Fission — chain reaction, critical mass, reactors (thermal vs fast): Worked examples

The scenario matrix

Think of the topic as a machine with a few dials. Each row below is a dial setting — a distinct type of problem. If we solve one example per row, you've seen the whole machine.

#	Cell (scenario class)	The core question it tests	Covered by
A	Balancing a fission equation	Do mass number $A$ and charge $Z$ conserve?	Ex 1
B	Energy from mass defect ( $Δ m$ given)	Turn missing mass into MeV	Ex 2
C	Energy from binding energy (BE/nucleon given)	Turn a binding-energy climb into MeV	Ex 3
D	Bulk / real-world energy (per kg, "how much coal?")	Scale one fission up to a fuel mass	Ex 4
E	$k < 1$ subcritical — reaction dies	Track neutrons over generations	Ex 5
F	$k > 1$ supercritical — reaction grows	Same maths, opposite sign of $(k - 1)$	Ex 6
G	$k = 1$ critical (degenerate) — steady state	The knife-edge: nothing grows or dies	Ex 6
H	Geometry / critical mass (surface vs volume)	Why size flips $k$ across 1	Ex 7
I	Breeder transmutation chain	Follow $Z, A$ through neutron capture + $β^{-}$	Ex 8
J	Exam twist — moderator collision physics	Why light nuclei slow neutrons best	Ex 9

The three signs of $k - 1$ (rows E, F, G) are the fission analogue of "every quadrant" — we cover negative, positive, and zero so no case is left out.

Ex 1 — Cell A: Balancing a fission equation

Worked example Fill in the missing fragment

$_{92}^{235} U +_{0}^{1} n ⟶_{56}^{144} Ba +_{?}^{?} X + 2_{0}^{1} n$ Find $A$ , $Z$ , and the element X.

Forecast: Guess the atomic number first — it must land near krypton ( $Z = 36$ ). Write your guess down.

Step 1 — Conserve mass number $A$ (the top numbers). Why this step? $A$ counts protons + neutrons (nucleons). In fission nucleons are only rearranged, never created or destroyed, so the totals on each side must match. $235 + 1 = 144 + A + 2 (1) \Rightarrow 236 = 146 + A \Rightarrow A = 90$

Step 2 — Conserve atomic number $Z$ (the bottom numbers). Why this step? $Z$ counts protons, i.e. total charge. Charge is conserved, so bottom numbers must balance too. $92 + 0 = 56 + Z + 2 (0) \Rightarrow Z = 36$

Step 3 — Name the element. $Z = 36$ is krypton, so $X =_{36}^{90} Kr$ .

Verify: Left totals $A = 236, Z = 92$ . Right totals $A = 144 + 90 + 2 = 236$ ✔, $Z = 56 + 36 + 0 = 92$ ✔. Both sides identical.

Ex 2 — Cell B: Energy straight from mass defect

The tool we reach for is $E=\Delta m\,c^2$. Why this tool and not simple arithmetic? Because the products weigh less than the reactants, and that missing mass $Δ m$ (call it the mass defect) is exactly the fuel — Einstein's equation is the only bridge from "grams" to "joules of energy."

Ex 3 — Cell C: Energy from a binding-energy climb

Now the data is different: instead of $Δ m$ we're handed binding energy per nucleon. This links to Binding Energy & Mass Defect. Why is this the right tool? Binding energy per nucleon is "how tightly glued each nucleon is." Products near the iron peak are more glued; the extra glue is the released energy — no need to look up masses.

Ex 4 — Cell D: Real-world word problem (energy per kg)

Worked example How much coal does 1 gram of

$^{235} U$ replace? Take $200 MeV$ per fission, and coal $\approx 3 \times 1 0^{7} J/kg$ . Constants: $N_{A} = 6.022 \times 1 0^{23}$ , $1 MeV = 1.602 \times 1 0^{- 13} J$ .

Forecast: Nuclear is millions of times denser than chemical fuel. Guess: 1 g $\approx$ tonnes of coal.

