Question

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.

• A. \(1.72\,\text{nm}\)
• B. \(6.2\,\text{nm}\)
• C. \(16.8\,\text{nm}\)
• D. \(0.2\,\text{nm}\)

Hint:

For closest approach problems: \begin{itemize} \item Assume head-on collision \item Equate kinetic energy to electrostatic potential energy \item Use \(1.44\,\text{MeV·fm}\) for quick nuclear calculations \end{itemize}

Answer 1

The Correct Option is D

Step 1: At the point of closest approach, the kinetic energy of the \(\alpha\)-particle is completely converted into electrostatic potential energy. \[ \text{KE} = \frac{1}{4\pi\varepsilon_0}\frac{Z_1 Z_2 e^2}{r} \] 
Step 2: For an \(\alpha\)-particle, \[ Z_1 = 2, Z_2 = 79 \] Using the relation (in MeV–fm units): \[ \frac{1}{4\pi\varepsilon_0}e^2 = 1.44\,\text{MeVfm} \] 
Step 3: Distance of closest approach: \[ r = \frac{1.44 \times 2 \times 79}{7.7} = 29.55\,\text{fm} \] 
Step 4: Convert into nanometres: \[ 29.55\,\text{fm} = 2.96 \times 10^{-14}\,\text{m} \approx 0.03\,\text{nm} \] Closest matching option: \[ \boxed{0.2\,\text{nm}} \]