Chapter 04 · Modern Physics · Full book

Nuclei

Core intuition: Chapter Roadmap

In the previous chapter, Rutherford's \(\alpha\)-particles ricocheting off gold foil revealed a tiny, dense core at the heart of every atom—the nucleus. We located it, measured its charge, and then walked away to study the electrons orbiting it. This chapter walks back. We open the nucleus itself and ask what holds it together, why it sometimes falls apart, and how a speck of matter smaller than a ten-thousandth of an atom powers stars and cities alike.

Composition, size, and density — protons and neutrons packed into a sphere of radius \(R = R_0 A^{1/3}\), with a density so enormous and so constant that every nucleus, light or heavy, is made of the same incompressible nuclear matter.
Mass–energy and the atomic mass unit — the bookkeeping of \(E = mc^2\) in units of \(\mathrm{u}\) and \(\mathrm{MeV}\), and the strange fact that a bound nucleus weighs less than its parts.
Binding energy and the curve of stability — the missing mass reappears as binding energy \(B = \Delta m\,c^2\); the binding energy per nucleon \(B/A\), peaking near iron, governs which way energy flows.
The nuclear force — short-ranged, saturating, and charge-independent: the strongest force in everyday matter, yet invisible beyond a few femtometres.
Radioactivity — the three classical decays \(\alpha\), \(\beta\), \(\gamma\), their displacement laws, and the \(Q\)-value that decides whether a decay can happen at all.
The law of radioactive decay — \(N = N_0 e^{-\lambda t}\), half-life, mean life, and activity: why a radioactive sample never quite reaches zero.
Nuclear energy — fission and fusion as two roads to the same peak of the \(B/A\) curve, and the source of the Sun's light.

Master this chapter and the periodic table stops at its true foundation: not the electron cloud, but the nuclear core that fixes an atom's identity and its fate.

Perplexing Questions

Perplexing questions

Before we begin, let these questions sit with you. Do not reach for formulas. You will return to each one—with full answers—before this chapter ends.

The Uniform-Density Puzzle: A uranium nucleus has \(238\) nucleons; a helium nucleus has \(4\). Yet if you could measure the density of each—mass per unit volume—you would get almost exactly the same staggering number, around \(2\times10^{17}\;\mathrm{kg\,m^{-3}}\). How can objects so different in size have identical densities, when ordinary matter from cork to lead does not?
The Missing Mass: Weigh two protons and two neutrons separately, add the numbers, then weigh a helium nucleus made of exactly those four particles. The helium nucleus comes out lighter—by about \(0.7\%\). The mass simply isn't there. Where did it go, and why does its disappearance bind the nucleus together?
The Two-Way Energy Source: Split a heavy uranium nucleus and you release energy—this powers reactors and bombs. Join two light hydrogen nuclei and you also release energy—this powers the Sun. How can both splitting and joining give out energy? Surely one of them must cost energy?
The Electron from Nowhere: In \(\beta^-\) decay, an electron shoots out of the nucleus at high speed. But the nucleus contains only protons and neutrons—no electrons at all. Where does the ejected electron come from?
The Sample That Never Empties: A radioactive sample loses half its atoms in one half-life, half of what remains in the next, and so on. After ten half-lives, less than a thousandth is left—but it is never zero. Does a radioactive sample ever truly finish decaying? And if each atom is identical, what decides which ones go first?
The Sun's Slow Fire: The Sun has shone for nearly five billion years and will shine for five billion more. No chemical fire could last a million. What fuel burns for aeons—and why does it release its energy as a steady glow rather than a single catastrophic flash?

Each question hides a property of the nucleus.
By the end of this chapter, every one of them will be transparent.

Composition, Size, and Density

◈ How big, how dense

live bench

Dial A and watch R = R₀A^(1/3) grow while the density refuses to budge — every nucleus packs the same 2.3×10¹⁷ kg/m³.

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↗ open in the Simulations page

The nucleus is built from two kinds of particle, collectively called nucleons: the positively charged proton and the electrically neutral neutron. Their masses are almost equal, and each is nearly two thousand times heavier than the electron: \[ m_p = 1.00728\;\mathrm{u} = 938.27\;\mathrm{MeV}/c^2, \qquad m_n = 1.00867\;\mathrm{u} = 939.57\;\mathrm{MeV}/c^2, \] against \(m_e = 0.00055\;\mathrm{u} = 0.511\;\mathrm{MeV}/c^2\). Almost all the mass of an atom—more than \(99.9\%\)—is locked in this nuclear core.

A specific nuclear species, or nuclide, is written \({}^{A}_{Z}X\), where

\(Z\), the atomic number, is the number of protons. It fixes the element—and hence the chemistry, since a neutral atom has \(Z\) electrons.
\(N\) is the number of neutrons.
\(A = Z + N\), the mass number, is the total number of nucleons.

Nuclides with the same \(Z\) but different \(N\) are isotopes (same element, different mass): \(\mathrm{{}^1H}\), \(\mathrm{{}^2H}\), \(\mathrm{{}^3H}\) are the three isotopes of hydrogen. Same \(N\), different \(Z\) are isotones; same \(A\) are isobars.

