Part 5 · Chapter 42

Nuclear Physics

Until now we have followed electrons through their quantum amusement park — trapped in wells, stacked into atoms — while quietly ignoring the speck at the very center. That speck, the nucleus, is where almost all of an atom's mass hides, packed into a volume a hundred thousand times smaller than the atom itself. In this chapter we ask how big it is, how tightly it is bound, and why most nuclei are not content to sit still: they decay, spitting out alpha particles, electrons, positrons, and neutrinos on a clock so reliable we can read the age of a wooden artifact or a rock from it. We will meet the binding-energy curve that explains both why the sun shines and why a reactor releases power, the exponential law that governs every radioactive sample, and the two complementary models — one treating the nucleus like a drop of liquid, the other like a tiny atom with shells of its own — that together make sense of the most concentrated matter we know.

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 100 min
i What you'll learn
  • A nuclide is labeled by its atomic number \(Z\) (protons), neutron number \(N\), and mass number \(A = Z + N\); nuclides with equal \(Z\) but different \(N\) are isotopes.
  • Nuclei are nearly spherical with radius \(r = r_0 A^{1/3}\) (\(r_0 \approx 1.2\,\mathrm{fm}\)), so nuclear density is roughly the same for all nuclei.
  • The mass defect \(\Delta m\) stores a binding energy \(\Delta E_{be} = \Delta m\,c^2\) (with \(1\,\mathrm{u} \leftrightarrow 931.494\,\mathrm{MeV}\)); the binding energy per nucleon \(\Delta E_{ben} = \Delta E_{be}/A\) peaks near iron.
  • Radioactive decay follows \(N = N_0 e^{-\lambda t}\), with activity \(R = \lambda N = R_0 e^{-\lambda t}\) and half-life \(T_{1/2} = \dfrac{\ln 2}{\lambda}\).
  • Alpha decay is barrier tunneling with disintegration energy \(Q = (m_X - m_Y - m_{\alpha})c^2\); beta decay (\(\beta^-,\ \beta^+\)) emits an electron or positron together with a (anti)neutrino.
  • Radioactive dating uses \(^{14}\mathrm{C}\) and \(^{40}\mathrm{K}\) as clocks; radiation dosage is measured as absorbed dose (gray) and dose equivalent (sievert).
  • The collective (liquid-drop) and independent-particle (shell) models, with magic numbers \(2, 8, 20, 28, 50, 82, 126\), combine into the modern picture of the nucleus.
Section 42-1

What Is Physics?

So far in our tour of the quantum world we have studied electrons confined in potential wells — including the atom — but we have left untouched the tiny object at the atom's center: the nucleus. For roughly a century a central goal of physics has been to work out the quantum behavior of nuclei, and for almost as long a central goal of engineering has been to put that physics to use, in applications stretching from radiation therapy against cancer to the detectors that warn of radon gas seeping into a basement. The nucleus is small, dense, and surprisingly lively, and understanding it begins with the experiment that revealed it existed at all.

Section 42-2

Discovering the Nucleus

In the early 1900s atoms were imagined as soft spheres of positive charge studded with electrons — the "plum-pudding" picture. To test it, Rutherford's group fired energetic alpha particles (helium nuclei, charge \(+2e\)) at a thin gold foil and watched where they went. Most passed nearly straight through, as the soft-pudding model predicted. But a tiny fraction bounced back through enormous angles, approaching \(180^\circ\) — something the diffuse-charge picture could never produce. Rutherford likened it to firing a shell at tissue paper and having it ricochet back at you.

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Why the big-angle scatter matters
A large deflection requires a large, concentrated force — only possible if the atom's positive charge and nearly all its mass are packed into a minute central nucleus, with the electrons occupying the vast space around it.

By analyzing how the deflection depended on the alpha energy and the target charge, Rutherford concluded the nucleus is some \(10^{4}\)\(10^{5}\) times smaller than the atom. The atom, it turns out, is almost entirely empty space wrapped around an astonishingly dense core.

