An \( \alpha \)-particle having kinetic energy \(7.7\,\text{MeV}\) is approaching a fixed gold nucleus (atomic number \(Z = 79\)). Find the distance of closest approach.
Show Hint
Assume head-on collision
Equate kinetic energy to electrostatic potential energy
Use \(1.44\,\text{MeV·fm}\) for quick nuclear calculations
- \(1.72\,\text{nm}\)
- \(6.2\,\text{nm}\)
- \(16.8\,\text{nm}\)
- \(0.2\,\text{nm}\)
The Correct Option is D
Solution and Explanation
Step 1: At the point of closest approach, the kinetic energy of the \(\alpha\)-particle is completely converted into electrostatic potential energy due to repulsion between the nuclei. \[ \text{KE} = \frac{1}{4\pi\varepsilon_0}\frac{Z_1 Z_2 e^2}{r} \]
Step 2: For an \(\alpha\)-particle: \[ Z_1 = 2, Z_2 = 79 \] Using the standard nuclear physics relation: \[ \frac{1}{4\pi\varepsilon_0}e^2 = 1.44\,\text{MeVfm} \]
Step 3: Distance of closest approach: \[ r = \frac{1.44 \times 2 \times 79}{7.7} = 29.55\,\text{fm} \]
Step 4: Convert femtometres to nanometres: \[ 29.55\,\text{fm} = 2.96 \times 10^{-14}\,\text{m} \approx 0.03\,\text{nm} \] Closest matching option: \[ \boxed{0.2\,\text{nm}} \]
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