Distance of Closest Approach Formula: r₀ = 2kZe²/K

The distance of closest approach formula is $r_0 = \frac{2Ze^2}{4\pi\varepsilon_0} \cdot \frac{1}{K}$ : it gives how near an alpha particle fired straight at a nucleus can get before the electric repulsion stops it. Here 2e is the alpha particle's charge, Ze is the nuclear charge and K is the kinetic energy the alpha particle starts with, far from the nucleus.

This result comes from the alpha particle scattering experiment in the Class 12 chapter on atoms, and it gives an upper limit on the size of a nucleus. It is pure energy conservation: all the kinetic energy turns into electric potential energy at the turning point. Revise the Coulomb side on the electrostatic force formula page, and see where the topic sits on the modern physics formula sheet.

Quick facts

Formula
$r_0 = \frac{2kZe^2}{K}$ , with $k = \frac{1}{4\pi\varepsilon_0}$
SI unit
metre (m); answers are usually quoted in femtometres, 1 fm = 10⁻¹⁵ m
Dimensional formula
[M⁰ L¹ T⁰]
Quick form
$r_0\,(\text{fm}) = \frac{2.88\,Z}{K\,(\text{MeV})}$
Condition
Head on approach (impact parameter zero), nucleus treated as fixed
Dependence
$r_0 \propto \frac{Z}{K}$ : double K and r₀ halves

Derivation of the distance of closest approach formula

Fire the alpha particle head on. Far away it has kinetic energy K and almost no potential energy. As it nears the nucleus, repulsion slows it. At the distance of closest approach it is momentarily at rest, so all of K has become electric potential energy.

Charges: alpha particle $q_1 = 2e$ , nucleus $q_2 = Ze$ .
Potential energy at separation r: $U = \frac{1}{4\pi\varepsilon_0}\,\frac{(2e)(Ze)}{r}$ .
Energy conservation at the turning point: $K = \frac{1}{4\pi\varepsilon_0}\,\frac{2Ze^2}{r_0}$ .
Rearrange: $r_0 = \frac{2Ze^2}{4\pi\varepsilon_0} \cdot \frac{1}{K}$ .

The nucleus is taken as fixed because a gold nucleus is about 50 times heavier than an alpha particle. The potential energy term is the same one that appears for any two point charges.

What each symbol stands for

Symbols and values
Symbol	Meaning	Value or unit
$r_0$	Closest distance between the centres	m (often fm)
$Z$	Atomic number of the target, 79 for gold	no unit
$e$	Size of the charge on one proton	1.602 × 10⁻¹⁹ C
$K$	Kinetic energy of the incoming alpha particle	J
$\frac{1}{4\pi\varepsilon_0}$	Coulomb's constant	8.99 × 10⁹ N m²/C² (board answers use 9 × 10⁹)

Questions usually give K in MeV. Multiply by 1.602 × 10⁻¹³ J per MeV before you substitute. A shortcut avoids the conversion: $\frac{e^2}{4\pi\varepsilon_0} = 1.44$ MeV fm, so $r_0 = \frac{2.88\,Z}{K}$ with K in MeV gives r₀ straight in femtometres.

Solved examples on closest distance of approach

Example 1: alpha particle on gold

A 5.0 MeV alpha particle moves head on toward a gold nucleus (Z = 79). Find the distance of closest approach.

Convert: K = 5.0 × 1.602 × 10⁻¹³ = 8.01 × 10⁻¹³ J.

$r_0 = \frac{8.99 \times 10^9 \times 2 \times 79 \times (1.602 \times 10^{-19})^2}{8.01 \times 10^{-13}}$

$r_0 = 4.55 \times 10^{-14}\ \text{m} \approx 46\ \text{fm}$

Check with the shortcut: 2.88 × 79 / 5.0 = 45.5 fm.

Example 2: speed given instead of energy

An alpha particle of mass 6.64 × 10⁻²⁷ kg moves at 2.0 × 10⁷ m/s toward a silver nucleus (Z = 47). Find r₀.

