The 'distance of closest approach' of an alpha-particle is '$d$' when it moves with a velocity $v$ head-on towards the target nucleus. If the velocity of alpha particle is halved, the new 'distance of closest approach' will be (A) $\dfrac{d}{2}$ (B) $2d$ (C) $\dfrac{d}{4}$ (D) $4d$

The 'distance of closest approach' of an alpha-particle is 'd' when it moves with a velocity v head-on towards

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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. …

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 …

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At closest approach, all kinetic energy converts to electrostatic potential energy. Since $KE \propto v^2$ , halving the velocity quarters the kinetic energy, which means the alpha-particle cannot penetrate as deeply — the distance of closest approach becomes $4d$ .

Why Distance of Closest Approach Depends on Kinetic Energy

When an alpha-particle ( $\alpha$ , carrying charge $+2e$ ) is fired head-on at a nucleus (charge $+Ze$ ), it slows down as electrostatic repulsion does negative work. At the distance of closest approach, the particle momentarily stops: all its initial kinetic energy has been converted into electrostatic potential energy.

The key insight is that the distance of closest approach is determined entirely by energy conservation. The greater the initial kinetic energy, the closer the alpha-particle can get before being turned back.

Step-by-Step Solution

Write the energy conservation equation at closest approach.

Initially, the alpha-particle has kinetic energy $KE = \frac{1}{2}m v^2$ and is far from the nucleus (so $PE \approx 0$ ). At closest approach (distance $d$ ), it has zero velocity and maximum potential energy:

\frac{1}{2}m v^2 = k \frac{(2e)(Ze)}{d}

where $k = \frac{1}{4\pi\epsilon_0}$ is Coulomb's constant.

Solve for the distance of closest approach $d$ .

Rearranging:

d = \frac{2kZe^2}{\frac{1}{2}m v^2} = \frac{4kZe^2}{m v^2}

This shows that $d \propto \frac{1}{v^2}$ .

Find the new distance when velocity is halved.

If the new velocity is $v' = \frac{v}{2}$ , the new distance $d'$ is: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set 55/1/11 markMCQ
$d$
›Reveal solutionSolution
At closest approach, all kinetic energy converts to electrostatic potential energy. Since $KE \propto v^2$ , halving the velocity quarters the kinetic energy, which means the alpha-particle cannot penetrate as deeply — the distance of closest approach becomes $4d$ .

Why Distance of Closest Approach Depends on Kinetic Energy

When an alpha-particle ( $\alpha$ , carrying charge $+2e$ ) is fired head-on at a nucleus (charge $+Ze$ ), it slows down as electrostatic repulsion does negative work. At the distance of closest approach, the particle momentarily stops: all its initial kinetic energy has been converted into electrostatic potential energy.

The key insight is that the distance of closest approach is determined entirely by energy conservation. The greater the initial kinetic energy, the closer the alpha-particle can get before being turned back.

Step-by-Step Solution

Write the energy conservation equation at closest approach.

Initially, the alpha-particle has kinetic energy $KE = \frac{1}{2}m v^2$ and is far from the nucleus (so $PE \approx 0$ ). At closest approach (distance $d$ ), it has zero velocity and maximum potential energy:

$\frac{1}{2}m v^2 = k \frac{(2e)(Ze)}{d}$
where $k = \frac{1}{4\pi\epsilon_0}$ is Coulomb's constant.

Solve for the distance of closest approach $d$ .

Rearranging:

$d = \frac{2kZe^2}{\frac{1}{2}m v^2} = \frac{4kZe^2}{m v^2}$
This shows that $d \propto \frac{1}{v^2}$ .

Find the new distance when velocity is halved.

If the new velocity is $v' = \frac{v}{2}$ , the new distance $d'$ is: …
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CBSE 2026Set ANNUAL1 mark
$Q. The perpendicular distance of the initial velocity vector of α-particle from the centre of the nucleus is termed as ________.$
›Reveal solutionSolution

This perpendicular distance is called the impact parameter (b); it determines how sharply an alpha particle is deflected in Rutherford scattering.

In Rutherford's alpha-particle scattering experiment, each incoming alpha particle travels toward the nucleus along a straight line in the absence of any deflecting force. The perpendicular distance between this initial (undeflected) line of approach and the centre of the target nucleus is defined as the impact parameter, b. A large impact parameter (the particle's path passes far from the nucleus) gives only a small deflection, while a very small impact parameter (a nearly head-on approach) produ …

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CBSE 2024Set 55/5/11 markMCQ
$Q. Assertion (A): An alpha particle is moving towards a gold nucleus. The impact parameter is maximum for the scattering angle of 180°. Reason (R): The impact parameter in an alpha particle scattering experiment does not depend upon the atomic number of the target nucleus. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is NOT the correct explanation of A. (C) A is true but R is false. (D) Both A and R are false.$
›Reveal solutionSolution

Impact parameter $b=\dfrac{Ze^2\cot(\theta/2)}{4\pi\epsilon_0 E}$ is minimum (zero) for a head-on collision ( $\theta=180^\circ$ ), not maximum -- Assertion (A) is false. The same formula shows $b$ depends directly on $Z$ (the atomic number), so Reason (R) is also false. Option (D).

