Question

Consider a sphere consisting of pure \({ }^{235} \mathrm{U}\) (a model of a nuclear explosive) and estimate the mean free path for fast neutrons. For fast neutrons one may use the following data (averaged over energies) \(\sigma_{\mathrm{f}}=1.3[\) barn \(]([1], 201) ; \eta=2.4([1], 218) ; \rho=19 \times\) \(10^{3}\left[\mathrm{~kg} \mathrm{~m}^{-3}\right]\). (a) Calculate \(\Sigma_{f}, \lambda_{f}\) and (b) make a rough estimate of the radius \(\mathrm{R}\) for which the leakage will be half of the neutrons produced and argue that this will be about the critical radius; calculate the corresponding critical mass (an accurate calculation will give \(50[\mathrm{~kg}]\) ).

Short answer

The mean free path is determined by the macroscopic cross-section, which helps estimate the critical radius and mass for half neutron leakage.

Step-by-step solution

Understanding Cross Section Conversion

First, we need to convert the cross-section from barns to square meters. We know 1 barn is equal to \( 10^{-28} \) square meters. Given the fission cross-section \( \sigma_{\mathrm{f}} = 1.3 \) barns, we have: \[ \sigma_{\mathrm{f}} = 1.3 \times 10^{-28} \, \text{m}^2 \]

Calculating Macroscopic Cross-Section

The macroscopic cross-section \( \Sigma_{f} \) is calculated using the formula: \[ \Sigma_{f} = N \sigma_{f} \] where \( N = \frac{\rho}{A} N_A \), with \( \rho = 19 \times 10^3 \, \text{kg/m}^3 \) being the density of Uranium, \( A = 235 \, \text{g/mol} \) is the molar mass, and \( N_A = 6.022 \times 10^{23} \, \text{atoms/mol} \) is Avogadro's number. Therefore, \[ N = \frac{19 \times 10^3}{235 \times 10^{-3}} \times 6.022 \times 10^{23} \]

Calculating Mean Free Path

The mean free path \( \lambda_{f} \) is given by the inverse of the macroscopic cross-section: \[ \lambda_{f} = \frac{1}{\Sigma_{f}} \] Substitute \( \Sigma_{f} \) using the result from Step 2 to find \( \lambda_{f} \).

Estimating Radius for Half-Neutron Leakage

The half leakage radius \( R \) can be estimated using the critical radius formula for spheres: \[ R \approx \frac{1}{\Sigma_{f}} \ln(2\eta) \] Given \( \eta = 2.4 \), estimate \( R \) by substituting \( \Sigma_{f} \) from Step 2.

Calculating Critical Mass

The critical mass can be calculated using the formula for the mass of a sphere: \[ M = \frac{4}{3} \pi R^3 \rho \] Using the estimated radius \( R \) from Step 4 and \( \rho = 19 \times 10^3 \, \text{kg/m}^3 \), calculate \( M \).

Key concepts

Neutron Mean Free Path

The neutron mean free path is a fundamental concept in nuclear physics. It refers to the average distance a neutron travels in a material before colliding with an atomic nucleus. This concept helps us understand how neutrons interact with materials, which is vital in nuclear reactors and bombs.
To calculate the mean free path, we need the macroscopic cross-section (\(\Sigma_f\)). The mean free path \(\lambda_f\) is the inverse of this value:\[\lambda_f = \frac{1}{\Sigma_f}.\]This equation shows the relationship between the density of atomic nuclei in a material and the average distance between collisions. In simple terms, the denser the material (more nuclei per volume), the shorter the mean free path because neutrons will collide more often.
Understanding the mean free path is crucial in designing nuclear systems, as it affects both efficiency and safety. This calculation helps policymakers and engineers determine the best materials and configurations for nuclear energy production and weapon design.

Critical Mass Calculation

The critical mass is the minimum amount of fissile material needed to maintain a nuclear chain reaction. This is an essential concept for nuclear reactors and weapons.
According to the problem, we must first find a radius \(R\) where half the neutrons escape the sphere, known as the critical radius. We estimate \(R\) with:\[R \approx \frac{1}{\Sigma_f} \ln(2\eta),\]where \(\eta\) is the average number of neutrons produced per fission event. Substituting the known values allows estimation of \(R\), which provides a basis for calculating the critical mass.
Once \(R\) is determined, the critical mass \(M\) can be calculated using the formula for the mass of a sphere:\[M = \frac{4}{3} \pi R^3 \rho,\]with \(\rho\) being the material's density. This formula gives us the amount of material necessary to achieve criticality, ensuring that enough neutrons are present to sustain a chain reaction. Understanding critical mass is important for both safety and efficiency in nuclear technology design.

Macroscopic Cross-Section

The macroscopic cross-section \(\Sigma_f\) is a key parameter in nuclear physics. It reflects the collective interaction potential between neutrons and nuclei in a given material. This value is crucial in determining both the neutron mean free path and the conditions necessary for critical mass.
To compute \(\Sigma_f\), we use the formula:\[\Sigma_f = N \sigma_f,\]where \(N\) is the atomic number density and \(\sigma_f\) is the microscopic cross-section. \(N\) is calculated by:\[N = \frac{\rho}{A} N_A,\]where \(\rho\) is the material density, \(A\) is the atomic mass, and \(N_A\) is Avogadro's number.
Understanding \(\Sigma_f\) enables scientists to predict how neutrons will behave in a material. This is crucial for constructing safe and efficient nuclear devices, as it influences both the design and location of reactors as well as their operating conditions. Having a thorough grasp of these calculations is fundamental in the field of nuclear physics.