Q.In Rutherford's α-scattering experiment, for what value of scattering angle, the impact parameter becomes minimum?

Rutherford Scattering Distance – From Intuition to Precision

Imagine you are firing a tiny, fast bullet at a large, heavy cannonball hidden inside a big cloud of cotton. Most bullets zip right through the cotton, barely slowing down. But a few bullets come very close to the cannonball itself. Those bullets get deflected sharply, sometimes even bouncing back.

The Rutherford scattering distance is the answer to this question: How close did that bullet get to the cannonball before it turned around?

In the real experiment, the "bullet" is an alpha particle (a helium nucleus, positively charged), the "cannonball" is the gold nucleus (also positively charged, and very heavy), and the "cotton" is the mostly empty space inside the gold atom. The alpha particle and the gold nucleus repel each other because both are positive. The closer the alpha particle gets, the stronger the repulsion.

The Intuitive Picture

Think of a ball rolling up a steep hill. The ball starts with some speed (kinetic energy). As it climbs, it slows down because gravity is pulling it back. At the very top of its climb, it stops for an instant — all its kinetic energy has been converted into gravitational potential energy. Then it rolls back down.

The alpha particle does the same thing, but with electric repulsion instead of gravity. It approaches the nucleus, slows down, stops at the closest possible point, and then flies back the way it came.

That closest point — the distance of closest approach — is the Rutherford scattering distance. It is the distance at which the alpha particle's initial kinetic energy is completely converted into electrostatic potential energy.

The Precise Statement

Let an alpha particle with charge $+2e$ and mass $m$ approach a gold nucleus with charge $+Ze$ (where $Z = 79$ for gold). The alpha particle starts from very far away with initial kinetic energy $K = \frac{1}{2} m v^2$.

At the distance of closest approach, call it $r_0$, the alpha particle's speed becomes zero. All its kinetic energy has become electrostatic potential energy:

Solving for $r_0$:

This is the Rutherford scattering distance (also called the distance of closest approach in a head-on collision).

What It Tells Us

• If the alpha particle hits the nucleus head-on, it comes exactly this close before reversing direction.
• If it misses slightly, it comes closer than $r_0$? No — it comes less close. The head-on collision gives the minimum possible distance of closest approach for a given initial energy. Any sideways motion means the particle never gets as close.
• If the initial kinetic energy is larger, $r_0$ becomes smaller — the alpha particle can punch closer to the nucleus before being stopped. …