University Physics — Year 3 · Bachelor Year 3

25Nuclear Physics

The high-school volume told the story’s outline: atoms have nuclei; some nuclei crumble, on schedules ranging from microseconds to billions of years; and grams of matter hide megatons of energy. This chapter reopens the nucleus with Year 3 tools. A droplet model with five terms predicts the binding of hundreds of nuclides to the percent; quantum tunnelling — the Year 2 volume’s barrier problem, promoted — explains how one formula spans twenty-five orders of magnitude in $\alpha$ -decay lifetimes; and the binding-energy curve’s quiet maximum at iron divides all of nuclear technology into two camps: split the heavy (reactors) or join the light (stars, and someday power plants). The chapter ends beside a reactor pool, glowing Cherenkov blue.

25.1 The nucleus: a saturated drop

Definition 25.1 (Nuclides, size, and density)

A nuclide $^{A}_{Z}\text{X}$ holds $Z$ protons and $N = A - Z$ neutrons, bound by the strong interaction: intense ( ${\sim}100\,\times$ Coulomb at contact), short-ranged ( ${\sim}1\,\mathrm{fm}$ , felt only by touching neighbours), and blind to charge (it grips protons and neutrons alike). Scattering experiments give every nucleus the same recipe: radius

R = r_0A^{1/3} , \qquad r_0 \approx 1.2\,\mathrm{fm} ,

so volume grows like $A$ and the density is a universal $\rho \approx 2.3 \times 10^{17}\,\mathrm{kg}/\mathrm{m}^{3}$ — a raindrop of it would weigh like a fleet of tankers, and Chapter 27 will meet whole stars at this density. Isotopes share $Z$ but differ in $N$ : same chemistry, different nuclear fates.

Proposition 25.2 (Mass defect and binding energy)

A bound nucleus weighs less than its parts; by Chapter 5, the missing mass is the binding energy,

B = \big(Zm_{\text{p}} + Nm_{\text{n}} - m_{\text{nucleus}}\big)c^2 ,

typically ${\sim}8\,\mathrm{MeV}$ per nucleon — a million times chemistry. Plotted per nucleon, $B/A$ climbs steeply through the light nuclei (surface effects fade), peaks at $\approx8.8\,\mathrm{MeV}$ near iron-56, then drifts down as Coulomb repulsion, long-ranged and unsaturating, taxes the heavyweights. Both slopes are energy mines: fusion climbs the left slope (the Sun’s business), fission rolls down the right (the reactor’s) — and iron, at the summit, is nuclear ash for both.

The binding-energy curve — arguably the most consequential graph in physics. Light nuclei gain by merging, heavy ones by splitting; per kilogram, the left slope pays about four times better, and the Sun sits on it.

Proposition 25.3 (The semi-empirical mass formula)

Model the nucleus as a charged liquid drop and the whole curve follows from five terms:

B = a_{\text{V}}A - a_{\text{S}}A^{2/3} - a_{\text{C}}\frac{Z^2}{A^{1/3}} - a_{\text{A}}\frac{(A - 2Z)^2}{A} + \delta(A) ,

with (in $\mathrm{MeV}$ ) $a_{\text{V}} = 15.8$ (each nucleon bonds its neighbours: volume), $a_{\text{S}} = 17.8$ (surface nucleons bond fewer: the drop’s “surface tension”), $a_{\text{C}} = 0.71$ (Coulomb, the long-range saboteur), $a_{\text{A}} = 23.7$ (asymmetry: by Chapter 19, protons and neutrons fill two separate Fermi ladders, cheapest when equally full), and $\delta$ a small pairing bonus for even numbers (Chapter 14). Minimising at fixed $A$ traces the valley of stability: $N \approx Z$ for light nuclei, bending neutron-rich ( $N/Z \to 1.5$ ) as Coulomb punishes protons — and off-valley nuclides roll back in by $\beta$ decay.

Proof. Admitted at this level. ∎

The valley of stability. Light nuclei balance their two Fermi ladders (N = Z); heavy ones dilute their Coulomb bill with extra neutrons. A nuclide off the valley floor -decays back toward it — and fission fragments, born far above the line, are furiously radioactive for exactly this reason. — The valley of stability. Light nuclei balance their two Fermi ladders ( $N = Z$ ); heavy ones dilute their Coulomb bill with extra neutrons. A nuclide off the valley floor $\beta$ -decays back toward it — and fission fragments, born far above the line, are furiously radioactive for exactly this reason.

