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Chapter 6 • Theory & Derivations

Neutron Chain Reactions, Multiplication & The Four/Six-Factor Formulas

Exhaustive treatment of self-sustaining nuclear fission chain reactions and critical multiplication physics: subcritical, critical, and supercritical operating regimes; infinite medium neutron life cycle and detailed derivation of the classical Four-Factor Formula k_infinity = epsilon * p * eta * f; the fast fission factor epsilon in uranium lattices; resonance escape probability p and Doppler resonance temperature dependence; thermal utilization factor f and competitive absorption balance; reproduction factor eta and the fissile figure of merit; finite core geometry leakage mechanics, fast non-leakage probability P_FNL, thermal non-leakage probability P_TNL, and the complete Six-Factor Formula k_eff = k_infinity * P_FNL * P_TNL; and homogeneous vs heterogeneous lattice design principles.

§6.1 The Self-Sustaining Nuclear Fission Chain Reaction: Critical States & Multiplications

1. The Multiplication Factor $k$

The central parameter governing any nuclear multiplying assembly is the effective multiplication factor $k_{\text{eff}}$ (or simply $k$), defined as the ratio of neutrons produced by fission in generation $n+1$ to the total neutrons lost by absorption and leakage in generation $n$: $$\mathbf{k_{\text{eff}} \equiv \frac{\text{Neutrons produced in generation } n+1}{\text{Neutrons lost (absorption + leakage) in generation } n} = \frac{\text{Production Rate}}{\text{Loss Rate}}}$$ Three distinct physical regimes exist:

  • Subcritical ($k_{\text{eff}} < 1$): Loss rate exceeds production rate. A chain reaction cannot sustain itself; neutron population decays exponentially to zero (or to a steady subcritical multiplier equilibrium supported by an external source).
  • Critical ($k_{\text{eff}} = 1.00000$): Production rate perfectly balances total losses. The neutron population and thermal power remain strictly constant in time. This is the normal operating state of all commercial nuclear power reactors.
  • Supercritical ($k_{\text{eff}} > 1$): Production exceeds losses. The neutron population grows exponentially in time.

2. Static Reactivity $\rho$

The fractional departure of a reactor from the exact critical state is the reactivity $\rho$: $$\mathbf{\rho \equiv \frac{k_{\text{eff}} - 1}{k_{\text{eff}}} = \frac{\Delta k}{k}}$$ Units of reactivity:

  • Dimensionless decimal ($\Delta k/k$).
  • Percent ($\% \Delta k/k$): $1\% = 10^{-2}$.
  • Percent Mille ($\text{pcm}$): $1\text{ pcm} = 10^{-5} = 0.001\%$. A change of $100\text{ pcm}$ is $0.1\% \Delta k/k$.
  • Dollars ($\$$) and Cents ($\cancel{\text{c}}$): Normalized by the effective delayed neutron fraction $\beta$: $$\rho\ [\$] \equiv \frac{\rho}{\beta_{\text{eff}}}, \qquad 1\$ = 100\cancel{\text{c}}$$

§6.2 Infinite Medium Neutron Life Cycle & The Four-Factor Formula

1. Tracking 1000 Neutrons Through One Generation

In an infinite multiplying medium (where leakage is zero by definition), every neutron lost is lost by absorption. Enrico Fermi and his team formulated the classic Four-Factor Formula to track the life cycle of neutrons from birth to the next generation: $$\mathbf{k_\infty = \epsilon \, p \, \eta \, f}$$ Let $N_0$ thermal neutrons be absorbed in fuel nuclei in generation $n$:

  1. Thermal Fission and Fast Birth: Absorbing $N_0$ neutrons in fuel produces $\eta N_0$ fast fission neutrons ($E \sim 2\text{ MeV}$).
  2. Fast Fission Boost ($\epsilon$): Before slowing down, some fast neutrons induce fast fission in fertile ${}^{238}\text{U}$, boosting the population by factor $\epsilon \ge 1$ to $\epsilon \eta N_0$ fast neutrons.
  3. Resonance Slowing Down ($p$): As these neutrons slow down through the epithermal resonance region, a fraction $1 - p$ are captured in ${}^{238}\text{U}$ resonances. The surviving population reaching thermal energy is $p \, \epsilon \eta N_0$.
  4. Thermal Absorption Competition ($f$): At thermal energy, neutrons are absorbed either in fuel or in non-fuel materials (moderator, clad, coolant, poisons). The fraction absorbed in fuel is $f$, yielding $f \, p \epsilon \eta N_0$ thermal neutrons absorbed in fuel in generation $n+1$.
Taking the ratio of absorbed fuel neutrons between generations: $$k_\infty = \frac{N_{n+1}}{N_n} = \frac{\epsilon \, p \, \eta \, f \, N_0}{N_0} = \mathbf{\epsilon \, p \, \eta \, f}$$

