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