25.2 Radioactivity, by the clock

Theorem 25.4 (The decay law)

An unstable nucleus has no memory and no aging: in every second it lives, it decays with the same probability $\lambda$ . For $N$ nuclei, $\dd N = -\lambda N\,\dd t$ , so

N(t) = N_0\,\eu^{-\lambda t} = N_0\,2^{-t/T_{1/2}} , \qquad T_{1/2} = \frac{\ln2}{\lambda} ,

and the activity $\mathcal A = \lambda N$ (decays per second, $\mathrm{Bq}$ ) fades on the same exponential. Half-lives run from microseconds to billions of years; the statistics of Chapter 16 guarantee that while one nucleus is perfectly unpredictable, a mole of them keeps exponential time better than any clock — the foundation of radiometric dating.

Proof. Constant per-second probability means $N(t + \dd t) = N(t)(1 - \lambda\dd t)$ ; integrate. Setting $N/N_0 = \tfrac12$ gives $T_{1/2}$ . ∎

Exponential decay: memoryless individuals, punctual population. Two half-lives leave a quarter; ten leave a thousandth — the ruler with which physicists date charcoal, glaciers, and the Earth.

Remark 25.5 (Three doors out of a nucleus)

$\boldsymbol\alpha$ : a heavy nucleus emits a $^4$ He cluster. Classically impossible — the Coulomb barrier tops $25\,\mathrm{MeV}$ and the $\alpha$ leaves with $5\,$ – $9\,\mathrm{MeV}$ — it happens by tunnelling (the Year 2 volume’s barrier, curved): Gamow’s exponential sensitivity turns a factor two in energy into twenty orders of magnitude in lifetime, exactly the Geiger–Nuttall pattern (Exercise 25.8). $\boldsymbol\beta$ : a neutron becomes a proton (or vice versa), emitting an electron (or positron) and a neutrino — the weak interaction at work, the valley-restoring force; the neutrino’s story waits for Chapter 26. $\boldsymbol\gamma$ : an excited nucleus sheds $\mathrm{MeV}$ photons, the nuclear analogue of Chapter 11’s spectral lines. Decays chain until the valley floor is reached: uranium’s ladder ends, fourteen rungs later, at stable lead.

25.3 Down the slopes: fission and fusion

Proposition 25.6 (Fission and the chain reaction)

A slow neutron absorbed by $^{235}$ U makes a wobbling drop that splits: two mid-mass fragments, two-to-three fresh neutrons, and $\approx200\,\mathrm{MeV}$ — the $B/A$ gap between uranium’s $7.6\,\mathrm{MeV}$ and the fragments’ $8.5\,\mathrm{MeV}$ , times 235. The freed neutrons can fission further nuclei: with multiplication factor $k$ (neutrons per neutron, one generation later), a population grows as $k^n$ — subcritical ( $k < 1$ ) dies out, critical ( $k = 1$ ) holds steady, supercritical explodes. A reactor moderates neutrons to slow speeds (fission’s preferred diet), leans on the 0.7 % of neutrons that arrive seconds late from fragment decays — the delay that makes $k \approx 1$ humanly adjustable — and lets control rods eat the surplus (Problem 25.1).

Proof. Admitted at this level. ∎

Liquid-drop fission: a slow neutron sets 235U wobbling; the deformed drop’s Coulomb repulsion beats its surface tension, and it tears into neutron-rich fragments plus the spare neutrons that make a chain possible. — Liquid-drop fission: a slow neutron sets $^{235}$ U wobbling; the deformed drop’s Coulomb repulsion beats its surface tension, and it tears into neutron-rich fragments plus the spare neutrons that make a chain possible.

Remark 25.7 (Fusion: the harder, better slope)

Joining light nuclei pays $\sim4\times$ more per kilogram than splitting heavy ones, with no long-lived fragments — but the reactants, both positive, must tunnel through their mutual Coulomb barrier, which demands temperatures of $10^7$ – $10^8\,\mathrm{K}$ (Exercise 25.11). Stars do it by being enormous (Chapter 27: the Sun turns $600\,$ million tonnes of hydrogen into helium each second); laboratories do it with magnetically bottled plasmas and, so far, an energy ledger still shy of its break-even ambitions.

Marie Curie (photograph by Henri Manuel, c. 1920, public domain): twice a Nobel laureate for opening this chapter’s subject — the gram of radium of was hers. — Marie Curie (photograph by Henri Manuel, c. 1920, public domain): twice a Nobel laureate for opening this chapter’s subject — the gram of radium of Exercise 25.4 was hers.

