nuclear binding energy and mass defect

A nucleus contains $Z$ protons and $N$ neutrons, yet its mass is smaller than the sum of the masses of the separate nucleons. The difference

$$\Delta m = Z m_p + N m_n - M_{\text{nuc}}$$

is called the mass defect. By Einstein's relation $E = mc^2$, this "missing" mass corresponds to the energy released when the nucleons bound together; it is called the nuclear binding energy:

$$E_{\text{b}} = \Delta m\,c^2.$$

In nuclear physics, masses are conveniently measured in atomic mass units and energies in electronvolts, with $1\ \text{u}\cdot c^2 = 931.494\ \text{MeV}$. In practice, tables list masses of neutral atoms, so it is convenient to use the atomic mass instead of the nuclear mass: the electrons are then accounted for automatically, and their binding energy (of order keV) is negligible compared with $E_{\text{b}}$ (of order MeV). Nuclear stability is characterized by the binding energy per nucleon $E_{\text{b}}/A$: for most nuclei it is about $8\ \text{MeV}$ per nucleon, the maximum lies near iron-56, and for light nuclei (deuteron, helium-4) it is noticeably smaller.

For the deuteron $^{2}\text{H}$, the proton mass is $m_p = 1.007276\ \text{u}$, the neutron mass is $m_n = 1.008665\ \text{u}$, and the nuclear mass is $M_{\text{nuc}} = 2.013553\ \text{u}$. Find the mass defect, the binding energy, and the binding energy per nucleon.

A table gives the mass of the neutral helium-4 atom: $M_{\text{atom}} = 4.002603\ \text{u}$. Using the hydrogen atom mass $m_{\text{H}} = 1.007825\ \text{u}$ and the neutron mass $m_n = 1.008665\ \text{u}$, calculate the binding energy of the $^{4}\text{He}$ nucleus and the binding energy per nucleon. Explain why atomic rather than nuclear masses may be used.

Consider the thermonuclear fusion reaction

$$^{2}\text{H} + {}^{3}\text{H} \rightarrow {}^{4}\text{He} + n.$$

Atomic masses: $M(^{2}\text{H}) = 2.014102\ \text{u}$, $M(^{3}\text{H}) = 3.016049\ \text{u}$, $M(^{4}\text{He}) = 4.002603\ \text{u}$, neutron mass $m_n = 1.008665\ \text{u}$. Determine whether energy is released or absorbed, and find its magnitude in MeV.

Build the binding-energy-per-nucleon curve yourself (data and simulation only, no radioactive samples).

1. Open an online simulation or interactive binding-energy curve, or a table of atomic masses (for example, AME). No radioactive materials are needed: the whole activity is computational and safe.
2. Write down the atomic masses for $^{2}\text{H}$, $^{4}\text{He}$, $^{12}\text{C}$, $^{56}\text{Fe}$, $^{238}\text{U}$ and the masses $m_{\text{H}} = 1.007825\ \text{u}$, $m_n = 1.008665\ \text{u}$.
3. For each nuclide, compute $\Delta m$, $E_{\text{b}} = \Delta m\,c^2$, and $E_{\text{b}}/A$.
4. Plot $E_{\text{b}}/A$ against mass number $A$ and mark the maximum.
5. State a conclusion: why fusion of light nuclei and fission of heavy nuclei release energy.
6. Check yourself: compare your values with reference ones ($^{2}\text{H}$: $2.22\ \text{MeV}$; $^{4}\text{He}$: $28.3\ \text{MeV}$; $^{56}\text{Fe}$: about $8.8\ \text{MeV}$ per nucleon).

Show how the binding energy per nucleon $E_{\text{b}}/A$ depends on mass number $A$ and where the maximum of nuclear stability lies.

A labeled sketch graph: horizontal axis — mass number $A$ (from 1 to 240), vertical axis — $E_{\text{b}}/A$ in $\text{MeV}$ per nucleon (from 0 to 9). Points are marked: $^{2}\text{H}$ ($1.11$), $^{4}\text{He}$ ($7.07$), $^{12}\text{C}$ (about $7.7$), $^{56}\text{Fe}$ (about $8.790$, near the maximum), $^{62}\text{Ni}$ (about $8.7945$, the true maximum), $^{238}\text{U}$ (about $7.6$). A horizontal dashed line is drawn at $8.7945\ \text{MeV}$; arrows show the direction of reactions: fusion from the left toward the maximum, fission from the right toward the maximum.

Every second the Sun fuses about $6\times10^{11}\ \mathrm{kg}$ of hydrogen into helium. A helium-4 nucleus is lighter than the four protons that built it: the mass defect is $\Delta m \approx 0.0304\ \mathrm{u} \approx 5.05\times10^{-29}\ \mathrm{kg}$, and the energy released per nucleus is $26.7\ \mathrm{MeV}$. Multiplying by the number of reactions needed to supply the Sun's luminosity $3.8\times10^{26}\ \mathrm{W}$ gives a mass loss of about $4.3\times10^9\ \mathrm{kg/s}$, i.e. 4.3 million tonnes per second.

This exercises the relativistic rest-energy relation $E=\Delta m c^2$: the missing mass $\Delta m$ is converted into the kinetic energy of the products and into gamma radiation. The $26.7\ \mathrm{MeV}$ per helium nucleus is exactly the rest-energy difference between the initial and final nuclei.

A typical ionization smoke detector contains about $0.3\ \mathrm{\mu g}$ of americium-241 with an activity of $33\ \mathrm{kBq}$. Each $^{241}\mathrm{Am}$ nucleus alpha-decays to $^{237}\mathrm{Np}$ with a half-life of 432 years. The masses do not balance: $m(^{241}\mathrm{Am}) - m(^{237}\mathrm{Np}) - m(^{4}\mathrm{He}) = 5.5\ \mathrm{MeV}/c^2 \approx 9.8\times10^{-30}\ \mathrm{kg}$. The emitted alpha particle carries a kinetic energy of $5.5\ \mathrm{MeV}$ ($8.8\times10^{-13}\ \mathrm{J}$).

The mass defect is the difference between the rest mass of the parent nucleus and the total rest mass of the daughter nucleus plus the alpha particle. Dividing this difference by $c^2$ gives the energy released, $Q = \Delta m c^2$; the alpha particle's kinetic energy and the recoil of the daughter nucleus share this energy, as required by energy and momentum conservation.

In a reactor, a thermal neutron splits $^{235}\mathrm{U}$ into two fragments, for example $^{141}\mathrm{Ba}$ and $^{92}\mathrm{Kr}$, plus three neutrons. The products together are lighter than the original nucleus plus neutron by $\Delta m \approx 0.2\ \mathrm{u} \approx 3.3\times10^{-28}\ \mathrm{kg}$, which corresponds to about $200\ \mathrm{MeV}$ per fission. A $1\ \mathrm{GW}$ reactor therefore consumes roughly $1\ \mathrm{kg}$ of $^{235}\mathrm{U}$ per day, since $1\ \mathrm{kg}$ releases $8.2\times10^{13}\ \mathrm{J}$.

This applies the definition of mass defect to a nuclear reaction: $\Delta m$ is the difference between the total rest mass before and after the reaction, and the released energy is $Q = \Delta m c^2$. The energy appears as kinetic energy of the fission fragments and neutrons, which heats the coolant.