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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 ($\Delta m$):

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

where $Z$ is the number of protons, $N$ the number of neutrons, $m_p$ the mass of a proton and $m_n$ 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 $^{14}\text{C}$: $$\prescript{14}{6}{\text{C}} \;\rightarrow\; \prescript{14}{7}{\text{N}} + \prescript{0}{-1}{\beta} + \bar{u}_e$$
• 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):
2. Calculate the mass of the separate nucleons: $$m_{\text{total}} = 2m_p + 2m_n = 2(1.007276) + 2(1.008665) = 4.031882\;\text{u}$$
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

Nucleus	Mass Number $A$	Binding Energy $E_{\text{b}}$ (MeV)	Binding Energy per Nucleon (MeV)
$^{2}\text{H}$ (Deuterium)	2	2.22	1.11
$^{4}\text{He}$ (Alpha particle)	4	28.30	7.07
$^{12}\text{C}$	12	92.2	7.68
$^{56}\text{Fe}$	56	492.3	8.80
$^{238}\text{U}$	238	1786	7.50

7. Practice Questions

1. Calculate the mass defect and binding energy of $^{12}\text{C}$ using the following masses: $m_p=1.007276\;\text{u}$, $m_n=1.008665\;\text{u}$, $m_{^{12}\text{C}}=12.000000\;\text{u}$.
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.