25.4 Exercises

Exercise 25.1 ★

Bookkeeping. (a) Give $Z$ , $N$ , $A$ for $^{12}$ C, $^{14}$ C, $^{235}$ U, $^{238}$ U. (b) Which pairs are isotopes? (c) Why do isotopes share chemistry but not nuclear stability? (d) $^{14}$ C and $^{14}$ N share $A = 14$ : what are such nuclides called, and what decay connects them?

Solution

Solution of Exercise 25.1.

(a) $^{12}$ C: $6, 6, 12$ ; $^{14}$ C: $6, 8, 14$ ; $^{235}$ U: $92, 143, 235$ ; $^{238}$ U: $92, 146, 238$ . (b) The carbons; the uraniums. (c) Chemistry is the electron cloud, fixed by $Z$ ; stability is the $N$ – $Z$ balance inside. (d) Isobars — connected by $\beta^-$ decay: $^{14}\text{C} \to {}^{14}\text{N} + \text{e}^- + \bar\nu$ , the radiocarbon clock’s tick.

Exercise 25.2 ★

The universal density. (a) From $R = r_0A^{1/3}$ , show the nuclear density is independent of $A$ and evaluate it. (b) Compute the mass of a teaspoon ( $5\,\mathrm{mL}$ ) of nuclear matter. (c) What fraction of an atom’s volume does its nucleus fill (take $R_{\text{atom}} = 1 \times 10^{-10}\,\mathrm{m}$ , $A = 60$ )? (d) What does the $A^{1/3}$ law itself say about the strong force’s range?

Solution

Solution of Exercise 25.2.

(a) $\rho = Am_{\text{u}}/\tfrac43\pi r_0^3A = 3m_{\text{u}}/4\pi r_0^3 \approx 2.3 \times 10^{17}\,\mathrm{kg}/\mathrm{m}^{3}$ : $A$ cancels. (b) $5\,\mathrm{mL} \to 1.2 \times 10^{12}\,\mathrm{kg}$ — a billion tonnes in a teaspoon. (c) $(R/R_{\text{atom}})^3 \approx (4.7\times10^{-5})^3 \approx 10^{-13}$ : the atom is empty space with a heavy speck. (d) Volume $\propto A$ means each nucleon claims fixed room and binds only its touching neighbours: the force saturates at $\mathrm{fm}$ range.

Exercise 25.3 ★

Weighing the glue. Nuclear masses: $m_{\text{p}} = 1.007\,276\,\mathrm{u}$ , $m_{\text{n}} = 1.008\,665\,\mathrm{u}$ , $m(^4\text{He}) = 4.001\,506\,\mathrm{u}$ ; $1\,\mathrm{u}\,c^2 = 931.5\,\mathrm{MeV}$ . (a) Compute $^4$ He’s mass defect and binding energy. (b) Give $B/A$ . (c) Compare $B$ with the $13.6\,\mathrm{eV}$ of Chapter 11: how many orders of magnitude separate nuclear from atomic glue? (d) Why does $^4$ He’s exceptional binding (the figure’s spike) make it the currency of $\alpha$ decay?

Solution

Solution of Exercise 25.3.

(a) $\Delta m = 2(1.007276 + 1.008665) - 4.001506 = 0.030\,376\,\mathrm{u}$ : $B = 28.3\,\mathrm{MeV}$ . (b) $B/A = 7.07\,\mathrm{MeV}$ . (c) $28.3\,\text{MeV}/13.6\,\text{eV} \approx 2\times10^6$ : six orders of magnitude — why nuclear fuel outclasses chemical by a million. (d) The $\alpha$ is a pre-assembled, exceptionally cheap parcel: emitting it exports four nucleons at a bargain $28.3\,\mathrm{MeV}$ rebate, which no single nucleon can match.

Exercise 25.4 ★

The first curie. Radium-226: $T_{1/2} = 1600$ years, molar mass $226\,\mathrm{g}/\mathrm{mol}$ . For one gram: (a) count the nuclei; (b) compute $\lambda$ ; (c) compute the activity and compare with the historic unit $1\,\text{Ci} = 3.7 \times 10^{10}\,\mathrm{Bq}$ (defined as exactly this gram); (d) how much of the gram survives after 4800 years?

Solution

Solution of Exercise 25.4.

(a) $N = N_{\text{A}}/226 = 2.66 \times 10^{21}\,$ . (b) $\lambda = \ln2/(1600\times3.156 \times 10^{7}\,\mathrm{s}) = 1.37 \times 10^{-11}\,\mathrm{s}^{-1}$ . (c) $\mathcal A = \lambda N = 3.7 \times 10^{10}\,\mathrm{Bq}$ : one curie, by construction — the Curies’ own gram. (d) Three half-lives: $0.125\,\mathrm{g}$ of radium (the rest now radon and its daughters, marching toward lead).

