Question

Step 1

The above problem can be understood as a case of energy conservation.

As the charges on both the alpha particle and the gold nucleus are postive and the gold nucleus being stationary, the alpha particle will eventually begin to slow down as it reaches near the gold nucleus due to repulsion.

Step 2

Initially when the alpha particle is very far from the gold nucleus the repulsive effect on it due to the stationary gold nucleus would be negligible and it can be assumed to possess total energy in the form of kinetic energy.

As the alpha particle begins to approach the gold nucleus, the alpha particle and the gold nucleus will together possess electrical potential energy due to the repulsion between them.

Step 3

Let the kinetic energy of the alpha particle at a very large distance from the nucleus and at the turning point close to the nucleus be denoted as *Ki *and *Kf* .

Let the electrical potential energy of the system of a...

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