Core intuition: The chemical symbol already tells you \(Z\)

Writing the proton number explicitly in \({}^{A}_{Z}X\) is redundant: the symbol \(X\) already names the element, and the element is \(Z\). Carbon is \(Z=6\) whether you decorate it or not. So \({}^{14}_{\ 6}\mathrm{C}\) and \(\mathrm{{}^{14}C}\) carry identical information. We keep the subscript only when balancing nuclear reactions, where lining up the \(Z\)'s makes the bookkeeping visible.

The size of the nucleus

Scattering experiments—fast electrons, neutrons, and \(\alpha\)-particles fired at nuclei—reveal that the nuclear radius grows with the cube root of the mass number: \[ R = R_0\,A^{1/3}, \qquad R_0 \approx 1.2\;\mathrm{fm} = 1.2\times10^{-15}\;\mathrm{m}. \] A femtometre (\(1\;\mathrm{fm} = 10^{-15}\;\mathrm{m}\), also called a fermi) is the natural ruler of the nucleus, roughly \(100{,}000\) times smaller than an atom. Thus aluminium (\(A=27\)) has \(R \approx 3.6\;\mathrm{fm}\), and uranium (\(A=238\)) only \(R \approx 7.4\;\mathrm{fm}\): a sixfold jump in nucleon count buys barely a doubling in radius.

The cube-root law is not arbitrary. If every nucleon occupies roughly the same little volume \(v_0\), and the nucleons pack together without gaps, then the total volume is \(A v_0\), and a sphere of that volume has radius proportional to \(A^{1/3}\). The nucleus behaves like a tightly packed droplet of incompressible fluid—an image we will use again when we meet the binding-energy curve and, later, nuclear fission.

The density of nuclear matter

The cube-root law has a startling consequence. Take a nucleus of mass number \(A\). Its mass is about \(A\) atomic mass units, and its volume is \(\tfrac{4}{3}\pi R^3 = \tfrac{4}{3}\pi R_0^3 A\). The mass and the volume are both proportional to \(A\)—so when we divide to get density, \(A\) cancels and vanishes.

Birth of the formula: Why Every Nucleus Has the Same Density

Model the nucleus as a uniform sphere. Its mass is \(m \approx A\,m_u\), where \(m_u = 1.66\times10^{-27}\;\mathrm{kg}\) is one atomic mass unit, and its volume is \[ V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi\bigl(R_0 A^{1/3}\bigr)^3 = \frac{4}{3}\pi R_0^3\,A. \] The density is therefore \[ \rho = \frac{m}{V} = \frac{A\,m_u}{\tfrac{4}{3}\pi R_0^3\,A} = \frac{m_u}{\tfrac{4}{3}\pi R_0^3}. \] The mass number \(A\) has cancelled top and bottom—the density does not depend on how big the nucleus is. Putting in numbers with \(R_0 = 1.2\;\mathrm{fm}\), \[ \rho = \frac{1.66\times10^{-27}} {\tfrac{4}{3}\pi\,(1.2\times10^{-15})^3} \approx 2.3\times10^{17}\;\mathrm{kg\,m^{-3}}. \] \[ \boxed{\;\rho_{\text{nuclear}} \approx 2.3\times10^{17}\;\mathrm{kg\,m^{-3}} \quad\text{(independent of }A\text{)}\;} \]

This number is almost impossible to picture. It is about \(10^{14}\) times the density of water: a matchbox of pure nuclear matter would weigh some \(10^{12}\;\mathrm{kg}\)—a thousand million tonnes. The same density turns up in neutron stars, which are essentially single gigantic nuclei held together by gravity. The constancy of \(\rho\) is the first hard fact about the nucleus: nuclear matter is incompressible, and packing in more nucleons makes the nucleus bigger, never denser. This answers Perplexing Question 1.

Misconception alert

“A bigger nucleus must be denser, since it has more particles crammed in.” This confuses amount with concentration. Adding nucleons increases the mass, but Equation shows the volume grows in exact step, because the radius rises as \(A^{1/3}\) and the volume as \(A\). The ratio—the density—stays fixed at \(\rho \approx 2.3\times10^{17}\;\mathrm{kg\,m^{-3}}\) for every nucleus from hydrogen to uranium. Heavier does not mean denser; it only means larger.

Worked Examples

Example 1 · Reading the mass number off a size

A certain nucleus is found to have a radius \(1.5\) times that of a \({}^{12}\mathrm{C}\) nucleus. Estimate its mass number, and name a nucleus that fits.

Solution. Given: \(R/R_{12} = 1.5\); carbon has \(A_{12}=12\).
Find: the mass number \(A\).
Setup: Since \(R = R_0 A^{1/3}\), the radius ratio is a pure cube-root of the mass-number ratio, and \(R_0\) cancels: \(A/A_{12} = (R/R_{12})^3\).
Solve: \[ A = A_{12}\left(\frac{R}{R_{12}}\right)^3 = 12\,(1.5)^3 = 12 \times 3.375 = 40.5 \approx 40. \] Answer: \(\boxed{A \approx 40}\) — calcium-40, \({}^{40}_{20}\mathrm{Ca}\).
Check: A \(50\%\) increase in radius should more than triple the volume (\(1.5^3 = 3.375\)), and volume \(\propto A\), so \(A\) should be about \(3.4\times12 \approx 41\) — consistent. The cube-root law is exactly why a nucleus \(59\) times heavier is only a few times wider. ✓

Example 2 · What it takes to see a nucleus

To resolve a structure you must probe it with a wavelength no larger than its size. What energy of electrons is needed to resolve nuclear detail on the scale of \(5\;\mathrm{fm}\)?