Exercise 25.5 ★★

The formula at work. With $a_{\text{V}} = 15.75$ , $a_{\text{S}} = 17.8$ , $a_{\text{C}} = 0.711$ , $a_{\text{A}} = 23.7$ ( $\mathrm{MeV}$ ) and pairing $+12/\sqrt A$ for even–even nuclei: (a) compute the four main terms of $B$ for $^{56}$ Fe ( $Z = 26$ ). (b) Sum (with pairing) and compare $B/A$ with the measured $8.79\,\mathrm{MeV}$ . (c) Which two terms fight hardest, and what does their balance set? (d) Why must the Coulomb term, alone, grow faster than $A$ ?

Solution

Solution of Exercise 25.5.

(a) Volume $+882$ ; surface $-261$ ; Coulomb $-0.711\times676/3.83 = -126$ ; asymmetry $-23.7\times16/56 = -6.8$ ( $\mathrm{MeV}$ ). (b) With pairing $+12/\sqrt{56} = +1.6$ : $B \approx 490\,\mathrm{MeV}$ , $B/A = 8.76$ against the measured $8.79$ — three per mille from a liquid drop. (c) Volume against surface $+$ Coulomb: their crossover in growth rates is what carves the maximum at iron. (d) Coulomb is long-ranged: every proton repels every other, $\propto Z^2$ , while the saturating strong force only ever pays $\propto A$ — the eventual defeat of the heavyweights.

Exercise 25.6 ★★

The valley floor. (a) Keeping only the $Z$ -dependent terms, minimise the mass at fixed $A$ and derive $Z^* = \dfrac{A/2}{1 + a_{\text{C}}A^{2/3}/4a_{\text{A}}}$ . (b) Evaluate for $A = 101$ and compare with ruthenium ( $Z = 44$ ). (c) Show the light- $A$ limit is $Z^* = A/2$ and interpret with the two Fermi ladders. (d) A fission fragment has $A = 95$ , $Z = 36$ : how far off the valley is it, and what sequence of decays follows?

Solution

Solution of Exercise 25.6.

(a) $\partial/\partial Z$ of the Coulomb and asymmetry terms gives $2a_{\text{C}}Z/A^{1/3} = 4a_{\text{A}}(A - 2Z)/A$ ; solve. (b) $Z^* = 50.5/1.163 = 43.4$ : nature’s choice at $A = 101$ is ruthenium, $Z = 44$ . (c) Without Coulomb, $Z^* = A/2$ : two Fermi ladders (Chapter 19), cheapest filled to equal height. (d) The fragment’s valley seat is $Z^* \approx 41$ : five protons short — a cascade of five $\beta^-$ decays climbs it back, each converting a neutron and emitting an electron and antineutrino.

Exercise 25.7 ★★

Radiocarbon. Living matter keeps $^{14}$ C ( $T_{1/2} = 5730$ years) topped up at 1 atom per $10^{12}$ carbons; death stops the refill. (a) A charcoal sample shows 30 % of the living activity: how old is the fire? (b) One gram of living carbon: compute its $^{14}$ C activity ( $\approx0.23\,\mathrm{Bq}$ expected). (c) Why is the method useless beyond $\sim50\,000\,$ years? (d) And why is it useless for dating rocks (which clocks take over)?

Solution

Solution of Exercise 25.7.

(a) $t = T_{1/2}\ln(1/0.3)/\ln2 = 5730\times1.74 \approx 10^{4}\,$ years: an end-of-ice-age fire. (b) $N_{^{14}\text{C}} = (N_{\text{A}}/12)\times10^{-12} = 5.0 \times 10^{10}\,$ ; $\lambda = 3.8 \times 10^{-12}\,\mathrm{s}^{-1}$ : $\mathcal A \approx 0.2\,\mathrm{Bq}$ — a dozen decays per minute per gram, the working count rate of every dating laboratory. (c) After nine half-lives the activity sinks below background: the clock still runs but can no longer be read. (d) Rocks never breathed atmospheric carbon; their clocks are the primordial ones — uranium–lead, potassium–argon — with half-lives of billions of years.

