Binding Energy and Mass Defect
Binding Energy and Mass Defect
The nucleus of an atom is held together by the strong nuclear force, the most powerful force in nature. This force binds protons and neutrons — collectively called nucleons — into a compact nucleus. But here is a remarkable fact: the mass of a nucleus is always slightly less than the sum of the masses of its individual protons and neutrons. This missing mass is called the mass defect, and it is the key to understanding nuclear energy.
Einstein's mass-energy equivalence, E equals mc squared, tells us that the missing mass has been converted into binding energy — the energy required to disassemble the nucleus into its individual nucleons. The larger the binding energy, the more tightly bound and stable the nucleus. Iron-56 has the highest binding energy per nucleon of any isotope, at about 8.79 MeV per nucleon. This is why iron is the endpoint of stellar nucleosynthesis — fusing lighter elements up to iron releases energy, but fusing heavier elements requires energy input.
The binding energy per nucleon curve is one of the most important graphs in nuclear physics. Light nuclei like hydrogen and helium have relatively low binding energy per nucleon, which means fusing them together releases enormous energy — this is nuclear fusion, the process that powers the sun. Heavy nuclei like uranium have lower binding energy per nucleon than mid-mass nuclei, which means splitting them apart also releases energy — this is nuclear fission, the process used in nuclear power plants.
Calculating binding energy is straightforward arithmetic: find the mass defect in atomic mass units, then convert to energy using the fact that 1 amu equals 931.5 MeV. This simple calculation reveals the immense energy locked inside every atomic nucleus.
The mass defect $\Delta m$ of a nucleus with $Z$ protons and $N$ neutrons is:
$$\Delta m = Z m_p + N m_n - m_{\text{nucleus}}$$
where $m_p = 1.007276$ u, $m_n = 1.008665$ u.
$$E_B = \Delta m \times 931.5 \text{ MeV/u}$$
Binding energy per nucleon:
$$\frac{E_B}{A} \text{ where } A = Z + N$$
Higher $E_B/A$ means a more stable nucleus. Maximum near $A \approx 56$ (iron).
Calculate the binding energy of Helium-4 ($^4_2$He). Nuclear mass = 4.001506 u.
Step 1: Find the mass of constituents.
$$2m_p + 2m_n = 2(1.007276) + 2(1.008665) = 4.031882 \text{ u}$$
Step 2: Find the mass defect.
$$\Delta m = 4.031882 - 4.001506 = 0.030376 \text{ u}$$
Step 3: Convert to energy.
$$E_B = 0.030376 \times 931.5 = 28.30 \text{ MeV}$$
Step 4: Binding energy per nucleon.
$$E_B/A = 28.30/4 = 7.07 \text{ MeV/nucleon}$$
Iron-56 has a nuclear mass of 55.92067 u ($Z = 26$, $N = 30$). Find $E_B/A$.
Step 1: Constituent mass: $26(1.007276) + 30(1.008665) = 26.18918 + 30.25995 = 56.44913$ u
Step 2: $\Delta m = 56.44913 - 55.92067 = 0.52846$ u
Step 3: $E_B = 0.52846 \times 931.5 = 492.3$ MeV
Step 4: $E_B/A = 492.3/56 = 8.79$ MeV/nucleon — the highest of any element.
In the fusion reaction $^2$H + $^3$H → $^4$He + n, how much energy is released? (Masses: D = 2.01410 u, T = 3.01605 u, He-4 = 4.00260 u, n = 1.00867 u)
Step 1: Total reactant mass: $2.01410 + 3.01605 = 5.03015$ u
Step 2: Total product mass: $4.00260 + 1.00867 = 5.01127$ u
Step 3: Mass converted to energy: $\Delta m = 5.03015 - 5.01127 = 0.01888$ u
Step 4: $E = 0.01888 \times 931.5 = 17.59$ MeV
This single fusion reaction releases 17.6 MeV — the basis of thermonuclear energy.
Practice Problems
Show Answer Key
1. $\Delta m = 1.007276 + 1.008665 - 2.01355 = 0.00239$ u. $E_B = 0.00239 \times 931.5 = 2.23$ MeV.
2. Constituents: $6(1.007276)+6(1.008665)=12.09565$ u. $\Delta m = 12.09565-11.99671=0.09894$ u. $E_B = 92.2$ MeV. $E_B/A = 7.68$ MeV/nucleon.
3. $E_B = 7.59 \times 235 = 1784$ MeV.
4. Fe-56 sits at the peak of the $E_B/A$ curve. Fusing lighter elements (going up the curve) releases energy. Fissioning heavier elements (going down toward Fe) also releases energy. Fe is the "ash" of both processes.
5. Reactants: $2(2.01410) = 4.02820$ u. Products: $3.01603+1.00867=4.02470$ u. $\Delta m = 0.00350$ u. $E = 3.26$ MeV.
6. $1$ u $=1.6605\times10^{-27}$ kg. $E=mc^2=1.6605\times10^{-27}\times(3\times10^8)^2=1.494\times10^{-10}$ J $=931.5$ MeV.
7. Constituents: $8(1.007276)+8(1.008665)=16.12753$ u. $\Delta m=0.13700$ u. $E_B=127.6$ MeV. $E_B/A=7.98$ MeV/nucleon.
8. Two He-4: $E_B=2\times28.30=56.60$ MeV. Be-8: $\Delta m=4(1.007276)+4(1.008665)-8.00531=0.06045$ u. $E_B=56.3$ MeV. Be-8 has less total binding energy than two separate He-4 nuclei, so it is unstable and splits instantly.
9. $0.030376/4.031882 \approx 0.754\%$.
10. $1$ GW $=10^9$ J/s. Each fission: $200\times1.602\times10^{-13}=3.204\times10^{-11}$ J. Fissions/s $= 10^9/3.204\times10^{-11}\approx 3.12\times10^{19}$.