Solution. Given: required \(\lambda \approx 5\;\mathrm{fm}\); probe particles are electrons.
Find: the electron energy.
Setup: By de Broglie \(\lambda = h/p\) (Chapter 11). At these wavelengths an electron is ultra-relativistic, so \(E \approx pc = hc/\lambda\), using the convenient constant \(hc = 1240\;\mathrm{MeV\,fm}\).
Solve: \[ E \approx \frac{hc}{\lambda} = \frac{1240\;\mathrm{MeV\,fm}}{5\;\mathrm{fm}} \approx 250\;\mathrm{MeV}. \] Answer: \(\boxed{E \sim 250\;\mathrm{MeV}}\).
Check: The energy is enormous compared with the few-eV photons of visible light, whose \(\sim\!500\;\mathrm{nm}\) wavelength is \(10^{8}\) times too coarse to see a nucleus. This is precisely why Hofstadter needed a \(\sim\!\mathrm{GeV}\) electron accelerator, not a microscope, to map nuclear charge distributions. ✓

✓ Checkpoint · Composition, Size, and Density

One nucleus has twice the radius of another. What is the ratio of their mass numbers?
\(R \propto A^{1/3}\), so \(A_1/A_2 = (R_1/R_2)^3 = 2^3 = 8\).
In one sentence, why is the density of every nucleus essentially the same?
Both mass and volume are proportional to \(A\) (since \(R \propto A^{1/3}\)), so \(\rho = m/V\) is independent of \(A\).
(Trap) “A \({}^{238}\mathrm{U}\) nucleus has about \(238\) times the volume of a single nucleon but only about \(6\) times the radius.” Is this consistent?
Yes. \(V \propto A\) gives \(238\times\); \(R \propto A^{1/3}\) gives \(238^{1/3} \approx 6.2\times\). Both statements describe the same cube-root law.

Mass, Energy, and the Atomic Mass Unit

◈ Mass defect → binding energy

live bench

Pick a nuclide and watch the ledger: loose nucleons weigh more than the bound nucleus, and the difference Δm × 931.5 is the energy that held it together.

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Nuclear masses are too small for the kilogram and too precise for casual rounding. The natural unit is the atomic mass unit (\(\mathrm{u}\)), defined so that one neutral carbon-12 atom weighs exactly \(12\;\mathrm{u}\): \[ 1\;\mathrm{u} = \frac{\text{mass of one } {}^{12}\mathrm{C}\text{ atom}}{12} = 1.66054\times10^{-27}\;\mathrm{kg}. \]

The decisive idea of nuclear physics is that mass and energy are interchangeable. Einstein's relation \(E = mc^2\) lets us quote any mass as an energy. Converting one atomic mass unit: \[ E = (1\;\mathrm{u})c^2 = (1.66054\times10^{-27})(2.998\times10^{8})^2 = 1.4924\times10^{-10}\;\mathrm{J} = 931.5\;\mathrm{MeV}. \] \[ \boxed{\;1\;\mathrm{u} = 931.5\;\mathrm{MeV}/c^2\;} \] This single conversion factor, \(931.5\;\mathrm{MeV}\) per atomic mass unit, will do nearly all the heavy lifting in this chapter. Whenever a nuclear process changes the total mass by \(\Delta m\) (in \(\mathrm{u}\)), it releases or absorbs an energy \(\Delta m \times 931.5\;\mathrm{MeV}\).

Core intuition: Mass is just very concentrated energy

The factor \(c^2 \approx 9\times10^{16}\;\mathrm{m^2\,s^{-2}}\) is enormous, so a minute mass change corresponds to a colossal energy. A single atomic mass unit—a speck \(1.7\times10^{-27}\;\mathrm{kg}\)—is worth \(931.5\;\mathrm{MeV}\), about a hundred million times the energy of a chemical bond. This is why nuclear processes, which shuffle masses at the level of a fraction of a \(\mathrm{u}\), dwarf chemical processes, which only rearrange electrons. The fuel is the same—energy—but the nucleus stores it a hundred million times more densely.

Atomic masses versus nuclear masses

Tables of masses almost always list atomic masses—the nucleus together with its full complement of \(Z\) electrons—because that is what a mass spectrometer actually measures. The atomic mass of hydrogen-1, for instance, is \[ m({}^1\mathrm{H}) = 1.00783\;\mathrm{u}, \] which is the proton (\(1.00728\;\mathrm{u}\)) plus one electron (\(0.00055\;\mathrm{u}\)), less a tiny electronic binding energy we may safely ignore. We will see in the next section that working with atomic masses, rather than bare nuclear masses, makes the electron bookkeeping cancel out almost automatically—provided we are careful.

Worked Examples

Example 3 · The energy hidden in one unit of mass

Express the energy equivalent of one atomic mass unit in \(\mathrm{MeV}\), and of one gram of any matter in joules.

