Question

An \( \alpha \)-particle of energy \(8\,\text{MeV}\) is directed towards a fixed copper nucleus \((Z = 29)\). Calculate the distance of closest approach.

• A. \(8.0\,\text{fm}\)
• B. \(9.6\,\text{fm}\)
• C. \(10.4\,\text{fm}\)
• D. \(12.0\,\text{fm}\)

Hint:

For distance of closest approach problems, directly use the energy form \[ r = \frac{1.44\,Z_1 Z_2}{E}\;(\text{in fm, if } E \text{ is in MeV}). \]

Answer 1

The Correct Option is C

Step 1: Recall the formula for distance of closest approach. 
For a head-on collision between an \( \alpha \)-particle and a nucleus, the distance of closest approach is given by: \[ r = \frac{1}{4\pi\varepsilon_0}\,\frac{Z_1 Z_2 e^2}{E}. \] In nuclear units: \[ \frac{e^2}{4\pi\varepsilon_0} = 1.44\,\text{MeVfm}. \] 
Step 2: Substitute known values. 
Charge of \( \alpha \)-particle: \[ Z_1 = 2, \] Charge of copper nucleus: \[ Z_2 = 29, \] Energy of \( \alpha \)-particle: \[ E = 8\,\text{MeV}. \] 
Step 3: Perform the calculation. 
\[ r = \frac{1.44 \times (2 \times 29)}{8} = \frac{1.44 \times 58}{8} = \frac{83.52}{8} = 10.44\,\text{fm}. \] 
Step 4: Final conclusion. 
The distance of closest approach is approximately: \[ \boxed{10.4\,\text{fm}}. \]