1. The stability of the neutron
a. Free neutron beta decay yield.
Neutron mass =  0.93956563 GeV/c^2
Proton mass =  0.93827231 GeV/c^2
Electron mass =  0.000511 GeV/c^2
Energy yield =   0.00078232 GeV/c^2 0.78232 MeV/c^2

b. deuterium beta decay threshold
deuterium mass =  1875.6000 MeV/c^2
two proton masses =  1876.5446 MeV/c^2
electron mass =  .5110 MeV/c^2
Energy yield =   -1.4556 MeV/c^2

c. alpha decay threshold estimate
alpha mass =   3727.400 MeV/c^2
three proton mass =   2814.817 MeV/c^2
neutron mass =   939.566 MeV/c^2
electron mass =   .511 MeV/c^2
Bind 4 nucleons at 5.3 MeV each =    21.200 MeV/c^2
Energy yield    -6.294 MeV/c^2

2. Neutron energy required comes from elastic collision.
For m1<<m2 and headon, the change in momentum for m1 is twice its original momentum
(I.e., it bounces straight back with essentially its original speed.)
.5*mO*vO^2 = 1.2 MeV = E, but mO*vO=2m(neutron)*v(neutron),
so KE neutron = Em(oxygen)/(4*m(neutron)) = (1.2 MeV)(1.48952E4 MeV)/(4*940MeV)
KE neutron = 4.753787234 MeV
 
3. Fission and fusion yields
a. uranium-235 fission
uranium-235 mass =  218896.8 MeV/c^2
xenon-143 mass =  132148.3031 MeV/c^2
strontium-90 mass =  83729.8 MeV/c^2
three neutrons =   2818.69689 MeV/c^2
Energy yield   200 MeV/c^2
Oops! My intention here was to get a net yield of 3 neutrons, or 4 neutrons out, but my
statement certainly didn't make that clear. Literally as stated, the yield would be
equal to 1139.56563 MeV, much too high a yield for a fission reaction.

b. deuterium-tritium fusion
tritium mass =  2808.9 MeV/c^2
deuterium mass =  1875.6000 MeV/c^2
alpha mass =   3727.400 MeV/c^2
neutron mass =   939.56563 MeV/c^2
Energy yield =   17.5344 MeV/c^2

c. The ratio of yield per event favors the fission with 200MeV/17.53 MeV =     11.40616971
However, the yield per unit mass of fuel favors the fusion reaction = (17.53/200)*(236/5)=      4.13708