# In Rutherford's scattering experiments, alpha particles

## Problem 4P Chapter 31

Physics with MasteringPhysics | 4th Edition

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Problem 4P

In Rutherford's scattering experiments, alpha particles (charge = +2e) were fired at a gold foil. Consider an alpha particle with an initial kinetic energy K heading directly for the nucleus of a gold atom (charge = +79e). The alpha particle will come to rest when all its initial kinetic energy has been converted to electrical potential energy. Find the distance of closest approach between the alpha particle and the gold nucleus for the case K = 3.0 MeV.

Step-by-Step Solution:

Solution 4P

Step 1 of 3</p>

Here, we have to calculate the distance of closest approach of alpha particle when it is moving towards the gold foil.

The alpha particle cannot move further beyond a specific point because of the strong electrostatic repulsion between the positively charged nucleus and the positively charged alpha particle.

Step 2 of 3</p>

The equation for the electrostatic potential,

Where,  - Charge of the gold nucleus

- Distance of closest approach

- Permittivity of free space

The electrostatic potential energy,

Where  - Charge of the alpha particle

We know that,

Provided,

Therefore, the electrostatic potential energy,

We know that,

Step 3 of 3

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