Impact parameter in Rutherford scattering

The impact parameter $b$ (perpendicular distance between the incident alpha particle's initial path and the nucleus) is related to the scattering angle $\theta$ by
$b=\frac{Ze^2\cot(\theta/2)}{4\pi\epsilon_0 E},$
where $Z$ is the atomic number of the target nucleus and $E$ is the alpha particle's kinetic energy.
$b=\frac{Ze^2}{4\pi\epsilon_0 E}\cot\left(\frac{\theta}{2}\right)$
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CBSE 2024Set 55/2/11 markMCQ
$d$
›Reveal solutionSolution
At closest approach, all kinetic energy converts to electrostatic potential energy; equating $K = \frac{kq_1q_2}{d}$ shows the distance of closest approach is inversely proportional to kinetic energy: $d \propto \frac{1}{K}$ .

The Geiger-Marsden experiment revealed the nuclear structure of the atom through alpha-particle scattering. When an alpha particle approaches a gold nucleus head-on, it experiences a repulsive Coulomb force that slows it down. At the distance of closest approach, the particle momentarily stops before reversing direction. This is a pure energy-conversion problem: kinetic energy transforms entirely into electrostatic potential energy.

The key insight is conservation of energy. Initially, the alpha particle has kinetic energy $K$ and negligible potential energy (it starts far away). At closest approach distance $d$ , the particle has zero kinetic energy and maximum potential energy.

Write the initial energy state. Far from the nucleus, the alpha particle has kinetic energy $K$ and potential energy $U_i \approx 0$ (taking $U = 0$ at infinity).

$E_{\text{initial}} = K + 0 = K$

Write the final energy state at closest approach. At distance $d$ , the particle stops momentarily, so kinetic energy is zero. The potential energy between the alpha particle (charge $q_\alpha = 2e$ ) and gold nucleus (charge $q_{\text{Au}} = Ze$ , where $Z = 79$ for gold) is:

$U_{\text{final}} = \frac{kq_\alpha q_{\text{Au}}}{d} = \frac{k(2e)(Ze)}{d} = \frac{2kZe^2}{d}$
$E_{\text{final}} = 0 + \frac{2kZe^2}{d}$ …
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CBSE 2024Set A11 markMCQ
$\alpha$
(a) both kinetic energy and potential energy are equal
(b) entire kinetic energy is converted into potential energy
(c) entire potential energy is converted into kinetic energy
(d) both kinetic energy and potential energy are zero
›Reveal solutionSolution

(b) entire kinetic energy is converted into potential energy. …

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CBSE 2019Set ANNUAL1 markMCQ
$r_0$
(a) $4r_0$
(b) $\dfrac{r_0}{4}$
(c) $2r_0$
(d) $\dfrac{r_0}{2}$
›Reveal solutionSolution

Because the distance of closest approach is inversely proportional to the square of the momentum, halving the momentum quadruples the distance of closest approach.

In Rutherford scattering, an alpha particle approaching a nucleus head-on is decelerated by the repulsive Coulomb force until, at the distance of closest approach $d$ , all of its kinetic energy has converted into electrostatic potential energy: $KE = \dfrac{1}{4\pi\epsilon_0}\dfrac{(2e)(Ze)}{d}$

Rearranging, $d = \dfrac{1}{4\pi\epsilon_0}\dfrac{2Ze^2}{KE}$ , so the distance of closest approach is inversely proportional to the kinetic energy: $d \propto \dfrac{1}{KE}$ .

The kinetic energy is related to momentum by $KE = \dfrac{p^2}{2m}$ , so for the same alpha particle (same mass $m$ ), $d \propto \dfrac{1}{KE} \propto \dfrac{m}{p^2} \propto \dfrac{1}{p^2}$ .

…

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Q.The 'distance of closest approach' of an alpha-particle is 'ddd' when it moves with a velocity vvv head-on towards the target nucleus. If the velocity of alpha particle is halved, the new 'distance of closest approach' will be (A) d2\dfrac{d}{2}2d​ (B) 2d2d2d (C) d4\dfrac{d}{4}4d​ (D) 4d4d4d

Why Distance of Closest Approach Depends on Kinetic Energy

Step-by-Step Solution

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Q.The 'distance of closest approach' of an alpha-particle is ' $d$ ' when it moves with a velocity $v$ head-on towards the target nucleus. If the velocity of alpha particle is halved, the new 'distance of closest approach' will be (A) $\dfrac{d}{2}$ (B) $2d$ (C) $\dfrac{d}{4}$ (D) $4d$

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