From our day-to-day experience of billiard-ball type collisions a high energy transfer collision can intui­tively be deduced to require a target nucleus of mass not too dissimilar to that of the impinging neutron.

On this basis it would be expected that hydrogen, whose single proton nucleus approximates well to the mass of neutron, would be the ‘best’ target nu­cleus and the heavier nuclei to be progressively less effective. This indeed is so. Applying Newton’s Laws of Motion to a target nucleus of mass number A it is possible to calculate the number of collisions required for a neutron to lose energy from some initial value E0, say, to a final value E. The mathe­matical expression is:

Number of scattering collisions = logn (E0/E)/£

where £, known as the logarithmic energy decrement, is given by

£ = [1 ~ (A — l)2/2A][logn (A + I)/(A — I)]

— 2/{A + 2/3)

for all but the very low mass number nuclei.

(In the case of hydrogen £ = 1.)

Thus to reduce neutron energy from 2 MeV to 0.025

eV requires

[logn (2 x 106)/(2.5 x 10"2)]£ = !8.2/£

In Table 1.6 the number of collisions required to thermalise 2 MeV neutrons are given for different nuclei. Also included are ordinary ‘light’ water FbO and ‘heavy’ water D2O where ‘D’ refers to deuterium, an isotope of hydrogen with a nucleus of one proton and one neutron. The smaller the moderator mass number the larger the value of £ or the fewer the scattering collisions to thermalise the neutrons.