§6.3 Fast Fission Factor Epsilon: Fission in Uranium-238

1. Fast Fission in Fertile Material

Although ${}^{238}\text{U}$ cannot fission from thermal neutrons, it possesses a significant fission cross section ($\sigma_f \approx 0.55\text{ b}$) for fast neutrons with energies above $E_{\text{th}} \approx 1.0\text{ MeV}$. Since approximately $69\%$ of prompt fission neutrons are born above $1.0\text{ MeV}$, some collide with ${}^{238}\text{U}$ before escaping the fuel rod, producing additional second-generation fast neutrons. The fast fission factor $\epsilon$ is defined as: $$\mathbf{\epsilon \equiv \frac{\text{Total fast neutrons born from all fission (thermal + fast)}}{\text{Fast neutrons born from thermal fission alone}}}$$

2. Quantitative Lattice Values

In a homogeneous mixture of natural uranium and graphite, $\epsilon \approx 1.000$ because fast neutrons immediately collide with carbon atoms and drop below $1\text{ MeV}$. In a heterogeneous reactor lattice with tightly packed fuel rods (such as a PWR assembly): $$\epsilon \approx 1.03\text{ to }1.08$$ This represents a crucial $3\%\text{ to }8\%$ boost to the core neutron population obtained completely free from fertile ${}^{238}\text{U}$!

§6.4 Resonance Escape Probability p: Resonance Capture & Doppler Broadening

1. Epithermal Capture in Uranium-238

As neutrons slow through the energy interval between $10\text{ keV}$ and $1\text{ eV}$, they encounter colossal capture resonances in ${}^{238}\text{U}$, such as the famous resonance at $E_0 = 6.67\text{ eV}$ where peak cross section exceeds $\sigma_\gamma > 20{,}000\text{ barns}$! The resonance escape probability $p$ is the fraction of fast neutrons that successfully escape resonance capture during moderation: $$\mathbf{p = \exp\left( - \frac{N_F}{\overline{\xi \Sigma_s}} I_{\text{eff}} \right)}$$ Typical values in thermal power reactors range from $p \approx 0.75\text{ to }0.90$.

2. Doppler Broadening and Inherent Safety

Thermal agitation of fuel atoms causes relative velocity motion between target nuclei and incoming neutrons. By Breit-Wigner theory, as fuel temperature increases:

  • The resonance peak height decreases.
  • The resonance width $\Gamma$ broadens (Doppler broadening).
  • Because resonances are self-shielded, broadening allows more neutrons to be absorbed in the resonance wings!
Consequently, an increase in fuel temperature decreases $p$: $$\frac{dp}{dT_{\text{fuel}}} < 0 \implies \alpha_D \equiv \frac{1}{k} \frac{dk}{dT_{\text{fuel}}} < 0$$ This negative Fuel Doppler Temperature Coefficient is the foundational passive safety mechanism of all commercial nuclear power reactors, guaranteeing that an accidental power excursion terminates itself within milliseconds without human or mechanical intervention!

§6.5 Thermal Utilization Factor f: Fuel vs Parasitic Absorption Balance

1. Definition of Thermal Utilization $f$

Once neutrons reach thermal equilibrium, they diffuse through the lattice until absorbed. The thermal utilization factor $f$ is the fraction of thermal neutrons absorbed in the nuclear fuel ($F$) relative to total absorption in all core materials: $$\mathbf{f \equiv \frac{\text{Thermal neutrons absorbed in fuel}}{\text{Total thermal neutrons absorbed in all core materials}} = \frac{\Sigma_a^F \phi_F V_F}{\Sigma_a^F \phi_F V_F + \Sigma_a^M \phi_M V_M + \Sigma_a^{\text{struct}} \phi_{\text{st}} V_{\text{st}}}}$$ where $F = \text{fuel}$, $M = \text{moderator}$, and $\text{struct} = \text{cladding, coolant, and structural poisons}$.

