Where does the missing mass go?

It was released as energy when the nucleus formed. Binding energy is the equivalent energy, via E = mc^2.

Why is iron the most stable nucleus?

It has the highest binding energy per nucleon, about 8.8 MeV, so neither fusion nor fission can release energy from it.

Why do nuclear reactions release far more energy than chemical reactions?

Nuclear binding energies are millions of electron-volts per nucleon; chemical bonds are a few electron-volts per atom.

Does mass change in chemical reactions too?

Yes, in principle, but the energy is so small that the mass change is far too tiny to measure.

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164–166Nuclear PhysicsStretch

Mass–Energy and Binding Energy

Weigh a helium nucleus and then weigh two protons and two neutrons separately: the nucleus is lighter. That missing mass, the $\textbf{mass defect}$ , is the energy that holds the nucleus together — the $\textbf{binding energy}$ — converted through $E = mc^{2}$ .

40 min Topic 11 · Nuclear Radiation

What you'll be able to do

Use $\Delta E = c^{2}\Delta m$ and convert between u, kg, J and MeV
Define mass defect and binding energy
Calculate binding energy per nucleon and interpret it as a measure of stability
Use the binding energy per nucleon curve to explain energy release in fission and fusion

Mass–energy equivalence

Einstein’s relation $\Delta E = c^{2}\Delta m$ says that any change in energy is accompanied by a change in mass. Because $c^{2} = 9.0\times10^{16}\ \mathrm{m^{2}\,s^{-2}}$ , a tiny mass corresponds to a vast energy.

Nuclear masses are measured in $\textbf{atomic mass units}$ : $1\ \mathrm{u} = 1.661\times10^{-27}$ kg, one-twelfth the mass of a carbon-12 atom.

It is convenient to know that $1\ \mathrm{u}$ is equivalent to $931.5$ MeV. So a mass difference in u, multiplied by 931.5, gives energy in MeV directly.

\Delta E = c^{2}\Delta m

Mass–energy equivalence.

1\ \mathrm{u} = 1.661\times10^{-27}\ \mathrm{kg} \equiv 931.5\ \mathrm{MeV}

Atomic mass unit.

Example 1Find the energy equivalent of 1.0 g of matter.

\Delta E = 9.0\times10^{16} \times 1.0\times10^{-3}

Answer

9.0\times10^{13}\ \mathrm{J}

— roughly the energy of a large power station running for a day.

Mass defect and binding energy

The $\textbf{mass defect}$ of a nucleus is the difference between the total mass of its separate protons and neutrons and the mass of the nucleus itself.

The $\textbf{binding energy}$ is the energy needed to separate a nucleus into its individual nucleons. Equivalently, it is the energy released when the nucleus forms from them — which is why the nucleus has less mass.

Binding energy $= \Delta m \times c^{2}$ , where $\Delta m$ is the mass defect.

\Delta m = Zm_{p} + (A - Z)m_{n} - m_{\text{nucleus}}

Mass defect.

Example 1A helium-4 nucleus has mass 4.00151 u. With

m_{p} = 1.00728

u and

m_{n} = 1.00867

u, find its binding energy in MeV.

1Separate nucleons:

2(1.00728) + 2(1.00867) = 4.03190

2Mass defect:

4.03190 - 4.00151 = 0.03039

3Binding energy:

0.03039 \times 931.5

Answer

28.3\ \mathrm{MeV}

, or

7.08

MeV per nucleon.

Tip — Keep five or six decimal places in u. The mass defect is a small difference of large numbers, and early rounding destroys it.

Binding energy per nucleon

Total binding energy grows with nucleus size, so to compare stability we divide by nucleon number: $\textbf{binding energy per nucleon}$ . The higher it is, the more tightly bound and stable the nucleus.

The graph of binding energy per nucleon against $A$ rises steeply for light nuclei, peaks at about 8.8 MeV near iron-56, then falls slowly for heavier nuclei.

Helium-4 sits well above its neighbours — an unusually stable arrangement, which is why alpha particles are emitted as a unit.

Energy release from the curve

Any nuclear reaction that moves products $\textbf{up}$ the binding energy per nucleon curve releases energy, because the products are more tightly bound than the reactants.

$\textbf{Fusion}$ joins light nuclei, left of the peak, into heavier ones with higher binding energy per nucleon. $\textbf{Fission}$ splits heavy nuclei, right of the peak, into medium ones with higher binding energy per nucleon.

The energy released equals the increase in total binding energy, which equals the decrease in total mass multiplied by $c^{2}$ .

E_{\text{released}} = \left(\sum m_{\text{before}} - \sum m_{\text{after}}\right)c^{2}

Energy from a mass decrease.

Tip — Binding energy is not energy stored in the nucleus. A higher binding energy means the nucleus has $\textit{lost}$ more energy on forming, so it has less mass.

Equation recap

\Delta E = c^{2}\Delta m

Mass–energy.

1\ \mathrm{u} \equiv 931.5\ \mathrm{MeV}

Conversion.

\text{B.E.} = \Delta m\, c^{2}

Binding energy.

Common mistakes to avoid

Saying the nucleus is heavier than its separate nucleons.

It is lighter by the mass defect.

Treating binding energy as energy stored in the nucleus.

It is the energy needed to pull it apart; more binding means less mass.

Comparing stability using total binding energy.

Use binding energy per nucleon.

Forgetting to convert u to kg before using

c^{2}

Use

1\ \mathrm{u} = 1.661\times10^{-27}

kg, or

931.5

MeV per u directly.

Key takeaways

$\Delta E = c^{2}\Delta m$ ; $1\ \mathrm{u} \equiv 931.5$ MeV.
Mass defect: separate nucleons minus nucleus.
Binding energy is the energy needed to separate a nucleus into nucleons.
Binding energy per nucleon peaks near iron-56.
Fusion and fission release energy by moving products up the curve.

Test yourself

Answer a few questions, then review anything you missed.

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