Question 9.2.Q3: Moderation (slowing down) of neutrons in nuclear reactors is......

(a) The simplest manner in which one can derive a recurrence equation to recursively define a sequence of events is to write the first few terms of the sequence and from a comparison of the terms deduce the recurrence equation. Moderation (slowing down) of neutrons by elastic scattering is a good example of a sequence that can be described by a recurrence equation and we derive the general equation by laying out the first few terms of the neutron slowing down process through elastic scattering.

For each elastic scattering event the mean energy \overline{\Delta E_{\mathrm{K}}} transferred from the incident neutron of kinetic energy \left(E_{\mathrm{K}}^{\mathrm{n}}\right)_{\mathrm{inc}} to the moderator is derived in (9.35) and given in (9.36). Consequently, we can write the kinetic energy of the neutron after each elastic scattering event as the difference between \left(E_{\mathrm{K}}^{\mathrm{n}}\right)_{\mathrm{inc}} \text { and } \overline{\Delta E_{\mathrm{K}}} as follows

\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_{\mathrm{after}}=\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_{\mathrm{inc}}-\overline{\Delta E_{\mathrm{K}}}=\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_{\mathrm{inc}}\left[1-\frac{2 A}{(1+A)^2}\right]=\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_{\mathrm{inc}}\left[\frac{1+A^2}{(1+A)^2}\right]            (9.37)

The first few elastic scattering events in the slowing down of a neutron in a sequence of elastic scattering events are now expressed as follows:

(1) We start with an incident neutron with kinetic energy \left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0 striking the moderator of atomic weight A. As suggested in (9.37), the kinetic energy \left(E_{\mathrm{K}}^{\mathrm{n}}\right)_1 of the neutron after the first elastic scattering interaction is given as

\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_1=\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0\left[\frac{1+A^2}{(1+A)^2}\right]               (9.38)

(2) Incident kinetic energy for the second scattering event is now \left(E_{\mathrm{K}}^{\mathrm{n}}\right)_1 so that, according to (9.37) in conjunction with (9.38), we write the neutron kinetic energy \left(E_{\mathrm{K}}^{\mathrm{n}}\right)_2 after the second scattering event as

\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_2=\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_1\left[\frac{1+A^2}{(1+A)^2}\right]=\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0\left[\frac{1+A^2}{(1+A)^2}\right]^2 .              (9.39)

(3) The third scattering event starts with neutron of kinetic energy \left(E_{\mathrm{K}}^{\mathrm{n}}\right)_2 and the neutron kinetic energy \left(E_{\mathrm{K}}^{\mathrm{n}}\right)_3 after the third scattering event with the help of (9.37) and (9.39) is

\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_3=\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_2\left[\frac{1+A^2}{(1+A)^2}\right]=\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0\left[\frac{1+A^2}{(1+A)^2}\right]^3            (9.40)

(4) The recurrence equation is now becoming obvious: after m elastic scattering interactions (with m an integer), neutron energy \left(E_{\mathrm{K}}^{\mathrm{n}}\right)_m can be written as follows

\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_m=\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0\left[\frac{1+A^2}{(1+A)^2}\right]^m               (9.41)

or

\frac{\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0}{\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_m}=\left[\frac{(1+A)^2}{1+A^2}\right]^m=\left[1+\frac{2 A}{1+A^2}\right]^m,            (9.42)

where \left(E_{\mathrm{K}}^{\mathrm{h}}\right)_0 is the kinetic energy of the incident neutron entering moderator A and starting the elastic scattering slowing down sequence.

(b) The process of slowing down energetic neutrons to thermal energy of 0.025 eV with elastic scattering in a moderator medium is referred to as thermalization of neutrons. The mean number m of elastic scattering interactions is easily determined by solving the recurrence equation (9.42) for m as follows

m=\frac{\ln \frac{\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0}{\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_{\mathrm{T}}}}{\ln \left[1+\frac{2 A}{1+A^2}\right]}             (9.43)

where

\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_\mathrm{T}  is the final kinetic energy of the neutron after thermalization through m scattering events.

\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0    is the initial kinetic energy of the neutron when it strikes the moderator

A    is the atomic mass number (atomic weight) of the moderator.

The number m of elastic scattering events for thermalization of \left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0 = 2 MeV neutrons in hydrogen (A = 1), carbon (A = 12), cadmium (A = 112), lead (A = 207), and uranium (A = 238), respectively, is determined as follows

m_{\mathrm{H}}=\frac{\ln \left[\left(2 \times 10^6\right) / 0.025\right]}{\ln \left[1+(2 \times 1) /\left(1+1^2\right)\right]} \approx 27,            (9.44)

m_{\mathrm{C}}=\frac{\ln \left[\left(2 \times 10^6\right) / 0.025\right]}{\ln \left[1+(2 \times 12) /\left(1+12^2\right)\right]} \approx 119,           (9.45)

m_{\mathrm{Cd}}=\frac{\ln \left[\left(2 \times 10^6\right) / 0.025\right]}{\ln \left[1+(2 \times 112) /\left(1+112^2\right)\right]} \approx 1029,           (9.46)

m_{\mathrm{Pb}}=\frac{\ln \left[\left(2 \times 10^6\right) / 0.025\right]}{\ln \left[1+(2 \times 207) /\left(1+207^2\right)\right]} \approx 1893,            (9.47)

m_{\mathrm{U}}=\frac{\ln \left[\left(2 \times 10^6\right) / 0.025\right]}{\ln \left[1+(2 \times 238) /\left(1+238^2\right)\right]} \approx 2175 .             (9.48)

In Fig. 9.4 we plot the mean number of elastic collisions m given by (9.43) against atomic weight A for three kinetic energies \left(E_K\right)_0 of incident neutrons: 100 keV, 2 MeV, and 10 MeV. The data points on the \left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0=2  \mathrm{MeV} line are data that were calculated for various target materials in (9.44) through (9.48).
For a given incident neutron energy \left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0 the plot of m against A is very close to a linear function even though (9.43) appears to be a complicated function. However, it turns out that ln[1 + 2A/(1 + A²)] can easily be expanded into a series as follows

\ln (1+x) \approx x-\frac{1}{2} x^2+\frac{1}{3} x^3-\frac{1}{4} x^4+\cdots \quad \text { for }-1<x<1             (9.49)

By setting x = 2A/(1 + A²) and using only the first two terms of the series (9.49) we obtain the following simplification for ln[1 + 2A/(1 + A²)] of (9.43)

\ln \left[1+\frac{2 A}{1+A^2}\right] \approx \frac{2 A}{1+A^2}\left[1-\frac{A}{1+A^2}\right] \approx \frac{\frac{2 A}{1+A^2}}{1+\frac{A}{1+A^2}}=\frac{2 A}{1+A+A^2}              (9.50)

Inserting (9.50) into (9.43) we get the following approximation for the mean number of elastic scattering events required to moderate a neutron from incident kinetic energy \left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0 of a few MeV down to thermal energy \left(E_{\mathrm{K}}^{\mathrm{n}}\right)_{\mathrm{T}} which typically amounts to 0.025 eV

m=\frac{\ln \frac{\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0}{\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_{\mathrm{T}}}}{\ln \left[1+\frac{2 A}{1+A^2}\right]} \approx \frac{1+A+A^2}{2 A} \ln \frac{\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_0}{\left(E_{\mathrm{K}}^{\mathrm{n}}\right)_{\mathrm{T}}}              (9.51)