First find K: $K = \frac{1}{2}mv^2 = 0.5 \times 6.64 \times 10^{-27} \times (2.0 \times 10^7)^2 = 1.328 \times 10^{-12}\ \text{J}$ .

$r_0 = \frac{8.99 \times 10^9 \times 2 \times 47 \times (1.602 \times 10^{-19})^2}{1.328 \times 10^{-12}} = 1.63 \times 10^{-14}\ \text{m}$

That is about 16 fm.

Example 3: a ratio question

The kinetic energy of the alpha particle is doubled. What happens to the distance of closest approach?

Since $r_0 \propto \frac{1}{K}$ , doubling K makes r₀ half its old value. If the speed is doubled instead, K becomes four times larger and r₀ falls to one quarter.

Impact parameter and the general case

Most alpha particles do not head straight for the nucleus. The impact parameter b is the perpendicular distance between the nucleus and the line of the alpha particle's original path. It links to the scattering angle θ by

$b = \frac{1}{4\pi\varepsilon_0}\,\frac{Ze^2 \cot(\theta/2)}{K} = \frac{r_0}{2}\cot\frac{\theta}{2}$

For b = 0 the angle is 180°, a straight bounce back, and the particle reaches the true r₀. A larger b means a smaller deflection and a turning point further out. Very few particles turned back in the experiment, which showed that the positive charge sits in a tiny core. The same energy method works for any charged projectile: replace 2e with its charge q to get $r_0 = \frac{kqZe}{K}$ , for example for a proton, with charge e.

Closest approach and the size of the nucleus

The alpha particle cannot get inside the nucleus without touching it, so r₀ is an upper limit on the nuclear radius. Nuclear radii follow $R = R_0 A^{1/3}$ with $R_0 \approx 1.2$ fm. For gold (A = 197) that gives about 7.0 fm, well inside the 46 fm found in Example 1. Only at much higher energies does the approach distance shrink to the nuclear surface, where the strong nuclear force starts to act and the simple Coulomb picture fails.

An r₀ smaller than about 10 fm for a heavy nucleus means the formula is being pushed beyond its range. Check the energy you used.

Common mistakes in exams

Using e instead of 2e for the alpha charge. The alpha particle carries two protons, so the numerator is 2Ze², not Ze².
Leaving K in MeV with SI constants. Either convert to joules or use the 1.44 MeV fm shortcut, never a mix.
Squaring the speed wrongly. When v is given, compute the kinetic energy K = ½mv² first and check the power of ten.
Applying it to an oblique path. The simple formula holds only for a head on approach.
Writing the answer without a unit. State metres or femtometres, and give 2 or 3 significant figures.

You can check your arithmetic with the calculator above by entering Z and K in joules.

Frequently asked questions

What is the formula of distance of closest approach?

For an alpha particle aimed straight at a nucleus of atomic number Z, r₀ = (1/4πε₀) × 2Ze² / K, where K is the alpha particle's initial kinetic energy. It comes from setting K equal to the electric potential energy at the turning point.

What is the distance of closest approach formula in Class 12?

Class 12 books write r₀ = 2Ze² / (4πε₀ K), sometimes with K written as ½mv². The chapter on atoms uses it to show that a nucleus is smaller than about 10⁻¹⁴ m.

How does the distance of closest approach depend on speed?

Since K = ½mv², r₀ is proportional to 1/v². Doubling the speed of the alpha particle makes the distance of closest approach four times smaller.

What is the difference between impact parameter and distance of closest approach?

The impact parameter is the sideways offset of the incoming path from the nucleus. The distance of closest approach is the smallest separation actually reached. For a head on shot the impact parameter is zero and the separation is r₀.

Is the formula for distance of closest approach valid for a proton?

Yes, with the charge changed. A proton has charge e, so r₀ = (1/4πε₀) × Ze² / K. In general use the product of the two charges divided by the kinetic energy.