Solution. Given: \(1\;\mathrm{u} = 1.66\times10^{-27}\;\mathrm{kg}\); \(c = 3\times10^{8}\;\mathrm{m\,s^{-1}}\).
Find: \(E(1\;\mathrm{u})\) in \(\mathrm{MeV}\) and \(E(1\;\mathrm{g})\) in \(\mathrm{J}\).
Setup: Both are \(E = mc^2\); convert units at the end.
Solve: \[ E(1\;\mathrm{u}) = (1.66\times10^{-27})(3\times10^{8})^2 = 1.49\times10^{-10}\;\mathrm{J} = 931.5\;\mathrm{MeV}, \] \[ E(1\;\mathrm{g}) = (10^{-3})(9\times10^{16}) = 9\times10^{13}\;\mathrm{J}. \] Answer: \(\boxed{1\;\mathrm{u} \leftrightarrow 931.5\;\mathrm{MeV}, \qquad 1\;\mathrm{g} \leftrightarrow 9\times10^{13}\;\mathrm{J}}\).
Check: \(9\times10^{13}\;\mathrm{J}\) is roughly the energy of a \(20\)-kilotonne fission bomb — from a single gram fully converted. That mass and energy are this tightly related is the entire reason a nucleus can power a city. ✓

Example 4 · Atomic mass is not nuclear mass

The atomic mass of \({}^{12}\mathrm{C}\) is exactly \(12.00000\;\mathrm{u}\) by definition. What is the mass of its nucleus alone?

Solution. Given: \(m_{\text{atom}}({}^{12}\mathrm{C}) = 12.00000\;\mathrm{u}\); \(Z=6\); \(m_e = 0.00055\;\mathrm{u}\).
Find: the nuclear mass \(m_{\text{nuc}}\).
Setup: An atomic mass includes \(Z\) electrons. Subtract them (the electronic binding energy, a few \(\mathrm{eV}\), is utterly negligible on the \(\mathrm{u}\) scale).
Solve: \[ m_{\text{nuc}} = m_{\text{atom}} - Z m_e = 12.00000 - 6(0.00055) = 11.99670\;\mathrm{u}. \] Answer: \(\boxed{m_{\text{nuc}} = 11.99670\;\mathrm{u}}\).
Check: The six electrons remove only \(0.0033\;\mathrm{u}\) — about \(0.03\%\). This tiny bookkeeping is exactly what makes the atomic-mass tables so convenient: in a \(\beta^-\) \(Q\)-value the daughter's extra orbital electron silently cancels the emitted one, so the electron masses never have to be tracked by hand. ✓

Binding Energy and the Curve of Stability

◈ The B/A curve — where the energy lives

live bench

Walk the binding-energy-per-nucleon curve nuclide by nuclide: everything drifts toward the iron peak, which is why light nuclei fuse and heavy ones split.

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Here is the experimental fact that everything turns on. Measure the mass of any nucleus and compare it to the sum of the masses of its separated constituents. The nucleus is always lighter. The difference is called the mass defect: \[ \Delta m = \bigl(Z\,m_p + N\,m_n\bigr) - m_{\text{nucleus}} \gt 0. \] For helium-4, four nucleons that separately weigh \(4.03188\;\mathrm{u}\) assemble into a nucleus of just \(4.00150\;\mathrm{u}\)—a deficit of \(0.0304\;\mathrm{u}\), about \(0.7\%\) of the total (Figure ). The mass is not lost in any mysterious sense: it was radiated away as energy when the nucleus formed. To pull the nucleus apart again, you would have to put that energy back.

From mass defect to binding energy

The energy equivalent of the mass defect is the binding energy \(B\): the energy you must supply to dismantle the nucleus into free, widely separated nucleons.

Birth of the formula: From Mass Defect to Binding Energy

Assembling \(Z\) protons and \(N\) neutrons into a nucleus releases energy, so the bound system is lighter by \(\Delta m\). By \(E = mc^2\), the energy released—and hence the energy needed to undo the assembly—is \[ B = \Delta m\,c^2 = \bigl[\,Z\,m_p + N\,m_n - m_{\text{nucleus}}\,\bigr]c^2. \] In practice we are handed atomic masses, which include electrons. Add \(Z\) electron masses to the protons (turning them into \(Z\) hydrogen atoms) and \(Z\) electron masses to the nucleus (turning it into the neutral atom). The same \(Z\) electrons appear on both sides, so they cancel: \[ \boxed{\;B = \bigl[\,Z\,m({}^1\mathrm{H}) + N\,m_n - m({}^{A}_{Z}\mathrm{X})\,\bigr]c^2\;} \] where \(m({}^1\mathrm{H})\) and \(m({}^{A}_{Z}\mathrm{X})\) are atomic masses. With masses in \(\mathrm{u}\), multiply the bracket by \(931.5\;\mathrm{MeV}/\mathrm{u}\). Worked check (helium-4). \(Z=2\), \(N=2\), \(m({}^4\mathrm{He}) = 4.00260\;\mathrm{u}\): \[ B = \bigl[\,2(1.00783) + 2(1.00867) - 4.00260\,\bigr](931.5) = (0.03038)(931.5) \approx 28.3\;\mathrm{MeV}. \] That is about \(7\;\mathrm{MeV}\) to remove each nucleon—roughly a million times the \(\sim\!10\;\mathrm{eV}\) that binds an atomic electron.

This answers Perplexing Question 2: the “missing” mass of a nucleus is the binding energy that holds it together, carried off when the nucleus formed. A nucleus is lighter than its parts precisely because it is bound.

Binding energy per nucleon

The total binding energy \(B\) grows steadily as nuclei get bigger—a uranium nucleus is far more strongly bound in total than a helium nucleus, simply because it has more nucleons. The physically revealing quantity is the binding energy per nucleon, \[ \frac{B}{A} = \frac{\text{total binding energy}}{\text{number of nucleons}}, \] which measures how tightly the average nucleon is held—and therefore how stable the nucleus is. Plotting \(B/A\) against \(A\) for all nuclei gives one of the most important graphs in physics (Figure ).