Distance of closest approach (head-on collision)
\[ K_i = \frac{1}{4\pi\varepsilon_0}\,\frac{q_\alpha q_{\text{nuc}}}{d} \qquad\Rightarrow\qquad d = \frac{1}{4\pi\varepsilon_0}\,\frac{(2e)(Ze)}{K_i} \]
In a head-on approach the alpha particle stops when all of its kinetic energy \(K_i\) has converted to electric potential energy. Solving for \(d\) gives an upper limit on the nuclear radius — typically a few tens of femtometers for a multi-MeV alpha, far smaller than an atom.
Section 42-3

Some Nuclear Properties

A nucleus is built from nucleons: positively charged protons and neutral neutrons. Three integers label any nuclide: the atomic number \(Z\) (number of protons, which fixes the chemical element), the neutron number \(N\), and the mass number \(A\).

Nucleon bookkeeping
\[ A = Z + N \]
A nuclide is written \(^{A}\!X\), e.g. \(^{197}\mathrm{Au}\) has \(Z = 79\), so \(N = 197 - 79 = 118\). Nuclides with the same \(Z\) but different \(N\) are isotopes; gold has 32 of them, but only \(^{197}\mathrm{Au}\) is stable.

Scattering experiments show that nuclear matter is packed at very nearly the same density everywhere, so the volume of a nucleus is roughly proportional to the number of nucleons it holds. Since volume scales as \(r^3\), the radius scales as \(A^{1/3}\).

Nuclear radius
\[ r = r_0\,A^{1/3}, \qquad r_0 \approx 1.2\,\mathrm{fm} \;\;(1\,\mathrm{fm} = 10^{-15}\,\mathrm{m}) \]
A useful "effective radius." Nuclei range from about \(1\,\mathrm{fm}\) for a single proton to roughly \(7\,\mathrm{fm}\) for the heaviest. Because \(r \propto A^{1/3}\), every nucleus has nearly the same enormous matter density, about \(2\times10^{17}\,\mathrm{kg/m^3}\) — a teaspoon would mass billions of tonnes.

Masses on the nuclear scale are reported in the atomic mass unit \(\mathrm{u}\), defined so that neutral \(^{12}\mathrm{C}\) has mass exactly \(12\,\mathrm{u}\), with \(1\,\mathrm{u} = 1.661\times10^{-27}\,\mathrm{kg}\). Through \(E = mc^2\) a mass and an energy are interchangeable, and the conversion factor is one worth memorizing.

Mass–energy conversion
\[ E = mc^2, \qquad c^2 = 931.494\,\mathrm{MeV/u} \]
So \(1\,\mathrm{u}\) of mass corresponds to \(931.5\,\mathrm{MeV}\) of energy. This single number turns any nuclear mass difference (in u) straight into an energy (in MeV) — the workhorse of every binding-energy and decay-energy calculation in the chapter.

Weigh a nucleus and you find it is less massive than the sum of its separate protons and neutrons. That missing mass — the mass defect \(\Delta m\) — is exactly the mass that was converted to energy and released when the nucleus assembled. To pull it apart again you would have to supply that same energy: the binding energy.

Binding energy and binding energy per nucleon
\[ \Delta E_{be} = \Big(\textstyle\sum m_{\text{nucleons}} - M_{\text{nucleus}}\Big)c^2, \qquad \Delta E_{ben} = \frac{\Delta E_{be}}{A} \]
A larger \(\Delta E_{ben}\) means a more tightly bound, more stable nucleus. Plotted against \(A\), the curve rises steeply, peaks near \(A \approx 56\) (iron–nickel) at about \(8.8\,\mathrm{MeV/nucleon}\), then falls gently. That single peak explains everything: fusing light nuclei up toward iron releases energy (the stars), and so does splitting heavy nuclei down toward iron (reactors) — the subject of the next chapter.
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The strong nuclear force
Protons repel one another electrically, yet nuclei hold together. A separate, far stronger nuclear force — attractive, charge-independent, and acting only over a range of a few femtometers — binds protons and neutrons alike.

Because it is short-ranged, a nucleon feels only its nearest neighbors, while the electric repulsion of the protons reaches across the whole nucleus. In heavy nuclei the accumulated repulsion eventually wins, which is why large nuclei need extra neutrons for stability and why the heaviest of all are radioactive.

