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From: Henry Spencer <>
Date: Wed, 10 Jul 1996 15:40:17 GMT

In article <4rtv8h$> (Alain Fournier) writes:
>Isn't the Helium3-Hydrogen reaction aneutronic and achievable at lower
>temperatures than the Deuterium-Tritium reaction? Helium-3 needs to be
>mined on the moon but...

Uh, no.  D-T has the lowest ignition temperature around; that's why the
controlled-fusion people are starting with it.  If 3He made things easier,
they would use it -- there isn't a lot of it around, but they don't need
much for experimental work.

D-D is next up the scale, but it too produces lots of neutrons.  3He-D is
third in line, and it's the one people are interested in 3He for; it is
*nearly* aneutronic (you still get some neutrons from D-D side reactions). 
I haven't heard of a useful 3He-H reaction, although I suppose it's

Truly aneutronic fusion tends to involve reactions like 11B+H, which is
a lot more difficult.
If we feared danger, mankind would never           |       Henry Spencer
go to space.                  --Ellison S. Onizuka |

From: (Paul F. Dietz)
Newsgroups: sci.physics.fusion
Subject: Re: Failing to see how hot fusion is expensive
Date: Fri, 15 Nov 1996 04:47:06 GMT

"Mike Asher" <> wrote:

>The most _likely_ reason fusion will never be economic is we never reach
>breakeven.  If we can, I wouldn't worry about the construction; build it
>and they will come.

Hah!  Breakeven is the least of the problems.  Lawrence Lidsky, a
nuclear engineer at MIT, observed in a (in)famous article in
Technology Review that the mainstream reactor concepts, even if they
could lead to "working" reactors, would not lead to reactors that a
utility would want to buy.

The basic problem is that DT fusion releases most of its energy as
neutrons.  These neutrons pass through the wall of the reactor.
There are limits on the power per unit area the first wall can
sustain.  As the reactor gets bigger, the square/cube law means the
volumetric power density of a fusion reactor must go down.  It's much
less than the power density of a fission reactor, so a fusion core of
equal power output will be larger.  It will also be much more
complicated, so it will be both more expensive and less reliable than
a fission reactor.

Concentrated efforts to optimize "paper" DT tokamak reactor designs
have not led to designs that are more than marginally competitive.
The electric utility research organization EPRI stopped all funding
for fusion some years ago.

     Paul Dietz

     "If you think even briefly about what the Federal
      budget will look like in 20 years, you immediately
      realize that we are drifting inexorably toward a
        -- Paul Krugman, in the NY Times Book Review

From: "Gordon D. Pusch" <>
Subject: Re: Pulsed fusion nuclear propulsion
Date: Sun, 26 Jan 1997 11:44:39 -0600

The following message is a courtesy copy of an article
that has been posted as well.

In article <> (AlmaMcCabe) writes:

>     Why does the D+D reaction not produce He-4. If it does, what
> percentage. If you're talking enough fuel to propel large spaceships,
> I think D+D is the only way to go. You can store it in heavy water
> and use the oxygen or use it to "cool" down the thrust.

It's about 18,769 times less likely than  D+D --> T+p or He3+n.

Nuclear reaction must conserve both momentum AND energy; this imposes
constraints on whether or not the reaction can ``go.'' In the case of
the dominant pure-D-fueled reactions, D+D --> T+p  and  D+D --> He3+n,
only rearangements of particles under the ``strong force'' are involved,
and the K.E. of the two outgoing particles can carry off the released
energy while conserving momentum.  For  D+D --> He4, there are =NO=
``extra'' particles to carry off the energy, so it can't go; one needs
at least one more particle in the outgoing channel. For the ``elctro-
magnetic'' de-excitation channel,  D+D --> He4+\gamma, the reaction-
probability would be supressed by a factor of the fine-structure
constant (~1/137+) --- except that it's a ``forbidden channel,''
because it violates parity.

SO, the first open E/M channel is D+D --> He4+2\gamma, which is
supressed by a factor of \alpha^2 (one for each gamma-photon) ---
so it'll occur about 18,769 time less often than the pure strong-force
reactions  D+D --> T+p or He3+n (the branching-ratio between these two
reactions is very close to 50%--50%, BTW).

--  Gordon D. Pusch   <>

Disclaimer:  I'm a consultant ---  I don't speak for ANL or the DOE,
and they *certainly* don't speak for =ME=  !!!

