164-167 - Nuclear Binding Energy, Fusion and Fission

Binding energy as a mass-energy difference

Imagine building a nucleus from separated protons and neutrons. Once the nucleons are close enough, the strong nuclear force binds them into a lower-energy arrangement. The surprising measurement is that the finished nucleus has slightly less mass than the same nucleons had when they were separated.

That missing mass has not disappeared. It has been transferred from the nuclear system as energy.

For one nucleus:

The Edexcel relationship is:

where:

• $\Delta E$ is the energy change in joules, J
• $\Delta m$ is the mass difference in kilograms, kg
• $c$ is the speed of light in a vacuum, $3.00 \times 10^{8}\,\mathrm{m\,s^{-1}}$

The equation matters because $c^{2}$ is enormous. A mass difference too small to notice on an ordinary balance can still correspond to a measurable nuclear energy.

A tempting wrong idea is that a large binding energy means energy is "stored inside" waiting to come out. The safer model is the opposite: a large binding energy means the nucleus is in a lower mass-energy, more stable state. Energy is released when a nuclear change moves the products to a lower total mass-energy state.

Calculating energy from a mass deficit

Nuclear masses are so small that the kilogram is awkward. Physicists often use the atomic mass unit, $u$.

For Edexcel calculations:

When a question gives masses in $\mathrm{u}$, the decision point is simple: convert the mass difference to kg before using $\Delta E = c^{2} \Delta m$, because $c$ is in SI units.

Here is a full route for a fusion reaction used in fusion research:

Suppose the masses are:

• deuterium atom: $2.014102\,\mathrm{u}$
• tritium atom: $3.016049\,\mathrm{u}$
• helium-4 atom: $4.002603\,\mathrm{u}$
• neutron: $1.008665\,\mathrm{u}$

The electron masses cancel in this balanced atomic-mass route: deuterium plus tritium have two electrons in total, and a helium atom also has two electrons.

First compare total initial mass with total final mass:

The initial mass is larger, so this reaction releases energy. Now convert the mass decrease:

Then use the mass-energy equation:

That is the energy from one reaction. It is tiny on everyday scales, but large for a single nuclear event. Dividing by $1.60 \times 10^{-19}\,\mathrm{J}$ per eV gives about $1.76 \times 10^{7}\,\mathrm{eV}$, or $17.6\,\mathrm{MeV}$, which is the scale used on nuclear binding-energy graphs.

For an Edexcel “calculate” item, the marking route normally needs the mass difference, the conversion to kg, substitution into $c^{2} \Delta m$, and a final energy with a unit. In a “show that” item, keep more figures during the working than the rounded value in the question.

The atomic mass unit is defined as one twelfth of the mass of a neutral carbon-12 atom. To express a mass in u, divide its value in kilograms by $1.66\times10^{-27}\,\mathrm{kg\,u^{-1}}$. A reverse problem can give binding energy and ask for nuclear mass: rearrange to $m_{\text{nucleus}}=Zm_p+(A-Z)m_n-E_{\text{binding}}/c^2$. Keep the input masses consistent: nuclear masses with proton masses, or atomic masses with matching electron totals. Do not subtract electron masses twice.

Reading the binding-energy-per-nucleon curve

Total binding energy tends to increase as nuclei get bigger, so it is not the best way to compare different nuclei. A fairer comparison is binding energy per nucleon:

This is the energy per nucleon required to separate the whole nucleus into free nucleons. It is not necessarily the energy needed to remove one particular nucleon; that separation energy depends on the original and residual nuclei. A larger value means the nucleons are, on average, more tightly bound and the nucleus is more stable.