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Mass Defect and Nuclear Binding Energy

Learning Objective

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

• Explain the concept of mass defect.
• Calculate the nuclear binding energy of a nucleus.
• Write simple nuclear reactions using nuclear notation.
• Interpret binding‑energy per nucleon trends.

1. What is Mass Defect?

The mass of a nucleus is always slightly less than the sum of the masses of its constituent protons and neutrons. This difference is called the mass defect (\$\Delta m\$):

\$\Delta m = \left(Zmp + Nmn\right) - m_{\text{nucleus}}\$

where \$Z\$ is the number of protons, \$N\$ the number of neutrons, \$mp\$ the mass of a proton and \$mn\$ the mass of a neutron.

2. From Mass Defect to Binding Energy

Einstein’s mass‑energy equivalence relates the mass defect to the energy required to separate the nucleus into its constituent nucleons:

\$E_{\text{b}} = \Delta m\,c^{2}\$

In practice we use the conversion factor \$1\;\text{u} = 931.5\;\text{MeV}/c^{2}\$, so the binding energy in mega‑electron‑volts is

\$E_{\text{b}}(\text{MeV}) = \Delta m(\text{u}) \times 931.5\$

3. Binding Energy per Nucleon

The average binding energy per nucleon is a useful indicator of nuclear stability:

\$\frac{E{\text{b}}}{A} = \frac{E{\text{b}}}{Z+N}\$

Plotting \$\frac{E_{\text{b}}}{A}\$ against mass number \$A\$ yields the familiar “binding‑energy curve”. Nuclei around \$A\approx 56\$ (e.g. \$^{56}\text{Fe}\$) have the highest values, indicating maximal stability.

4. Writing Nuclear Equations

A nuclear reaction is written in the form

\$\prescript{A}{Z}{\text{X}} + \prescript{a}{z}{\text{y}} \;\rightarrow\; \prescript{A'}{Z'}{\text{X'}} + \prescript{b}{b}{\text{z}}\$

where the superscript denotes the mass number and the subscript the atomic number.

Examples:

• Alpha decay of \$^{238}\text{U}\$:

  \$\prescript{238}{92}{\text{U}} \;\rightarrow\; \prescript{234}{90}{\text{Th}} + \prescript{4}{2}{\alpha}\$

• Beta‑minus decay of \$^{14}\text{C}\$:

  \$\prescript{14}{6}{\text{C}} \;\rightarrow\; \prescript{14}{7}{\text{N}} + \prescript{0}{-1}{\beta} + \bar{\nu}_e\$

• Fusion of deuterium and tritium:

  \$\prescript{2}{1}{\text{H}} + \prescript{3}{1}{\text{H}} \;\rightarrow\; \prescript{4}{2}{\text{He}} + \prescript{1}{0}{\text{n}}\$

5. Sample Calculation – Binding Energy of \$^{4}\text{He}\$

1. Masses (in atomic mass units, u):

   • \$m_{\alpha}=4.002603\,\text{u}\$ (mass of \$^{4}\text{He}\$ nucleus)
   • \$m_p=1.007276\,\text{u}\$ (proton)
   • \$m_n=1.008665\,\text{u}\$ (neutron)

2. Calculate the mass of the separate nucleons:

   \$m{\text{total}} = 2mp + 2m_n = 2(1.007276) + 2(1.008665) = 4.031882\;\text{u}\$

3. Mass defect:

   \$\Delta m = m{\text{total}} - m{\alpha} = 4.031882 - 4.002603 = 0.029279\;\text{u}\$

4. Binding energy:

   \$E_{\text{b}} = 0.029279 \times 931.5 = 27.2\;\text{MeV}\$

5. Binding energy per nucleon:

   \$\frac{E_{\text{b}}}{A} = \frac{27.2}{4} = 6.8\;\text{MeV/nucleon}\$

6. Table of Typical Binding Energies

7. Practice Questions

1. Calculate the mass defect and binding energy of \$^{12}\text{C}\$ using the following masses: \$mp=1.007276\;\text{u}\$, \$mn=1.008665\;\text{u}\$, \$m_{^{12}\text{C}}=12.000000\;\text{u}\$.
2. Write the nuclear equation for the beta‑plus decay of \$^{11}\text{C}\$.
3. Explain why energy is released in the fusion of two deuterium nuclei but not in the fission of \$^{4}\text{He}\$.

8. Summary

The mass defect quantifies the loss of mass when nucleons bind together, and via \$E=mc^{2}\$ it gives the nuclear binding energy. Binding energy per nucleon provides a clear measure of nuclear stability, explaining why heavy nuclei release energy by fission and light nuclei by fusion. Mastery of nuclear notation allows clear communication of these processes.