Practice question
Question
The kinetic energy of an alpha-particle is 7.7 MeV. If it approaches a nucleus with atomic number 79,
what is its distance of closest approach? (Use constants: \( \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \,
\text{N·m}^2/\text{C}^2 \), \( e = 1.6 \times 10^{-19} \, \text{C} \), 1 MeV = \( 1.6 \times 10^{-13} \,
\text{J} \))
Explanation
**Energy in hydrogen** total E_n = -13.6/n² eV, kinetic K = +13.6/n² eV, potential U = -27.2/n² eV, ratio K/U = -1/2, total negative indicates bound, zero at ionization, negative total means electron bound, requires energy to free. For n=4 E=-0.85 eV, U=-1.7 eV, K=0.85 eV, potential energy twice total negative. d = (2Ze²/4πepsilon₀ K) . K = 7.7 × 1.6 × 10⁻¹³ = 1.232 × 10⁻¹² J . d = (2 × 79 × (1.6 × 10⁻¹⁹)² × 9 × 10⁹/1.232 × 10⁻¹²) . d = (3.641 × 10⁻²⁸/1.232 × 10⁻¹²) ≈ 2.95 × 10⁻¹⁴ m ≈ 30 fm . Using E_n = -13.6/n²
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