Section 42-4

Radioactive Decay

Most known nuclides are radioactive: each spontaneously emits a particle and transforms into a different nuclide. You cannot predict when any single nucleus will decay — only the probability per unit time, which is the same for every nucleus of a given species and unaffected by temperature, pressure, or chemistry. With huge numbers of nuclei, this fixed probability produces an exact exponential law.

The decay law
\[ \frac{dN}{dt} = -\lambda N \qquad\Longrightarrow\qquad N = N_0\,e^{-\lambda t} \]
The number of undecayed nuclei \(N\) falls exponentially from its initial value \(N_0\). The disintegration constant \(\lambda\) (units \(\mathrm{s^{-1}}\)) is the decay probability per nucleus per second — large for short-lived species, tiny for long-lived ones.
Activity (decay rate) of a sample
\[ R = -\frac{dN}{dt} = \lambda N = R_0\,e^{-\lambda t} \]
The activity is how many decays occur per second; it falls off with the same exponential as \(N\). Its SI unit is the becquerel (\(1\,\mathrm{Bq} = 1\) decay/s); the older curie is \(1\,\mathrm{Ci} = 3.7\times10^{10}\,\mathrm{Bq}\).
Half-life and mean (average) life
\[ T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}, \qquad \tau = \frac{1}{\lambda} \]
The half-life \(T_{1/2}\) is the time for either \(N\) or \(R\) to fall to half its value; after \(n\) half-lives a fraction \((1/2)^n\) remains. The mean life \(\tau = 1/\lambda\) is the average survival time, a bit longer than \(T_{1/2}\). Half-lives span an extraordinary range, from fractions of a second to billions of years.
Section 42-5

Alpha Decay

In alpha decay a heavy nucleus emits a tightly bound \(^{4}\mathrm{He}\) nucleus (two protons, two neutrons), so \(Z\) drops by 2 and \(A\) by 4. The classic example is uranium decaying to thorium:

Example reaction and its disintegration energy
\[ ^{238}\mathrm{U} \;\longrightarrow\; ^{234}\mathrm{Th} + \,^{4}\mathrm{He}, \qquad Q = \big(m_X - m_Y - m_{\alpha}\big)c^2 \]
The disintegration energy \(Q\) is the mass that vanishes, expressed as energy; it appears as kinetic energy shared by the products. A positive \(Q\) means the decay is energetically allowed. For \(^{238}\mathrm{U}\), \(Q \approx 4.25\,\mathrm{MeV}\), most of it carried by the light alpha particle.
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Why such wildly different half-lives?
An alpha particle inside the nucleus is trapped behind a Coulomb potential barrier it lacks the energy to climb. It escapes only by quantum-mechanical barrier tunneling — a rare leak whose probability is fantastically sensitive to the barrier's height and width.

This sensitivity explains the famous puzzle: \(^{238}\mathrm{U}\) has a half-life of \(4.5\times10^{9}\,\mathrm{y}\) while \(^{228}\mathrm{U}\) lives only about \(9\) minutes — a factor of \(10^{14}\) in half-life — even though their \(Q\) values differ by less than a factor of two. A small change in barrier transparency produces an immense change in lifetime.

Section 42-6

Beta Decay

In beta decay the mass number \(A\) stays fixed but a neutron and a proton interconvert, shifting \(Z\) by one. In beta-minus (\(\beta^-\)) decay a neutron becomes a proton, emitting an electron; in beta-plus (\(\beta^+\)) decay a proton becomes a neutron, emitting a positron.

The two beta processes
\[ n \;\to\; p + e^- + \bar{\nu}, \qquad\qquad p \;\to\; n + e^+ + \nu \]
Each emission is accompanied by a nearly massless, chargeless neutrino \(\nu\) (or antineutrino \(\bar{\nu}\)). The neutrino was proposed precisely to rescue energy and momentum conservation, and decades later detected directly.
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Why beta electrons have a spread of energies
Alpha particles emerge with one sharp energy, but beta electrons come out with every energy from zero up to a maximum equal to \(Q\). The reason is the unseen neutrino: it shares the released energy \(Q\) with the electron in any proportion.

