 # Rates of Neutron Reactions

Consider a mono-energetic neutron beam having a neutron density n (the number of neutrons per m3). If the neutron speed is v m/s, then nv is the number of neutrons falling on 1 m2 of the target material per second. If о m2 is the effective area per single nucleus, for a given reaction (or reactions) and neutron energy, we have the effective area ? of all the nuclei per m3 of the target. Hence, the product of effective area per single nucleus and the effective area, that is, ?nv give the number of interactions (between neutrons and nuclei) per m3 of target material per second. This is an important result as it gives the number of neutrons per second involved in any interaction (or interactions) with 1 m3 of material for which ? is the macroscopic cross section. Sometimes, it may be written in a slightly different form by replacing the neutron flux in place of the neutron density. The neutron flux is defined as the product of the neutron density and the velocity, that is, It is expected in units of neutrons/m2 ? s, which is equal to the total distance in metres travelled in 1 s by all neutrons present in 1 m3. Sometimes, it is referred to as the track length. Substituting ф for nv in Equation 2.9, it follows that This flux is a special case of the angular flux, which adds a spatial angular dependency to the neutron density. The angular flux is a scalar. For the present discussion, we neglect the angular dependency by assuming that the integrated effect is negligible.

Fick's law of diffusion states: 'If the concentration of a solute in one region is greater than in another of a solution, the solute diffuses from the region of higher concentration to the region of lower concentration' (Lamarsh 1983). Using Fick's law, the diffusion approximation in reactor theory is discussed next.

Let us assume the following:

• 1. Consider an infinite medium.
• 2. The cross sections are constants, which are independent of positions and implying a uniform medium.
• 3. Scattering of neutrons are isotropic in the laboratory system.
• 4. The neutron flux function slowly varies with respect to the position.
• 5. We use a one-speed system where the neutron density is not a function of time.
• 6. We use a steady-state system where the neutron density is not a function of time.
• 7. There is no fission source in the system.

Some of these assumptions will be relaxed afterwards. For instance, the diffusing medium will be taken as finite in size rather than infinite.

We shall attempt to calculate the current density at the centre of the coordinate system. The vector J is given by so we must evaluate the components Jx , Jy , Jz.

These net current components can be written in terms of the partial axial currents as Let us investigate the estimation of one single component Jz crossing the element of area dSz at the origin of the coordinate system in the negative z direction.

Due to the scattering collision of the neutrons, every neutron passes through the area dSz in the x-y plane. Neutron scattering above x-y plane will thus flow downwards through the area dSz.

Consider the volume element shown in Figure 2.2, which may be represented as The number of scattering collisions occurring per unit time in the volume element dV is represented as where ф(г) and Es are called the particle flux and macroscopic scattering cross section. FIGURE 2.2

Geometry for the neutron current.

Since scattering is isotropic in LAB system (Glasstone and Sesonke 2004) and the fraction of scattering neutron arriving to dSz is subtended by the solid angle dQ, which may be written as The number of neutrons scattered per unit time in dV reaching dSz and the exponential factor which is attenuated in the medium are The partial current J- can now be written as Since ф(г) is an unknown function, so we expand it in a Taylor series assuming that it varies slowly with position, which will be Writing x, y and z in spherical coordinates, Equation 2.19 becomes Substituting Equation 2.20 into Equation 2.18, we have Integrating the terms containing cos ф and sin ф to zero over the interval ф e [0, 2n], we get The first term may be evaluated as follows: The second term becomes Thus, Equation 2.22 transforms to Similarly, Substituting Equations 2.23 and 2.24 into Equation 2.13, we get Substituting Equation 2.12, we get the expression for the current density after dropping the evaluation at the origin notation. Since the origin of the coordinates is arbitrary, we have where  