Birth of the formula: The Curve of Binding Energy per Nucleon

Computing \(B/A\) from measured atomic masses for a handful of nuclei traces the shape of the curve:

Nuclide	\(A\)	\(B\) (MeV)	\(B/A\) (MeV)
\({}^{2}\mathrm{H}\)	\(2\)	\(2.22\)	\(1.11\)
\({}^{4}\mathrm{He}\)	\(4\)	\(28.3\)	\(7.07\)
\({}^{12}\mathrm{C}\)	\(12\)	\(92.2\)	\(7.68\)
\({}^{16}\mathrm{O}\)	\(16\)	\(127.6\)	\(7.98\)
\({}^{56}\mathrm{Fe}\)	\(56\)	\(492.3\)	\(8.79\)
\({}^{235}\mathrm{U}\)	\(235\)	\(1783.9\)	\(7.59\)

The pattern is unmistakable. \(B/A\) rises steeply for the lightest nuclei, climbs through carbon and oxygen, and reaches a broad maximum of about \(8.8\;\mathrm{MeV}\) near \(A \approx 56\) (the iron–nickel region). Beyond the peak it declines slowly, falling to about \(7.6\;\mathrm{MeV}\) for uranium. \[ \boxed{\;\left(\frac{B}{A}\right)_{\max} \approx 8.8\;\mathrm{MeV} \ \text{at } A \approx 56\ \text{(iron peak)}\;} \]

Two features of this curve deserve emphasis, because between them they explain nuclear energy itself.

First, across the broad middle of the chart—from about \(A=30\) to \(A=120\)—the curve is nearly flat at \(\sim\!8.5\;\mathrm{MeV}\) per nucleon. The binding energy per nucleon barely changes, which tells us each nucleon is bound only to its immediate neighbours, not to the whole nucleus. This is the property of saturation, and it is our first clue about the nuclear force (Section ).

Second, the curve has a peak. Iron sits at the bottom of the energy valley: it is the most tightly bound, most stable nucleus per nucleon. Everything else—lighter or heavier—is less tightly bound, and can release energy by moving toward the peak. Light nuclei reach the peak by fusing together; heavy nuclei reach it by splitting apart. Both processes climb the same hill from opposite sides, and both give out the binding energy difference as usable energy. This answers Perplexing Question 3—we develop it fully in Section .

Misconception alert

“Uranium has more binding energy than iron, so uranium is more stable.” This conflates total binding energy \(B\) with binding energy per nucleon \(B/A\). Uranium does have a far larger total \(B\) (\(\approx 1784\;\mathrm{MeV}\) versus iron's \(\approx 492\;\mathrm{MeV}\))—but only because it has many more nucleons to add up. What measures stability is how tightly each individual nucleon is held, i.e. \(B/A\). By that measure iron (\(8.79\;\mathrm{MeV}\)) beats uranium (\(7.59\;\mathrm{MeV}\)), and uranium is the less stable nucleus—which is exactly why uranium can release energy by fissioning toward the iron peak, and iron cannot. When a problem asks about stability or about energy release, always reason with \(B/A\), never with \(B\) alone.

Misconception alert

“In a nuclear reaction, mass is destroyed and turned into energy out of nothing.” Mass is not destroyed; mass-energy is conserved. The total energy before equals the total energy after. What changes is the split between rest-mass energy and kinetic energy: when nucleons bind more tightly, some of their rest-mass energy is converted into the kinetic energy of the products (and into radiation). The bound nucleus weighs less than its parts because that energy has left the nucleus, not because mass has vanished into nonexistence. Mass and energy are two accounts of the same quantity, and the books always balance.

◇ Figure 1

Fig 1. The mass of a bound nucleus is less than the summed mass of its free constituents. The deficit \(\Delta m\) (exaggerated here) is the mass equivalent of the binding energy, \(B = \Delta m\,c^2\). For \({}^4\mathrm{He}\), \(\Delta m \approx 0.030\;\mathrm{u}\) gives \(B \approx 28\;\mathrm{MeV}\).

◇ Figure 2

Fig 2. Binding energy per nucleon \(B/A\) versus mass number \(A\). The curve rises steeply, peaks near \(A \approx 56\) (iron) at about \(8.8\;\mathrm{MeV}\), then declines slowly. Light nuclei release energy by fusing toward the peak; heavy nuclei by splitting toward it.

Worked Examples

Example 5 · Binding energy of oxygen-16

Find the total binding energy and the binding energy per nucleon of \({}^{16}_{\ 8}\mathrm{O}\), given \(m({}^{16}\mathrm{O}) = 15.99491\;\mathrm{u}\).

