Rutherford’s experiment told us the nucleus is tiny — but how tiny, exactly? The trick is beautifully simple. Fire a positive alpha particle straight at a nucleus and it slows, stops, and reverses, like a ball rolled up a hill. At the point where it momentarily stops, all its kinetic energy has become electrical potential energy. Set those two equal and you can calculate exactly how close it got — the distance of closest approach — which puts an upper limit on the nuclear radius. From there, one elegant formula gives the size of any nucleus.
📘 What you need to know
A head-on alpha particle slows to a momentary stop at the distance of closest approach d
At that point, all its kinetic energy has become electric potential energy: Ek = Ep
Electric potential energy: Ep = kQq/d, with alpha charge Q = 2e and nucleus charge q = Ze
Rearranging gives the distance of closest approach: d = k(2Ze2)/Ek
This d is an upper limit on the nuclear radius
Nuclear radius depends on nucleon number: R = R0A1/3, with the Fermi radiusR0 = 1.20 × 10−15 m
Nuclear density is constant for all nuclei (~1017 kg m−3) — nucleons are evenly packed
Nuclear density ≫ atomic density, confirming the atom is mostly empty space
The distance of closest approach
Imagine firing an alpha particle dead-on at a gold nucleus. Both are positive, so the nucleus repels the alpha, slowing it down. The alpha keeps going until it stops completely for an instant, then gets pushed straight back the way it came. At that turning point, its speed — and so its kinetic energy — is zero. All of it has been converted into electric potential energy.
The alpha comes in with all its energy kinetic, slows as it’s repelled, and stops at distance d where all its energy is now potential. Then it’s flung back. Setting Ek = Ep gives d.
Setting the initial kinetic energy equal to the electric potential energy at the closest point:
Energy conservation at closest approachEk = Ep = kQq/dalpha: Q = 2e • nucleus: q = Ze → Ek = k(2Ze2)/d
Rearrange for the distance of closest approachd = k(2Ze2) / EkZ = proton number of the nucleus • Ek = initial kinetic energy of the alpha
All kinetic Ek
converts to
All potential Ep
solve for
Closest approach d
Remember the charge of the alpha is 2e — it’s a helium nucleus, two protons. That’s where the “2” in d = k(2Ze2)/Ek comes from. And the value of d you get is an upper limit on the nuclear radius, not the exact radius: the alpha stops when repulsion balances its energy, which is at — or just outside — the nuclear surface. Fire a faster alpha and it gets closer, tightening the limit.
WE 1
An alpha particle with kinetic energy 6.0 MeV is fired directly at a gold nucleus (Z = 79). Calculate the distance of closest approach. (k = 8.99 × 109 N m2 C−2, e = 1.60 × 10−19 C, 1 MeV = 1.60 × 10−13 J)
Step 1 — kinetic energy in joulesEₖ = 6.0 × (1.60 × 10⁻¹³) = 9.6 × 10⁻¹³ JStep 2 — use d = k(2Ze²)/Eₖd = (8.99 × 10⁹)(2)(79)(1.60 × 10⁻¹⁹)² / (9.6 × 10⁻¹³)d = 3.8 × 10⁻¹⁴ m (38 fm)This is the closest the 6.0 MeV alpha gets — an upper limit on the gold nuclear radius. It’s much bigger than gold’s true radius (~7 fm) because a 6 MeV alpha doesn’t have enough energy to reach the surface. Push the energy higher and d shrinks toward the real radius.
The nuclear radius formula
Do this for many nuclei and a clean pattern emerges. The more nucleons a nucleus has, the bigger it is — and the relationship is a cube root, because volume grows in proportion to the number of nucleons.
Nuclear radiusR = R0A1/3R = nuclear radius (m) • A = nucleon number • R0 = Fermi radius = 1.20 × 10−15 m
The Fermi radiusR0 is the radius of a single-proton nucleus (hydrogen, A = 1). Because R ∝ A1/3, plotting R against A1/3 gives a straight line through the origin with gradient R0.
