Nuclear Radius Formula

Equivalent forms

V \propto A, \rho \approx const.

Constant nuclear density across the chart of nuclides is the headline fact behind the cube root.

Symbol	Name	SI unit	Dimension	Range
$R$	Nuclear radiusoutput Effective nuclear radius	m	L	0 – 1e-14
$R₀$	Nuclear radius constant Empirical constant ~1.2 fm	m	L	1e-15 – 2e-15
$A$	Mass number Total number of nucleons	1	1	1 – 250

Unit systems

Symbol	SI	CGS	Natural
$R$	m	cm	fm
$R₀$	m	cm	fm
$A$	1	1	1

Where it holds

Good

to \sim 10

% for spherical nuclei

A \ge 12

. Deformed nuclei (rare earths, actinides) require multipole corrections.

Dimensional analysis

R \to [L]

R_{0} \cdot A^(1/3) \to [L]\cdot [1] = [L]

Length units.

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Discovery

Robert Hofstadter · 1956

Hofstadter's electron-scattering experiments at Stanford mapped charge distributions of nuclei and confirmed the A^(1/3) scaling, earning him the 1961 Nobel Prize.

Try this

How big is a uranium nucleus compared to a hydrogen one?

Using R = R₀ A^(1/3) with R₀ = 1.2 fm, compute the radii of H-1 and U-238 and compare.

Research status: stable

Real-world applications

Cross section estimates $\sigma \approx \pi R^{2} for$ nuclear reactions
Coulomb barrier height in alpha decay and fusion
Heavy-ion collision geometry (RHIC, LHC)
Mean-field nuclear models (Hartree-Fock, density functional)

Common misconceptions

$R_{0}$ varies slightly depending on the probe (charge radius $\sim 1.2\,\mathrm{fm} vs$ nuclear-matter radius $\sim 1.25\,\mathrm{fm})$ .
The formula gives the half-density radius; the surface is diffuse $(\sim 2\,\mathrm{fm}$ thick), not sharp.
Nuclei are not always spherical — many are prolate/oblate; A^(1/3) gives an effective average.

Experimental verification

Confirmed by elastic electron scattering (Hofstadter), muonic atom X-ray spectra, and proton-nucleus scattering form factors across

A=4

to 238 with

R_{0} in

the range 1.1–1.4 fm.

Derivation

Assume constant nuclear matter density

\rho \approx 0.17

nucleons/

fm^{3}

.

Volume

V = (4/3)\pi R^{3} = A/\rho

.

Solving

R = (3/(4\pi \rho ))^(1/3)

A^

(1/3) = R_{0} A^(1/3)

with

R_{0} \approx 1.2\,\mathrm{fm}

.

Limiting cases

A = 1

⟶

R = R_{0} \approx 1.2\,\mathrm{fm}

Single nucleon — sets the baseline scale.

A = 238

⟶

R \approx 7.4\,\mathrm{fm}

Uranium-238: about

6\times the

proton radius.

A \to \infty

⟶

R \to \infty \propto A^(1/3)

No upper bound in this formula; in reality, nuclei become unstable past

A\sim 300

.

What if…

What if nuclear density weren't constant?

R would not scale as A^(1/3); instead, heavier nuclei could be denser or more diffuse, changing the chart of nuclides entirely.

What if

R_{0}

were 2 fm?

All nuclei would be larger, Coulomb barriers lower, and alpha decay rates much faster.

What if we used neutron skin?

Neutron-rich nuclei show R_n > $R_p by \sim 0.2\,\mathrm{fm}$ ; precise probes (PREX) test the EOS of neutron matter.

1

Uranium-238 radius

Given ·

R₀:: 1.2e-15
A:: 238

Find · R

Steps

Step 1: A^ $(1/3) = 238^(1/3) \approx 6.196$ .
Step 2: $R = 1.2 \times 6.196\,\mathrm{fm} \approx 7.44\,\mathrm{fm}$ .
Step 3: That's about 6 times the proton radius.

Result ·

R = 1.2e-15 \times 238^(1/3) = 1.2e-15 \times 6.196 \approx 7.44e-15\,\mathrm{m}

2

Density of nuclear matter

Given ·

R₀:: 1.2e-15
A:: 12

Find ·

\rho

Steps

Step 1: $R = 1.2e-15 \times 12^(1/3) \approx 2.74e-15\,\mathrm{m}$ .
Step 2: $V = (4/3)\pi R^{3} \approx 8.62e-44\,\mathrm{m}^{3}$ .
Step 3: Mass $= 12 \times 1.67e-27 = 2.0e-26\,\mathrm{kg}$ .
Step 4: $\rho \approx 2.0e-26 / 8.62e-44 \approx 2.3e17\,\mathrm{kg/m}^{3}$ — same for all nuclei.

Result ·

\rho = A m_p / V \approx 12 \times 1.67e-27 / [(4/3)\pi (2.74e-15)^{3}] \approx 2.3e17\,\mathrm{kg/m}^{3}

Semi-Empirical Mass Formula→rutherford scattering cross sectionnuclear density