2. Homogeneous vs Heterogeneous Expressions

For an intimate homogeneous mixture where flux is spatially uniform ($\phi_F = \phi_M$): $$\mathbf{f = \frac{\Sigma_a^F}{\Sigma_a^F + \Sigma_a^M + \Sigma_a^{\text{other}}} = \frac{N_F \sigma_a^F}{N_F \sigma_a^F + N_M \sigma_a^M}}$$ Typical values: $f \approx 0.85\text{ to }0.95$ in enriched reactors. There is a fundamental design trade-off between $p$ and $f$:

  • Adding more moderator increases $p$ (more moderation means faster crossing of resonances), but decreases $f$ (more moderator absorbs more thermal neutrons parasitically).
  • Plotting $k_\infty = \epsilon p \eta f$ versus moderator-to-fuel ratio $(V_M / V_F)$ reveals a distinct maximum—the optimum moderation pitch.

§6.6 Thermal Reproduction Factor Eta: Fission Neutrons Per Fuel Absorption

1. Microscopic Definition of $\eta$

The thermal neutron reproduction factor $\eta$ is the average number of fast fission neutrons emitted per thermal neutron absorbed in the fuel material: $$\mathbf{\eta \equiv \nu \, \frac{\Sigma_f^F}{\Sigma_a^F} = \nu \, \frac{\sigma_f^F}{\sigma_a^F} = \nu \, \frac{\sigma_f^F}{\sigma_f^F + \sigma_c^F} = \frac{\nu}{1 + \alpha}}$$ where $\nu$ is the average number of neutrons emitted per fission, $\sigma_c$ is radiative capture $(n, \gamma)$ without fission, and $\alpha \equiv \sigma_c / \sigma_f$ is the capture-to-fission ratio.

2. Pure Fissile Isotopes at 2200 m/s

Fissile Isotope $\nu$ $\sigma_f\ (\text{b})$ $\sigma_c\ (\text{b})$ $\alpha = \sigma_c/\sigma_f$ $\eta = \nu / (1+\alpha)$
Uranium-233 (${}^{233}\text{U}$) $2.49$ $531$ $45.5$ $0.086$ $\mathbf{2.29}$ (Highest thermal $\eta$)
Uranium-235 (${}^{235}\text{U}$) $2.43$ $585$ $99$ $0.169$ $\mathbf{2.08}$
Plutonium-239 (${}^{239}\text{Pu}$) $2.88$ $748$ $269$ $0.360$ $\mathbf{2.12}$

3. $\eta$ in Enriched and Natural Uranium Fuel

For a fuel mixture containing both fissile ${}^{235}\text{U}$ ($5$) and fertile ${}^{238}\text{U}$ ($8$): $$\mathbf{\eta = \frac{\nu_5 \, N_5 \, \sigma_{f, 5}}{N_5 \sigma_{a, 5} + N_8 \sigma_{a, 8}} = \nu_5 \frac{\sigma_{f, 5}}{\sigma_{a, 5} + \frac{N_8}{N_5} \sigma_{a, 8}}}$$ For natural uranium ($N_8/N_5 = 99.28 / 0.72 \approx 138$): $$\sigma_{a, 8} \approx 2.7\text{ b} \implies 138 \times 2.7 = 372.6\text{ b}$$ $$\sigma_{a, 5} = 684\text{ b}, \quad \sigma_{f, 5} = 585\text{ b}, \quad \nu_5 = 2.43$$ $$\eta_{\text{NatU}} = 2.43 \times \frac{585}{684 + 372.6} = 2.43 \times \frac{585}{1056.6} \approx \mathbf{1.345}$$ Because $\eta_{\text{NatU}} = 1.345$ is so close to $1.0$, a natural uranium reactor has a tiny margin for neutron losses: the product $\epsilon p f$ must exceed $1 / 1.345 = 0.743$, which is impossible in light water!

§6.7 Finite Reactor Non-Leakage Probabilities & The Six-Factor Formula

1. Fast and Thermal Neutron Leakage

In a finite real reactor core, neutrons escape across the boundary into the surroundings. We define two non-leakage probabilities:

  • Fast Non-Leakage Probability ($P_{\text{FNL}}$): The probability that a fast fission neutron slows down to thermal energy without leaking from the core during moderation: $$\mathbf{P_{\text{FNL}} = \exp\left( - B_g^2 \tau \right) \approx \frac{1}{1 + B_g^2 \tau}}$$ where $B_g^2$ is the geometric buckling of the core and $\tau$ is Fermi age.
  • Thermal Non-Leakage Probability ($P_{\text{TNL}}$): The probability that a thermal neutron does not leak out while diffusing, but is absorbed inside the core: $$\mathbf{P_{\text{TNL}} = \frac{1}{1 + B_g^2 L^2}}$$ where $L^2$ is the thermal diffusion area.

