Physics 2Quantum Physics and the NucleusHS-PS1-8

Mass-Energy and Binding Energy

E = mc²: a nucleus has less mass than its separate protons and neutrons, and the missing mass is its binding energy.

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In Fission and Fusion you saw that nuclear reactions release energy as a little mass disappears. Einstein's $E = mc^{2}$ says how much: mass is a form of energy, with $c^{2}$ as the exchange rate. Energy is conserved, as in Conservation of Mechanical Energy, once you count rest energy too. Nuclear energies are millions of times bigger than chemical ones, so we measure them in MeV: $1\ \text{MeV} = 10^{6}\ \text{eV}$, and an electron volt is the energy one electron gains crossing $1\ \text{V}$ (Electric Potential and Voltage).

Key idea

$E = mc^{2}$. A nucleus has less mass than its separate protons and neutrons. The missing mass, the mass defect $\Delta m$, times $c^{2}$ is the binding energy, the energy needed to pull the nucleus apart. In atomic units, $1\ \text{u} = 931.5\ \text{MeV}/c^{2}$.

Why it works

When protons and neutrons snap together, the strong force pulls them into a deep energy well and the binding energy leaves as gamma rays and kinetic energy. The bound nucleus then has less total energy than the parts had apart, and since $E = mc^{2}$, less mass. To separate them again you must put that energy back. The more tightly bound, the bigger the mass defect per nucleon.
Binding energy per nucleon peaks near iron and nickel. Fusing light nuclei or splitting heavy ones both move toward the peak, releasing energy.

Common mistake

Thinking a nucleus should weigh more than its parts because it "holds" energy. The binding energy was released when it formed, so the nucleus has less energy and less mass than its separate parts. Check: a nucleus that weighed more than its parts would fly apart on its own and give off energy.

Worked example: Helium-4

Find the binding energy of helium-4 (atomic mass $4.002603\ \text{u}$). Use hydrogen-1 atoms ($1.007825\ \text{u}$, to include the electrons) and neutrons ($1.008665\ \text{u}$).
  1. Parts: $$2(1.007825) + 2(1.008665) = 4.032980\ \text{u}.$$
  2. Mass defect: $$\Delta m = 4.032980 - 4.002603 = 0.030377\ \text{u}.$$
  3. $$E_B = (0.030377)(931.5\ \text{MeV}) = 28.3\ \text{MeV},$$ or $7.07\ \text{MeV}$ per nucleon.

Answer: $28.3\ \text{MeV}$

Worked example: Energy from fusion

In the fusion reaction $^{2}\text{H} + {}^{3}\text{H} \to {}^{4}\text{He} + n$, the masses are $2.014102$, $3.016049$, $4.002603$ and $1.008665\ \text{u}$. How much energy is released?
  1. Before: $2.014102 + 3.016049 = 5.030151\ \text{u}$. After: $4.002603 + 1.008665 = 5.011268\ \text{u}$.
  2. Mass lost: $0.018883\ \text{u}$.
  3. $Q = (0.018883)(931.5) = 17.6\ \text{MeV}$, mostly carried off by the neutron.

Answer: $17.6\ \text{MeV}$

Worked example: Mass into joules

How much energy is $1.0\ \text{g}$ of mass?
  1. $E = mc^{2} = (1.0 \times 10^{-3})(3.00 \times 10^{8})^{2}$.
  2. $E = 9.0 \times 10^{13}\ \text{J}$, roughly the energy of $20$ thousand tonnes of TNT.

Answer: $9.0 \times 10^{13}\ \text{J}$

Practice

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Sources and further reading

  • OpenStax College Physics 2e, Section 31.6: Binding Energy
  • OpenStax College Physics 2e, Section 28.6: Relativistic Energy
  • Atomic masses: the 2020 Atomic Mass Evaluation (AME2020), as tabulated by NIST
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