For every decay the electron's energy plus the neutrino's energy equals the same \(Q\). Before the neutrino was understood, this continuous spectrum looked like a violation of energy conservation — the neutrino was the hidden bookkeeper that balanced the accounts. The same logic governs the positron spectrum in \(\beta^+\) decay.

Section 42-7

Radioactive Dating

Because radioactive decay runs on a clock no external condition can alter, a known half-life turns a measured activity into an age. The best-known method is radiocarbon dating. Cosmic rays keep the atmosphere stocked with a steady trace of \(^{14}\mathrm{C}\) (half-life \(5730\,\mathrm{y}\)), which living things absorb. When an organism dies it stops taking in carbon, and its \(^{14}\mathrm{C}\) decays at a fixed rate.

Age from the surviving activity
\[ R = R_0\,e^{-\lambda t} \qquad\Longrightarrow\qquad t = \frac{1}{\lambda}\ln\!\frac{R_0}{R} = \frac{T_{1/2}}{\ln 2}\,\ln\!\frac{R_0}{R} \]
Here \(R_0\) is the activity of a living (or fresh) sample and \(R\) is the activity measured today. Comparing the two yields the elapsed time. Radiocarbon reaches back tens of thousands of years; for rocks, the much longer-lived \(^{40}\mathrm{K}\) (\(T_{1/2} \approx 1.25\times10^{9}\,\mathrm{y}\)) and \(^{238}\mathrm{U}\) serve as clocks for the age of the Earth itself.
Section 42-8

Measuring Radiation Dosage

Radiation from cosmic rays, the rocks beneath us, and medical and industrial sources deposits energy in living tissue, and how much damage it does depends both on the energy absorbed and on the kind of radiation. Two quantities capture this, and it is important to keep them straight.

Absorbed dose
\[ \text{Absorbed dose} = \frac{\text{energy absorbed}}{\text{mass of tissue}}, \qquad 1\,\mathrm{Gy} = 1\,\mathrm{J/kg} \]
The SI unit is the gray (Gy); the older unit is the rad, with \(1\,\mathrm{rad} = 0.01\,\mathrm{Gy}\). This measures pure energy deposited, regardless of biological effect.
Dose equivalent
\[ \text{Dose equivalent} = \text{absorbed dose} \times \mathrm{RBE} \]
Equal energies of alpha radiation and X-rays do unequal biological harm, so the absorbed dose is weighted by a relative biological effectiveness (RBE) factor. The SI unit is the sievert (Sv); the older unit is the rem, with \(1\,\mathrm{rem} = 0.01\,\mathrm{Sv}\). A typical person receives a few millisieverts per year from natural background.
Section 42-9

Nuclear Models

No single picture captures the nucleus completely, so physicists use two complementary models that emphasize opposite extremes of how strongly the nucleons interact.

The collective (liquid-drop) model
Bohr imagined the nucleons interacting so strongly and randomly that the nucleus behaves like a drop of incompressible liquid — its surface can oscillate, and a struck nucleus shares the incoming energy among all nucleons. This model beautifully describes nuclear reactions and, especially, fission, where a deformed drop splits in two.
The independent-particle (shell) model
At the opposite extreme, each nucleon moves nearly freely in the average potential of all the others, occupying quantized energy levels just as atomic electrons do. Filled nucleon shells make a nucleus unusually stable, which shows up at the magic numbers \(2, 8, 20, 28, 50, 82, 126\) of protons or neutrons — the nuclear analog of the noble-gas electron counts.
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The combined model
The two pictures are reconciled by letting a few outer nucleons move in shell-model orbits around a core that itself behaves like a liquid drop and can be deformed by those outer nucleons.

This unified view accounts for the magic-number stabilities, the collective vibrations and rotations of the drop, and the spins and magnetic moments of real nuclei — a single coherent description of the most concentrated matter we know.

Worked Examples

Putting It to Work

1 Distance of closest approach

Problem. An alpha particle (charge \(+2e\)) with kinetic energy \(K_i = 5.30\,\mathrm{MeV}\) heads straight at a gold nucleus (\(Z = 79\)). Find its distance of closest approach \(d\).

