Q.Answer the following questions, which help you understand the difference between Thomson's model and Rutherford's model better.

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.

A Quick Numerical Feel

For a typical alpha particle from radioactive decay (kinetic energy about $5\ \text{MeV}$) and a gold nucleus ($Z = 79$):

That's about 45 femtometers — roughly 10,000 times smaller than the atom itself. This tiny number was the first direct evidence that the positive charge in an atom is concentrated in an incredibly small nucleus.

The Key Takeaway

Rutherford scattering distance is the distance at which an alpha particle, approaching a nucleus head-on, comes to a complete stop and reverses direction. It is given by:

It is the minimum possible distance of closest approach for a given initial kinetic energy, and it revealed that the atom's positive charge is packed into a volume far smaller than the atom itself.

The distance of closest approach in Rutherford scattering is a classic numerical from the NCERT Class 12 Physics chapter on Atoms, regularly appearing in "Rutherford scattering distance of closest approach formula" searches and in JEE Main/NEET important-questions compilations on atomic structure. Mastering this derivation also builds the groundwork for later problems on nuclear size and the scale of the atom covered in the same chapter.