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: $$\Delta m = \left(Z m_p + N m_n\right) - m_{\text{nucleus}}$$ where $Z$ is the number of protons, $N$ the number of neutrons, $m_p$ the mass of a proton, $m_n$ 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 ($m_p = 1.007276\;\text{u}$) and a neutron ($m_n = 1.008665\;\text{u}$).
3. Obtain the experimental nuclear mass $m_{\text{nucleus}}$ from a table of atomic masses.
4. Calculate the total mass of the separated nucleons: $Z m_p + N m_n$.
5. Determine the mass defect: $\Delta m = (Z m_p + N m_n) - m_{\text{nucleus}}$.
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 ($^{4}\text{He}$)

Nucleus	Mass (u)	Mass of nucleons (u)	Mass defect $\Delta m$ (u)	Binding energy $E_b$ (MeV)
$^{4}\text{He}$	4.002603	$2m_p + 2m_n = 2(1.007276) + 2(1.008665) = 4.031882$	$4.031882 - 4.002603 = 0.029279$	$0.029279 \times 931.5 \approx 27.2$

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.