Binding Energy
Energy holding a nucleus together.
Formula
Variables
How it works
A bound nucleus has less mass than the total mass of its separate protons and neutrons; that missing mass is the mass defect, \(\Delta m\). In relativity, mass and energy are equivalent, so the energy released when the nucleus formed, or the energy required to pull it apart again, is \(E_b = \Delta m c^2\). This works because the nucleus sits in a lower-energy state than the free nucleons, and the lost rest-mass shows up as nuclear binding energy.
When to use it
Use this when you need to find how tightly a nucleus is bound, compare nuclear stability, or calculate the energy released in nuclear reactions and decays. It is also used in nuclear physics and chemistry problems involving atomic masses, mass defects, and reaction Q-values.
Worked example, step by step
A helium-4 nucleus has a mass defect of \(\Delta m = 0.0304\,\text{u}\). Convert mass defect to kilograms: \(\Delta m = 0.0304 \times 1.6605\times10^{-27} = 5.04\times10^{-29}\,\text{kg}\) Use \(E_b = \Delta m c^2\): \(E_b = (5.04\times10^{-29})(3.00\times10^8)^2\) Square the speed of light: \((3.00\times10^8)^2 = 9.00\times10^{16}\) Multiply: \(E_b = 5.04\times10^{-29} \times 9.00\times10^{16} = 4.54\times10^{-12}\,\text{J}\) Convert to MeV: \(E_b = \frac{4.54\times10^{-12}}{1.602\times10^{-13}} \approx 28.3\,\text{MeV}\) Final answer: the binding energy is \(4.54\times10^{-12}\,\text{J}\) or about \(28.3\,\text{MeV}\).
Common mistakes to avoid
- Using the nucleus mass directly instead of the mass defect
- Forgetting to convert atomic mass units to kilograms before using \(c\) in SI units
- Confusing binding energy with energy per nucleon or with the total rest energy of the nucleus
Did You Know?
A helium nucleus weighs slightly less than its parts, and that missing mass is the energy powering the Sun.
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