The big idea: The nucleus sits at the centre of the atom and is fantastically small — about 10⁻¹⁵ m across (a femtometre, fm).
Size, and the HL step
- It is roughly 100 000 times smaller than the whole atom.
- At HL you go further: the radius of a nucleus is not random.
- It depends only on how many nucleons (A) it contains, through a simple cube-root law.
What A means here: A is the nucleon number (also called the mass number) — the total count of protons + neutrons in the nucleus. A bigger A means a heavier, larger nucleus.
Because every nucleon takes up roughly the same amount of space, packing more of them in grows the nucleus in a very predictable way — like adding marbles to a ball of clay. Double the nucleons and the volume roughly doubles, so the radius grows as the cube root.
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Radius grows as the cube root of A: Experiments (high-energy scattering) show the nuclear radius follows a single law: R = R₀A1/3, where R₀ ≈ 1.2 × 10⁻¹⁵ m is a constant the same for every nucleus.
- nuclear radius (m)
- constant = 1.2 × 10⁻¹⁵ m (same for all nuclei)
- nucleon number (protons + neutrons, no unit)
Worked example — radius of a nucleus with A = 64
Estimate the radius of a nucleus that contains A = 64 nucleons. Take R₀ = 1.2 × 10⁻¹⁵ m.
Solution
- Write the given radius law first:
- Find the cube root of A: ∛64 = 4.0, then substitute:
- Multiply out — keep the unit:
Final answer
R = 4.8 × 10⁻¹⁵ m (about 4.8 fm).
Cube roots without panicking: Look for the perfect cube hiding in A: 8 = 2³, 27 = 3³, 64 = 4³, 125 = 5³, 216 = 6³. Then A1/3 is just that whole number, no calculator gymnastics needed.
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Same density for every nucleus: Every nucleus has the same density — about 10¹⁷ kg m⁻³, regardless of the element.
Why
- Because R³ ∝ A, the volume (∝ R³) grows in exact step with the number of nucleons A.
- Mass ∝ A and volume ∝ A.
- So density = mass ÷ volume is the same for every nucleus.
This is the key HL insight: nuclei are not 'fluffier' or 'denser' for big atoms versus small atoms. They are all packed to the same incredibly high density, like identical marbles glued together. A bigger nucleus is simply a bigger ball of the same stuff.
A small nucleus (A = 8)
- 8 nucleons
- R = 1.2 × 10⁻¹⁵ × 2.0 = 2.4 × 10⁻¹⁵ m
- Smaller ball
- Same packing density
A larger nucleus (A = 64)
- 64 nucleons (8 × as many)
- R = 4.8 × 10⁻¹⁵ m (2 × the radius)
- 2³ = 8 × the volume
- Same packing density
Density against nucleon number: Asked for a graph of nuclear density against nucleon number A? It is a horizontal line: every nucleus has the same density.
Worked example — same density, 8× the volume
Compare a nucleus with A = 8 to one with A = 64. Show that the larger one has 8 times the volume, and explain why both have the same density.
Solution
- Radius of the A = 8 nucleus (∛8 = 2.0):
- Radius of the A = 64 nucleus (∛64 = 4.0):
- Volume ratio (volume ∝ R³, and 4.8/2.4 = 2):
- Mass is 8 × larger (8 × the nucleons) and volume is 8 × larger, so:
Final answer
The A = 64 nucleus has 8 × the volume of the A = 8 nucleus, but also 8 × the mass — so both have the same density.
An energy question: Head-on, the alpha's kinetic energy turns into electric potential energy until it stops. At that moment it is as close as it can get: the distance of closest approach d.
- initial kinetic energy of the alpha particle (J)
- Coulomb constant, 8.99 × 10⁹ N m² C⁻²
- charges of the alpha particle and of the nucleus (C)
- distance of closest approach (m)
Worked example — an alpha particle and a silver nucleus
An alpha particle with kinetic energy 6.0 MeV is fired head-on at a silver nucleus (Z = 47). Find the distance of closest approach.
Solution
- Energy in joules:
- All of it becomes electric potential energy at d:
- Work it out:
Final answer
d ≈ 2.3 × 10⁻¹⁴ m: about four times the radius of a silver nucleus (R = 1.2 × 10⁻¹⁵ × 1081/3 ≈ 5.7 × 10⁻¹⁵ m).
Why Rutherford's model fails at high energy: Rutherford's prediction assumes the alpha feels only the electric force. Since d ∝ 1/Ek, a fast enough alpha gets within the range of the strong nuclear force, and the number scattered back (near 180°) no longer fits the prediction.
The energy E₀ at which this first happens gives an estimate of the nuclear radius: put E₀ = k(2e)(Ze)/R and solve for R.
Isotopes give the same d: d depends on the charge Ze, not on the nucleon number. Isotopes of an element have the same Z, so an alpha of the same energy stops at the same distance, even though their radii differ slightly.
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Why fire electrons at a nucleus?: You cannot put a ruler on a nucleus. Instead you fire a beam of high-energy electrons at it and look at how they scatter/diffract.
The rings give the size: The first dark ring in the diffraction pattern reveals the diameter of the nucleus — exactly how a small object diffracts light reveals its size.
The wavelength must match the target: To resolve an object you need a probe whose wavelength is about the same size as the object — here ≈ 10⁻¹⁵ m. A particle's de Broglie wavelength λ = h/p shrinks as its momentum p grows, so you need very high momentum, i.e. very high energy, electrons to get λ down to nuclear size.
Electrons are chosen because they are point-like and feel only the electromagnetic force (not the messy strong nuclear force), so the scattering pattern is clean and easy to interpret. Speeding them up to high energy shortens their de Broglie wavelength until it matches the femtometre scale of the nucleus.
The logic chain
- Nucleus is ≈ 10⁻¹⁵ m across
- To resolve it, need probe wavelength λ ≈ 10⁻¹⁵ m
- λ = h/p ⇒ small λ needs large momentum p
- Large p ⇒ high-energy electrons
- Diffraction-ring angle ⇒ nuclear diameter
Where it shows up: Nuclear size, density and closest approach are HL only (E.1). It comes up in Paper 1A and Paper 2.
| Paper | What they ask |
|---|---|
| Paper 1A | A one-step 'estimate R for this A', 'how does R depend on A?', 'which graph shows nuclear density against A?' or an expression for the distance of closest approach. |
| Paper 2 | Calculate a closest-approach distance or the proton number from one, outline why scattering deviates from Rutherford's model at high energy and how that estimates R, show d compared with R = R₀A1/3. |
Three easy marks: (1) Quote R = R₀A1/3 and use R₀ = 1.2 × 10⁻¹⁵ m. (2) 'Density constant' comes from volume ∝ R³ ∝ A, same as mass ∝ A. (3) High energy ⇒ large p ⇒ small de Broglie λ ⇒ can resolve a 10⁻¹⁵ m nucleus.
IB-style question — two nuclei, same density
Nucleus X has A = 27 and nucleus Y has A = 216. (a) Calculate the radius of each. (b) Hence state the ratio of their volumes and comment on their densities.
Solution
- Use the given law for X (∛27 = 3.0):
- And for Y (∛216 = 6.0):
- Volume ratio (volume ∝ R³, 7.2/3.6 = 2):
- Mass ∝ A is also 8 × larger, so density is unchanged:
Final answer
RX = 3.6 × 10⁻¹⁵ m, RY = 7.2 × 10⁻¹⁵ m; VY is 8 × VX (= the mass ratio), so both nuclei have the same density.