Question

An \(\alpha\)-particle accelerated by potential difference of V volt strikes with a nucleus (atomic no. = Z). If r be the nearest distance of the particle to reach the nucleus then prove that : \(r = 14.4 \frac{Z}{V} \AA\)

Hint:

For the alpha particle problem, energy conservation is the central idea. The conversion of kinetic energy to potential energy is a common theme in collision and scattering problems in both mechanics and electromagnetism.

Answer 1

Step 1: Understanding the Concept:
The proof is based on the principle of conservation of energy. An \(\alpha\)-particle (charge \(q_1 = +2e\)) is accelerated by a potential difference V, gaining kinetic energy. As it approaches a positively charged nucleus (charge \(q_2 = +Ze\)), the electrostatic repulsive force slows it down. The distance of closest approach (\(r\)) is the point where the \(\alpha\)-particle momentarily stops and turns back. At this point, all of its initial kinetic energy has been converted into electrostatic potential energy of the system.

Step 2: Key Formula or Approach:
1. Kinetic Energy (KE) gained by the \(\alpha\)-particle: \(KE = q_1 V = (2e)V\)
2. Electrostatic Potential Energy (PE) at separation \(r\): \(PE = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r} = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r}\)
By conservation of energy: \(KE_{\text{initial}} = PE_{\text{final}}\)

Step 3: Detailed Explanation (Derivation):
Equating the initial kinetic energy with the final potential energy: \[ (2e)V = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{r} \] We can cancel \(2e\) from both sides: \[ V = \frac{1}{4\pi\epsilon_0} \frac{Ze}{r} \] Now, we solve for the distance of closest approach, \(r\): \[ r = \left(\frac{1}{4\pi\epsilon_0}\right) \frac{Ze}{V} \] To get the numerical value, we substitute the known constants: \begin{itemize} \item Coulomb's constant, \(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\) N·m\(^2\)/C\(^2\). \item Elementary charge, \(e = 1.6 \times 10^{-19}\) C. \end{itemize} \[ r = \frac{(9 \times 10^9) \times Z \times (1.6 \times 10^{-19})}{V} \] \[ r = \frac{14.4 \times 10^{-10} \times Z}{V} \text{(in meters)} \] Since 1 Angstrom (\(\AA\)) = \(10^{-10}\) meters, we can express the distance in Angstroms: \[ r = \frac{14.4 \times Z}{V} \times 10^{-10} \text{ m} = 14.4 \frac{Z}{V} \AA \] This proves the required relation.