The averaged fractional energy lost per elastic scattering event of neutron moderation process is shown in the following formulation
\frac{\overline{\Delta E}}{E} = \frac{1}{2}(1 - \alpha)
where $\overline{\Delta E} = E - \overline{E'}$, which is referring to the averaged energy lost after an elastic scattering event; $\alpha = \left(\frac{A - 1}{A + 1}\right)^2$, where A = M/m refers to the mass ratio of the target nuclide and a neutron.
If the energy before scattering E = 5 × 10? eV, while the desired energy after \"moderation\" is 1 eV, how many collisions does a neutron required to encounter for such process, assuming this neutron always encounter elastic scattering interaction? Calculate the result assuming the target nuclides are ¹H and ¹²C. (atomic mass of ¹H is 1.007825; atomic mass of ¹²C is 12.00000).