Worked examples — Fission — chain reaction, critical mass

Before we start, three symbols you must own (each is just plain counting):

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The scenario matrix

Every fission problem is one of these cells. The examples below are tagged with the cell they hit.

Cell	What varies	Example
A — energy per event	one fission, $Q=\Delta m{c}^2$	Ex 1
B — many events / power	$N$ fissions/s from power $P$	Ex 2
C — subcritical $k<1$	reaction dies, count decays	Ex 3
D — critical $k=1$	steady state, count constant	Ex 3 (limit)
E — supercritical $k>1$	count grows, doubling time	Ex 4
F — degenerate $k=0$	every neutron lost, one-shot	Ex 5
G — geometry & threshold	critical radius $R_c$, sign of $k-1$	Ex 6
H — limiting behaviour	$R\to \infty$ and $R\to 0$	Ex 6 (limits)
I — real-world word problem	fuel burn-up over a day	Ex 7
J — exam twist	density/compression, $M_c\propto 1/{\rho }^2$	Ex 8

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Example 1 — Cell A: energy of a single fission

Step 1. Use the conversion $1u\cdot c^2=931.5$ MeV. $Q=\Delta m{c}^2=0.2146\times 931.5\text{MeV}\approx 200\text{MeV}.$

Why this step? Mass and energy are the same currency (Mass-Energy Equivalence E=mc^2); the "exchange rate" for one atomic mass unit is exactly $931.5$ MeV, so we just multiply.

Step 2. Convert MeV to joules using $1\text{MeV}=1.6\times 10^{-13}$ J. $Q=200\times 1.6\times 10^{-13}\text{J}=3.2\times 10^{-11}\text{J}.$

Why this step? Joules are needed the moment we mix in power (watts $=$ joules/second) in Ex 2.

Verify: $0.2146\times 931.5=199.9\approx 200$ MeV. Sanity: the parent note derived $\approx 200$ MeV from the Binding Energy per Nucleon Curve ($235\times 0.9$ MeV). Independent route, same answer. ✓

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Example 2 — Cell B: from power to fissions per second

Step 1. Fissions per second $=\frac{P}{Q}$. $\frac{P}{Q}=\frac{3000\times 10^6\text{J/s}}{3.2\times 10^{-11}\text{J}}=9.375\times 10^{19}\text{fissions/s}.$

Why this step? Total energy rate ÷ energy per event $=$ events per rate. Units cancel to $1/\text{s}$.

Step 2. Fissions per day $=9.375\times 10^{19}\times 86400\text{s}=8.1\times 10^{24}$.

Why this step? One day $=86400$ s; multiplying a per-second rate by seconds gives a raw count.

Step 3. Each fissioned nucleus weighs $235u=235\times 1.66\times 10^{-27}$ kg $=3.90\times 10^{-25}$ kg. Mass burned: $m=8.1\times 10^{24}\times 3.90\times 10^{-25}\text{kg}\approx 3.16\text{kg}.$

Why this step? Count × mass-per-nucleus $=$ total mass. About 3 kg of uranium a day for a big reactor — a genuinely useful sanity anchor.

Verify: $9.375\times 10^{19}$ is far below $6\times 10^{23}$, so under one mole of nuclei splits per second. Mass $\approx 3.2$ kg/day is the textbook figure for a $\sim 3$ GW plant. ✓

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Example 3 — Cells C & D: subcritical and critical bookkeeping

Step 1 (a). Use $N_n=N_0{k}^n$ (each generation multiplies by $k$). $N_{200}=10^6\times 0.98^{200}.$

Why this step? This is the parent's exponential-growth law; with $k<1$ it becomes exponential decay.

Step 2 (a). Compute $0.98^{200}=e^{200\ln 0.98}=e^{200\times (-0.02020)}=e^{-4.04}\approx 0.0176$. $N_{200}\approx 10^6\times 0.0176\approx 1.76\times 10^4.$

Why this step? Logs turn a big power into an easy exponential — see Ex 4 for the same trick used forward.

Step 3 (b). With $k=1$: $N_n=10^6\times 1^n=10^6$ for every $n$.

Why this step? This is exactly the critical state: production balances loss, so the population is frozen — a reactor's normal operating point.

Verify: $0.98^{200}=0.01758$, giving $N_{200}=17,580$ — a drop by a factor $\approx 57$, i.e. subcritical dies away. For $k=1$, $1^{200}=1$, count unchanged. ✓

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Example 4 — Cell E: supercritical growth and doubling time

Step 1 (a). Need $k^n=1000$. Take logs (this is why we use logarithms — they extract an exponent): $n=\frac{\ln 1000}{\ln k}=\frac{6.908}{\ln 1.05}=\frac{6.908}{0.04879}\approx \mathrm{141.6.}$

Why this step? $n$ is trapped in an exponent; the only tool that pulls it out is the logarithm — the function that undoes exponentiation.

Step 2 (b). Real time $=n\tau =141.6\times 10^{-8}\text{s}\approx 1.42\times 10^{-6}$ s.

Why this step? Generations × time-per-generation $=$ elapsed time. About a microsecond — no human or machine can intervene.

Step 3 — contrast. In a reactor the effective $\tau$ is stretched to $\sim 0.1$ s by delayed neutrons, so the same growth would take $\sim 14$ s: controllable. (See Nuclear Reactor.)

Why this step? It shows the same math with a different $\tau$ separates a bomb from a power plant.

