Slowing down spectrometer

When the value of $\xi$ is small, the number of collisions necessary for slowing down a neutron of 2 MeV, for example to thermal energies becomes very large. For lead, for example it requires 760 collisions. Since the mean free path in lead is around 3cm, the neutron travels as much as 23 meters before being slowed down. It is clear that this process will take a long time, hence the possibility to use this slowing down time to characterize the neutron. We give a schematic derivation of the relation between the neutron velocity and energy and the slowing down time[37]. The average velocity loss at each collision is $\frac{\Delta v}{v}=\frac{\xi}{2}$. The velocity after n collisions is simply

$$v_{n}=v_{0}(1\ -\ \frac{\xi}{2})^{n}$$ (4.56)

Assuming a constant scattering cross-section and mean free path $\lambda,$ the average time of collision n is, therefore:

$$t_{n}=\ {\frac{\lambda}{v_{0}}}\left( 1\ +\ {\frac{1}{1-\frac... ...ght) ^{n+1}}{\frac{\xi}{2}\left( 1-\frac{\xi}{2}\right) ^{n}}}$$ (4.57)

$$t_{n}=\ {\frac{\lambda}{v_{0}}}\ {\frac{1-\left( 1-\frac{\xi}... ...ht) ^{n+1}}{\frac{\xi}{2}\left( 1-\frac{\xi}{2}\right) ^{n}}}\$$ (4.58)

Using equations 3.61 and 3.59 and dropping the index n one gets

$$v=\frac{v_{0}\left( 1-\frac{\xi}{2}\right) }{\frac{\frac{\xi}{2}v_{0} }{\lambda}t+\left( 1-\frac{\xi}{2}\right) }$$ (4.59)

and

$$E=\frac{E_{0}\left( 1-\frac{\xi}{2}\right) ^{2}}{\left( \frac... ...\frac{2E_{0}}{m}}t+\left( 1-\frac{\xi}{2}\right) \right) ^{2}}$$

which is currently writen

$$E=\frac{K}{\left( t+t_{0}\right) ^{2}}$$ (4.60)

with
$$\begin{align*}K & =\frac{2m\lambda^{2}\left( 1-\frac{\xi}{2}\right) ^{2}}{\xi^{2... ...1-\frac{\xi}{2}\right) \frac{2\lambda}{\xi}\sqrt{\frac {m}{2E_{0}}} \end{align*}$$
For a lead scatterer the TARC experiment[36] provided:

$$E_{keV}=\ {\frac{172}{\ \left( t_{\mu s}\ +\ 0.3\right) ^{2}}}$$ (4.61)

in good agreement with Slovacek et al.[37] who calculated K=165.