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

To calculate the distance of closest approach for an \( \alpha \)-particle approaching a copper nucleus, we use the principle of energy conservation. For an \( \alpha \)-particle (He nucleus, \( Z = 2 \)), the initial kinetic energy is converted into potential energy due to electrostatic forces at the point of closest approach.

Concept and Formula Used: 

The potential energy at the distance of closest approach (\( r_0 \)) is given by the electrostatic potential energy between two charges:

\(U = \frac{k \cdot Z_1 \cdot Z_2 \cdot e^2}{r_0}\)

where:

• \(k = \frac{1}{4\pi\varepsilon_0}\) is Coulomb's constant, approximately \( 8.9875 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
• \(Z_1 = 2\) (atomic number of alpha particle)
• \(Z_2 = 29\) (atomic number of copper nucleus)
• \(e = 1.6 \times 10^{-19} \, \text{C}\) is the charge of an electron
• \(r_0\) is the distance of closest approach

At the point of closest approach, the initial kinetic energy of the alpha particle equals its potential energy:

\(\text{KE} = \frac{1}{2} m v^2 = U\)

Given \(\text{K.E.} = 8\,\text{MeV}\), convert this energy into joules:

\(8\,\text{MeV} = 8 \times 1.6 \times 10^{-13}\,\text{J} = 1.28 \times 10^{-12} \,\text{J}\)

Calculation:

\(1.28 \times 10^{-12} = \frac{k \cdot 2 \cdot 29 \cdot (1.6 \times 10^{-19})^2}{r_0}\)

Plugging in the constants:

\(r_0 = \frac{8.9875 \times 10^9 \cdot 2 \cdot 29 \cdot (1.6 \times 10^{-19})^2}{1.28 \times 10^{-12}}\)

Calculate \(r_0\):

\(r_0 = 10.4 \times 10^{-15} \,\text{m} = 10.4\,\text{fm}\)

Conclusion:

The distance of closest approach by the alpha particle to the copper nucleus is \( 10.4\,\text{fm} \), which matches the correct option.