Verify: $\ln 1000/\ln 1.05=141.6$; $141.6\times 10^{-8}=1.416\times 10^{-6}$ s. Microsecond regime confirmed. ✓

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Example 5 — Cell F: the degenerate case $k=0$

Step 1. Generation-0 neutrons cause up to $500$ fissions, but $k=0$ means generation 1 has $N_1=N_0\times k=500\times 0=0.$

Why this step? $k=0$ is the boundary of the formula $N_n=N_0{k}^n$: the chain is severed after one step. This is the no-chain limit.

Step 2. Total fissions $\approx N_0=500$ (one generation only), then it stops dead.

Why this step? With no surviving neutrons there is no generation 1 — the reaction is a single flash, not a chain. This is what "subcritical" degenerates into at its extreme.

Verify: $500\times 0^1=0$ and ${\sum }_{n\ge 1}{N}_0{k}^n=0$ — geometric series with ratio $0$ sums to nothing beyond generation 0. The lump is inert. ✓

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Example 6 — Cells G & H: critical radius and both limits (with figure)

Step 1 (a). Critical means $k=1$. Set $\frac{a}{b}{R}_c=1$: $R_c=\frac{b}{a}=\frac{12}{1}=12\text{cm}.$

Why this step? $k=1$ is the razor's edge between growing and dying; solving for $R$ there gives the threshold — the geometric origin of critical mass.

Step 2 (b). $k(8)=\frac{1}{12}\times 8=0.667<1$ → subcritical (look at the blue point left of the dashed line in the figure). $k(20)=\frac{1}{12}\times 20=1.667>1$ → supercritical (green point, right of the line).

Why this step? The sign of $k-1$ is the entire classification. Left of $R_c$: dies. Right: explodes.

Step 3 (c) — limits.

• As $R\to \infty$: $k=\frac{a}{b}R\to \infty$. The leakage fraction $\propto R^2/R^3=1/R\to 0$; production totally wins. (In reality $k$ saturates once absorption caps it, but it never drops back below 1.)
• As $R\to 0$: $k\to 0$. All surface, no volume — every neutron escapes instantly. This is Ex 5's degenerate $k=0$ case, recovered as a geometric limit.

Why this step? Checking both extremes proves the model is monotonic in $R$: there is exactly one crossing at $R_c$, so critical mass is unique — not magic, just geometry.

Verify: $R_c=12$; $k(8)=0.667$, $k(20)=1.667$; ${\lim }_{R\to 0}k=0$ matches Ex 5. ✓

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Example 7 — Cell I: real-world word problem (burn-up)

Step 1. Fissions per second $=\frac{150\times 10^6}{3.2\times 10^{-11}}=4.6875\times 10^{18}$/s.

Why this step? Same power-÷-energy logic as Ex 2; "steady" means $k=1$, so power is constant.

Step 2. Seconds in a year $=3.156\times 10^7$. Total fissions $=4.6875\times 10^{18}\times 3.156\times 10^7=1.479\times 10^{26}$.

Why this step? Rate × total time $=$ total count. "Continuous" is the key word — no downtime factor.

Step 3. Mass $=1.479\times 10^{26}\times 3.90\times 10^{-25}\text{kg}\approx 57.7$ kg.

Why this step? Count × mass-per-nucleus. So under 60 kg of uranium powers a submarine for a year — the whole point of nuclear energy density.

Verify: $4.6875\times 10^{18}\times 3.156\times 10^7\times 3.90\times 10^{-25}\approx 57.7$ kg. Order of magnitude (tens of kg/year) is the standard textbook result. ✓

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Example 8 — Cell J: exam twist (compression and $M_c\propto 1/{\rho }^2$)

Step 1. Since $M_c\propto {\rho }^{-2}$, the ratio of critical masses is $\frac{M_c^{\prime }}{M_c}={\left(\frac{\rho }{{\rho }^{\prime }}\right)}^2={\left(\frac{1}{2.5}\right)}^2=\mathrm{0.16.}$

Why this step? Compressing shrinks $R$ (less leakage) and packs nuclei closer (neutrons hit sooner). Both effects push the threshold down; the combined scaling is the inverse-square law quoted in the parent.

Step 2. $M_c^{\prime }=0.16\times 10\text{kg}=1.6$ kg.

Why this step? The compressed metal only needs $1.6$ kg to go critical.

Step 3. The lump still contains all $10$ kg of plutonium — now far above its new $1.6$ kg threshold. So it is massively supercritical ($k>1$): this is exactly how an implosion device detonates a sub-critical mass.

Why this step? Same atoms, same total mass — only geometry/density changed. Critical mass is about arrangement, echoing Ex 6's geometric threshold.

Verify: $(1/2.5{)}^2=0.16$; $0.16\times 10=1.6$ kg $<10$ kg → supercritical. ✓

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Recall Self-check: match each to its cell

A single split releases 200 MeV — which cell? ::: Cell A (Ex 1) The number that decides bomb vs reactor vs dud ::: $k$; $k>1$, $=1$, $<1$ As $R\to 0$, what does $k$ become and which earlier example does that reproduce? ::: $k\to 0$, the degenerate one-shot of Ex 5 (Cell F) Why does compressing plutonium make a subcritical lump explode? ::: $M_c\propto 1/{\rho }^2$, so higher density lowers the threshold below the mass present (Cell J) A reactor burns roughly how much ${}^{235}$U per day at 3 GW? ::: About 3 kg (Ex 2)

Related: Nuclear Reactor · Neutron Cross-section · Radioactive Decay and Half-life · Nuclear Fusion · Binding Energy per Nucleon Curve