Step 1 — Count atoms in 1 g. Why this step? Each U-235 atom is one potential fission; we need how many we have. $N = \frac{1}{235} \times 6.022 \times 1 0^{23} = 2.563 \times 1 0^{21} atoms$

Step 2 — Joules per fission. Why this step? Convert the per-fission energy out of MeV into SI joules so it can meet coal's J/kg. $200 MeV = 200 \times 1.602 \times 1 0^{- 13} = 3.204 \times 1 0^{- 11} J$

Step 3 — Total energy from 1 g. $E = 2.563 \times 1 0^{21} \times 3.204 \times 1 0^{- 11} = 8.21 \times 1 0^{10} J$

Step 4 — Convert to an equivalent mass of coal. $m_{coal} = \frac{8.21 \times 1 0 ^{10}}{3 \times 1 0 ^{7}} \approx 2.74 \times 1 0^{3} kg \approx 2.7 tonnes$

Verify: Units: $J \div (J/kg) = kg$ ✔. So 1 g of U-235 ≈ 2.7 tonnes of coal — the "million-fold" density we forecast. ✔

Ex 5 — Cell E: Subcritical, $k < 1$ (reaction dies)

The tool is the parent's generation formula $N_{n} = N_{0} k^{n}$ , where $k$ = multiplication factor = (neutrons causing fission this generation) ÷ (last generation). Why powers of $k$ ? Because each generation multiplies by the same factor $k$ — repeated multiplication is an exponent. Compare with Radioactive Decay & Half-life, where a fixed fraction survives each half-life; same "repeated fraction" logic.

Ex 6 — Cells F & G: Supercritical ( $k > 1$ ) and Critical ( $k = 1$ )

Same formula, opposite sign of $(k - 1)$ . We do both signs and the zero case in one shot — the "all quadrants" requirement for $k$ .

Worked example Two cores from the same

$N_{0} = 1000$ : (a) $k = 1.02$ , (b) $k = 1$ . Compare after $n = 100$ generations. Forecast: (a) grows a little each round — but 100 rounds compounds. (b) stays put forever. Guess (a)'s final number.

Step 1 (case F, $k > 1$ ) — apply the formula. Why this step? $k > 1$ is supercritical growth; compounding over 100 generations can be dramatic even for tiny excess. $N_{100} = 1000 \times 1.0 2^{100}$

Step 2 — Evaluate. $1.0 2^{100} \approx 7.245 \Rightarrow N_{100} \approx 7245$ A mere 2% excess per generation multiplies the population $\sim$ 7-fold in 100 steps — this is why runaway is so fast.

Step 3 (case G, $k = 1$ ) — the degenerate knife-edge. Why this step? $k = 1$ is the reactor's operating point; test what "steady" means mathematically. $N_{100} = 1000 \times 1^{100} = 1000$ Nothing grows, nothing dies — exactly steady.

Verify: Sign logic — $k > 1 \Rightarrow N$ up ( $7245 > 1000$ ) ✔; $k = 1 \Rightarrow N$ unchanged ( $= 1000$ ) ✔; and Ex 5's $k < 1 \Rightarrow N$ down. All three signs of $(k - 1)$ behave as defined. ✔

Ex 7 — Cell H: Geometry — surface, volume, and critical mass

This is the geometric cell, so it gets figures. The idea: fissions (neutron production) fill the volume $\propto r^{3}$ ; leakage happens through the surface $\propto r^{2}$ . The ratio surface/volume $\propto 1/ r$ decides whether $k$ crosses 1.

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

Worked example Compare neutron leakage for a small sphere (

$r = 1$ ) and a big one ( $r = 2$ ) Use the surface-to-volume ratio $\frac{S}{V} = \frac{4 π r ^{2}}{\frac{4}{3} π r ^{3}} = \frac{3}{r}$ as a leakage proxy.

Forecast: Bigger sphere → relatively less surface → less leakage. Guess by how much the ratio drops when $r$ doubles.

Step 1 — Derive the ratio. Why this step? Only the ratio (leakage vs production), not the raw areas, tells us if enough neutrons stay inside. $\frac{S}{V} = \frac{4 π r ^{2}}{\frac{4}{3} π r ^{3}} = \frac{3}{r}$

Step 2 — Plug both radii. $r = 1 : \frac{S}{V} = 3.0 r = 2 : \frac{S}{V} = 1.5$

Step 3 — Interpret. Doubling $r$ halves the leakage-per-fission proxy. Grow the lump enough and leakage falls below production → $k$ climbs past 1 → the critical mass is reached.