Exercise 25.8 ★★

Gamow’s cliff. Polonium-212 emits an $8.78\,\mathrm{MeV}$ $\alpha$ ; its barrier tops $\approx26\,\mathrm{MeV}$ . (a) Show classical escape is impossible, and classical entry too — yet it decays in $0.3\,\text{µ}\mathrm{s}$ . (b) Uranium-238’s $\alpha$ carries $4.2\,\mathrm{MeV}$ and its half-life is $4.5\times10^9$ years: compute the ratio of the two decay constants. (c) Explain, with the tunnelling exponential, how a factor two in energy buys $\sim10^{24}$ in rate. (d) Why does the same physics set the Sun’s core temperature requirement?

Solution

Solution of Exercise 25.8.

(a) The $\alpha$ ’s $8.78\,\mathrm{MeV}$ is far under the $26\,\mathrm{MeV}$ rim: classically it can neither leave nor have entered — yet the half-life is $0.3\,\text{µ}\mathrm{s}$ . (b) $\lambda_1 = \ln2/3 \times 10^{-7}\,\mathrm{s} = 2.3 \times 10^{6}\,\mathrm{s}^{-1}$ ; $\lambda_2 = \ln2/1.4 \times 10^{17}\,\mathrm{s} = 4.9 \times 10^{-18}\,\mathrm{s}^{-1}$ : ratio $\approx 5\times10^{23}$ . (c) The tunnelling rate is $\eu^{-2G}$ with the Gamow exponent $\propto Z/\sqrt E$ : halving $E$ raises $2G$ by some fifty-five units, and $\eu^{55} \approx 10^{24}$ — Geiger–Nuttall’s outrageous straight line, from one exponential. (d) Solar protons also tunnel: the core temperature must be just high enough for the Maxwell tail times the Gamow factor to sustain the burn — same cliff, climbed from the other side.

Exercise 25.9 ★★

Two hundred million electron-volts. (a) Justify the $200\,\mathrm{MeV}$ per fission from the $B/A$ curve. (b) Compute the energy in one kilogram of $^{235}$ U fully fissioned, in joules and in tonnes of oil equivalent ( $42\,\mathrm{GJ}/\mathrm{t}$ ). (c) A $1\,\mathrm{GW}_{e}$ power plant at 33 % efficiency: how many kilograms of $^{235}$ U per year? (d) Why do the fragments, not the neutrons, carry most of the $200\,\mathrm{MeV}$ , and into what form does it immediately go?

Solution

Solution of Exercise 25.9.

(a) The split moves ${\sim}235$ nucleons from $7.6\,$ to ${\sim}8.5\,\mathrm{MeV}$ of binding: $235\times0.9 \approx 200\,\mathrm{MeV}$ . (b) $N = 2.56 \times 10^{24}\,$ nuclei: $E = 8.2 \times 10^{13}\,\mathrm{J} \approx 2000$ tonnes of oil — per kilogram. (c) Thermal $3\,\mathrm{GW}$ for a year is $9.5 \times 10^{16}\,\mathrm{J}$ : about $1.2\,\mathrm{t}$ of $^{235}$ U. (d) The two positive fragments spring apart under their own Coulomb repulsion, carrying ${\sim}170\,\mathrm{MeV}$ as kinetic energy that becomes heat within micrometres — a reactor is a kettle whose flame is electrostatic recoil.

Exercise 25.10 ★★★

Taming $k$ . In a reactor, prompt neutrons reproduce in $\ell \approx 10^{-4}\,\mathrm{s}$ . (a) With $k = 1.001$ on prompt neutrons alone, compute the power multiplication in one second — and the verdict on mechanical control. (b) A fraction $\beta = 0.7\,\%$ of neutrons is delayed by $\sim10\,\mathrm{s}$ : explain why, for $k - 1 < \beta$ , the chain’s effective clock becomes seconds. (c) Compute the same one-second multiplication with the effective generation time $\sim0.1\,\mathrm{s}$ . (d) State in one sentence what Chernobyl’s operators lost when their reactor went prompt-supercritical.

Solution

Solution of Exercise 25.10.

(a) $10^4$ generations: $(1.001)^{10^4} = \eu^{10} \approx 22000$ — megawatts to tens of gigawatts inside a second; no motor moves a rod that fast. (b) With $k - 1 < \beta$ the chain cannot reproduce on prompt neutrons alone: every growth step waits for the ${\sim}10\,\mathrm{s}$ stragglers, so the effective generation time is a fraction of a second or more. (c) $(1.0005)^{10} \approx 1.005$ : half a percent per second — rod-and-operator territory. (d) Driving $k - 1$ past $\beta$ put the reactor on the $10^{-4}\,\mathrm{s}$ prompt clock: they lost the delayed-neutron grace period that makes reactors steerable.

