Question

4. A neutron with a kinetic energy of 2 MeV undergoes a series of elastic scattering collisions with 1°F to slow down to a kinetic energy of 0.025 eV. For this to occur, determine (a) the minimum number of collisions, and (b) the mean number of collisions.

Answer 1

Answer: 86 collisions

In the head-on collision, the neutron will bounce back from the more massive carbon nucleus. Considering a general nucleus of atomic mass number A the conservation of momentum and energy can be expressed as

mun = AmVA - mvn

m(un)2/2 = Am(VA)2/2 + m(vn)2/2

where a simple approximation has been made in terms of an average nucleon mass m. Canceling this out and slightly rearranging

vn = AVA - un

(un)2 = A(VA)2 + (vn)2

Thus un = VA(A + 1)/2

The energy loss is just the recoil kinetic energy of the nucleus

Am(VA)2/2 = 2Am(vn)2/(A + 1)2 = 0.568 MeV

This expression can be used to find the ratio of neutron energy before to neutron energy after the collision - this is

Ratio of energies = [(A - 1)/(A + 1)]2

After q collisions, the neutron energy is reduced by a factor

[(A - 1)/(A + 1)]2q

Thus to reduce from 2 MeV to 0.025 eV requires about 86 collisions.