1 markMCQSection A

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 −

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 −

**Correct option: (D) $4d$** At closest approach, the alpha-particle's kinetic energy is fully converted to electrostatic potential energy: - $\frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0}(Ze\cdot 2e)/d$, so **$d \propto 1/v^2$**. - Halving v (→ $v/2$) multiplies d by $1/(1/2)^2$ = **4**, giving the new distance = **$4d$**.

Q: Does the mass of the alpha-particle affect the distance of closest approach?

Yes. From $d = \frac{2kq_1q_2}{mv^2}$, d is also inversely proportional to the mass m for a given speed; a heavier particle at the same speed approaches closer.

Atoms

Distance of Closest Approach (Rutherford Scattering)

Options

(A)

d/2

(B)

2d

(C)

d/4

(D)

4d

Official Answer

Correct option: (D) $4d$

At closest approach, the alpha-particle's kinetic energy is fully converted to electrostatic potential energy:

$\frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0}(Ze\cdot 2e)/d$ , so $d \propto 1/v^2$ .
Halving v (→ $v/2$ ) multiplies d by $1/(1/2)^2$ = 4, giving the new distance = $4d$ .

distance of closest approachalpha particled proportional to 1/v^2kinetic energyelectrostatic potential energyRutherford scatteringhead-on collision4d

Marking Scheme

11 mark: correct option (D) $4d$ .
2Key reasoning: $d \propto 1/v^2$ from $\frac{1}{2}mv^2 = \frac{kq_1q_2}{d}$ , so halving v gives $4d$ .

Hint

Equate kinetic energy $\frac{1}{2}mv^2$ to Coulomb potential energy — notice d depends on $v^2$ , not v.

Quick Oral Answer

Since the alpha-particle's kinetic energy fully converts to electrostatic potential energy at the closest point, the distance is inversely proportional to the square of the speed, so halving the velocity makes the distance four times larger.

Analysis & Explanation

This tests the energy-conservation derivation of the distance of closest approach in Rutherford's alpha-scattering experiment.

Concept: For a head-on collision the alpha-particle momentarily stops when all its kinetic energy becomes potential energy:

$\frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0}\cdot\frac{2Ze^2}{d} \implies d = \frac{1}{4\pi\varepsilon_0}\cdot\frac{4Ze^2}{mv^2}$ .
Thus d is inversely proportional to $v^2$ (for fixed nucleus and charge).

Why (D) is correct: Replacing v by $v/2$ gives $d' \propto \frac{1}{(v/2)^2} = \frac{4}{v^2}$ , i.e. $d' = 4d$ .

Why the others are wrong:

(A) $d/2$ and (C) $d/4$ wrongly assume d increases with speed or scales linearly with v.
(B) $2d$ comes from taking $d \propto 1/v$ (linear) instead of the correct $1/v^2$ dependence.

Exam trap: Students often forget the square in the kinetic-energy term, giving a factor of 2 instead of 4.

Common Mistakes

1Using $d \propto 1/v$ (linear) instead of $d \propto 1/v^2$ , giving $2d$ rather than the correct $4d$ .
2Assuming the distance decreases when speed decreases — a slower particle is repelled sooner, so it stops farther away.
3Forgetting that at closest approach the velocity is momentarily zero, so all $KE = PE$ .

Interesting Facts

Rutherford's 1911 analysis of the closest approach (a few $\times 10^{-14}$ m) first revealed the tiny, dense atomic nucleus, overturning Thomson's plum-pudding model.

The distance of closest approach gives an upper estimate of nuclear size; for typical alpha energies it is about $10^4$ times smaller than the atom itself.

Spotted a mistake or something unclear?

Tell us — we fix reported answers fast.

Frequently Asked Questions

Why is the distance of closest approach inversely proportional to $v^2$ ?

At the closest point the alpha-particle stops, so its entire kinetic energy $\frac{1}{2}mv^2$ equals the Coulomb potential energy $\frac{kq_1q_2}{d}$ . Solving for d gives $d = \frac{2kq_1q_2}{mv^2}$ , which is inversely proportional to v².

Does the mass of the alpha-particle affect the distance of closest approach?

Yes. From $d = \frac{2kq_1q_2}{mv^2}$ , d is also inversely proportional to the mass m for a given speed; a heavier particle at the same speed approaches closer.