2. The Six-Factor Formula

Combining the infinite multiplication factor with both non-leakage probabilities yields the complete Six-Factor Formula for effective multiplication: $$\mathbf{k_{\text{eff}} = k_\infty \, P_{\text{FNL}} \, P_{\text{TNL}} = \left( \epsilon \, p \, \eta \, f \right) \cdot P_{\text{FNL}} \cdot P_{\text{TNL}}}$$ In the modified one-group diffusion approximation, since $B_g^2 \tau \ll 1$ and $B_g^2 L^2 \ll 1$: $$P_{\text{FNL}} P_{\text{TNL}} \approx \frac{1}{1 + B_g^2 (L^2 + \tau)} = \mathbf{\frac{1}{1 + M^2 B_g^2}}$$ Thus: $$\mathbf{k_{\text{eff}} = \frac{k_\infty}{1 + M^2 B_g^2}}$$ where $M^2 \equiv L^2 + \tau$ is the migration area!

ADVANCED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, quantitative calculations, and step-by-step examination solutions for Unit 1.

Solved Problem Example 6.1: Detailed Four-Factor Formula Evaluation for Natural Uranium / Heavy Water

A homogeneous thermal reactor system consists of natural uranium metal dissolved in pure heavy water ($\text{D}_2\text{O}$) with an atomic ratio $N_D / N_U = 120$. Thermal microscopic parameters:

  • For $^{235}\text{U}$: $\sigma_a = 680\text{ b}, \sigma_f = 582\text{ b}, \nu = 2.43$
  • For $^{238}\text{U}$: $\sigma_a = 2.71\text{ b}, \sigma_f = 0\text{ b}$
  • For Deuterium (${}^2\text{H}$): $\sigma_a = 0.00053\text{ b}, \sigma_s = 3.4\text{ b}, \xi = 0.725$
  • For Oxygen (${}^{16}\text{O}$): $\sigma_a = 0.00020\text{ b}, \sigma_s = 3.8\text{ b}$

Natural uranium contains $0.720\%\ {}^{235}\text{U}$ and $99.280\%\ {}^{238}\text{U}$. Fast fission factor $\epsilon = 1.005$. The effective resonance integral for this mixture is $I_{\text{eff}} = 24.5\text{ b}$. (a) Calculate the reproduction factor $\eta$ for natural uranium. (b) Calculate the thermal utilization factor $f$. (c) Calculate the resonance escape probability $p$. (d) Compute the infinite multiplication factor $k_\infty = \epsilon \, p \, \eta \, f$ and determine whether an infinite critical reactor is possible.

(a) Reproduction Factor $\eta$: For natural uranium:

$$\frac{N_5}{N_U} = 0.00720, \qquad \frac{N_8}{N_U} = 0.99280$$
$$\bar{\sigma}_{f, U} = 0.00720 \times 582\text{ b} = 4.1904\text{ b}$$
$$\bar{\sigma}_{a, U} = 0.00720 \times 680\text{ b} + 0.99280 \times 2.71\text{ b} = 4.896 + 2.6905 = 7.5865\text{ b}$$
$$\eta = \nu \frac{\bar{\sigma}_{f, U}}{\bar{\sigma}_{a, U}} = 2.43 \times \frac{4.1904\text{ b}}{7.5865\text{ b}} \approx \mathbf{1.3421}$$

(b) Thermal Utilization Factor $f$: Thermal absorption in heavy water per uranium atom ($N_D/N_U = 120$, $N_O/N_U = 60$):

$$\frac{\Sigma_a^M}{N_U} = 120 \times \sigma_{a, D} + 60 \times \sigma_{a, O} = 120(0.00053) + 60(0.00020) = 0.0636 + 0.0120 = 0.0756\text{ b}$$

Thermal absorption in fuel:

$$\frac{\Sigma_a^F}{N_U} = \bar{\sigma}_{a, U} = 7.5865\text{ b}$$

Thermal utilization:

$$f = \frac{\Sigma_a^F}{\Sigma_a^F + \Sigma_a^M} = \frac{7.5865}{7.5865 + 0.0756} = \frac{7.5865}{7.6621} \approx \mathbf{0.9901}$$

