Exercises — Fission — chain reaction, critical mass, reactors (thermal vs fast)

WHAT we do: conserve mass number (top) and charge (bottom) — the two quantities that cannot change in any nuclear reaction. Mass number: $235+1=236$. On the right so far $140+A+x$. Charge: $92+0=92$. On the right $54+Z+0$, so $Z=92-54=38$ (that is indeed strontium, Sr — consistency check). Strontium's most common fission fragment here is ${}^{94}\text{Sr}$, so try $A=94$: then $140+94+x=236\Rightarrow x=2$. Answer: ${}_{38}^{94}\text{Sr}$ and $x=2$ neutrons. WHY it looks right: two mid-sized fragments plus 2 neutrons — a textbook fission split.

$k<1$ ⇒ subcritical. Each generation has fewer fission-neutrons than the last ($N_0{k}^n$ with $k<1$ shrinks toward 0), so the chain dies out.

(a) moderator → slows neutrons to thermal speed. (b) control rod → absorbs excess neutrons (tunes $k$). (c) coolant → carries heat away to make steam. (d) fuel → undergoes fission.

WHY this tool: the missing mass becomes energy via $E=\Delta m{c}^2$; the shortcut $1\text{u}\to 931.5\text{MeV}$ saves us from writing $c^2$ explicitly. $E=0.2148\times 931.5\approx 200.1\text{MeV}$ Answer: $\approx 200\text{MeV}$ — the canonical value.

WHAT: $100\text{W}=100\text{J/s}$. Convert one fission to joules, then divide. One fission $=200\text{MeV}=200\times 1.602\times 10^{-13}\text{J}=3.204\times 10^{-11}\text{J}$. $\text{rate}=\frac{100}{3.204\times 10^{-11}}\approx 3.12\times 10^{12}\text{fissions/s}$ WHY it matters: a few trillion fissions per second is tiny on the atomic scale — that is why a fingernail of fuel lasts a long time.

$N=N_0{k}^n=1\times 2^{10}=1024$ WHY: each generation doubles the count — this is the avalanche that $k>1$ produces. Ten steps already gives over a thousand.

Step 1 — count atoms. Moles $=\frac{1000\text{g}}{235\text{g/mol}}=4.255\text{mol}$. Atoms $=4.255\times 6.022\times 10^{23}=2.563\times 10^{24}$. Why: each nucleus is one potential fission. Step 2 — joules per fission. $200\text{MeV}=3.204\times 10^{-11}\text{J}$. Step 3 — multiply. $E=2.563\times 10^{24}\times 3.204\times 10^{-11}\approx 8.21\times 10^{13}\text{J}$ Answer: $\approx 8.2\times 10^{13}\text{J/kg}$ — millions of times a chemical fuel.

WHY this ratio: fissions (neutron birth) fill the volume $\propto r^3$; leakage happens through the surface $\propto r^2$. Leakage-per-fission scales like $\frac{r^2}{r^3}=\frac{1}{r}$.

• Small sphere: $\frac{S}{V}=\frac{4\pi r^2}{\frac{4}{3}\pi r^3}=\frac{3}{r}$.
• Big sphere: $\frac{3}{2r}$ — exactly half the small one. Conclusion: the bigger sphere leaks relatively less, so its $k$ is higher — it is closer to critical. Doubling the radius halves fractional leakage. See figure.

A cross-section is the effective "target area" a nucleus presents to a neutron — big cross-section means a hit is likely. For ${}^{235}\text{U}$ the fission cross-section is hundreds of times larger for slow (thermal, ~0.025 eV) neutrons than for fast (~2 MeV) ones. So slowing neutrons dramatically raises the chance each one causes fission, letting the chain sustain with only ~3–5% enrichment. The cost is carrying a moderator (light nuclei: H, D, C) that slows neutrons by elastic collisions.

