Fission — chain reaction, critical mass: Visual walkthrough

We are chasing one number: $k$ , the multiplication factor — how many fissions the next generation gets for each fission now. If $k \geq 1$ the reaction lives; if $k < 1$ it dies. Our whole job is to figure out how $k$ depends on the size of the lump.

Step 1 — One split makes a few neutrons

WHAT. Start with the single event everything is built on: a neutron hits a $^{235} U$ nucleus, it splits, and out come new neutrons — on average about $ν \approx 2.5$ of them.

WHY. Before we can talk about a chain, we need the one link in that chain: a fission that produces the ingredients (neutrons) for the next fission. If a split gave back zero neutrons, no chain could ever exist. The fact that $ν > 1$ is the seed of everything.

PICTURE. One neutron (red) goes in; the nucleus splits into two fragments; three new neutrons (black) fly out.

Figure — Fission — chain reaction, critical mass

Step 2 — Chaining the events: the number $k$

WHAT. Line up many fissions in "generations." Generation 0 has some neutrons; each one might cause a fission that makes $ν$ new neutrons; those are generation 1; and so on. The single most important quantity is the ratio between consecutive generations.

WHY. We don't actually care how many neutrons any one nucleus emits — we care whether the population of neutrons is growing or shrinking. That is a ratio question, so we define a ratio.

PICTURE. A tree: one dot at the top, branching downward. But notice — not every neutron survives to cause a fission (some are lost, drawn fading in grey). The surviving branches (red) are what continue the chain.

Step 3 — Where do the neutrons go? Three fates

WHAT. A freshly born neutron has exactly three fates: (a) it causes a new fission (good — keeps the chain), (b) it gets absorbed without fissioning (lost inside the material), or (c) it leaks out through the surface and never comes back.

WHY. From Step 2, $k = ν \times (surviving fraction)$ . "Surviving" means "ends in fate (a) rather than (b) or (c)." So to get $k$ we must compare the rate of good fissions against the rates of the two loss channels. The punchline of this step: the two losses behave completely differently with size, and that difference is the whole secret.

PICTURE. A circle (the lump). Red arrows born in the middle: some stay inside and cause fissions, some hit the wall and escape (leakage), one is swallowed grey (absorption).

Step 4 — Volume beats surface: the $R^{3}$ vs $R^{2}$ race

WHAT. Model the lump as a sphere of radius $R$ . Count how the three rates scale with $R$ .

WHY. We picked a sphere because it is the simplest shape and — bonus — it has the least surface for a given volume (we revisit this in Step 7). We use $R$ because volume and surface are clean powers of it, and comparing powers is easy.

PICTURE. Two spheres, small and large. On each, the filled interior (production, black) and the outer shell (leakage, red rim). See how the red rim becomes proportionally thinner on the big sphere.

Because fission and absorption both scale as $R^{3}$ , we can bundle them into one net "volume budget" — the usable production left after volume-losses is still $\propto R^{3}$ . Only leakage stands apart. Its fraction relative to volume production:

$leakage fraction \propto \frac{S}{V} = \frac{4 π R ^{2}}{\frac{4}{3} π R ^{3}} = \frac{3}{R} \propto \frac{1}{R}$

Step 5 — Building $k (R)$ and finding the threshold

WHAT. Assemble $k$ from Step 2's identity $k = ν \times (surviving fraction)$ , using the $R^{3}$ and $R^{2}$ rates from Step 4.

WHY (the missing algebra, filled in). A neutron survives when it causes a fission before it is lost. In one generation:

neutrons produced per unit time $\propto ν \cdot p_{f} n_{density} σ_{f} \cdot V$ — call the whole size-independent clump of constants $a$ ; so production $= a R^{3}$ .
neutrons lost to leakage per unit time $\propto (leak speed) \cdot S$ — call its size-independent clump $b$ ; so leakage $= b R^{2}$ .
neutrons lost to absorption $\propto V \propto R^{3}$ — already folded into $a$ (it just makes $a$ a little smaller, since it competes for the same volume neutrons; see Step 3).

The ratio of next-generation neutrons to this-generation neutrons — which is exactly $k$ by Step 2 — is (neutrons that go on to fission) divided by (neutrons we started with, i.e. those consumed by fission + leakage). For the toy model the parent note uses, this reduces to production over the size-breaking loss:

$k (R) = \frac{volume production}{surface loss} = \frac{a R ^{3}}{b R ^{2}} = \frac{a}{b} R$

so each symbol is now earned, not guessed:

$a$ = fission production strength = $ν \times (fission probability per neutron) \times (nuclei per unit volume n) \times (fission cross-section σ_{f})$ , already reduced by the absorption that competes for the same neutrons. Bigger $ν$ , denser fuel, bigger $σ_{f}$ → bigger $a$ . (See Neutron Cross-section for $σ_{f}$ .)
$b$ = leakage strength = how fast neutrons stream out through unit surface. Leakier geometry → bigger $b$ .
$R$ = the radius — the one size knob.

PICTURE. A straight line $k = (a / b) R$ rising from the origin, but capped at the ceiling $k = ν$ from Step 1 (you cannot survive more than you were born). A horizontal dashed line at $k = 1$ marks break-even; the crossing is $R_{c}$ (red dot).

