[abcs12000]

avg of mean free path for neutrons
can be found with analogy to gamma mean free path which is equal to

1/(macro scopic cross section


we know that macro scobic cross section


N#*microscopic cross section=

A(Cd)=112.411 (gm/mol) amu

note that N is the number of atoms per unit volume

=8.64(gm/cm^3)/112.411gm/mol

=.0768mol /cm^3
0.0768mol*Av no.=4.6286*10^22 atom of Cd
ratio of Cd -113=12%
so our sample contain .88 of Cd-112
when we calculate microscopic crosss section we took the total cs.
ie. there is no importance to these ratios
what you need is to multiply
4.6286*10^22*2*10^4*10^-24=#
1/# is the mean free path
dont care about ratios