Fission — chain reaction, critical mass: Exercises

Before we begin, one figure fixes the geometry that half these problems lean on — the tug-of-war between neutron production (grows with volume) and neutron leakage (grows with surface).

Figure — Fission — chain reaction, critical mass

Look at the two coloured curves: the burnt-orange (production, $\propto R^{3}$ ) starts below the teal (leakage $\propto R^{2}$ ) for a small ball, then overtakes it. Where they cross, $k = 1$ — that crossing radius is the critical radius.

Level 1 — Recognition

Problem 1.1

State, in one line each, what happens to a lump of fissile material when (a) $k = 0.8$ , (b) $k = 1.0$ , (c) $k = 1.3$ .

Recall Solution

$k$ = (neutrons in the new generation) ÷ (neutrons in the old generation).

(a) $k = 0.8 < 1$ : each generation is smaller → the chain dies out (subcritical).
(b) $k = 1.0$ : every generation the same size → steady, self-sustaining (critical — a reactor's normal state).
(c) $k = 1.3 > 1$ : each generation bigger → exponential growth (supercritical — bomb / reactor start-up).

Problem 1.2

A fission event releases about $200$ MeV. Roughly how many neutrons does one fission emit on average, and why does that number (not the energy) decide whether a chain reaction is possible?

Recall Solution

On average $ν \approx 2.5$ neutrons per fission (the reaction releases $2$ – $3$ ). The energy heats things up, but it is the emitted neutrons that go on to trigger the next fission. A chain needs, on average, at least one surviving neutron per fission to hit a new nucleus. So survival of neutrons — not joules released — is what sustains the chain.

Level 2 — Application

Problem 2.1

A reactor runs at $P = 300$ MW of thermal power. Each fission releases $200$ MeV. How many fissions happen each second?

Recall Solution

Step 1 — one fission in joules. Why? Power is joules per second, so energy-per-event must also be in joules. $E = 200 MeV = 200 \times 1.6 \times 1 0^{- 13} J = 3.2 \times 1 0^{- 11} J .$ Step 2 — divide. Why? (total energy rate) ÷ (energy per event) = events per second. $fissions/s = \frac{300 \times 1 0 ^{6}}{3.2 \times 1 0 ^{- 11}} = 9.375 \times 1 0^{18} s^{- 1} .$

Problem 2.2

Starting with $N_{0} = 1000$ neutrons and $k = 1.02$ , how many neutrons remain after $50$ generations?

Recall Solution

Step 1 — the growth law. Each generation multiplies by $k$ , so after $n$ generations $N_{n} = N_{0} k^{n}$ . Step 2 — plug in. Why raise to a power? Repeated multiplication by the same factor is exponentiation. $N_{50} = 1000 \times 1.0 2^{50} = 1000 \times 2.6916 \approx 2692 neutrons .$

Problem 2.3

$1$ kg of $^{235} U$ is completely fissioned. How much energy (in joules) is released? (Take $200$ MeV per fission, molar mass $235$ g/mol.)

Recall Solution

Step 1 — how many nuclei? Why? Energy = (number of fissions) × (energy each), and every nucleus fissions once. $N = \frac{1000 g}{235 g/mol} \times 6.022 \times 1 0^{23} = 2.5626 \times 1 0^{24} nuclei .$ Step 2 — total energy. $E = N \times 3.2 \times 1 0^{- 11} J = 2.5626 \times 1 0^{24} \times 3.2 \times 1 0^{- 11} \approx 8.20 \times 1 0^{13} J .$ That is about $20$ kilotonnes of TNT from one kilogram — the scale of a fission weapon.

Level 3 — Analysis

Problem 3.1

The mass of one $^{235} U$ atom plus the incoming neutron exceeds the total mass of the products by $Δ m = 0.2154 u$ . Find the energy $Q$ released and compare with the " $200$ MeV" rule of thumb.