Solution. At closest approach all kinetic energy has become electric potential energy, so \(K_i = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(2e)(79e)}{d}\). Using \(\dfrac{e^2}{4\pi\varepsilon_0} = 1.44\,\mathrm{eV\cdot nm} = 1.44\times10^{-9}\,\mathrm{eV\cdot m}\).

d = (1/4πε₀)(2·79·e²)/K_i
\[ d = \frac{(2)(79)(1.44\times10^{-9}\,\mathrm{eV\cdot m})}{5.30\times10^{6}\,\mathrm{eV}} \approx 4.3\times10^{-14}\,\mathrm{m} = 43\,\mathrm{fm} \]

The alpha stops about \(43\,\mathrm{fm}\) from the center — well outside the gold nucleus's \(\sim 7\,\mathrm{fm}\) radius — so it reverses without ever touching the nuclear surface, feeling only the Coulomb repulsion. This is exactly how Rutherford set an upper bound on nuclear size.

2 Nuclear size and density

Problem. Find the radius of a \(^{197}\mathrm{Au}\) nucleus, and estimate its matter density.

Solution. Use \(r = r_0 A^{1/3}\) with \(r_0 = 1.2\,\mathrm{fm}\) and \(A = 197\); then density is mass over volume, with mass \(\approx A\,m_p\).

r = r₀ A^(1/3); ρ = m / (4/3 π r³)
\[ r = (1.2\,\mathrm{fm})(197)^{1/3} \approx (1.2)(5.82)\,\mathrm{fm} \approx 7.0\,\mathrm{fm} \]
\[ \rho \approx \frac{(197)(1.67\times10^{-27}\,\mathrm{kg})}{\tfrac{4}{3}\pi(7.0\times10^{-15}\,\mathrm{m})^3} \approx 2\times10^{17}\,\mathrm{kg/m^3} \]

The radius is about \(7\,\mathrm{fm}\), and the density is roughly \(2\times10^{17}\,\mathrm{kg/m^3}\). Repeat the calculation for any nuclide and you get nearly the same density — the cancellation of \(A\) in mass and volume is precisely why nuclear matter is so uniform.

3 Binding energy per nucleon

Problem. The \(^{4}\mathrm{He}\) nucleus (2 protons, 2 neutrons) has a mass \(0.0304\,\mathrm{u}\) less than its separated nucleons. Find its binding energy and binding energy per nucleon.

Solution. Convert the mass defect to energy with \(c^2 = 931.5\,\mathrm{MeV/u}\), then divide by \(A = 4\).

ΔE = Δm·c²; ΔE_ben = ΔE / A
\[ \Delta E_{be} = (0.0304\,\mathrm{u})(931.5\,\mathrm{MeV/u}) \approx 28.3\,\mathrm{MeV}, \qquad \Delta E_{ben} = \frac{28.3}{4} \approx 7.1\,\mathrm{MeV/nucleon} \]

Each nucleon is bound by about \(7\,\mathrm{MeV}\) — already high, which is why the alpha particle is so stable and so readily emitted whole in alpha decay. The very heaviest nuclei sit lower on the binding curve (\(\sim 7.6\,\mathrm{MeV}\)), so splitting them toward the iron peak releases energy.

4 Decay constant, half-life, and activity

Problem. A radioactive source contains \(N = 2.0\times10^{15}\) nuclei of a nuclide with half-life \(T_{1/2} = 14.3\,\mathrm{d}\) (phosphorus-32). Find its disintegration constant and current activity.

Solution. First \(\lambda = (\ln 2)/T_{1/2}\) in \(\mathrm{s^{-1}}\), then \(R = \lambda N\). Convert \(14.3\,\mathrm{d} = 1.236\times10^{6}\,\mathrm{s}\).

λ = ln2 / T½; R = λN
\[ \lambda = \frac{0.693}{1.236\times10^{6}\,\mathrm{s}} \approx 5.6\times10^{-7}\,\mathrm{s^{-1}} \]
\[ R = \lambda N = (5.6\times10^{-7})(2.0\times10^{15}) \approx 1.1\times10^{9}\,\mathrm{Bq} \approx 0.030\,\mathrm{Ci} \]

The sample undergoes about \(1.1\times10^{9}\) decays per second, roughly \(30\,\mathrm{mCi}\). After one half-life (14.3 days) both \(N\) and \(R\) drop to half; after ten half-lives only about a thousandth remains.