Solution. Given: \(Z=8\), \(N=8\); \(m({}^1\mathrm{H})=1.00783\), \(m_n=1.00867\), \(m({}^{16}\mathrm{O})=15.99491\;\mathrm{u}\).
Find: \(B\) and \(B/A\).
Setup: Using atomic masses, \(B = [\,Z m({}^1\mathrm{H}) + N m_n - m_{\text{atom}}\,]c^2\); the eight electron masses cancel between the \(Z\) hydrogens and the oxygen atom.
Solve: \[ \Delta m = 8(1.00783) + 8(1.00867) - 15.99491 = 16.13200 - 15.99491 = 0.13709\;\mathrm{u}, \] \[ B = 0.13709 \times 931.5 = 127.7\;\mathrm{MeV}, \qquad \frac{B}{A} = \frac{127.7}{16} = 7.98\;\mathrm{MeV}. \] Answer: \(\boxed{B \approx 127.7\;\mathrm{MeV},\quad B/A \approx 7.98\;\mathrm{MeV}}\).
Check: \(B/A \approx 8\;\mathrm{MeV}\) places \({}^{16}\mathrm{O}\) right on the broad plateau of the stability curve, just below the iron peak — as expected for a stable, doubly-magic light nucleus. ✓

Example 6 · Why splitting a heavy nucleus pays

A heavy nucleus of \(A = 240\) sits at \(B/A = 7.6\;\mathrm{MeV}\). It splits into two fragments of \(A = 120\), each at \(B/A = 8.5\;\mathrm{MeV}\). Estimate the energy released.

Solution. Given: parent \(A=240\), \(B/A=7.6\); fragments at \(B/A=8.5\;\mathrm{MeV}\).
Find: energy released \(Q\).
Setup: Energy released \(=\) (final total binding) \(-\) (initial total binding), since a more tightly bound system has less mass. Nucleon number is conserved, so \(Q = A\,\big[(B/A)_{\text{final}} - (B/A)_{\text{initial}}\big]\).
Solve: \[ Q = 240\,(8.5 - 7.6) = 240 \times 0.9 \approx 216\;\mathrm{MeV}. \] Answer: \(\boxed{Q \approx 216\;\mathrm{MeV}}\).
Check: This matches the textbook “\(\sim\!200\;\mathrm{MeV}\) per fission” to within the crudeness of reading the curve, and it is \(\sim\!10^{7}\) times the few-\(\mathrm{eV}\) yield of a chemical reaction per atom — the whole case for nuclear power in one subtraction. ✓

Example 7 · Binding the last neutron: separation energy

How much energy is needed to remove one neutron from \({}^{13}\mathrm{C}\)? Use \(m({}^{12}\mathrm{C}) = 12.00000\), \(m({}^{13}\mathrm{C}) = 13.00335\;\mathrm{u}\).

Solution. Given: the two carbon masses and \(m_n = 1.00867\;\mathrm{u}\).
Find: the neutron separation energy \(S_n\).
Setup: \(S_n\) is the \(Q\) to undo \({}^{13}\mathrm{C} \to {}^{12}\mathrm{C} + n\): \(S_n = [\,m({}^{12}\mathrm{C}) + m_n - m({}^{13}\mathrm{C})\,]c^2\). (Atomic masses; the six electrons match on both sides.)
Solve: \[ S_n = (12.00000 + 1.00867 - 13.00335)\times 931.5 = 0.00532 \times 931.5 \approx 4.95\;\mathrm{MeV}. \] Answer: \(\boxed{S_n \approx 4.95\;\mathrm{MeV}}\).
Check: This is well below the average \(B/A \approx 7.7\;\mathrm{MeV}\) of \({}^{13}\mathrm{C}\): pulling off one specific nucleon costs less than the per-nucleon average, because the average is dominated by the tightly bound \({}^{12}\mathrm{C}\) core. Total binding and last-nucleon binding are different questions — a distinction examiners love. ✓

✓ Checkpoint · Mass Defect and Binding Energy

Define the mass defect and state its relation to the binding energy.
\(\Delta m = (Zm_p + Nm_n) - m_{\text{nucleus}}\); the binding energy is \(B = \Delta m\,c^2\).
Why is \(B/A\), not the total \(B\), the right measure of stability?
Total \(B\) grows simply because heavy nuclei have more nucleons; \(B/A\) measures how tightly the average nucleon is held, which is what stability means.
(Trap) “The most stable nucleus is the one with the largest total binding energy.” Identify the flaw.
False. Stability is set by \(B/A\), which peaks at iron (\(A\approx56\)). Uranium has a larger total \(B\) but a smaller \(B/A\), and is less stable.

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · Size of a heavy nucleus

Estimate the radius of a \({}^{238}_{\ 92}\mathrm{U}\) nucleus, and compare it with that of a \({}^4_2\mathrm{He}\) nucleus.

Solution. Given: \(A_{\mathrm U} = 238\), \(A_{\mathrm{He}} = 4\), \(R_0 = 1.2\;\mathrm{fm}\).
Find: \(R_{\mathrm U}\) and the ratio \(R_{\mathrm U}/R_{\mathrm{He}}\).
Setup: Use \(R = R_0 A^{1/3}\); the ratio depends only on the mass numbers.
Solve: \[ R_{\mathrm U} = 1.2\,(238)^{1/3} = 1.2 \times 6.20 \approx 7.4\;\mathrm{fm}, \qquad \frac{R_{\mathrm U}}{R_{\mathrm{He}}} = \left(\frac{238}{4}\right)^{1/3} = (59.5)^{1/3} \approx 3.9. \] Answer: \(\boxed{R_{\mathrm U} \approx 7.4\;\mathrm{fm}}\); the uranium nucleus is only about \(3.9\) times wider than helium, despite having \(59\) times as many nucleons.
Check: Cube-root scaling: \(59^{1/3} \approx 3.9\), so a \(59\)-fold jump in nucleon number buys under a fourfold jump in radius — the signature of constant-density packing. ✓

Example 2 · Density of nuclear matter

Estimate the density of nuclear matter, treating a nucleus of mass number \(A\) as a uniform sphere.

