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 establish the relationship between the stopping distance \( d \) of an alpha particle and its initial kinetic energy \( K \), principles of energy conservation and Coulomb force are applied. 1. Initial Kinetic Energy: The alpha particle begins with a kinetic energy denoted by \( K \). 2. Potential Energy at Distance \( d \): Upon halting, the alpha particle's kinetic energy is fully converted into electrostatic potential energy. The potential energy \( U \) between the alpha particle (charge \( +2e \)) and the gold nucleus (charge \( +79e \)) at distance \( d \) is calculated as: \[ U = \frac{1}{4\pi\epsilon_0} \cdot \frac{(2e)(79e)}{d} \] Here, \( \epsilon_0 \) represents the permittivity of free space. 3. Energy Conservation: At the point of cessation, the alpha particle's kinetic energy is zero, and its potential energy is equivalent to the initial kinetic energy: \[ K = \frac{1}{4\pi\epsilon_0} \cdot \frac{158e^2}{d} \] 4. Solving for \( d \): The equation is rearranged to isolate \( d \): \[ d = \frac{1}{4\pi\epsilon_0} \cdot \frac{158e^2}{K} \] This equation demonstrates an inverse proportionality between \( d \) and \( K \): \[ d \propto \frac{1}{K} \] Consequently, the correct answer is: \[ \boxed{C} \]