Recall Solution

Step 1 — mass-energy equivalence. Why $E = m c^{2}$ ? Fission converts a tiny lost mass (the "mass defect") into energy; see Mass-Energy Equivalence E=mc^2. $Q = Δ m \cdot c^{2} = 0.2154 \times 931.5 MeV \approx 200.6 MeV .$ Step 2 — compare. This matches the $\approx 200$ MeV estimate we got from the Binding Energy per Nucleon Curve ( $0.9$ MeV/nucleon $\times 235$ ). Two independent routes — mass defect and binding-energy difference — agree, confirming the number.

Problem 3.2

A reactor sits exactly at $k = 1$ . An operator withdraws a control rod, jumping $k$ to $1.01$ . The effective generation time (dominated by delayed neutrons) is $τ = 0.1$ s. How long until the neutron population doubles?

Recall Solution

Step 1 — generations to double. From $N_{n} = N_{0} k^{n}$ , set the ratio to $2$ : $k^{n} = 2 \Rightarrow n = \frac{l n 2}{l n k} = \frac{0.6931}{l n 1.01} = \frac{0.6931}{0.009950} \approx 69.7 generations .$ Why take logs? The unknown $n$ sits in the exponent; the logarithm is the tool that pulls an exponent down to solve for it. Step 2 — convert to time. Why multiply by $τ$ ? Each generation lasts $τ$ seconds. $t = n τ \approx 69.7 \times 0.1 \approx 7.0 s .$ Seven seconds is comfortably controllable — that is the gift of delayed neutrons. With prompt-only neutrons ( $τ \sim 1 0^{- 8}$ s) the same $k$ would double in under a microsecond. See Nuclear Reactor.

Problem 3.3

Using the toy model $k (R) = \frac{a}{b} R$ (production $\propto R^{3}$ , surface leakage $\propto R^{2}$ ), a sphere of radius $R = 8$ cm has $k = 0.8$ . Find the critical radius $R_{c}$ .

Recall Solution

Step 1 — pin down the constant. Why? $k$ is linear in $R$ in this toy model, so one data point fixes the slope $a / b$ . $\frac{a}{b} = \frac{k}{R} = \frac{0.8}{8 cm} = 0.1 cm^{- 1} .$ Step 2 — set $k = 1$ . Why? Critical means production exactly balances leakage. $1 = \frac{a}{b} R_{c} = 0.1 R_{c} \Rightarrow R_{c} = 10 cm .$ Our $8$ cm ball is below $R_{c}$ , consistent with its subcritical $k = 0.8$ .

Level 4 — Synthesis

Problem 4.1

Critical mass scales as $M_{c} \propto 1/ ρ^{2}$ with density $ρ$ . An implosion device compresses a plutonium core to $2.5 \times$ its normal density. By what factor does its critical mass drop, and why does this turn a subcritical lump supercritical?

Recall Solution

Step 1 — apply the scaling. Why $1/ ρ^{2}$ ? Higher density packs nuclei closer, so a neutron travels a shorter distance before it hits one (smaller mean free path); the geometry that sets leakage shrinks in both the smaller radius and the tighter packing — two factors of $ρ$ . $\frac{M _{c} ( compressed )}{M _{c} ( normal )} = \frac{1}{2. 5 ^{2}} = \frac{1}{6.25} = 0.16.$ So the critical mass falls to $16%$ of its original value. Step 2 — why it goes supercritical. The actual mass of the core is unchanged. If that fixed mass was just below the old critical mass, then slashing the critical mass to $16%$ leaves the fixed mass now sitting well above the new threshold → $k > 1$ → supercritical. This is the whole trick of an implosion bomb.

Problem 4.2

A sample of $^{235} U$ also undergoes ordinary radioactive decay with half-life $7.0 \times 1 0^{8}$ years. Suppose instead of decaying, each nucleus underwent chain fission with an effective doubling time of $1 0^{- 8}$ s in a supercritical assembly. Contrast the two timescales by finding how many doublings occur in one microsecond, versus the fraction of nuclei that decay by ordinary radioactivity in that same microsecond.

