Neutron energy after one elastic collision

[knoximator]

ok, this is the question:

neutrons scatter elastically at 1.0MeV. after one scattering collision, determine what fraction of neutrons will have energy of less than 0.5 MeV if they scatter from:

a. hydrogen
b. Deuterium
c. Carbon-12
d. Uranium-238


solution process...

the basic equation to be used: n=\frac{1}{ζ}*ln\frac{E_{0}}{E_{n}}
n= number of collisions
ζ=depends on atomic mass of target≈ \frac{2}{A+\frac{2}{3}} (A= atomic mass)
for A=1, ζ=1!
E_{0}= original energy of neutron before collision
E_{n}= energy of neutron after n collisions

so, inputting n=1, i get the equation E_{1}=E_{0}*e^{-ζ}

and subsequently, i get the following energies:

a. E_{1}=0.367*E_{0}
b. E_{1}=0.472*E_{0}
c. E_{1}=0.0853*E_{0}
d. E_{1}=0.9916*E_{0}

and that is where i get stuck, i have no clue on how to continue and get a fraction out of the information i got.
in the book, there's a probability equation presented, but i can't see any use of it to my question


Edit

ok, so after some deep book delving session, i might have found my problem.
basically, i don't think i need the equation above, but should rely more on the neutron cross section σ tables for the elements mention above and the specific energies.

for example: \frac{σ_{s}(E)}{σ_{t}(E)} is the probability of a neutron to scatter for a certain energy E

my question is, how to use this relation, and which energies to use?

 

[mathman]

Your answer for c is incorrect, the leading 0 shouldn't be there.

General approach to the problem is basically a classical mechanics problem - for one collision. First transform the particles (neutron and stationary target) to the center of mass coordinate frame. Next assume the elastic collision is isotropic. Transform back to the lab frame, getting the velocity vector of the neutron.
Here you can calculate the neutron energy as a function of scattering angle. Find the angle where the energy is 0.5 Mev. and get the probability that the scattering angle will be greater.

Hint: for U-238 and C-12, the probability will be 0. For H1 (from what I remember) it will be ~ .5, while for H2 it will smaller - I don't remember exactly.