Nuclear Size & Density

PHYNuclei — Structure & Binding Energy

Experiments show that a nucleus behaves like a tiny sphere whose radius grows with mass number as $R = R_0 A^{1/3}$ , with $R_0 \approx 1.2\,\text{fm}$ . The cube-root law has a striking consequence: because volume $\propto R^3 \propto A$ and mass $\propto A$ , the nuclear density comes out the same for every nucleus — an enormous $\sim 2.3\times 10^{17}\,\text{kg m}^{-3}$ .

ISC examines this because it tests one clean idea (the $A^{1/3}$ scaling) in three guises: finding a single nucleus's radius, comparing two nuclei, and proving density is independent of $A$ .

Nuclear radius (cube-root law)

R = R_0 A^{1/3}, \qquad R_0 \approx 1.2\ \text{fm}

R

is the nuclear radius,

A

the mass number,

R_0 = 1.2\,\text{fm} = 1.2\times 10^{-15}\,\text{m}

the empirical constant. For aluminium (

A=27

A^{1/3}=3

, so

R = 3.6\,\text{fm}

Ratio of two nuclear radii

\dfrac{R_1}{R_2} = \left(\dfrac{A_1}{A_2}\right)^{1/3}

R_1, R_2

are the radii and

A_1, A_2

the mass numbers. For

A_1:A_2 = 27:64

, the ratio is

(27/64)^{1/3} = 3/4

. The constant

R_0

cancels, so only the cube root of the mass-number ratio matters.

Nuclear density (independent of A)

\rho = \dfrac{\text{mass}}{\text{volume}} = \dfrac{A\,m_N}{\tfrac{4}{3}\pi R_0^{3} A} = \dfrac{3\,m_N}{4\pi R_0^{3}}

m_N \approx 1.67\times 10^{-27}\,\text{kg}

is the mass of one nucleon,

R_0 = 1.2\,\text{fm}

. The mass number

A

cancels top and bottom, so

\rho

is constant (

\approx 2.3\times 10^{17}\,\text{kg m}^{-3}

) for all nuclei.

The radius grows only as the cube root of $A$ : doubling $A$ increases $R$ by just $2^{1/3} \approx 1.26$ , so even heavy nuclei are only a few times larger than light ones.
In radius ratios the constant $R_0$ cancels, so you never need its value — just take the cube root of the mass-number ratio.
Because volume $\propto R^3 \propto A$ and mass $\propto A$ , the density is independent of $A$ : this is the key conceptual result and a favourite Assertion-Reason item.
Nuclear density ( $\sim 10^{17}\,\text{kg m}^{-3}$ ) is about $10^{14}$ times ordinary matter density, showing how tightly mass is packed into the nucleus.
Use femtometres (fermi): $1\,\text{fm} = 10^{-15}\,\text{m}$ . Keep all lengths in metres when computing volume so the density comes out in $\text{kg m}^{-3}$ .
Treat the nucleon mass as roughly equal for protons and neutrons ( $\approx 1.67\times 10^{-27}\,\text{kg}$ ); the small proton/neutron mass difference is negligible for a density estimate.
The model assumes a uniform spherical nucleus; the constant density reflects the short-range, saturating nature of the strong nuclear force.

Where the marks go

Forgetting the cube root: writing $R \propto A$ instead of $R \propto A^{1/3}$ , which wrongly makes large nuclei enormous and density fall with $A$ .
Taking the wrong root for radius ratios — use $(A_1/A_2)^{1/3}$ , not $(A_1/A_2)$ or its square root.
Leaving $R$ in fm while computing volume, giving a density off by $(10^{15})^3$ . Convert fm to metres first.
Claiming density depends on $A$ : the $A$ in mass and the $A$ in volume cancel, so all nuclei share the same density.

How the board asks it

Numericalthe cube-root law $R = R_0 A^{1/3}$
Given $R_0 = 1.2\,\text{fm}$ , calculate the radius of the nucleus of $_{79}^{197}\text{Au}$ , expressing your answer in metres.
Numericalratio of nuclear radii with $R_0$ cancelling
Find the ratio of the nuclear radii of $_{13}^{27}\text{Al}$ and $_{52}^{125}\text{Te}$ .
Derive / provenuclear density independent of $A$
Using the relation $R = R_0 A^{1/3}$ , show that the density of a nucleus is independent of its mass number $A$ , and hence calculate its value taking $R_0 = 1.2\,\text{fm}$ and nucleon mass $\approx 1.67\times 10^{-27}\,\text{kg}$ .
Assertion–Reasondensity independent of $A$
Assertion: All nuclei have nearly the same density irrespective of their mass number. Reason: The nuclear radius is proportional to $A^{1/3}$ , so nuclear volume is proportional to $A$ while nuclear mass is also proportional to $A$ . Select the correct option.
Define / statethe cube-root law and constant $R_0$
State how the radius of a nucleus depends on its mass number, write the relation between them, and name the constant involved.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.