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:  
The problem involves an alpha particle being projected towards a fixed gold nucleus. The kinetic energy of the particle is given, and the goal is to find the diameter of the nucleus. The situation involves the interaction between the alpha particle and the gold nucleus, which is described using electrostatic repulsion. 
Step 1: Use the energy conservation principle. 
The kinetic energy of the alpha particle is completely converted into electrostatic potential energy at the point where the particle is just able to touch the nucleus (i.e., at the closest distance of approach). The potential energy between the alpha particle and the nucleus is given by: \[ U = \frac{1}{4 \pi \epsilon_0} \frac{Z_1 Z_2 e^2}{r} \] where: - \(Z_1 = 2\) (atomic number of the alpha particle), - \(Z_2 = 79\) (atomic number of gold), - \(e\) is the elementary charge, - \(r\) is the distance of closest approach (which corresponds to the radius of the nucleus), - \(\epsilon_0\) is the permittivity of free space. 
Step 2: Set up the equation. 
The initial kinetic energy of the alpha particle is given as 7.9 MeV. At the point of closest approach, this energy is equal to the potential energy, so: \[ K.E. = U \] \[ 7.9 \, \text{MeV} = \frac{1}{4 \pi \epsilon_0} \frac{Z_1 Z_2 e^2}{r} \] 
Step 3: Substitute values and solve for \(r\). 
Converting the given energy into joules and substituting the constants: \[ 7.9 \, \text{MeV} = 7.9 \times 10^6 \times 1.6 \times 10^{-13} \, \text{J} \] Substituting the known values of constants: \[ \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \] The equation becomes: \[ 7.9 \times 10^6 \times 1.6 \times 10^{-13} = 9 \times 10^9 \times \frac{(2)(79)(1.6 \times 10^{-19})^2}{r} \] Solving for \(r\), we get: \[ r \approx 57.6 \, \text{fm} \] 
Step 4: Conclusion. 
The diameter of the nucleus is approximately: \[ \boxed{57.6 \, \text{fm}} \]