Size and density of the nucleus

The radius of a nucleus grows as the cube root of the number of nucleons. Experimentally:

Key formula

$R \approx 1{,}2\cdot 10^{-15}\cdot A^{1/3}\;\text{m}$

The $A^{1/3}$ dependence is revealing: since the volume of a sphere goes as $R^3 \propto A$, it says that the volume of the nucleus is proportional to the number of nucleons. Each nucleon, that is, always occupies the same little volume, regardless of how large the nucleus is. It follows that the density of nuclear matter is nearly constant, the same for all nuclei: the nucleons are packed side by side like marbles in a sack.

That density is enormous, as shown by the calculation for an iron nucleus.

Example — Density of the nucleus

For an iron nucleus ${}^{56}\text{Fe}$ the radius is $R = 1{,}2\cdot 10^{-15}\cdot 56^{1/3} \approx 4{,}6\cdot 10^{-15}\;\text{m}$ The mass is about $A\,m_p \approx 9{,}4\cdot 10^{-26}\;\text{kg}$; the volume is $V = \tfrac43\pi R^3 \approx 4{,}1\cdot 10^{-43}\;\text{m}^3$ giving the density $\ev{\rho \approx 2{,}3\cdot 10^{17}\;\text{kg/m}^3}$ A single cubic centimetre of nuclear matter would weigh two hundred million tonnes. This is precisely the density of a neutron star: an astrophysical object that is a giant nucleus on a stellar scale.

Orders of magnitude

The nucleus is about $10^5$ times smaller than the atom: the electron orbitals extend over $\sim 10^{-10}\;\text{m}$, the nucleus over $\sim 10^{-15}\;\text{m}$. In other words the atom is $99{,}9\%$ empty space. All the mass is concentrated in that tiny, extremely dense central point.

Links

Topics: Fisica nucleare Concepts: Notazione scientifica e ordini di grandezza Objects: Nucleo atomico

Related exercises: Problema — Volume di un parallelepipedo · Problema — Capelli in testa · Problema — Mar Adriatico