1 MeV/s = 1.6021917 * 10^-13 W

So for a fusion reaction that releases X MeV and Y MeV as neutrons (total X+Y)
to break even it must output

1 neutron per (X+Y) = Z MeV total release

1.6/Z * 10^13 neutrons each second

for every watt input, if it outputs neutrons with 100% certainity for each
reaction.

For normal D+T this is
	D+T -> He4 (3.5 MeV) + n (14.1 MeV)
	Z = 17.6
	Neutrons per second: 9.09 * 10^11 * W_in

For D+D, only half of the chain leads to neutron emission. Therefore, on
average, between two neutrons there will be both one neutronic reaction and
an aneutronic reaction, which leads us to have to sum both reactions. These
are
	D+D -> T (1.01 MeV) + p (3.02 MeV) (50%)
	    -> He3 (0.82 MeV) + n (2.45 MeV) (50%)

	1.01 + 3.02 + 0.82 + 2.45 = 7.3

	Neutrons per second: 2.19 * 10^12 * W_in

	(stray reactions from D+He3 notwithstanding.)

(D + Li6 looks promising. It is aneutronic and outputs a 22.4 MeV (!!)
 He4 nucleus. Of course, IANANP, and bremsstrahlung could severely
 mess it up for fusors)

-----------------------------------------

n + Li6 -> He4 (2.1 MeV) + T (2.7 MeV)
Total = 4.8 MeV.

So this is only worth it if the neutron carries less than 4.8 MeV of energy.
It doesn't do that in D+T, but it does in D+D. If we're able to extract
the ions' energies perfectly by EHD (utopian, yes, but AFAIK the true ranges
are ~85%), we get
	1.01 + 3.02 + 0.82 + 4.8 = 9.65
	and
	Neutrons per second: 1.65 * 10^12 * W_in
			     2.07 * 10^12 * W_in for eff = 0.8

These values refer to the flux before the Li coat/gas/whatnot itself.

Note that if you don't protect this, you'll get an output equivalent to
1 gray/sec if close to the machine, so stay away or shield it!


Perfect beryllium multiplication (Be9 + n -> Be8 + 2n) would double the
Li6 stage output. However, this is only in theory, assuming the neutron
always hits a Be8 then Li6. 

Also, obviously this isn't true because then a solid block of Be9 would be 
supercritical, each neutron from a Be9 hitting two more Be9 atoms and so on, 
resulting in an enormous neutron flux and the solid beryllium floating away as 
gas (4He). 

7.5% of all natural lithium is Li6.
100% of all natural beryllium is Be9. Be8 decays extremely quickly (on the
	order of 7 * 10^-17 s) to two alphas at 0.05 MeV each.

--------------------------------------------

Bremsstrahlung losses ("Fundamental Limitations on Plasma Fusion Systems Not
in Thermodynamic Equilibrium", Rider):

fuel	P_brem/P_fus	Efficiency
D-T	0.007		99.3 %
D-He3	0.19		84.0 %
D-D	0.35		74.0 %
He3-He3	1.39		41.8 %
p-B11	1.74		36.5 %
p-Li6	4.81		17.2 %

-

From this, D-T seems to be the best choice. The neutron carries a
disproportionate amount of the output energy though; how do we get our
paws on it? (It may ultimately be shown that D+D even with its brems losses
gets higher efficiency output due to MHD versus thermal energy capturing
methods)

-

A neutron decays after 10.3 minutes (into a proton, an electron, and neutrino).
The neutron's mass is 1.6749 * 10^-27 kg, and at 14 MeV, its kinetic energy is
2.2430 * 10^-12 Joule.

The relativistic velocity is then 0.17c = 5.118 * 10^7 m/s.
(http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/releng.html#c5)

Gamma is 1.01477, so due to time dilation, we have to wait about 10.45 minutes
for the decay. This is equal to 627 sec, and during this time (unless impeded)
it will have traveled 5.118 * 10^7 * 627 = 3.208 * 10^10 m.

As a comparison, the earth's equatorial diameter is 1.276 * 10^7 m (which is
less than what the neutron travels in a second), so using decay to get the 
energy from neutrons is clearly infeasible.