define and use the terms mass defect and binding energy

Mass Defect and Nuclear Binding Energy

Learning Objective

By the end of this lesson you should be able to:

• Define mass defect and binding energy.
• Calculate the mass defect of a nucleus from atomic masses.
• Convert mass defect to binding energy using Einstein’s relation.
• Interpret the significance of binding energy per nucleon.

Key Definitions

Mass defect (\$\Delta m\$) is the difference between the sum of the masses of the individual nucleons (protons and neutrons) that would make up a nucleus and the actual mass of the nucleus:

\$\Delta m = \left(Z mp + N mn\right) - m_{\text{nucleus}}\$

where \$Z\$ is the number of protons, \$N\$ the number of neutrons, \$mp\$ the mass of a proton, \$mn\$ the mass of a neutron and \$m_{\text{nucleus}}\$ the measured nuclear mass.

Binding energy (\$E_b\$) is the energy required to separate a nucleus into its constituent protons and neutrons. It is obtained from the mass defect via Einstein’s mass‑energy equivalence:

\$E_b = \Delta m\,c^{2}\$

In nuclear physics it is convenient to use the conversion \$1\;\text{u}c^{2}=931.5\;\text{MeV}\$, so

\$E_b\;(\text{MeV}) = \Delta m\;(\text{u}) \times 931.5.\$

Step‑by‑Step Calculation

1. Write down the number of protons (\$Z\$) and neutrons (\$N\$) for the nucleus.
2. Find the atomic mass of a proton (\$mp = 1.007276\;\text{u}\$) and a neutron (\$mn = 1.008665\;\text{u}\$).
3. Obtain the experimental nuclear mass \$m_{\text{nucleus}}\$ from a table of atomic masses.
4. Calculate the total mass of the separated nucleons: \$Z mp + N mn\$.
5. Determine the mass defect: \$\Delta m = (Z mp + N mn) - m_{\text{nucleus}}\$.
6. Convert \$\Delta m\$ to binding energy using \$E_b = \Delta m \times 931.5\;\text{MeV}\$.
7. Optionally, find the binding energy per nucleon: \$E_b/A\$, where \$A = Z+N\$.

Example: Helium‑4 (\$^{4}\text{He}\$)

Thus the binding energy per nucleon for \$^{4}\text{He}\$ is \$27.2\;\text{MeV} / 4 \approx 6.8\;\text{MeV}\$.

Why Binding Energy Matters

The binding energy per nucleon indicates the stability of a nucleus. Nuclei with higher \$E_b/A\$ are more tightly bound and less likely to undergo spontaneous fission or decay. The curve of binding energy per nucleon peaks around iron (\$^{56}\text{Fe}\$), explaining why both fission of heavy nuclei and fusion of light nuclei release energy.

Practice Questions

1. Calculate the mass defect and binding energy of \$^{12}\text{C}\$ using \$m_{\text{nucleus}} = 12.000000\;\text{u}\$.
2. Explain why the binding energy per nucleon of \$^{238}\text{U}\$ is lower than that of \$^{56}\text{Fe}\$.
3. Given a mass defect of \$0.0189\;\text{u}\$ for a nucleus, determine its binding energy in MeV.

Summary

• Mass defect is the “missing mass” when nucleons bind to form a nucleus.
• Binding energy quantifies the energy equivalent of this missing mass.
• Using \$E = \Delta m c^{2}\$ (or \$931.5\;\text{MeV/u}\$) we can convert mass defect to binding energy.
• Binding energy per nucleon provides insight into nuclear stability and the energy released in fission and fusion.