5 Disintegration energy of \(^{238}\mathrm{U}\)

Problem. For \(^{238}\mathrm{U}\to{}^{234}\mathrm{Th}+{}^{4}\mathrm{He}\), the atomic masses are \(238.05079\,\mathrm{u}\), \(234.04363\,\mathrm{u}\), and \(4.00260\,\mathrm{u}\). Find \(Q\).

Solution. The mass lost is \(\Delta m = m_X - m_Y - m_{\alpha}\); using atomic masses, the electrons cancel. Then \(Q = \Delta m\,c^2\).

Q = (m_U − m_Th − m_He) c²
\[ \Delta m = 238.05079 - 234.04363 - 4.00260 = 0.00456\,\mathrm{u} \]
\[ Q = (0.00456\,\mathrm{u})(931.5\,\mathrm{MeV/u}) \approx 4.25\,\mathrm{MeV} \]

A positive \(Q \approx 4.25\,\mathrm{MeV}\) confirms the decay is allowed; nearly all of it goes to the light alpha as kinetic energy, with a small recoil for the thorium. Yet the half-life is \(4.5\times10^{9}\,\mathrm{y}\) — the barrier, not the energy balance, sets the pace.

6 Radiocarbon age of an artifact

Problem. A wooden tool shows a \(^{14}\mathrm{C}\) activity that is \(0.25\) times that of fresh wood. With \(T_{1/2} = 5730\,\mathrm{y}\), how old is it?

Solution. Use \(t = \dfrac{T_{1/2}}{\ln 2}\ln\dfrac{R_0}{R}\) with \(R/R_0 = 0.25 = (1/2)^2\).

t = (T½ / ln2) · ln(R₀/R)
\[ t = \frac{5730\,\mathrm{y}}{0.693}\,\ln\!\frac{1}{0.25} = \frac{5730}{0.693}(1.386) \approx 1.15\times10^{4}\,\mathrm{y} \]

Since \(0.25\) is exactly two halvings, the answer is simply two half-lives, about \(11{,}500\) years — a quick sanity check that matches the formula. For arbitrary ratios the logarithm does the work.

Review

Chapter Summary

The nucleus

Rutherford scattering revealed a tiny, dense, positive core. Nucleons = protons + neutrons; \(A = Z + N\); same \(Z\), different \(N\) → isotopes.

Size and density

\(r = r_0 A^{1/3}\) with \(r_0 \approx 1.2\,\mathrm{fm}\), so all nuclei share a density near \(2\times10^{17}\,\mathrm{kg/m^3}\).

Binding energy

Mass defect → \(\Delta E_{be} = \Delta m\,c^2\) (\(931.5\,\mathrm{MeV/u}\)); \(\Delta E_{ben}\) peaks near iron, driving both fusion and fission.

Decay law

\(N = N_0 e^{-\lambda t}\), activity \(R = \lambda N\), half-life \(T_{1/2} = (\ln 2)/\lambda\); units Bq and Ci.

Alpha decay

\(Z\!-\!2,\ A\!-\!4\); energy \(Q = (m_X-m_Y-m_\alpha)c^2\); escape by barrier tunneling sets the half-life.

Beta decay

\(n\to p+e^-+\bar\nu\) or \(p\to n+e^++\nu\); electron energy spreads from 0 to \(Q\) because the neutrino shares it.

Dating & dosage

\(^{14}\mathrm{C}\) and \(^{40}\mathrm{K}\) as clocks; absorbed dose in grays, dose equivalent (dose × RBE) in sieverts.

Nuclear models

Collective (liquid drop) + independent-particle (shell), with magic numbers \(2,8,20,28,50,82,126\), combine into one picture.

