Nuclear Binding Energy & Reactions

Weigh a nucleus and something is missing

The last lesson treated decay as bookkeeping in time: a nucleus has a chance per second of falling apart, and you followed the population with $N = N_0 e^{-\lambda t}$. It never asked the more basic question. Why does a nucleus hold together at all, and why does one sometimes pay to break?

Here is the experiment that answers it. Take a helium-4 nucleus and put it on the most precise scale ever built. Then weigh two free protons and two free neutrons separately and add those four numbers up. The two results are not the same. The assembled nucleus is lighter: by about half a percent, far more than any measurement error.

Half a percent of a nucleus is enormous by the standards of chemistry, and it is not an accounting slip. Mass went somewhere. What happened is that when those four nucleons snapped together, the strong force pulled them into a bound state and a burst of energy left the system. By $E = mc^2$, energy leaving means mass leaving. The nucleus you are holding is the parts minus what escaped.

That missing mass is the whole subject. It tells you how tightly a nucleus is held, which nuclei are stable, and how much energy comes out of a reactor or a star.

Mass defect and binding energy

Write the arithmetic down. For a nucleus with $Z$ protons and $A$ nucleons total, so $A - Z$ neutrons, the mass defect is how much lighter it is than its scattered parts:

$\Delta m = Z m_p + (A - Z) m_n - m_{\text{nucleus}}$

Multiply by $c^2$ and you have the binding energy:

$E_B = \Delta m\, c^2$

Read it in the direction that is easier to trust. $E_B$ is the energy you would have to supply to pull that nucleus completely apart into free nucleons at rest. Big $E_B$ means hard to take apart. It is not stored in the nucleus waiting to come out; it is energy already gone.

Two practical points make this arithmetic bearable. First, nuclear masses come in atomic mass units, defined so that a neutral carbon-12 atom weighs exactly $12$ u. The conversion you will use constantly is

$1\ \text{u} = 931.5\ \text{MeV}/c^2$

so a mass defect quoted in u becomes an energy in MeV by one multiplication.

Second, published tables give atomic masses (nucleus plus its $Z$ electrons) not bare nuclear masses. Rather than subtracting electrons, use $m_H = 1.007825$ u, the mass of a whole hydrogen atom, in place of the proton. Then the $Z$ electrons on the left cancel the $Z$ electrons carried along inside the atomic mass on the right:

$E_B = \left[Z m_H + (A - Z) m_n - m_{\text{atom}}\right] c^2$

The electron binding energies you are ignoring here are a few eV against millions, so the cancellation is exact for any purpose you have.

Worked example 1: helium-4. Its atomic mass is $4.002603$ u. With $Z = 2$ and $A - Z = 2$:

$Z m_H + N m_n = 2(1.007825) + 2(1.008665) = 2.015650 + 2.017330 = 4.032980\ \text{u}$

Subtract the real mass: $4.032980 - 4.002603 = 0.030377$ u. That is the defect. Now convert:

$E_B = 0.030377 \times 931.5 = 28.30\ \text{MeV}$

Four nucleons, $28.3$ MeV to separate them. Per nucleon that is $28.3/4 = 7.07$ MeV. For contrast, ripping an electron off a hydrogen atom costs $13.6$ eV. Nuclear energies run about a million times larger, and that single fact is why a gram of uranium outruns a tonne of coal.

The curve that runs everything

Total binding energy is not a fair way to compare nuclei. Uranium beats helium easily, but only because it has sixty times as many nucleons. Divide it out. The quantity worth plotting is the binding energy per nucleon, $E_B/A$, in MeV: how tightly the average nucleon is held.

Plot that against $A$ and you get a shape you should carry in your head. It climbs steeply from the deuteron at $1.1$ MeV per nucleon, passes $7.07$ at helium-4 and about $7.9$ at oxygen-16, flattens into a broad maximum of $8.79$ MeV per nucleon near iron-56, then sags slowly back down to $7.57$ MeV at uranium-238.

Two competing forces make that shape, and you can reason both out. The strong nuclear force is powerful but very short ranged, reaching barely past a nucleon's nearest neighbours. So each nucleon binds to a fixed number of partners no matter how big the nucleus gets, and total binding grows roughly in proportion to $A$: that is the flat top. Nucleons on the surface have fewer neighbours, which is pure loss, and the surface fraction is largest for small nuclei: that is why the left-hand side climbs as the surface-to-volume ratio drops. Coulomb repulsion, by contrast, has unlimited range: every proton pushes on every other one, so that penalty grows like $Z^2$ and eventually wins. That is the long decline at the heavy end.