Solution. Given: mass \(\approx A m_u\) with \(m_u = 1.66\times10^{-27}\;\mathrm{kg}\); \(R = R_0 A^{1/3}\), \(R_0 = 1.2\times10^{-15}\;\mathrm{m}\).
Find: \(\rho\).
Setup: \(\rho = m/V\) with \(V = \tfrac{4}{3}\pi R^3\); the factor \(A\) cancels.
Solve: \[ \rho = \frac{A m_u}{\tfrac{4}{3}\pi R_0^3 A} = \frac{m_u}{\tfrac{4}{3}\pi R_0^3} = \frac{1.66\times10^{-27}}{\tfrac{4}{3}\pi(1.2\times10^{-15})^3} \approx 2.3\times10^{17}\;\mathrm{kg\,m^{-3}}. \] Answer: \(\boxed{\rho \approx 2.3\times10^{17}\;\mathrm{kg\,m^{-3}}}\), independent of \(A\).
Check: This is \(\sim\!10^{14}\) times the density of water — a matchbox of it would weigh \(\sim\!10^{12}\;\mathrm{kg}\). The cancellation of \(A\) is exactly why the number is the same for every nucleus. ✓

Example 3 · Counting the constituents

For a neutral atom of gold, \({}^{197}_{\ 79}\mathrm{Au}\), state the number of protons, neutrons, and electrons.

Solution. Given: \(Z = 79\), \(A = 197\).
Find: numbers of \(p\), \(n\), \(e\).
Setup: protons \(= Z\); neutrons \(= A - Z\); in a neutral atom, electrons \(= Z\).
Solve: \(p = 79\); \(n = 197 - 79 = 118\); \(e = 79\).
Answer: \(\boxed{79\ \text{protons},\ 118\ \text{neutrons},\ 79\ \text{electrons}}\).
Check: \(p + n = 79 + 118 = 197 = A\), and \(e = Z\) makes the atom neutral — both bookkeeping checks pass. ✓

Example 4 · Energy locked in mass

Find the energy equivalent of the electron's rest mass, \(m_e = 0.00055\;\mathrm{u}\), in \(\mathrm{MeV}\).

Solution. Given: \(m_e = 0.00055\;\mathrm{u}\), \(1\;\mathrm{u} = 931.5\;\mathrm{MeV}/c^2\).
Find: \(E = m_e c^2\).
Setup: multiply the mass in \(\mathrm{u}\) by \(931.5\;\mathrm{MeV/u}\).
Solve: \(E = 0.00055 \times 931.5 \approx 0.51\;\mathrm{MeV}\).
Answer: \(\boxed{E \approx 0.51\;\mathrm{MeV}}\) — the familiar \(0.511\;\mathrm{MeV}\) electron rest energy.
Check: Twice this, \(2m_e c^2 = 1.02\;\mathrm{MeV}\), is precisely the surcharge a \(\beta^+\) channel must pay — the same number that reappears in the \(Q\)-value rules. ✓

Example 5 · Mass defect of the deuteron

The deuteron \({}^2_1\mathrm{H}\) has atomic mass \(2.01410\;\mathrm{u}\). Find its mass defect and binding energy.

Solution. Given: \(m({}^2\mathrm{H}) = 2.01410\;\mathrm{u}\), one proton and one neutron.
Find: \(\Delta m\) and \(B\).
Setup: use atomic masses, \(\Delta m = [m({}^1\mathrm{H}) + m_n] - m({}^2\mathrm{H})\); then \(B = \Delta m \times 931.5\;\mathrm{MeV}\).
Solve: \[ \Delta m = (1.00783 + 1.00867) - 2.01410 = 0.00240\;\mathrm{u}, \qquad B = 0.00240 \times 931.5 \approx 2.22\;\mathrm{MeV}. \] Answer: \(\boxed{\Delta m \approx 0.0024\;\mathrm{u},\quad B \approx 2.22\;\mathrm{MeV}}\).
Check: \(B/A = 2.22/2 = 1.1\;\mathrm{MeV}\) per nucleon — far below the \(\sim\!8\;\mathrm{MeV}\) plateau, correctly marking the deuteron as the most weakly bound stable nucleus. ✓

Problem bank

Five questions from this chapter’s 50-question bank, free — attempt each one before you reveal the answer. The rest of the bank, and the timed test that draws on all of it, are in the full book.