Exercise 25.11 ★★★

The Sun’s arithmetic. Overall, $4\,^1\text{H} \to {}^4\text{He}$ converts $0.71\,\%$ of the mass to energy ( $26.7\,\mathrm{MeV}$ , neutrinos included). (a) From $L_\odot = 3.8 \times 10^{26}\,\mathrm{W}$ , compute the mass converted per second, and the hydrogen consumed per second. (b) With $2 \times 10^{30}\,\mathrm{kg}$ of sun, a tenth of it burnable core hydrogen (take the hydrogen fraction $\approx0.75$ ): estimate the lifetime. (c) Two protons at $1.5 \times 10^{7}\,\mathrm{K}$ : compare $k_{\text{B}}T$ with their Coulomb barrier at $1\,\mathrm{fm}$ and conclude who does the crossing (Exercise 25.8). (d) Why does fusion, unlike fission, leave essentially no radioactive ash?

Solution

Solution of Exercise 25.11.

(a) $\dot m = L_\odot/c^2 = 4.2 \times 10^{9}\,\mathrm{kg}/\mathrm{s}$ of pure mass; hydrogen throughput $\dot m/0.0071 \approx 6 \times 10^{11}\,\mathrm{kg}/\mathrm{s}$ — six hundred million tonnes a second. (b) Burnable hydrogen ${\sim}2\times10^{30}\times0.1\times0.75 = 1.5 \times 10^{29}\,\mathrm{kg}$ : $t \approx 2.5 \times 10^{17}\,\mathrm{s} \approx 8$ billion years — the Sun is middle-aged. (c) Barrier ${\sim}1.4\,\mathrm{MeV}$ against $k_{\text{B}}T \approx 1.3\,\mathrm{keV}$ : a thousandfold deficit — only the Maxwell tail’s fastest protons, tunnelling (Exercise 25.8), ever fuse; hence the Sun burns for gigayears instead of exploding. (d) The ash is $^4$ He — stable, doubly magic contentment; there are no neutron-rich fragments to $\beta$ -decay for centuries.

Exercise 25.12 ★★★

Nuclear medicine. (a) Technetium-99m ( $T_{1/2} = 6\,\mathrm{h}$ , $\gamma$ of $140\,\mathrm{keV}$ ) is medicine’s workhorse: compute the number of nuclei, and their mass, behind an injected $500\,\mathrm{MBq}$ . (b) Explain why a six-hour half-life and a pure $\gamma$ are each exactly what a diagnostic wants. (c) PET: a positron from $^{18}$ F annihilates with an electron — use Chapter 5 to give the energy and geometry of the two photons, and hence the detector’s principle. (d) Radiotherapy delivers $2\,\mathrm{Gy}$ ( $\mathrm{J}/\mathrm{kg}$ ) to a tumour: compare the energy with a sip of warm water, and reconcile the harmlessness of the joules with the lethality of the ionisation.

Solution

Solution of Exercise 25.12.

(a) $\lambda = \ln2/21\,600\,\mathrm{s} = 3.2 \times 10^{-5}\,\mathrm{s}^{-1}$ : $N = \mathcal A/\lambda = 1.6 \times 10^{13}\,$ nuclei — $2.6\,\mathrm{ng}$ . Medicine by the nanogram. (b) Six hours outlives the scan but not the weekend: the dose switches itself off; a pure $140\,\mathrm{keV}$ $\gamma$ leaves the body to the camera without the tissue-burning $\alpha$ / $\beta$ toll. (c) Electron and positron at rest annihilate into two $511\,\mathrm{keV}$ photons, back to back (energy and momentum, Chapter 5): each coincident pair defines a line through the tumour, and thousands of lines triangulate it — tomography by conservation law. (d) $2\,\mathrm{J}/\mathrm{kg}$ warms tissue half a millikelvin — thermally a sip of tepid water; but delivered as ${\sim}10^{17}$ ionisations per kilogram, each snipping $\mathrm{eV}$ -scale chemical bonds, it is chemistry-lethal to a dividing cell. Joules measure heat; grays measure sabotage.

The reactor pool of the weekend problem: fragment electrons outrunning light in water wrap the core in Cherenkov blue — the shielding, visibly at work. — The reactor pool of the weekend problem: fragment $\beta$ electrons outrunning light in water wrap the core in Cherenkov blue — the shielding, visibly at work.

25.5 Problem: Open day at the research reactor

Problem 25.1

Open day at the research reactor. The national laboratory opens its $20\,\mathrm{MW}$ pool-type research reactor to visitors, and you have talked your way onto the balcony: below, through eight metres of limpid water, the core glows an unearthly blue. The guide — a retiring reactor physicist — has agreed to let you do the arithmetic.