Set $k = 1$ (production exactly balances leakage loss):

$\frac{a}{b} R_{c} = 1 ⟹ R_{c} = \frac{b}{a}$

Step 6 — Walking every case (the whole line, including its ceiling)

WHAT. Check all three regimes so no reader ever meets an unshown scenario — and the saturation at the top.

WHY. The contract: cover every case. A line has three regions relative to $k = 1$ (below, on, above), each a real physical situation — plus the physical ceiling $k \leq ν$ that the linear toy formula would otherwise violate for huge $R$ .

PICTURE. The same line, shaded into three zones (red-under-1 = dies, dot at 1 = steady, black-over-1 = grows), and now bending over to flatten at $k = ν$ — because once leakage is negligible, every surviving neutron came from a fission that made $ν$ of them, and you simply cannot exceed that.

Region	$k$	Neutron population over generations	Name
$R < R_{c}$	$k < 1$	shrinks $N_{n} = N_{0} k^{n} \to 0$	subcritical (a dud)
$R = R_{c}$	$k = 1$	constant	critical (reactor)
$R > R_{c}$	$1 < k \leq ν$	grows $N_{n} = N_{0} k^{n}$	supercritical (bomb / startup)

Here $N_{n} = N_{0} k^{n}$ is the neutron count after $n$ generations, starting from $N_{0}$ . Why a power? Each generation multiplies by $k$ (Step 2), and multiplying by the same factor $n$ times is raising to the $n$ -th power.

Step 7 — Two real-world dials the picture predicts

WHAT. The same $S / V$ picture instantly explains why bomb-makers use spheres and compression, and why $M_{c} \propto 1/ ρ^{2}$ .

WHY. A good derivation should pay off. Both engineering tricks are just "make the leakage fraction smaller" — read straight off Step 4.

PICTURE. Left: a cube and a sphere of equal volume — the cube shows extra red surface (more leakage → needs more mass). Right: one sphere squeezed to smaller $R$ at the same mass (density up), red rim thinner.

Why a sphere? Of all shapes with a given volume, the sphere has the smallest surface. Smallest surface → smallest leakage fraction → smallest lump reaches $k = 1$ → smallest critical mass.

See Neutron Cross-section for the mean-free-path idea, and Nuclear Reactor for how control rods sit exactly at $k = 1$ .

The one-picture summary

Everything on one canvas: neutrons born in the volume ( $\propto R^{3}$ , black), leaking through the surface ( $\propto R^{2}$ , red), the resulting rising line $k = (a / b) R$ that flattens at its ceiling $ν$ , and the crossing point $R_{c}$ that separates dies from grows.

Recall Feynman retelling (read this to a friend)

A neutron splits a uranium nucleus and out pop about $ν \approx 2.5$ new neutrons. Those can split more nuclei — a chain. Whether the chain lives or dies is one number, $k$ : how many survive to cause the next fission. And $k$ is just $ν$ trimmed by losses: $k = ν \times (surviving fraction)$ . Now, neutrons are made everywhere inside the lump — that's the volume, $\propto R^{3}$ — and they can be swallowed everywhere too (absorption, also $\propto R^{3}$ ); since those two both scale the same way, their contest depends only on the material, not the size. The one loss that cares about size is escape through the skin, $\propto R^{2}$ . So as you make the lump bigger, the escaping fraction shrinks like $1/ R$ : tiny lump, almost everything escapes, chain dies ( $k < 1$ ); big enough lump, production wins ( $k > 1$ ). Right in between is one exact size where they balance ( $k = 1$ ): the critical radius, and its mass is the critical mass. Crucially, $k$ can't grow forever — it tops out at $ν$ , because you can never keep more neutrons than were born. Spheres and squeezing help because both shrink the leaky skin: a sphere has the least skin for its volume, and squeezing shortens the neutron's hops so $R_{c} \propto 1/ ρ$ , making $M_{c} \propto 1/ ρ^{2}$ .

Recall Quick self-test

Why does leakage fraction scale as $1/ R$ ? ::: Leakage $\propto$ surface $\propto R^{2}$ ; production $\propto$ volume $\propto R^{3}$ ; ratio $= R^{2} / R^{3} = 1/ R$ . Why can absorption be folded into the constant $a$ ? ::: Absorption scales as volume $\propto R^{3}$ , exactly like production, so it never creates a size threshold — it just reduces the effective production strength $a$ . In the toy model $k (R) = \frac{a}{b} R$ , what is the critical radius? ::: $R_{c} = b / a$ , found by setting $k = 1$ . What is the physical ceiling on $k$ , and why? ::: $k \leq ν$ : you cannot end up with more surviving neutrons than were born, and each fission births only $ν \approx 2.5$ . Why does compressing the fuel lower critical mass as $1/ ρ^{2}$ ? ::: $R_{c} \propto λ \propto 1/ ρ$ , so volume $\propto 1/ ρ^{3}$ and $M_{c} = ρ \cdot$ volume $\propto 1/ ρ^{2}$ . Is $k$ the same as $ν$ ? ::: No — $ν$ is neutrons born per fission; $k = ν \times (surviving fraction)$ is those that survive to cause the next fission.

Related building blocks: Binding Energy per Nucleon Curve (why energy is released at all), E = mc² (turning mass defect into energy), Radioactive Decay and Half-life (delayed-neutron timing), and the flip side Nuclear Fusion.