Q values: what a reaction pays

For any nuclear reaction, the energy released is the mass that vanished:

$Q = \left(\sum m_{\text{initial}} - \sum m_{\text{final}}\right) c^2$

Positive $Q$ means energy comes out, carried off as kinetic energy of the products and gamma rays. Negative $Q$ means you must feed energy in and the reaction will not go by itself. Use atomic masses throughout and the electrons take care of themselves, with one exception noted below.

Worked example 2: D-T fusion. Deuterium plus tritium gives helium-4 plus a neutron. Atomic masses: $^2$H $= 2.014102$, $^3$H $= 3.016049$, $^4$He $= 4.002603$, n $= 1.008665$ u.

Initial: $2.014102 + 3.016049 = 5.030151$ u. Final: $4.002603 + 1.008665 = 5.011268$ u. Difference: $5.030151 - 5.011268 = 0.018883$ u.

$Q = 0.018883 \times 931.5 = 17.59\ \text{MeV}$

Five nucleons in, $17.6$ MeV out, which is $3.52$ MeV per nucleon. Compare that with fission's roughly $0.85$ MeV per nucleon and you see why fusion is the richer fuel, and why stars use it.

There is a second way to get the same number without any masses. Helium-4 has $E_B = 28.3$ MeV; the deuteron has $2.22$ MeV and the triton $8.48$ MeV. The free neutron has none. So

$Q = \sum E_B(\text{final}) - \sum E_B(\text{initial}) = 28.3 - (2.22 + 8.48) = 17.6\ \text{MeV}$

Note the flip: with masses you subtract final from initial, with binding energies you subtract initial from final. Both say the same thing, because becoming more tightly bound is becoming lighter. Use whichever the problem hands you.

Fission and fusion are the same idea

Both directions run downhill on the same curve. Anything that moves nucleons closer to iron pays out.

Fission takes a heavy nucleus at about $7.6$ MeV per nucleon and splits it into mid-mass fragments at about $8.5$. The gain is about $0.9$ MeV per nucleon, and with $236$ nucleons involved that is roughly $200$ MeV per event, most of it as kinetic energy of the two fragments flying apart under their mutual Coulomb repulsion.

Worked example 3: is splitting uranium-238 in half worth it? Suppose $^{238}$U, at $7.57$ MeV per nucleon, split into two identical $^{119}$Pd fragments at $8.50$ MeV per nucleon. Work with totals, not per-nucleon figures, or you will get lost.

Parent: $238 \times 7.57 = 1801.7$ MeV of binding. Fragments: $2 \times 119 \times 8.50 = 2023.0$ MeV.

The products are bound more tightly by $2023.0 - 1801.7 = 221$ MeV, so that much energy comes out. The reaction is allowed as far as energy is concerned. It does not happen appreciably anyway, because the nucleus has to stretch through a badly bound intermediate shape first, and that barrier is what a slow neutron helps a $^{235}$U nucleus over. Energetically allowed and actually happening are different questions, and this one is the clearest example of the gap.

Fusion climbs the steep left-hand side instead, and the gain per nucleon is several times larger. The price is getting two positive nuclei close enough for the strong force to reach, against their Coulomb repulsion. That barrier height is straight electrostatics:

$U = \frac{k q_1 q_2}{r} = \frac{Z_1 Z_2 (k e^2)}{r}$

and the useful constant is $ke^2 = 1.44$ MeV·fm. Two deuterons that must reach $3.00$ fm need

$U = \frac{(1)(1)(1.44)}{3.00} = 0.48\ \text{MeV}$

which, as a thermal energy, corresponds to billions of kelvin. Stars run cooler than that and fuse anyway, by quantum tunnelling and by the lucky tail of the speed distribution.

Turning MeV into watts

Most of the harder problems are one energy calculation wrapped in unit conversion. Three numbers do all of it: $1$ MeV $= 1.602 \times 10^{-13}$ J, $1$ u $= 1.6605 \times 10^{-27}$ kg, and $N_A = 6.022 \times 10^{23}$.

Worked example 4: how long can the Sun burn hydrogen? The core holds about $10\%$ of the Sun's $2.0 \times 10^{30}$ kg as usable hydrogen, hydrogen burning converts $0.71\%$ of the fuel's rest mass to energy, and the luminosity is $3.9 \times 10^{26}$ W.