(c) Resonance Escape Probability $p$: Slowing down power per uranium atom:

$$\frac{\Sigma_s}{N_U} = 120 \times 3.4\text{ b} + 60 \times 3.8\text{ b} + 1 \times 8.3\text{ b} = 408 + 228 + 8.3 = 644.3\text{ b}$$

Effective decrement:

$$\bar{\xi} \approx \frac{120(3.4)(0.725) + 60(3.8)(0.120)}{644.3} = \frac{295.8 + 27.36}{644.3} \approx 0.5015$$
$$\frac{\xi \Sigma_s}{N_U} = 0.5015 \times 644.3 \approx 323.1\text{ b}$$

Resonance escape probability:

$$p = \exp\left( - \frac{I_{\text{eff}}}{\frac{\xi \Sigma_s}{N_U}} \right) = \exp\left( - \frac{24.5\text{ b}}{323.1\text{ b}} \right) = \exp(-0.07583) \approx \mathbf{0.9270}$$

(d) Infinite Multiplication Factor $k_\infty$:

$$k_\infty = \epsilon \cdot p \cdot \eta \cdot f = 1.005 \times 0.9270 \times 1.3421 \times 0.9901$$
$$k_\infty = 1.005 \times 0.9270 \times 1.3288 \approx \mathbf{1.238}$$

Because $k_\infty = 1.238 > 1$, this natural uranium / heavy water system is robustly supercritical with $+23.8\%$ excess multiplication, easily overcoming finite reactor leakage!

Solved Problem Example 6.2: Heterogeneous Lattice Advantage: Resonance Self-Shielding & Spatial Flux Depression

A reactor designer replaces a homogeneous natural uranium / graphite mixture with a square heterogeneous lattice of solid uranium fuel rods ($d = 2.5\text{ cm}$) surrounded by graphite blocks. In the homogeneous mixture: $p_{\text{hom}} = 0.690$, $f_{\text{hom}} = 0.920$, $\epsilon_{\text{hom}} = 1.000$, $\eta = 1.340$. In the heterogeneous lattice:

  • Resonance self-shielding increases resonance escape to $p_{\text{het}} = 0.885$.
  • Fast fission inside the dense fuel rod increases $\epsilon_{\text{het}} = 1.035$.
  • Spatial flux depression (thermal flux inside rod is lower than moderator, $\phi_F / \phi_M = 0.82$) reduces thermal utilization to $f_{\text{het}} = 0.875$.

(a) Calculate $k_{\infty, \text{hom}}$ for the homogeneous mixture and show that it cannot achieve criticality. (b) Calculate $k_{\infty, \text{het}}$ for the heterogeneous lattice. (c) Calculate the net reactivity gain $\Delta \rho = \frac{k_{\text{het}} - k_{\text{hom}}}{k_{\text{het}}}$ in pcm.

(a) Homogeneous Multiplication Factor $k_{\infty, \text{hom}}$:

$$k_{\infty, \text{hom}} = \epsilon \cdot p \cdot \eta \cdot f = 1.000 \times 0.690 \times 1.340 \times 0.920 \approx \mathbf{0.8506}$$

Because $k_{\infty} = 0.8506 \ll 1$, an infinite homogeneous mixture of natural uranium and graphite cannot become critical under any circumstances.

(b) Heterogeneous Multiplication Factor $k_{\infty, \text{het}}$:

$$k_{\infty, \text{het}} = \epsilon \cdot p \cdot \eta \cdot f = 1.035 \times 0.885 \times 1.340 \times 0.875$$

Calculating step by step:

$$\epsilon \times p = 1.035 \times 0.885 = 0.915975$$
$$\eta \times f = 1.340 \times 0.875 = 1.1725$$
$$k_{\infty, \text{het}} = 0.915975 \times 1.1725 \approx \mathbf{1.0740}$$

By lumping the fuel into discrete rods, $k_\infty$ jumps from $0.851$ to $1.074$, allowing a critical graphite-moderated reactor (such as Enrico Fermi's historic Chicago Pile-1)!

(c) Net Reactivity Gain:

$$\Delta \rho = \frac{k_{\text{het}} - k_{\text{hom}}}{k_{\text{het}}} = \frac{1.0740 - 0.8506}{1.0740} = \frac{0.2234}{1.0740} \approx 0.2080$$

In pcm ($1\text{ pcm} = 10^{-5}$):

$$\Delta \rho = 0.2080 \times 10^5 = \mathbf{+20{,}800\text{ pcm}} = \mathbf{+20.8\%\ \Delta k/k}$$

Lumping the fuel into a heterogeneous lattice yields a colossal $+20{,}800\text{ pcm}$ reactivity improvement, turning an impossible reactor into a functioning power plant!

