Energy from the Nucleus

The short version

A nucleus has less mass than the protons and neutrons that make it. The difference is the mass defect $\Delta m$, and it corresponds to the binding energy $E_B=\Delta m{c}^2$: the energy needed to pull the nucleus apart into free nucleons.

Dividing by the number of nucleons gives the binding energy per nucleon, a measure of how tightly each nucleon is held. It rises steeply for light nuclei, peaks near iron-56 at about $8.8$ MeV, and falls slowly for heavy nuclei.

Any change that makes nucleons more tightly bound releases energy. Fusion of light nuclei and fission of heavy nuclei both move towards iron, so both release energy. The energy released is

$E=\left(m_{\text{reactants}}-m_{\text{products}}\right)c^2$

With masses in atomic mass units, multiply the mass lost by $931.5$ MeV per u.

A nucleus weighs less than its parts

A helium-4 nucleus contains two protons and two neutrons. Add up their masses and you get $4.031882$ u. Weigh a helium-4 nucleus and you get $4.001506$ u. About $0.75$ per cent of the mass is missing.

That missing mass is the mass defect. It is not a measuring error, and it has not gone anywhere mysterious. When the nucleons came together, the strong nuclear force pulled them in and energy was released, as it is whenever something falls into a well. By $E=m{c}^2$, a system that has lost energy $E$ has lost mass $E/c^2$.

To separate the nucleus again you would have to put that energy back. It is called the binding energy.

Working in atomic mass units

Nuclear masses are tabulated in unified atomic mass units, where $1$ u $=1.661\times 10^{-27}$ kg. Converting to kilograms every time is slow, so use the energy equivalent of one mass unit:

$1\text{u}\times c^2=1.66054\times 10^{-27}\times (2.99792\times 10^8{)}^2=1.49242\times 10^{-10}\text{J}=931.5\text{MeV}$

The extra digits matter here: with $c=3.00\times 10^8$ the product comes out near $933$ MeV, which is enough to throw off a calculation built on differences in the fourth decimal place.

So a mass defect in u, multiplied by $931.5$, gives an energy in MeV. Never multiply by $931.5$ and by $c^2$: the $c^2$ is already inside the $931.5$.

Tables usually list atomic masses, which include the electrons. That is fine as long as the electrons balance. For binding energy, use hydrogen-1 atoms instead of bare protons, so the $Z$ electrons appear on both sides and cancel.

The binding energy curve

Plot binding energy per nucleon against mass number and one of the most important graphs in physics appears.

Three features matter.

• The steep rise. In a small nucleus many nucleons sit on the surface with few neighbours, so each is weakly held. Adding nucleons gives each more neighbours to bond with through the short-range strong force.
• The peak near iron. Around $A=56$ to $62$, nucleons are as tightly bound as they can be, at about $8.8$ MeV each. Nuclei here are the most stable.
• The slow fall. In large nuclei the strong force still only reaches nearest neighbours, but every proton repels every other proton. The repulsion grows faster than the attraction, so each nucleon is held a little less tightly.

Helium-4 sits well above its neighbours. It is an exceptionally tight package, which is why heavy nuclei shed whole alpha particles rather than single protons.

Now use the sliders. Take $A=236$ and split it into pieces of about $94$ and $142$: both pieces sit higher on the curve than the parent, and over $200$ MeV is released. Then take $A=24$ and split it into two $12$s: the pieces sit lower, so splitting costs energy, and running it backwards, fusing them, releases it.

Energy released in any nuclear change

The same mass bookkeeping works for every decay and reaction. First balance the equation, so nucleon number and charge are conserved. Then compare the total mass on each side.

Mass on its own is not conserved in a nuclear reaction: the products are lighter. What is conserved is mass–energy. The rest mass that disappears reappears as the kinetic energy of the products and as gamma rays.

Alpha decay

Beta decay

In beta-minus decay a neutron becomes a proton, and an electron and an antineutrino are emitted. With atomic masses, the daughter atom has one extra electron, which accounts for the emitted one, so the atomic masses can be subtracted directly. The energy is shared between the electron and the antineutrino, which is why beta particles emerge with a spread of energies up to a maximum.

Fission

A slow neutron absorbed by uranium-235 makes an unstable uranium-236 nucleus, which splits into two medium-sized fragments and two or three neutrons. The fragments are further up the binding energy curve, so energy is released, mostly as their kinetic energy.

Fusion

In fusion two light nuclei join. Because the left side of the binding energy curve is so steep, the product is much more tightly bound per nucleon, and a large share of the mass is released.

The difficulty is getting there. Both nuclei are positive, and the strong force only acts over about $10^{-15}$ m, so they must approach at enormous speed to overcome their electrostatic repulsion. That needs temperatures of millions of kelvin, as in the core of the Sun, and confinement to keep the hot plasma together.

Where the curve comes from

The smooth curve on the graph is the liquid drop model, which treats the nucleus like a drop of incompressible fluid. It adds up a few effects:

• A volume term: each nucleon bonds with its neighbours, so binding grows in proportion to $A$.
• A surface term: nucleons on the surface have fewer neighbours, which subtracts in proportion to the surface area, $A^{2/3}$. This dominates for small nuclei and causes the steep rise.
• A Coulomb term: every pair of protons repels, subtracting roughly $Z^2/A^{1/3}$. This dominates for heavy nuclei and causes the slow fall.
• An asymmetry term: nuclei with very unequal numbers of protons and neutrons are less stable, which is why heavy nuclei need more neutrons than protons but not too many more.

With only four fitted numbers it reproduces measured binding energies to within about one per cent for all but the lightest nuclei. Helium-4, carbon-12 and oxygen-16 sit above the curve because they are built from complete shells of nucleons, an effect the drop model ignores.

Why fusion wins per kilogram

Deuterium–tritium fusion releases $17.6$ MeV from $5$ nucleons; a uranium fission releases about $180$ MeV from $236$. Per kilogram of fuel, fusion gives roughly five times as much energy. It also produces no long-lived fission fragments, and deuterium can be extracted from seawater. The obstacle is engineering: holding a plasma above $100$ million kelvin long enough, and densely enough, for more energy to come out than went in.

Where the elements came from

Hydrogen, helium and a little lithium formed in the first few minutes after the Big Bang. Stars then fuse lighter elements into heavier ones up to iron, releasing energy at each step. Beyond iron, fusion absorbs energy, so the heavier elements, including every atom of gold and uranium, were made in supernovae and neutron star collisions, where huge amounts of energy were available.