Fuel mass: $0.10 \times 2.0 \times 10^{30} = 2.0 \times 10^{29}$ kg. Mass actually converted: $0.0071 \times 2.0 \times 10^{29} = 1.42 \times 10^{27}$ kg. Energy available:

$E = mc^2 = (1.42 \times 10^{27})(3.00 \times 10^8)^2 = 1.28 \times 10^{44}\ \text{J}$

Divide by the power: $t = 1.28 \times 10^{44} / 3.9 \times 10^{26} = 3.3 \times 10^{17}$ s. In years, $3.3 \times 10^{17} / 3.16 \times 10^7 \approx 1 \times 10^{10}$: ten billion years, which is the right order for a star like this one. The same chain run backwards gives a reactor's fuel burn: divide the required power by the energy per reaction to get reactions per second, then multiply by the mass per reaction.

Predicting the curve: the semi-empirical mass formula

The reasoning about volume, surface and Coulomb can be made quantitative. Add a term punishing any imbalance between protons and neutrons, and one small correction for pairing, and you get the semi-empirical mass formula:

$B = a_v A - a_s A^{2/3} - a_c \frac{Z(Z-1)}{A^{1/3}} - a_{\text{sym}} \frac{(A - 2Z)^2}{A} \pm \frac{12}{\sqrt{A}}$

with $a_v = 15.75$, $a_s = 17.8$, $a_c = 0.711$, $a_{\text{sym}} = 23.7$ MeV. Term by term: volume binding proportional to $A$; a surface correction proportional to the area $A^{2/3}$; Coulomb repulsion over all $Z(Z-1)/2$ proton pairs, divided by the radius which goes as $A^{1/3}$; a symmetry penalty that vanishes when $Z = A/2$; and a pairing term, $+$ for even-even nuclei, $-$ for odd-odd, zero otherwise.

Worked example 5: calcium-40, with $Z = 20$, $A = 40$, even-even.

Volume: $15.75 \times 40 = 630.0$. Surface: $40^{2/3} = 11.70$, so $17.8 \times 11.70 = 208.2$. Coulomb: $40^{1/3} = 3.420$, and $Z(Z-1) = 380$, so $0.711 \times 380/3.420 = 79.0$. Symmetry: $A - 2Z = 0$, so nothing. Pairing: $+12/\sqrt{40} = +1.90$.

$B = 630.0 - 208.2 - 79.0 - 0 + 1.9 = 344.7 \approx 345\ \text{MeV}$

That is $8.6$ MeV per nucleon, within a percent of the measured value from a formula with five constants.

The formula also tells you which nucleus is the stable one at a given $A$. Only the Coulomb and symmetry terms depend on $Z$, so set $\partial B/\partial Z = 0$ and solve. Doing that gives

$Z_{\min} = \frac{A}{2 + (a_c/2a_{\text{sym}})A^{2/3}}$

For $A = 125$: $125^{2/3} = 25.0$, and $a_c/2a_{\text{sym}} = 0.711/47.4 = 0.0150$, so the denominator is $2 + 0.0150 \times 25.0 = 2.375$ and $Z_{\min} = 125/2.375 = 52.6$. The real stable nuclide at $A = 125$ is tellurium, $Z = 52$. Notice what the formula says: for light nuclei the denominator is close to $2$ and $Z \approx A/2$, equal numbers of each. For heavy nuclei the Coulomb term pushes the ratio toward more neutrons, which is why the valley of stability bends away from the $N = Z$ line.

Recognising the technique

You are in binding-energy territory whenever a problem hands you masses to more than four decimal places. That precision is only ever there so you can subtract and find a defect. The routine follows from what the problem gives you:

Given atomic masses and one nuclide, you want $E_B$: sum $Z m_H + N m_n$, subtract the tabulated mass, multiply by $931.5$. Divide by $A$ if it asks per nucleon.

Given masses on both sides of a reaction, you want $Q$: initial minus final, times $931.5$. This covers alpha decay, beta decay endpoints and every fusion reaction alike.

Given binding energies rather than masses, flip the direction: final minus initial. Separation energies are the same move with one nucleon removed: taking a neutron out of helium-4 costs $28.3 - 7.72 = 20.6$ MeV, and the photon that does it needs at least that energy.

Given a power and an energy per event, you are converting units: watts to events per second, or grams per day, through $1$ MeV $= 1.602 \times 10^{-13}$ J.

Given two nuclei that must touch, it is not binding energy at all, it is $ke^2 Z_1 Z_2/r$ with $ke^2 = 1.44$ MeV·fm.

Every one of these is the same conservation law you have used since mechanics. Mass is simply another place energy can sit. Next you go below the nucleus, to what protons and neutrons are actually made of, in Intro to Particle Physics.