A straight line through the origin: nuclear radius grows as the cube root of nucleon number, and the gradient gives the Fermi radius R0 = 1.20 fm.
WE 2
Estimate the radius of a copper nucleus, which has a nucleon number of 64. (R0 = 1.20 × 10−15 m)
Step 1 — use R = R₀A^(1/3)A^(1/3) = 64^(1/3) = 4R = (1.20 × 10⁻¹⁵) × 4R = 4.8 × 10⁻¹⁵ m (4.8 fm)64 is a perfect cube (4³), so the cube root is a clean 4 — a favourite exam number. For messier A values, use the cube-root button on your calculator. A few fm is the typical size of a medium nucleus.
Nuclear density
Now combine the radius formula with the mass to find the density. Something remarkable falls out: the A cancels, so every nucleus has the same density.
Nuclear density (A cancels out)ρ = m/V = Au / (&frac43;πR03A) = 3u / (4πR03)
Because A disappears, the density is a constant — the same for a tiny helium nucleus and a huge uranium one. This tells us nucleons are evenly packed throughout every nucleus, regardless of size.
WE 3
Determine the value of nuclear density, taking the Fermi radius as 1.20 fm. (u = 1.661 × 10−27 kg)
Step 1 — use ρ = 3u / (4πR₀³)ρ = 3(1.661 × 10⁻²⁷) / [4π(1.20 × 10⁻¹⁵)³]ρ = 2.3 × 10¹⁷ kg m⁻³Step 2 — what it tells us
The same for all nuclei, and far greater than atomic density.
nucleons are evenly packed; the atom is mostly empty spaceAround 10¹⁷ kg m⁻³ is the order of magnitude to remember — a sugar-cube of nuclear matter would weigh billions of tonnes. Because it’s so much denser than an atom, almost all the atom’s mass really is crammed into that tiny nucleus.
🔬 Working a nuclear-size question
Closest approach? Set Ek = Ep, then d = k(2Ze2)/Ek. Alpha charge = 2e.
Convert MeV to J (× 1.60 × 10−13) before substituting.
Nuclear radius?R = R0A1/3. Use the cube-root button.
Straight-line graph? Plot R vs A1/3; gradient is R0.
Nuclear density?ρ = 3u/(4πR03) — the same ~1017 kg m−3 for all nuclei.
d is an upper limit on the radius, not the exact value.
💡 Top tips
Alpha charge is 2e (helium nucleus) — that’s the “2” in the closest-approach formula.
Convert MeV to J with × 1.60 × 10−13.
The distance of closest approach is an upper limit on the nuclear radius.
R = R0A1/3 — a plot of R vs A1/3 is a straight line, gradient R0.
Nuclear density is constant (~1017 kg m−3) because A cancels.
Constant density → nucleons are evenly packed; the atom is mostly empty.
⚠ Common mistakes
Using e instead of 2e for the alpha charge
Leaving kinetic energy in MeV instead of joules
Treating d as the exact nuclear radius — it’s an upper limit
Forgetting the cube root in R = R0A1/3
Cubing R0 wrongly (or forgetting the R03) in the density formula
Thinking bigger nuclei are denser — density is the same for all
Quick recap: A head-on alpha particle stops at the distance of closest approach, where Ek = Ep = k(2Ze2)/d, giving d = k(2Ze2)/Ek as an upper limit on the nuclear radius. The radius of any nucleus follows R = R0A1/3 with R0 = 1.20 fm, so R vs A1/3 is a straight line. Combining these gives a constant nuclear density of ~1017 kg m−3, confirming nucleons are evenly packed and the atom is mostly empty space.
This whole method assumes the only force at play is electrostatic repulsion between the two positive charges — Rutherford’s original picture. But push the alpha to very high energies and it gets close enough for something new to take over. The experimental results start to peel away from Rutherford’s prediction, and that deviation is our first evidence of the strong nuclear force. Next page: Deviations from Rutherford Scattering.
Nuclear size calculations not landing?
Book a free meeting and we’ll work through closest approach, R = R0A1/3, and why nuclear density is constant.
