understand the equivalence between energy and mass as represented by E = mc2 and recall and use this equation

Mass Defect and Nuclear Binding Energy

Learning Objective

Students will be able to:

• Explain the equivalence of mass and energy as expressed by Einstein’s equation \$E = mc^{2}\$.
• Define mass defect and nuclear binding energy.
• Calculate the mass defect of a nucleus from atomic masses.
• Convert mass defect into binding energy using \$E = mc^{2}\$.
• Apply these ideas to solve typical A‑Level exam questions.

1. The Principle \$E = mc^{2}\$

Einstein’s famous relation links the energy \$E\$ of a system to its mass \$m\$ through the speed of light \$c\$:

\$E = mc^{2}\$

where \$c = 3.00 \times 10^{8}\ \text{m s}^{-1}\$. In nuclear physics we usually work with electron‑volts (eV) as the energy unit and atomic mass units (u) as the mass unit. The conversion factor is:

\$1\ \text{u} = 931.5\ \text{MeV}/c^{2}\$

Thus a mass change \$\Delta m\$ (in u) corresponds to an energy change \$\Delta E\$ (in MeV) given by:

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

2. What Is Mass Defect?

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

\$\Delta m = Zm{p} + Nm{n} - M_{\text{nucleus}}\$

where:

• \$Z\$ = number of protons, \$m_{p}\$ = mass of a proton.
• \$N\$ = number of neutrons, \$m_{n}\$ = mass of a neutron.
• \$M_{\text{nucleus}}\$ = measured atomic mass of the nuclide (including electron masses, which are subtracted when necessary).

3. Nuclear Binding Energy

The energy required to separate a nucleus into its individual nucleons is the binding energy \$E_{b}\$. It is obtained by converting the mass defect into energy:

\$E_{b} = \Delta m\,c^{2} = \Delta m \times 931.5\ \text{MeV}\$

Binding energy per nucleon is a useful indicator of nuclear stability:

\$\frac{E_{b}}{A} = \frac{\text{total binding energy}}{\text{mass number }A}\$

4. Example Calculation – Helium‑4

Given data:

Step 1 – Sum of separate nucleon masses:

\$M_{\text{separate}} = 2(1.007276) + 2(1.008665) = 4.031882\ \text{u}\$

Step 2 – Mass defect:

\$\Delta m = M{\text{separate}} - M{\text{nucleus}} = 4.031882 - 4.002603 = 0.029279\ \text{u}\$

Step 3 – Binding energy:

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

Step 4 – Binding energy per nucleon:

\$\frac{E_{b}}{A} = \frac{27.2\ \text{MeV}}{4} = 6.8\ \text{Me \cdot per nucleon}\$

5. Common Pitfalls

• For atomic masses, remember to subtract the electron masses if you are using nuclear masses only.
• Use the correct number of significant figures – the conversion factor 931.5 MeV u⁻¹ is usually given to four significant figures.
• Do not forget to convert the mass defect from atomic mass units to energy using the factor 931.5 MeV u⁻¹.

6. Practice Questions

1. Calculate the mass defect and binding energy of \$^{12}\text{C}\$ using the following data: \$m{p}=1.007276\ \text{u}\$, \$m{n}=1.008665\ \text{u}\$, \$M(^{12}\text{C})=12.000000\ \text{u}\$.
2. A certain isotope has a binding energy per nucleon of \$8.5\ \text{MeV}\$. If its mass number is \$A=56\$, find the total binding energy and the mass defect.
3. Explain why the binding energy per nucleon peaks around iron (Fe) and what this implies for nuclear fission and fusion.

7. Summary

Mass defect and nuclear binding energy are direct consequences of the mass–energy equivalence \$E = mc^{2}\$. By measuring atomic masses we can determine how much mass is “missing” when nucleons bind together, and by converting this missing mass to energy we obtain the binding energy, a key quantity for understanding nuclear stability, fission, and fusion processes.