Solved Problem Example 6.3: Six-Factor Critical Multiplication Factor and Thermal Leakage in a Bare Reactor

A bare cubical research reactor has side length $\tilde{a} = 280\text{ cm}$ (including extrapolation distance). The core material parameters are:

  • $k_\infty = 1.0720$
  • Thermal diffusion area $L^2 = 145\text{ cm}^2$
  • Fermi age $\tau = 45\text{ cm}^2$

(a) Calculate the geometric buckling $B_g^2$ of the cubical core in $\text{cm}^{-2}$. (b) Calculate the fast non-leakage probability $P_{\text{FNL}}$ and thermal non-leakage probability $P_{\text{TNL}}$. (c) Determine the effective multiplication factor $k_{\text{eff}}$ and calculate the net reactivity $\rho$ in pcm. (d) What side length $\tilde{a}_{\text{crit}}$ would make the reactor precisely critical ($k_{\text{eff}} = 1.0000$)?

(a) Geometric Buckling of Cubical Core: For a cube of dimension $\tilde{a}$:

$$B_g^2 = 3 \left( \frac{\pi}{\tilde{a}} \right)^2 = 3 \left( \frac{3.14159265}{280\text{ cm}} \right)^2 = 3 \times (0.011220)^2 \approx \mathbf{3.7766 \times 10^{-4}\text{ cm}^{-2}}$$

(b) Non-Leakage Probabilities:

  • Fast Non-Leakage Probability $P_{\text{FNL}}$:
$$P_{\text{FNL}} = \frac{1}{1 + B_g^2 \tau} = \frac{1}{1 + (3.7766 \times 10^{-4} \times 45)} = \frac{1}{1 + 0.016995} \approx \mathbf{0.98329}$$

Fast neutron leakage fraction is $1 - 0.98329 = 1.67\%$.

  • Thermal Non-Leakage Probability $P_{\text{TNL}}$:
$$P_{\text{TNL}} = \frac{1}{1 + B_g^2 L^2} = \frac{1}{1 + (3.7766 \times 10^{-4} \times 145)} = \frac{1}{1 + 0.05476} \approx \mathbf{0.94808}$$

Thermal neutron leakage fraction is $1 - 0.94808 = 5.19\%$.

(c) Effective Multiplication Factor $k_{\text{eff}}$ and Reactivity:

$$k_{\text{eff}} = k_\infty \cdot P_{\text{FNL}} \cdot P_{\text{TNL}} = 1.0720 \times 0.98329 \times 0.94808 \approx \mathbf{0.99938}$$

Reactivity:

$$\rho = \frac{k_{\text{eff}} - 1}{k_{\text{eff}}} = \frac{0.99938 - 1.00000}{0.99938} = \frac{-0.00062}{0.99938} \approx -6.20 \times 10^{-4} = \mathbf{-62\text{ pcm}}$$

The reactor is slightly subcritical with a reactivity deficit of $-62\text{ pcm}$.

(d) Critical Dimensions $\tilde{a}_{\text{crit}}$: At criticality, $k_{\text{eff}} = 1.0000 \implies k_\infty = 1 + M^2 B_{g, \text{crit}}^2$: Migration area:

$$M^2 = L^2 + \tau = 145 + 45 = 190\text{ cm}^2$$
$$B_{g, \text{crit}}^2 = \frac{k_\infty - 1}{M^2} = \frac{1.0720 - 1}{190\text{ cm}^2} = \frac{0.0720}{190} \approx 3.7895 \times 10^{-4}\text{ cm}^{-2}$$

For a cube:

$$3 \left( \frac{\pi}{\tilde{a}_{\text{crit}}} \right)^2 = B_{g, \text{crit}}^2 \implies \frac{\pi}{\tilde{a}_{\text{crit}}} = \sqrt{\frac{3.7895 \times 10^{-4}}{3}} = \sqrt{1.2632 \times 10^{-4}} \approx 0.011239\text{ cm}^{-1}$$
$$\tilde{a}_{\text{crit}} = \frac{\pi}{0.011239} \approx \mathbf{279.5\text{ cm}}$$

Decreasing the side by just $0.5\text{ cm}$ (or tightening the core) restores exact criticality!