach? (Use \( \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 \), \(
e = 1.6 \times 10^{-19} \, \text{C} \), 1 MeV = \( 1.6 \times 10^{-13} \, \text{J} \))
Explanation
**Angular momentum** L_n = n h/2π, h=6.6×10⁻³⁴ J·s, L₂=2×6.6×10⁻³⁴/2π=2.11×10⁻³⁴ J·s, speed v_n = e²/(2 ε₀ h) ×1/n ≈2.2×10⁶/n m/s, kinetic energy ½ m v² =13.6/n² eV, illustrating Bohr model predictions for hydrogen-like atoms. d = (2Ze²/4πepsilon₀ K) . K = 4.0 × 1.6 × 10⁻¹³ = 6.4 × 10⁻¹³ J . d = (2 × 79 × (1.6 × 10⁻¹⁹)² × 9 × 10⁹/6.4 × 10⁻¹³) . d = (3.641 × 10⁻²⁸/6.4 × 10⁻¹³) ≈ 5.69 × 10⁻¹⁴ m ≈ 57 fm . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and
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