Recall Solution

Step 1 — chain doublings in $1 μ$ s. Why divide? Number of doublings = (elapsed time) ÷ (doubling time). $\frac{1 0 ^{- 6} s}{1 0 ^{- 8} s} = 100 doublings \Rightarrow growth factor 2^{100} \approx 1.27 \times 1 0^{30} .$ Step 2 — ordinary decay in $1 μ$ s. Why the exponential-decay law? Radioactivity follows $N = N_{0} e^{- λ t}$ with $λ = ln 2/ T_{1/2}$ ; see Radioactive Decay and Half-life. $λ = \frac{0.6931}{7.0 \times 1 0 ^{8} \times 3.156 \times 1 0 ^{7} s} \approx 3.14 \times 1 0^{- 17} s^{- 1} .$ Fraction decayed $\approx λ t = 3.14 \times 1 0^{- 17} \times 1 0^{- 6} \approx 3.1 \times 1 0^{- 23}$ — utterly negligible. Conclusion: In the same microsecond, a chain reaction multiplies by $\sim 1 0^{30}$ while natural decay barely touches one part in $1 0^{23}$ . Chain fission is stupendously faster — that is why an assembled critical mass releases its energy essentially instantaneously.

Level 5 — Mastery

Problem 5.1

A "gun-type" weapon slams two subcritical $^{235} U$ pieces together to form a supercritical mass with $k = 1.05$ and prompt generation time $τ = 1 0^{- 8}$ s. Starting from a single stray neutron, estimate the time for the population to reach $1 0^{20}$ neutrons (enough to fission a significant fraction of the fuel).

Recall Solution

Step 1 — real time from the growth law. In real time, $N (t) = N_{0} k^{t / τ}$ . Set $N_{0} = 1$ , $N (t) = 1 0^{20}$ : $1 0^{20} = 1.0 5^{t / τ} .$ Step 2 — take logs to free the exponent. Why logs again? Same reason as L3: the unknown is trapped in the exponent. $\frac{t}{τ} = \frac{l n 1 0 ^{20}}{l n 1.05} = \frac{20 \times 2.3026}{0.04879} = \frac{46.05}{0.04879} \approx 943.8 generations .$ Step 3 — convert. $t = 943.8 \times 1 0^{- 8} s \approx 9.44 \times 1 0^{- 6} s \approx 9.4 μ s .$ Under ten microseconds from one neutron to full-scale energy release — no control system on Earth could intervene. This is exactly why a bomb deliberately uses fast (unmoderated) neutrons and $k > 1$ , while a reactor leans on delayed neutrons and holds $k = 1$ .

Problem 5.2

Design-reasoning question. You are handed $60$ kg of $90%$ -enriched $^{235} U$ (bare-sphere critical mass $\approx 52$ kg). Explain, using surface-to-volume and cross-section reasoning, three independent ways to make this assembly more reactive without adding fuel, and one way to make it safely subcritical.

Recall Solution

More reactive (raise $k$ ):

Compress it (raise density). $M_{c} \propto 1/ ρ^{2}$ : squeezing lowers the critical mass, so our fixed $60$ kg sits further above threshold. Physically, a shorter mean free path means neutrons find nuclei before leaking (Problem 4.1).
Wrap it in a neutron reflector / tamper. A tamper bounces would-be-escaping neutrons back into the core, cutting the leakage term ( $\propto R^{2}$ ). Fewer neutrons lost at the surface → higher $k$ , lower effective critical mass.
Shape it as a sphere. A sphere has the smallest surface area for its volume, minimising leakage; any non-spherical shape (slab, rod) has more surface per unit volume and leaks more. See the crossing point in the figure above.

(Cross-section angle:) slowing neutrons with a moderator raises the fission Neutron Cross-section of $^{235} U$ , another route to more fissions per neutron — though a fast weapon deliberately avoids this to stay prompt.

Safely subcritical:

Split the mass into separated sub-pieces (or flatten it into a thin slab). Breaking the sphere into two distant lumps, each below $52$ kg, drives the surface-to-volume ratio of each piece way up → leakage dominates → $k < 1$ for each. This is precisely how fissile material is stored.

Why it all reduces to one idea: every lever here is either raising production (density, cross-section) or cutting leakage (reflector, spherical shape) — or, for safety, deliberately maximising leakage (splitting, flattening). Reactivity is the ledger of production versus loss.