Question

An alpha particle approaches a gold nucleus in Geiger-Marsden experiment with kinetic energy \( K \). It momentarily stops at a distance \( d \) from the nucleus and reverses its direction. Then \( d \) is proportional to:

• A. \( \frac{1}{\sqrt{K}} \)
• B. \( \sqrt{K} \)
• C. \( \frac{1}{k} \)
• D. \( K \)

Hint:

The distance at which a charged particle stops due to Coulomb's interaction is inversely proportional to the square root of its kinetic energy.

Answer 1

The Correct Option is A

To determine the relationship between the distance \( d \) at which the alpha particle stops and its initial kinetic energy \( K \), we can use the principles of energy conservation and the Coulomb force. 1. Initial Kinetic Energy: The alpha particle has an initial kinetic energy \( K \). 2. Potential Energy at Distance \( d \): When the alpha particle stops, all its kinetic energy is converted into electrostatic potential energy. The potential energy \( U \) between the alpha particle (charge \( +2e \)) and the gold nucleus (charge \( +79e \)) at a distance \( d \) is given by: \[ U = \frac{1}{4\pi\epsilon_0} \cdot \frac{(2e)(79e)}{d} \] where \( \epsilon_0 \) is the permittivity of free space. 3. Energy Conservation: At the point where the alpha particle stops, its kinetic energy is zero, and its potential energy equals the initial kinetic energy: \[ K = \frac{1}{4\pi\epsilon_0} \cdot \frac{158e^2}{d} \] 4. Solving for \( d \): Rearrange the equation to solve for \( d \): \[ d = \frac{1}{4\pi\epsilon_0} \cdot \frac{158e^2}{K} \] This shows that \( d \) is inversely proportional to \( K \): \[ d \propto \frac{1}{K} \] Therefore, the correct answer is: \[ \boxed{C} \]