Nuclear binding energy

Here is the surprising fact from which all nuclear energy arises: the mass of a nucleus is less than the sum of the masses of its constituents taken separately. If we weigh $Z$ free protons and $N$ free neutrons and compare them with the nucleus they form, some mass is missing. That missing mass, the mass defect, has not disappeared: it has been converted into energy according to $E = mc^2$ and was released when the nucleons bound together. To reconstruct the free nucleons, that energy would have to be given back to the nucleus.

Principle — Nuclear binding energy

$\ev{B(Z,A) = \bigl[\,Z\,m_p + (A-Z)\,m_n - m_\text{nucleo}\,\bigr]\,c^2}$

$B$ is the energy that must be spent to completely dismantle the nucleus into its free nucleons; equivalently, it is the energy that was released in assembling it. The typical energy per nucleon is of the order of $\sim 8\;\text{MeV}$, a value a million times larger than the energies involved in chemical bonds (eV): this is why a nuclear reaction releases, for the same mass, millions of times more energy than a chemical combustion.

It is convenient to reason in terms of the binding energy per nucleon $B/A$: it is the “cohesion budget” of each single nucleon, and it allows nuclei of different sizes to be compared.

Key formula

$B/A = \text{binding energy per nucleon}$ Peak $\approx 8{,}8\;\text{MeV/nucleon}$ near ${}^{56}\text{Fe}$. Conversion factor: $1\;\text{u of mass} \leftrightarrow 931{,}5\;\text{MeV}$.

The factor $931{,}5\;\text{MeV}$ per atomic mass unit ($1\;\text{u} = 1{,}66054\cdot 10^{-27}\;\text{kg}$) is the working tool of all nuclear physics: it converts a mass defect, measured in u, directly into the energy released in MeV.

Example — Binding energy of the deuteron

Deuterium ${}^2$H has mass $m_d = 2{,}013553\;\text{u}$. The sum of the masses of a free proton ($m_p = 1{,}007276\;\text{u}$) and a free neutron ($m_n = 1{,}008665\;\text{u}$) is $m_p + m_n = 2{,}015941\;\text{u}$. The mass defect is $\Delta m = 2{,}015941 - 2{,}013553 = 0{,}002388\;\text{u}$ and the binding energy $\ev{B = \Delta m \cdot 931{,}5 \approx 2{,}22\;\text{MeV}}$ A mass change of barely one thousandth is enough to hold the nucleus together (Halliday, Resnick and Walker 2014).

Collegamenti

Argomenti: Fisica nucleare Concetti: Energia di legame nucleare · Equivalenza massa-energia Oggetti: Nucleo atomico

Esercizi collegati: Perché la fissione conviene · Difetto di massa del trizio · Energia di legame del ferro