One hemisphere: it has less mass and a worse (higher) surface-to-volume ratio than the full sphere — a hemisphere exposes a large flat cut face. More fractional leakage ⇒ $k<1$ ⇒ subcritical, chain dies. Slammed together: the two halves reform (nearly) one sphere. Surface-to-volume drops back to the low value of a full ball ⇒ leakage falls ⇒ $k$ jumps above 1 ⇒ supercritical, runaway growth. This is the gun-type bomb assembly. WHY it works: geometry alone changes $k$ — same total material, different shape, different leakage.

Step 1 — total energy. $1.0\text{GW}=1.0\times 10^9\text{J/s}$. One day $=86400\text{s}$. $E=1.0\times 10^9\times 86400=8.64\times 10^{13}\text{J}$ Step 2 — divide by energy per kg. $m=\frac{8.64\times 10^{13}}{8.2\times 10^{13}}\approx 1.05\text{kg}$ Answer: about $1.05\text{kg}$ of ${}^{235}\text{U}$ per day — roughly a kilogram, a striking contrast with the tonnes of coal a similar plant burns.

Capture: ${}_{92}^{238}\text{U}+{}_0^{1}n\to {}_{92}^{239}\text{U}$ (mass $238+1=239$; charge unchanged). First ${\beta }^{-}$: a neutron becomes a proton, so $Z$ rises by 1, $A$ stays: ${}_{92}^{239}\text{U}\stackrel{{\beta }^{-}}{\to }{}_{93}^{239}\text{Np}$. Second ${\beta }^{-}$: again $Z+1$: ${}_{93}^{239}\text{Np}\stackrel{{\beta }^{-}}{\to }{}_{94}^{239}\text{Pu}$. WHY ${\beta }^{-}$: each decay conserves $A$ and raises $Z$ by 1 (recall Radioactive Decay & Half-life) — the exact rule that walks U → Np → Pu. Fast neutrons are used because moderating them wastes the spare neutrons needed to breed.

Fission per nucleon: $\frac{200}{235}\approx 0.85\text{MeV/nucleon}$. Fusion per nucleon: $\approx 3.4\text{MeV/nucleon}$. Ratio: $\frac{3.4}{0.85}=4.0$ — fusion delivers about 4× more energy per nucleon. WHY: both processes climb the binding-energy curve toward the iron peak; the slope is much steeper on the light side (fusion) than the shallow heavy side (fission), so light nuclei gain far more binding per nucleon.

Neutrons causing fission $=100-30-25=45$. Those $45$ fissions each release $2.5$ neutrons ⇒ next generation is born with $45\times 2.5=112.5$ neutrons. Define $k$ by fission-causing neutrons: we compare this generation's fission-causers to the previous one. If the previous generation had produced the $100$ born here through its own fissions, that previous generation had $100/2.5=40$ fission-causers. $k=\frac{45}{40}=1.125$ Verdict: $k>1$ ⇒ supercritical; control rods must be pushed in further to raise absorption until $k=1$. WHY the bookkeeping matters: producing 2.5 neutrons per fission does not guarantee growth — but here losses were small enough that it did. This is the L2 misconception laid bare with numbers.

Set up. Born $=100$; leak $=30$; let non-fission absorption $=x$. Fission-causers $=100-30-x=70-x$. For $k=1$ the next generation must again be born with $100$ neutrons: $(70-x)\times 2.5=100$. $70-x=40\Rightarrow x=30$ Answer: absorb $30$ neutrons without fission (control rods soak up 5 more than in L5.2). Then exactly $40$ cause fission, reproducing the $100$-neutron generation — steady, self-sustaining, $k=1$. WHY this is the operator's job: control rods are the adjustable term $x$; sliding them tunes $k$ to exactly $1$, which is a reactor's whole point (contrast Nuclear Reactor Safety & Waste, where losing this control is the danger).

Energy per U-235 fission ::: ~200 MeV Energy per kg of U-235 ::: ~8.2 × 10¹³ J How S/V scales with radius ::: proportional to 1/r (leakage falls as r grows) k for subcritical / critical / supercritical ::: k<1 / k=1 / k>1 Fusion vs fission energy per nucleon ::: fusion ≈ 4× larger (~3.4 vs ~0.85 MeV/nucleon)