8 min readConcept sheet1-mark MCQ, 2-mark proof that nuclear density is the same for all nuclei, 3-mark numericals; also a NEET favourite

Size and density of the nucleus

Nuclear radius grows as R = R₀ ∛A, with R₀ = 1.2 fm. So volume is proportional to A, and nuclear density is the same for every nucleus.

Nuclear radiusR = R₀ ∛A (R₀ times the cube root of A), with R₀ = 1.2 × 10⁻¹⁵ m = 1.2 fm

Femtometre (fermi)1 fm = 10⁻¹⁵ m

Nuclear densityAbout 2.3 × 10¹⁷ kg m⁻³, the same for all nuclei

How R was measuredBy scattering fast electrons from nuclei; alpha scattering gave only an upper limit

Why R depends on the cube root of A

Alpha-particle scattering showed that a gold nucleus is smaller than about 4 × 10⁻¹⁴ m. Later, scattering of fast electrons gave exact sizes. They show that nucleons are packed together like marbles in a bag. Each nucleon takes up about the same space, so the volume of a nucleus is proportional to A. Since volume = (4/3)πR³, we get R³ ∝ A, so R = R₀ ∛A with R₀ = 1.2 fm.

Proof: nuclear density does not depend on A

Let m be the mass of one nucleon (about 1.66 × 10⁻²⁷ kg).

1
Mass of the nucleus: M ≈ A × m.
2
Volume of the nucleus: V = (4/3)πR³ = (4/3)π(R₀ ∛A)³ = (4/3)πR₀³A.
3
Density ρ = M/V = Am ÷ (4/3)πR₀³A.
4
A cancels: ρ = 3m/(4πR₀³).
5★
So, ρ depends only on constants (m and R₀), not on A: all nuclei have nearly the same density.

★ key step

Worked example: the value of nuclear density

Problem

Estimate the density of nuclear matter. (m = 1.66 × 10⁻²⁷ kg, R₀ = 1.2 × 10⁻¹⁵ m, π = 3.14)

Working

Formula: ρ = 3m/(4πR₀³).
R₀³ = (1.2 × 10⁻¹⁵)³ = 1.728 × 10⁻⁴⁵ m³.
4πR₀³ = 4 × 3.14 × 1.728 × 10⁻⁴⁵ = 2.17 × 10⁻⁴⁴ m³.
ρ = 3 × 1.66 × 10⁻²⁷ ÷ 2.17 × 10⁻⁴⁴ = 4.98 × 10⁻²⁷ ÷ 2.17 × 10⁻⁴⁴ ≈ 2.3 × 10¹⁷ kg m⁻³.
Compared with water (10³ kg m⁻³), this is about 2 × 10¹⁴ times denser.

Answer

ρ ≈ 2.3 × 10¹⁷ kg m⁻³

Worked example: comparing two nuclei

Problem

Find the radii of ²⁷Al and ¹²⁵Te nuclei, and the ratio of their volumes. (R₀ = 1.2 fm)

Working

Formula: R = R₀ ∛A.
²⁷Al: ∛27 = 3, so R = 1.2 × 3 = 3.6 fm.
¹²⁵Te: ∛125 = 5, so R = 1.2 × 5 = 6.0 fm.
Ratio of volumes = (3.6/6.0)³ = (3/5)³ = 27/125, the same as the ratio of mass numbers.

Answer

3.6 fm and 6.0 fm; volumes in the ratio 27 : 125

Picture the scale

A nucleus is about 10⁻¹⁵ to 10⁻¹⁴ m across; the atom is about 10⁻¹⁰ m.
A teaspoon (5 cm³) of nuclear matter would have a mass of about 10¹² kg, more than a hundred crore tonnes.
Neutron stars are giant balls of matter with roughly nuclear density.

In real life

Neutron stars, the collapsed cores of giant stars, are about as dense as a nucleus. Indian astronomers study them with AstroSat, India's space observatory launched by ISRO in 2015.

Exam angle

2 marks: “Show that the density of nuclear matter is the same for all nuclei.” M = Am, V = (4/3)π(R₀∛A)³ = (4/3)πR₀³A. ρ = M/V = 3m/(4πR₀³), which has no A, so it is the same for all nuclei.

2 marks: “The radius of a ²⁷Al nucleus is 3.6 fm. Find the radius of a ¹²⁵Te nucleus.” R ∝ ∛A. R(Te) = 3.6 × ∛(125/27) = 3.6 × 5/3 = 6.0 fm.

1 marks: “What is the ratio of the nuclear densities of ¹⁶O and ²⁰⁸Pb?” 1 : 1, since nuclear density does not depend on A.

Remember it

Radius goes as cube root, density goes nowhere: R = 1.2 ∛A fm; ρ ≈ 2.3 × 10¹⁷ kg m⁻³ for every nucleus.