If the average logarithmic energy decrement of a material is 0.2 , calculate the number of elastic collisions needed to reduce the neutron energy from \(10 \mathrm{MeV}\) to \(10 \mathrm{eV}\).

\(\Delta=M-A\) is the "mass excess," where \(M\) is the mass of a nuclide and \(A\) its mass number. These data are given in the Nuclear Wallet cards, and elsewhere. The data given below may be useful in solving some of the problems.

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Nuclide n H 2H 4He 23 Na 24Na A (MeV) 8.071 7.289 13.136 2.425 -9.530 -8.418 56Fe -60.601 Nuclide 231 Th 232 Th 235U 236U 238U A (MeV) 33.811 35.444 40.914 42.441 47.304