E4-2. Fission Chain Reactions and Critical Mass

1. The Mechanics of a Chain Reaction

Definition

  • When a U-235 nucleus undergoes induced fission, it releases 2 or 3 fast-moving neutrons.
  • If these newly born neutrons strike other U-235 nuclei, they induce further fission events, which release even more neutrons. This geometrically multiplying sequence is a nuclear chain reaction.
Uncontrolled Exponential Fission Chain Reaction Generation 1 Generation 2 Generation 3 U U U U U U U Uncontrolled Exponential Growth

The Multiplication Factor ($k$)

  • The effective neutron multiplication factor ($k$) is the ratio of fissions produced in one generation to the number of fissions in the preceding generation.
  • Subcritical ($k < 1$): The neutron population decays; the reaction dies out.
  • Critical ($k = 1$): The reaction is steady and self-sustaining (the goal of a nuclear power plant).
  • Supercritical ($k > 1$): The reaction grows exponentially (the mechanics of a nuclear weapon).

2. Critical Mass

  • Neutrons are created throughout the entire volume of a piece of uranium, but they can be lost by escaping through the surface into the environment.
  • Since Volume scales as $r^3$ and Surface Area scales as $r^2$, a physically larger sphere has a significantly lower Surface-to-Volume ratio than a small sphere.
  • Critical Mass: The minimum mass of fissile material required to sustain a chain reaction ($k \ge 1$). If the mass is below critical, too high a percentage of neutrons leak out through the relatively large surface area before they can strike another nucleus.

3. Examples

Example 1: Exponential Kinetics of a Supercritical Mass

Problem: In a supercritical mass of U-235, the multiplication factor is slightly above one, $k = 1.002$. The average time between one fission event and the neutrons causing the next generation of fission is $t_g = 1.0 \times 10^{-4}\text{ s}$ (the neutron generation time).
(a) If the reaction begins with a baseline of $N_0 = 100$ fissions, how many fission events will occur in a single generation exactly $1.0\text{ second}$ later?
(b) Why is thermal reactor control physically dependent on "delayed neutrons" (neutrons emitted seconds later by fission fragments) rather than these prompt neutrons?

Solution:

Part (a): Exponential Growth
Calculate the number of generations ($g$) that elapse in $1.0\text{ s}$: $$g = \frac{t_{\text{total}}}{t_g} = \frac{1.0\text{ s}}{1.0 \times 10^{-4}\text{ s/gen}} = 10,000\text{ generations}$$ The number of fissions $N$ after $g$ generations follows a geometric progression: $$N = N_0 \times k^g = 100 \times (1.002)^{10000}$$ Using a calculator: $1.002^{10000} \approx 4.79 \times 10^8$. $$N = 100 \times 4.79 \times 10^8 \approx \mathbf{4.8 \times 10^{10}\text{ fissions/generation}}$$

Part (b): The Necessity of Delayed Neutrons
If a reactor relied solely on prompt neutrons (which trigger the next fission in $\sim 10^{-4}\text{ s}$), human operators and mechanical control rods could not possibly move fast enough to stop a supercritical surge. Luckily, about $0.65\%$ of neutrons are emitted by radioactive fission fragments seconds or minutes after the initial split. By keeping the reactor strictly subcritical for prompt neutrons alone, but exactly critical when including delayed neutrons, the effective generation time is artificially extended to several seconds, allowing mechanical control rods ample time to safely throttle the power level.

Example 2: Volume-to-Surface Leakage Dynamics

Problem: In reactor design, neutron production is proportional to volume $V = 4\pi r^3/3$ and neutron leakage is proportional to surface area $A = 4\pi r^2$. A pure spherical chunk of Plutonium-239 will achieve criticality if the ratio of Volume to Surface Area ($V/A$) exceeds a critical threshold of $0.015\text{ m}$. Calculate the critical radius ($r_c$) of this sphere in cm, and compute the absolute critical mass if the density of Plutonium is $19,800\text{ kg/m}^3$.

Solution:

Step 1: Relate the $V/A$ ratio algebraically to the radius.
$$\frac{V}{A} = \frac{\frac{4}{3}\pi r^3}{4\pi r^2} = \frac{r}{3}$$ We are given that the critical ratio is $0.015\text{ m}$. $$\frac{r_c}{3} = 0.015\text{ m} \implies r_c = 3 \times 0.015 = 0.045\text{ m} = \mathbf{4.5\text{ cm}}$$

Step 2: Calculate the Critical Mass.
First, compute the geometric volume at the critical radius. $$V = \frac{4}{3}\pi (0.045)^3 = \frac{4}{3}\pi (9.1125 \times 10^{-5}) \approx 3.817 \times 10^{-4}\text{ m}^3$$ Now, multiply by the density to find the physical mass: $$M = \rho \times V = 19800\text{ kg/m}^3 \times (3.817 \times 10^{-4}\text{ m}^3) \approx \mathbf{7.56\text{ kg}}$$