✎ Free practice questions

Radius of a copper nucleus.
Estimate the radius of a \({}^{64}_{29}\mathrm{Cu}\) nucleus (\(R_0 = 1.2\;\mathrm{fm}\)).
\(R = 1.2\,(64)^{1/3} = 1.2\times4 = 4.8\;\mathrm{fm}\).
Radius ratio.
By what factor is the radius of a nucleus with \(A = 216\) larger than one with \(A = 27\)?
\((216/27)^{1/3} = 8^{1/3} = 2\). Twice as large.
Composition.
State the number of protons and neutrons in \({}^{235}_{\ 92}\mathrm{U}\).
\(92\) protons, \(235-92 = 143\) neutrons.
Binding of oxygen-16.
Given \(m({}^{16}\mathrm{O}) = 15.99491\;\mathrm{u}\), find \(B\) and \(B/A\).
\(\Delta m = 8(1.00783)+8(1.00867)-15.99491 = 0.13701\;\mathrm{u}\); \(B = 127.6\;\mathrm{MeV}\), \(B/A \approx 7.98\;\mathrm{MeV}\).
Is lead's nucleus denser?
Compute the matter density of a \({}^4\mathrm{He}\) nucleus and a \({}^{208}\mathrm{Pb}\) nucleus (\(R_0 = 1.2\;\mathrm{fm}\)) and settle the question.
\(\rho = \dfrac{A\,m_u}{\frac{4}{3}\pi (R_0 A^{1/3})^3} = \dfrac{m_u}{\frac{4}{3}\pi R_0^3}\) — the \(A\) cancels. Both give \(\rho \approx 2.3\times10^{17}\;\mathrm{kg\,m^{-3}}\). Lead is \(52\times\) more massive but \(52\times\) larger in volume: same density. The trap is forgetting that \(R\propto A^{1/3}\).

Chapter test

A paper drawn at random from this chapter's bank. Choose the exam you are training for — the marking scheme, pace and difficulty mix follow the real pattern. Work on paper; when you finish (or the clock runs out), the answers are revealed and you mark yourself honestly.

Pattern

Questions

Clock

Free practice questions

Which climbs faster toward iron?
Fission of \({}^{235}\mathrm{U}\) (\(B/A \approx 7.6\;\mathrm{MeV}\)) yields fragments with \(B/A \approx 8.5\;\mathrm{MeV}\); fusion of deuterium (\(B/A \approx 1.11\;\mathrm{MeV}\)) yields \({}^4\mathrm{He}\) (\(B/A \approx 7.07\;\mathrm{MeV}\)). A student says “fission releases \(200\;\mathrm{MeV}\) and fusion only a few, so fission wins.” Compare the energy released per nucleon and expose the trap.
Per nucleon: fission gains \(\Delta(B/A)\approx 8.5-7.6 = 0.9\;\mathrm{MeV}\); fusion gains \(7.07-1.11 \approx 6.0\;\mathrm{MeV}\) — about \(7\times\) more. The trap is comparing total \(Q\) (uranium simply has more nucleons) instead of energy density. Per unit mass fusion is the richer source.
The missing \(2m_e\).
\({}^{22}\mathrm{Na}\) decays by \(\beta^+\) to \({}^{22}\mathrm{Ne}\). Using atomic masses \(M({}^{22}\mathrm{Na}) = 21.994437\;\mathrm{u}\), \(M({}^{22}\mathrm{Ne}) = 21.991385\;\mathrm{u}\), a student writes \(Q = (M_{\mathrm{Na}}-M_{\mathrm{Ne}})c^2\). Give that value, then the correct \(Q\).
Naive: \((21.994437-21.991385)(931.5) = 2.84\;\mathrm{MeV}\) — the trap. For \(\beta^+\) the atomic-mass formula carries a \(2m_e c^2\) penalty (a positron is created and the daughter has one fewer orbital electron): \(Q_{\beta^+} = (M_{\mathrm{Na}}-M_{\mathrm{Ne}}-2m_e)c^2 = 2.84 - 1.02 = \boxed{1.82\;\mathrm{MeV}}\).
“Safe after two half-lives.”
A source reads \(8000\;\mathrm{Bq}\) with \(t_{1/2} = 6.0\;\mathrm{h}\). A technician assumes it is “essentially gone” after \(2\) half-lives and handles it at \(t = 12\;\mathrm{h}\). What is the actual activity then, and after how long does it truly fall below \(10\;\mathrm{Bq}\)?
\(\mathcal{A}(12\,\mathrm h) = 8000\,(1/2)^2 = 2000\;\mathrm{Bq}\) — a quarter remains, not zero. For \(\mathcal{A}\lt 10\): \((1/2)^n \lt 10/8000 \Rightarrow n \gt \log_2 800 = 9.6\), so \(10\) half-lives, \(t = \boxed{60\;\mathrm{h}}\).
Longer-lived, therefore hotter?
Compare \(1\;\mathrm{mg}\) of \({}^{226}\mathrm{Ra}\) (\(t_{1/2} = 1600\;\mathrm{yr}\)) with \(1\;\mathrm{mg}\) of \({}^{131}\mathrm{I}\) (\(t_{1/2} = 8.0\;\mathrm{d}\)). Which is the more active source, and by roughly what factor?
\(\mathcal{A} = \lambda N = (\ln2/t_{1/2})(m/M)N_A\). Radium: \(\approx 3.7\times10^{7}\;\mathrm{Bq}\) (\(\approx 1\;\mathrm{mCi}\)). Iodine: \(\approx 4.6\times10^{12}\;\mathrm{Bq}\). The iodine is \(\sim\!10^5\) times more active despite the same mass — a shorter half-life means more decays per second. “Long-lived” and “dangerous now” are opposite tendencies.

Nuclei

Perplexing Questions

Composition, Size, and Density

The size of the nucleus

The density of nuclear matter

Worked Examples

Mass, Energy, and the Atomic Mass Unit

Atomic masses versus nuclear masses

Worked Examples

Binding Energy and the Curve of Stability

From mass defect to binding energy

Binding energy per nucleon

Worked Examples

Solved examples

Problem bank

Chapter test

The chapter continues.