Part I — The energy ledger.

Each fission of $^{235}$ U releases about $200\,\mathrm{MeV}$ . Convert to joules, and compute the fissions per second sustaining $20\,\mathrm{MW}$ .
Convert to grams of $^{235}$ U consumed per day.
The same $20\,\mathrm{MW}$ from coal ( $30\,\mathrm{MJ}/\mathrm{kg}$ ): tonnes per day? Form the mass ratio and connect it to the $\mathrm{MeV}$ -versus- $\mathrm{eV}$ scales of this chapter and chemistry.
The fragments are born at $\sim8.5\,\mathrm{MeV}$ per nucleon binding, uranium at $7.6\,\mathrm{MeV}$ : verify $235\times0.9 \approx 200\,\mathrm{MeV}$ is consistent.
The fragments stop within micrometres of their birthplace. Into what does their $200\,\mathrm{MeV}$ convert, and on what timescale does it reach the cooling water?
Why are the fragments (e.g. $A = 95$ , $Z = 36$ ) inevitably radioactive? Use the valley of stability.

Part II — Keeping $k = 1$ .

Define the multiplication factor $k$ and state what $k = 0.999$ , $1.000$ , $1.001$ mean for the neutron population.
The pool water moderates. Explain in two sentences why slowing neutrons to thermal speeds helps fission $^{235}$ U, and why hydrogen is the best moderator (recall elastic collisions from the Year 1 volume).
Control rods are boron steel. What does boron do, and how does raising or lowering the rods steer $k$ ?
With prompt neutrons alone ( $\ell = 10^{-4}\,\mathrm{s}$ ) and $k = 1.0005$ , compute the power growth over one second and conclude.
The 0.7 % delayed neutrons stretch the effective generation time to $\sim0.1\,\mathrm{s}$ : recompute, and state the design rule that keeps $k - 1$ safely below the delayed fraction.
The water is also the coolant. Explain the intrinsic safety feature: what happens to moderation — and hence to $k$ — if the water boils away?
Why does the reactor still need emergency cooling after shutdown? (Name the heat source that control rods cannot touch.)

Part III — The blue light.

The glow is Cherenkov radiation: light’s speed in water is $c/n$ with $n = 1.33$ . What must a charged particle do to emit it?
Compute the threshold speed, and with Chapter 5 the threshold kinetic energy for an electron.
Which reactor particles qualify? Trace the chain from fission fragment to fast $\beta$ electron in the water.
The spectrum’s intensity grows toward short wavelengths: why does the pool glow blue rather than red?
The guide says: “the glow is the shielding working.” Explain — what does eight metres of water do to the core’s radiation, and roughly why is the balcony safe?
After shutdown the blue dims over minutes but does not vanish for days. Connect to Part I’s question 6.

Part IV — What the reactor is for.

This reactor’s day job is making molybdenum-99 ( $T_{1/2} = 66\,\mathrm{h}$ ), parent of medicine’s technetium-99m. Why must the world’s hospitals be resupplied weekly, and why can no warehouse stockpile it?
A $100\,\mathrm{GBq}$ Mo-99 shipment leaves on Monday; the hospital elutes Tc-99m on Friday (96 hours later). What activity of the parent remains?
Neutron beams from the core also feed the diffraction instruments of Chapter 23: why are reactor neutrons, once thermalised, born with the right wavelength?
The guide’s parting problem: a fuel element spends five years in the pool before shipment. Using the fragments’ mix of half-lives, explain the logic — what has five years of pool time bought?
Estimate the decay heat one hour after shutdown ( $\approx1\,\%$ of $20\,\mathrm{MW}$ ) and compare it with a household’s electric heater: is passive pool cooling plausible?
Close the ledger in four lines: $B/A$ pays $200\,\mathrm{MeV}$ a split; delayed neutrons lend the seconds that make $k$ steerable; water moderates, cools, shields, and glows; and the day’s real product leaves in medical vials, not megawatts.

Solution

Solution of Problem 25.1.

