Question

An \(\alpha\)-particle is projected towards a fixed gold nucleus (\(Z=79\)) with kinetic energy \(7.9\,\text{MeV}\). If the particle is just able to touch the nuclear boundary, find the diameter of the nucleus.

• A. \(57.6\,\text{fm}\)
• B. \(45.6\,\text{fm}\)
• C. \(36.6\,\text{fm}\)
• D. \(20.6\,\text{fm}\)

Hint:

For closest approach problems:
Use head-on collision assumption
Equate kinetic energy to Coulomb potential energy
Remember the constant \( \dfrac{e^2}{4\pi\varepsilon_0} = 1.44\,\text{MeV·fm} \)

Answer 1

The Correct Option is A

Concept: 
When an \(\alpha\)-particle approaches a heavy nucleus head-on, it is slowed down by the electrostatic (Coulomb) repulsion. At the distance of closest approach, the initial kinetic energy of the \(\alpha\)-particle is completely converted into electrostatic potential energy. For a head-on collision: \[ \text{K.E.} = \text{Electrostatic P.E.} \] The Coulomb potential energy between two charges is: \[ U = \frac{1}{4\pi\varepsilon_0}\frac{Z_1 Z_2 e^2}{r} \] 
Step 1: Identify the charges. 
Charge of \(\alpha\)-particle: \(Z_1 = 2\) 
Charge of gold nucleus: \(Z_2 = 79\) 
Step 2: Write the energy balance equation. Given kinetic energy: \[ K = 7.9\,\text{MeV} \] At closest approach \(r = r_{\min}\): \[ 7.9 = \frac{1}{4\pi\varepsilon_0}\frac{(2)(79)e^2}{r_{\min}} \] Using the standard nuclear physics constant: \[ \frac{e^2}{4\pi\varepsilon_0} = 1.44\text{MeVfm} \] 
Step 3: Substitute numerical values. \[ 7.9 = \frac{2 \times 79 \times 1.44}{r_{\min}} \] \[ 7.9 = \frac{227.52}{r_{\min}} \] \[ r_{\min} = \frac{227.52}{7.9} = 28.8\,\text{fm} \] 
Step 4: Find the diameter of the nucleus. Since the \(\alpha\)-particle just touches the nuclear boundary: \[ \text{Radius of nucleus} = r_{\min} \] \[ \text{Diameter} = 2r_{\min} = 2 \times 28.8 = 57.6\,\text{fm} \] \[ \boxed{\text{Diameter of nucleus} = 57.6\,\text{fm}} \]