Weigh a helium-4 atom and compare it with two hydrogen atoms plus two neutrons: the atom is lighter. The missing mass became the energy that holds the nucleus together, by $E = mc^2$. This connects stability with the energy of fission and fusion, and uses scientific notation throughout.
Key idea
Mass defect: $\Delta m = Z\,m(^1\mathrm{H}) + N\,m_n - m(\text{atom})$. Binding energy: $E_b = \Delta mc^2$, which is $931.5\ \text{MeV}$ for every $1\ \text{u}$ of mass defect. Binding energy per nucleon, $E_b/A$, measures how tightly a nucleus is held; it peaks near iron and nickel at about $8.8\ \text{MeV}$.
Why it works
Bound systems have less energy than their separate parts: energy is released when they form, and must be supplied to pull them apart. Relativity says energy has mass, so the bound nucleus weighs less. Using hydrogen-1 atoms (a proton plus an electron) instead of bare protons means the electrons are counted on both sides, so atomic masses from tables can be used directly. Because $E_b/A$ peaks near $A = 56$–62, fusing light nuclei moves up the curve and releases energy, and so does splitting heavy ones.
Worked examples
Worked example: Helium-4
Find the mass defect, binding energy and binding energy per nucleon of helium-4 (atomic mass $4.002603\ \text{u}$). Use $m(^1\mathrm{H}) = 1.007825\ \text{u}$ and $m_n = 1.008665\ \text{u}$.
- Parts: $2(1.007825) + 2(1.008665) = 4.032980\ \text{u}$.
- $\Delta m = 4.032980 - 4.002603 = 0.030377\ \text{u}$.
- $E_b = (0.030377)(931.5) = 28.30\ \text{MeV}$; per nucleon, $28.30/4 = 7.07\ \text{MeV}$.
Answer: $0.030377\ \text{u}$; $28.3\ \text{MeV}$; $7.07\ \text{MeV}$ per nucleon.
Worked example: In joules and per mole
Express helium-4's binding energy in joules per nucleus and in kJ/mol ($1\ \text{u} = 1.6605 \times 10^{-27}\ \text{kg}$, $c = 3.00 \times 10^8\ \text{m/s}$).
- Mass in kg: $$(0.030377)(1.6605 \times 10^{-27}) = 5.044 \times 10^{-29}\ \text{kg}$$
- $$E = mc^2 = (5.044 \times 10^{-29})(3.00 \times 10^8)^2 = 4.54 \times 10^{-12}\ \text{J}$$ per nucleus.
- Per mole: $$(4.54 \times 10^{-12})(6.022 \times 10^{23}) = 2.73 \times 10^{12}\ \text{J/mol} = 2.73 \times 10^{9}\ \text{kJ/mol}$$ millions of times a chemical bond energy.
Answer: $4.54 \times 10^{-12}\ \text{J}$; $2.73 \times 10^{9}\ \text{kJ/mol}$.
Worked example: Iron-56
Iron-56 has atomic mass $55.934936\ \text{u}$. Find its binding energy per nucleon.
- Parts: $$26(1.007825) + 30(1.008665) = 56.463400\ \text{u}$$; $\Delta m = 0.528464\ \text{u}$.
- $E_b = (0.528464)(931.5) = 492.3\ \text{MeV}$; per nucleon $492.3/56 = 8.79\ \text{MeV}$, near the top of the curve.
Answer: $8.79\ \text{MeV}$ per nucleon.
Common mistake
Thinking a nucleus weighs more than its separate protons and neutrons, or using $E = mc^2$ with the mass still in atomic mass units. The nucleus is always lighter, and $c^2$ in $\text{m}^2/\text{s}^2$ needs mass in kilograms to give joules. Check: a binding energy in joules per nucleus should be around $10^{-12}$ to $10^{-10}\ \text{J}$.