Practice

Problems

Take \(c^2 = 931.5\,\mathrm{MeV/u}\), \(r_0 = 1.2\,\mathrm{fm}\), \(N_A = 6.02\times10^{23}\,\mathrm{mol^{-1}}\), \(\dfrac{e^2}{4\pi\varepsilon_0} = 1.44\,\mathrm{eV\cdot nm}\), proton/neutron mass \(\approx 1.67\times10^{-27}\,\mathrm{kg}\). Use \(r = r_0 A^{1/3}\), \(N = N_0 e^{-\lambda t}\), \(R = \lambda N\), and \(T_{1/2} = (\ln 2)/\lambda\).

  1. The nuclide \(^{197}\mathrm{Au}\) has \(Z = 79\). State its number of protons, neutrons, and nucleons, and write the symbol for an isotope with two more neutrons.
  2. Find the radius of a \(^{56}\mathrm{Fe}\) nucleus and of a \(^{238}\mathrm{U}\) nucleus, and the ratio of their volumes.
  3. Show, using \(r = r_0 A^{1/3}\), that the mass density of nuclear matter is essentially independent of \(A\), and estimate its value.
  4. An alpha particle with \(K_i = 7.0\,\mathrm{MeV}\) is fired head-on at an aluminum nucleus (\(Z = 13\)). Find its distance of closest approach.
  5. The deuteron \(^{2}\mathrm{H}\) has a mass defect of \(0.00239\,\mathrm{u}\). Find its binding energy and binding energy per nucleon, and compare with \(^{4}\mathrm{He}\).
  6. Why does the binding-energy-per-nucleon curve peak near \(A \approx 56\), and what does that single feature imply about both fusion and fission?
  7. A nuclide has half-life \(8.0\,\mathrm{d}\). What fraction of an initial sample remains after (a) \(8.0\,\mathrm{d}\), (b) \(24\,\mathrm{d}\), and (c) \(40\,\mathrm{d}\)?
  8. A source has activity \(4.0\times10^{8}\,\mathrm{Bq}\) and half-life \(6.0\,\mathrm{h}\). Find its disintegration constant and the number of radioactive nuclei present.
  9. Show that the mean life \(\tau = 1/\lambda\) is longer than the half-life, and find the ratio \(\tau/T_{1/2}\).
  10. For the alpha decay \(^{226}\mathrm{Ra}\to{}^{222}\mathrm{Rn}+{}^{4}\mathrm{He}\), the atomic masses are \(226.02540\), \(222.01757\), and \(4.00260\,\mathrm{u}\). Find \(Q\).
  11. Explain why two alpha emitters with nearly equal \(Q\) can have half-lives differing by many orders of magnitude.
  12. In the beta-minus decay \(^{14}\mathrm{C}\to{}^{14}\mathrm{N}+e^-+\bar\nu\), the maximum electron energy is the full \(Q\). Why do most emitted electrons carry less than this maximum?
  13. A sample of charcoal from a fire pit has a \(^{14}\mathrm{C}\) activity \(0.60\) times that of living wood. Using \(T_{1/2} = 5730\,\mathrm{y}\), find its age.
  14. A tumor absorbs \(0.20\,\mathrm{J}\) of gamma radiation in a mass of \(0.50\,\mathrm{kg}\). Find the absorbed dose in grays, and the dose equivalent in sieverts if \(\mathrm{RBE} = 1\).
  15. List the proton (or neutron) magic numbers, and explain why a doubly magic nuclide such as \(^{208}\mathrm{Pb}\) (\(Z = 82,\ N = 126\)) is exceptionally stable.
Tip: three habits tame this whole chapter. First, every energy question — binding energy, alpha \(Q\), beta \(Q\) — is the same calculation: find the mass that disappears and multiply by \(931.5\,\mathrm{MeV/u}\); with atomic masses the electrons usually cancel for you, so resist the urge to track them separately. Second, anything involving time is the one exponential \(N = N_0 e^{-\lambda t}\) and its twin \(R = \lambda N\) — convert the half-life to \(\lambda = (\ln 2)/T_{1/2}\) first, keep units consistent, and when the surviving fraction is a clean power of \(\tfrac12\) just count half-lives instead of reaching for a calculator. Third, separate what is allowed from how fast it happens: a positive \(Q\) says a decay can occur, but the barrier (for alpha) or the weak interaction (for beta) sets the half-life — which is why energy balance and lifetime are almost unrelated.