In Physics 2 you saw that a nucleus weighs less than the protons and neutrons it is made of, and that the missing mass, times $c^2$, is its binding energy. Here we look at the whole family of nuclei: how big they are, and which ones are held most tightly. The answer explains why the Sun shines by fusing light nuclei and why reactors run by splitting heavy ones.
Key idea
A nucleus with $A$ nucleons has radius $$R = r_0A^{1/3},\qquad r_0 \approx 1.2\ \text{fm}\ (1\ \text{fm} = 10^{-15}\ \text{m}),$$ so every nucleus has about the same density, $2.3 \times 10^{17}$ kg/m³. Its binding energy, from atomic masses, is $$E_B = \left[Zm(^1\text{H}) + Nm_n - M(^A\text{X})\right]c^2,\qquad 1\ \text{u}\cdot c^2 = 931.5\ \text{MeV}.$$ The binding energy per nucleon, $E_B/A$, rises steeply for light nuclei, peaks near iron and nickel at about 8.8 MeV, and falls slowly for heavy nuclei.
Why it works
If nucleons pack together like marbles, each taking the same volume, the nucleus's volume is proportional to $A$, and its radius to $A^{1/3}$. Constant density means the strong force holds each nucleon to its near neighbors only (it has a short range). That explains the shape of the curve. A nucleon inside a big nucleus has a full set of neighbors, but in a small nucleus many nucleons sit at the surface with fewer neighbors, so light nuclei are loosely bound. Protons also repel each other electrically, and that repulsion reaches across the whole nucleus, growing roughly like $Z^2$; in heavy nuclei it wins back some of the binding. The best compromise is near $A \approx 56$ to 62. Using atomic masses (with $^1$H, not a bare proton) keeps the electrons balanced: $Z$ hydrogen atoms bring the same $Z$ electrons as the neutral atom.
Worked examples
Worked example: How big is a nucleus?
Find the radii of iron-56 and uranium-238.
- Fe-56: $R = 1.2\ \text{fm} \times 56^{1/3} = 1.2 \times 3.83 = 4.6$ fm.
- U-238: $R = 1.2 \times 238^{1/3} = 1.2 \times 6.20 = 7.4$ fm.
- U-238 has 4.25 times as many nucleons but is only $4.25^{1/3} = 1.62$ times as wide.
Answer: About 4.6 fm and 7.4 fm, while an atom is about 100,000 fm across
Worked example: Helium-4
Find the binding energy of helium-4 and its binding energy per nucleon. Atomic masses: $^4$He 4.002603 u, $^1$H 1.007825 u, neutron 1.008665 u.
- Parts: $2(1.007825) + 2(1.008665) = 4.032980$ u.
- Mass defect: $4.032980 - 4.002603 = 0.030377$ u.
- $E_B = 0.030377 \times 931.5 = 28.3$ MeV, and $E_B/A = 28.3/4 = 7.07$ MeV.
Answer: $28.3$ MeV, or 7.07 MeV per nucleon: unusually tight for so light a nucleus
Worked example: Iron-56
Repeat for iron-56 (26 protons, 30 neutrons, atomic mass 55.934936 u).
- Parts: $26(1.007825) + 30(1.008665) = 56.463400$ u.
- Mass defect: $56.463400 - 55.934936 = 0.528464$ u.
- $E_B = 0.528464 \times 931.5 = 492.3$ MeV, and $E_B/A = 8.79$ MeV.
Answer: $492$ MeV, or 8.79 MeV per nucleon, near the top of the curve
Common mistake
Thinking bigger nuclei are always more tightly bound because they have more binding energy in total. Total binding grows with $A$, but what decides stability and energy release is the binding per nucleon, and that peaks near iron. Uranium's 7.6 MeV per nucleon is less than iron's 8.8, which is why splitting uranium releases energy. Check the direction: energy comes out when nucleons end up more tightly bound per nucleon, whether by fusion or by fission.