Suppose an alpha particle accelerated by a potential of $V$ volt is allowed to collide with a nucleus whose atomic number is $Z$, then the distance of closest approach of alpha particle to the nucleus is (a) $14.4\dfrac{Z}{V}$ Å (b) $14.4\dfrac{V}{Z}$ Å (c) $1.44\dfrac{Z}{V}$ Å (d) $1.44\dfrac{V}{Z}$ Å

Suppose an alpha particle accelerated by a potential of V volt is allowed to collide with a nucleus whose atom

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V

(a)

14.4\dfrac{Z}{V}

(b)

14.4\dfrac{V}{Z}

(c)

1.44\dfrac{Z}{V}

(d)

1.44\dfrac{V}{Z}

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Concept understanding — Rutherford Scattering Distance

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.

Note

This distance is not the radius of the nucleus. It is the distance at which the alpha particle would just touch the nucleus if the nucleus were a point charge. In reality, the alpha particle never actually reaches the nucleus — it turns around before that.

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:

K = \frac{1}{4\pi\varepsilon_0} \cdot \frac{(2e)(Ze)}{r_0}

Solving for $r_0$ :

r_0 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2Ze^2}{K}

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.

Watch out

Do not confuse this with the impact parameter (the perpendicular distance from the nucleus to the initial line of motion). The impact parameter is a different quantity — it tells you how "off-center" the collision is. The Rutherford scattering distance $r_0$ is the actual closest distance achieved during the collision, which depends on both the impact parameter and the initial energy.

A Quick Numerical Feel

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

r_0 \approx \frac{9 \times 10^9 \cdot 2 \cdot 79 \cdot (1.6 \times 10^{-19})^2}{5 \times 10^6 \cdot 1.6 \times 10^{-19}} \approx 4.5 \times 10^{-14}\ \text{m}

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:

\boxed{r_0 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2Ze^2}{K}}

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.

Why this formula?

Rutherford Scattering: Why the Distance of Closest Approach Formula Works

The distance of closest approach — often denoted $d_0$ or $r_0$ — is the minimum separation between an alpha particle and the nucleus in a head-on collision. It's a beautiful example of energy conservation doing all the heavy lifting.

The Physical Picture

Imagine an alpha particle (charge $+2e$ ) fired straight at a gold nucleus (charge $+Ze$ ). As it approaches, the Coulomb repulsion slows it down. At the point of closest approach, the alpha particle's radial velocity becomes zero — it stops moving toward the nucleus, and is about to turn around and fly back.

At that instant, all the kinetic energy it had at infinity has been converted into electrostatic potential energy. No other forces are at play (gravity is negligible, and we're far from the nuclear force range).

The Derivation in One Step

Let the alpha particle have initial kinetic energy $K = \frac{1}{2} m v^2$ at a large distance (where potential energy is zero). At the distance of closest approach $r_0$ , its speed is zero, so kinetic energy is zero. Energy conservation gives:

\frac{1}{2} m v^2 = \frac{1}{4\pi\epsilon_0} \cdot \frac{(2e)(Ze)}{r_0}

r_0 = \frac{1}{4\pi\epsilon_0} \cdot \frac{2Ze^2}{K}

That's it. The formula is a direct consequence of energy conservation in a pure Coulomb field.

Why This Makes Physical Sense

Higher kinetic energy → the alpha particle can push closer before being stopped → $r_0$ is smaller.
Higher nuclear charge $Z$ → stronger repulsion → the alpha stops farther away → $r_0$ is larger.
The factor $2Ze^2$ comes from the product of charges: $(2e)(Ze) = 2Ze^2$ .

Note

This is the head-on distance. For non-head-on collisions (nonzero impact parameter), the distance of closest approach is larger because some energy remains in the perpendicular component of motion. The general formula involves the impact parameter $b$ and scattering angle $\theta$ , but the head-on case gives the absolute minimum possible approach.

A Common Misconception

Students sometimes think the alpha particle "hits" the nucleus at $r_0$ . It doesn't — it's turned around purely by the electric field. The fact that Rutherford's experiment did see some alpha particles bounce back at large angles (implying they got very close) is what told him the nucleus must be extremely small — smaller than $r_0$ for those particles. If the nucleus were larger, the alpha would have hit it and the scattering pattern would have been different.

Watch out

The formula $r_0 = \frac{2Ze^2}{4\pi\epsilon_0 K}$ assumes the nucleus is point-like and stationary. In reality, the nucleus recoils slightly, so the reduced mass should technically be used. But for gold ( $A \approx 197$ ) vs alpha ( $A=4$ ), the correction is tiny — less than 2%.

The Deeper Insight

Rutherford didn't just measure $r_0$ — he used it to set an upper limit on nuclear size. By observing that alpha particles with kinetic energy $K$ were still being scattered (not absorbed), he knew the nucleus must be smaller than the corresponding $r_0$ . This gave the first experimental evidence that the atom's positive charge is concentrated in a region less than $10^{-14}$ m across — a thousand times smaller than the atom itself.

That's why this simple energy-conservation formula is historically monumental: it opened the door to the nuclear age.

Step 1. An alpha particle carries charge $2e$ . Accelerated through potential $V$ , its kinetic energy is $E_k=2eV$ .

Step 2. At the distance of closest approach $r_0$ (head-on collision), all of this kinetic energy has converted to electrostatic potential energy between the alpha particle (charge $2e$ ) and the nucleus (charge $Ze$ ):

2eV=\frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{r_0}\ \Rightarrow\ r_0=\frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{eV}=\frac{1}{4\pi\varepsilon_0}\frac{Ze}{V}

Step 3. Using the standard constant $\dfrac{e}{4\pi\varepsilon_0}$ evaluated in the units this classic result is conventionally quoted in gives the numerical coefficient $1.44$ , so

r_0=1.44\,\frac{Z}{V}\ \text{Å}

Step 4. Checking the other options: (a) and (b) have the wrong power balance between $Z$ and $V$ or the wrong numerical prefactor (10 times too large); (d) inverts the correct $Z/V$ dependence - a larger accelerating potential $V$ should let the alpha particle approach closer, i.e. $r_0$ must decrease as $V$ increases, so any option with $V$ in the numerator is physically wrong.

✓Final answer

1.44\dfrac{Z}{V}

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Q.Suppose an alpha particle accelerated by a potential of VVV volt is allowed to collide with a nucleus whose atomic number is ZZZ, then the distance of closest approach of alpha particle to the nucleus is

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Q.Suppose an alpha particle accelerated by a potential of $V$ volt is allowed to collide with a nucleus whose atomic number is $Z$ , then the distance of closest approach of alpha particle to the nucleus is

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