1. $200\,\mathrm{MeV} = 3.2 \times 10^{-11}\,\mathrm{J}$ : $20\times10^6/3.2\times10^{-11} = 6.3 \times 10^{17}\,$ fissions per second. 2. $5.4\times10^{22}$ fissions per day $\times 235/N_{\text{A}}$ : about $21\,\mathrm{g}$ of $^{235}$ U a day. 3. Coal: $1.7 \times 10^{12}\,\mathrm{J}/3 \times 10^{7}\,\mathrm{J}/\mathrm{kg} \approx 58$ tonnes a day — a mass ratio of $\sim3\times10^{6}$ , which is the $\mathrm{MeV}$ -to- $\mathrm{eV}$ ratio of nuclear to chemical bonds. 4. $235\times(8.5 - 7.6) \approx 210\,\mathrm{MeV}$ : consistent, the small change bookkept by neutrons and neutrinos. 5. Fragment kinetic energy $\to$ ionisation $\to$ heat, within micrometres and microseconds; conduction hands it to the water — the reactor is a fragment-stopping kettle. 6. They inherit uranium’s $N/Z \approx 1.55$ , far above the valley floor at their $A$ : each must run a chain of $\beta^-$ decays back down — built-in radioactivity, by geometry of the valley. 7. $k$ = neutrons begotten per neutron, one generation on: $0.999$ dies out, $1.000$ holds steady (an operating reactor), $1.001$ grows. 8. Slow neutrons linger near the nucleus and $^{235}$ U’s fission appetite grows steeply at thermal energies. Elastic collisions shed energy fastest onto equal masses — and hydrogen matches the neutron’s mass: water is moderator made to order. 9. Boron-10 devours neutrons ( $\text{n} + {}^{10} \text{B} \to \alpha + {}^{7}\text{Li}$ ): rods in, neutrons eaten, $k$ down; rods out, $k$ up — the throttle. 10. $(1.0005)^{10^4} = \eu^{5} \approx 150$ : powers of a hundred and fifty per second — hopeless for machinery. 11. On the $0.1\,\mathrm{s}$ delayed clock: $(1.0005)^{10} \approx 1.005$ , half a percent per second. Rule: keep $|k - 1|$ well below $\beta = 0.007$ , so the delayed neutrons always hold the casting vote. 12. Boiling removes the moderator: neutrons stay fast, fission starves, $k$ falls — the water-moderated design throttles itself (a negative feedback its graphite cousins lacked). 13. Decay heat: the fragments’ radioactivity — Part I question 6 — obeys half-lives, not control rods, and still yields megawatts just after shutdown. 14. Outrun light in the water: $v > c/n$ . 15. $v > 0.752c$ : $\gamma = 1.51$ , so $E_{\text{k}} > 0.51\times511\,\mathrm{keV} \approx 0.26\,\mathrm{MeV}$ . 16. $\beta$ electrons from fragment decays carry $\mathrm{MeV}$ s, and core $\gamma$ s Compton-kick electrons to similar energies: both sail past $0.26\,\mathrm{MeV}$ — the pool is full of qualifying electrons. 17. The Cherenkov spectrum strengthens toward short wavelengths ( $\propto1/\lambda^2$ in intensity): the eye is handed blue and violet — the colour is the spectrum’s slope. 18. Water attenuates $\gamma$ s and neutrons exponentially; eight metres is dozens of halving-lengths, so the balcony sits at background dose. The glow is the water absorbing the radiation’s energy — visible proof the shield is eating it. 19. The short-lived fragments die within minutes (the dimming); the longer-lived tail — the same decay heat of question 13 — keeps a faint glow and a real heat load for days. 20. With $T_{1/2} = 66\,\mathrm{h}$ the parent loses a factor ${\sim}6$ per week: stockpiles decay themselves away, so hospitals live on a weekly “technetium cow” delivered from reactors like this one. 21. $96/66 = 1.45$ half-lives: $2^{-1.45} \approx 0.36$ — about $36\,\mathrm{GBq}$ remain. 22. Thermalised to room temperature, $\lambda = h/\sqrt{3m_{\text{n}}k_{\text{B}}T} \approx 1.5\,\text{Å}$ (Exercise 23.5): born matched to lattice spacings. 23. Five years kills every half-life up to months: activity and decay heat fall by orders of magnitude, until the element can ride in a shielded cask instead of a swimming pool. 24. ${\sim}200\,\mathrm{kW}$ — a hundred domestic heaters into a hundred-tonne pool: tens of kelvin per day at worst, comfortably removed by natural convection — passive safety by sheer heat capacity. 25. $B/A$ pays $200\,\mathrm{MeV}$ a split and $21\,\mathrm{g}$ a day; delayed neutrons lend the seconds that make $k$ steerable; the water moderates, cools, shields — and glows blue precisely because it is working; and the product that matters most leaves in vials for Monday’s hospitals.

Terms defined in this chapter

Nuclides, size, and density

See all 431 terms in the glossary