Verify: $\frac{3}{r}$ at $r = 1$ is $3$ , at $r = 2$ is $1.5$ ; ratio of ratios $= 2$ ✔ — leakage proxy exactly halves as $r$ doubles. This is why a sphere below critical size ( $k < 1$ ) dies but a larger one ( $k \geq 1$ ) sustains. ✔

Ex 8 — Cell I: Breeder transmutation chain

Links to Nuclear Reactor Safety & Waste and the parent's breeder box. Why track $Z$ and $A$ separately? Because a neutron capture and a $β^{-}$ decay change them in different, predictable ways — capture adds a nucleon, $β^{-}$ swaps a neutron for a proton.

Worked example Follow

$_{92}^{238} U$ absorbing one neutron, then two $β^{-}$ decays. End nucleus? Recall $β^{-}$ decay: a neutron becomes a proton, emitting an electron — so $Z$ increases by 1, $A$ unchanged (see Radioactive Decay & Half-life).

Forecast: Start at $Z = 92$ (U). Two $β^{-}$ steps push $Z$ up by 2. Guess the final element.

Step 1 — Neutron capture. Why this step? Adding one neutron raises $A$ by 1, leaves $Z$ alone (neutrons carry no charge). $_{92}^{238} U +_{0}^{1} n \to_{92}^{239} U$

Step 2 — First $β^{-}$ . $Z : 92 \to 93$ , $A$ stays $239$ . $_{92}^{239} U β^{-}_{93}^{239} Np$

Step 3 — Second $β^{-}$ . $Z : 93 \to 94$ , $A$ stays $239$ . $_{93}^{239} Np β^{-}_{94}^{239} Pu$

Verify: Mass number never changes after the initial capture: $239$ throughout ✔. Charge climbs $92 \to 93 \to 94$ ✔. End product $_{94}^{239} Pu$ — fissile plutonium, more fuel than we started with, which is exactly what a breeder does. ✔

Ex 9 — Cell J: Exam twist — moderator collision physics

The parent claimed a moderator uses light nuclei to slow neutrons best, "like billiard balls." Let's prove that number, borrowing head-on elastic-collision physics.

Worked example A neutron (mass

$m$ ) hits a stationary nucleus (mass $M$ ) head-on and elastically. What fraction of energy does it lose to hydrogen ( $M = m$ ) vs carbon ( $M = 12 m$ )? The fraction of kinetic energy transferred in a head-on elastic collision is $f = \frac{4 m M}{( m + M ) ^{2}}$ Forecast: A neutron should dump the most energy into a nucleus close to its own mass. Guess which of H or C strips more.

Step 1 — Hydrogen, $M = m$ . Why this step? Equal masses is the "billiard ball hits billiard ball" limit — expect maximum transfer. $f_{H} = \frac{4 m \cdot m}{( m + m ) ^{2}} = \frac{4 m ^{2}}{4 m ^{2}} = 1$ A neutron can lose all its energy to a proton in one head-on hit.

Step 2 — Carbon, $M = 12 m$ . Why this step? A heavier target barely recoils — like hitting a bowling ball — so transfer should be small. $f_{C} = \frac{4 m \cdot 12 m}{( m + 12 m ) ^{2}} = \frac{48 m ^{2}}{169 m ^{2}} = \frac{48}{169} \approx 0.284$

Step 3 — Compare. $f_{H} = 1 ≫ f_{C} \approx 0.28$ . Light nuclei slow neutrons fastest, confirming why $H_{2} O$ and $D_{2} O$ are excellent moderators while heavy nuclei are useless.

Verify: $f = \frac{4 m M}{( m + M ) ^{2}}$ is maximised at $M = m$ where $f = 1$ ✔; $f_{C} = 48/169 \approx 0.284$ ✔. Matches the parent's billiard-ball intuition. ✔

Wrap-up

Recall Did every cell get covered?

Balancing (Ex1) ::: Cell A ✔ Energy from $Δ m$ (Ex2) ::: Cell B ✔ Energy from BE/nucleon (Ex3) ::: Cell C ✔ Per-kg / coal comparison (Ex4) ::: Cell D ✔ $k < 1$ , $k > 1$ , $k = 1$ (Ex5, Ex6) ::: Cells E, F, G ✔ Surface/volume & critical mass (Ex7) ::: Cell H ✔ Breeder chain (Ex8) ::: Cell I ✔ Moderator collision (Ex9) ::: Cell J ✔

Related deep reading: Binding Energy & Mass Defect · Nuclear Fusion · Neutron Cross-section · Einstein Mass-Energy Equivalence · back to the parent topic.