From: (Henry Spencer)
Subject: Re: Deuterium
Date: Tue, 6 Apr 1999 20:49:34 GMT

In article <7ed137$9v1$>,
theresa knott <> wrote:
>I don't know the details of how Li or B are produced [ I expect others will
>be able to tell you] but Carbon is produced in the triple alpha burning
>(4)He + (4) He ---> (8)Be + photon
>(8)Be + (4) He ---> (12)C + photon

No, it doesn't go in two stages like that.  8Be is very, very, very unstable;
its half-life is a fraction of a picosecond, and it fissions back to a pair
of 4He's.  The reaction is

(4)He + (4)He + (4)He ---> (12)C + photon

Such a three-particle fusion is much less probable than two-particle
reactions, which is why the step from hydrogen burning to helium burning
requires a great increase of temperature and pressure in a star.

Li, Be, and B are all extremely unstable in thermonuclear conditions, and
probably are not formed in stars at all.  Just how they are formed is
still slightly uncertain; the current best guess is that they are formed
by cosmic-ray hits on ordinary atoms.  They are all relatively rare.
The good old days                   |  Henry Spencer
weren't.                            |      (aka

From: "Paul F. Dietz" <>
Newsgroups: sci.physics.fusion,sci.physics.particle,
Subject: Re: Deuterium
Date: Thu, 08 Apr 1999 21:59:16 -0500

Ian Bannister wrote:
>  D will react
> quite readily but He(3) + He(4) probably has to wait until things hot up a
> bit in the stars later life.

Actually, 3He + 4He --> 7Be + photon is an
important side reaction of the pp chain.  The 7Be
then either

(1) decays by electron capture to 7Li, which reacts
  with protons to make helium, or

(2) itself captures a proton to make 8B, which decays
  to 8Be, which then immediately splits into helium.

These are important reactions because their neutrinos
(especially from 8B) are energetic and more easily
detected on earth than the pp neutrinos.  Segue to
the solar neutrino problem...


From: "Paul F. Dietz" <>
Subject: Re: future of spacetravel
Date: Tue, 18 Jan 2000 06:49:14 -0600

James Hunter Heinlen wrote:

> > Fusing ordinary hydrogen (protons) is not easy.  It is
> > extremely hard.  It is not at all clear it can be done
> > in objects less massive than stars.
> Look into spherical inertial-electrostatic-confinement fusion.  There are
> companies (Daimler-Chrysler is one: that
> make desktop fusion system for commercial applications.

It's doubtful this would work for even DT fusion, due to various
loss mechanisms.  For PP fusion, it fails bt a large number of
orders of magnitude.

Consider:  under the conditions in the center of the sun, deuterium
fuses with hydrogen in about a second.  If you had pure deuterium
at the number density and temperature of the center of the sun,
it would fuse in much less than a second.  Just protons, under those
conditions, burns slowly for billions of years.  The PP reaction rate
is extremely small.  We would not exist if it were not.


From: "Paul F. Dietz" <>
Subject: Re: He-3 fusion radioactivity (was Re: Water as a radiation shield?)
Date: Mon, 26 Jun 2000 19:46:26 -0500

Jonathan A Goff wrote:

> What about He3-He3 reactions?  I imagine they would be almost
> as clean as a Boron-Hydrogen reaction, but would be a wee bit
> easier.

Actually, 1H-11B is easier.  The energy barrier is a bit higher,
but the reduced mass is much lower, so tunneling through it
is easier.  Also, 1H-11B would go through a resonance (in 12C).

Having said that, it's unlikely that anything more advanced
that D-3He could be burned in a fusion reactor (and even that
would dump at least 5% of its energy into neutrons.)  This
includes colliding beam, electrostatic devices, etc.  A PhD
student at MIT a few years back showed the advanced fuels
don't work, under rather general assumptions, I understand.


From: "Paul F. Dietz" <>
Subject: Re: He-3 fusion radioactivity (was Re: Water as a radiation shield?)
Date: Sat, 01 Jul 2000 14:45:01 -0500

Tomorrow Calling wrote:

> Do you have a reference to this, I always thought the more exotic ones
> like the CNO cyclc were unlikely but that p-B was considered at least
> possible

This summary and critique of Todd Rider's thesis is online:


From: (Henry Spencer)
Subject: Re: Dueterium as rocket fuel?
Date: Mon, 24 Jul 2000 23:33:03 GMT

In article <>,
Ryan Healey  <> wrote:
>> Correct.  Given that, and the slightly poorer performance, there's just
>> no good reason to do it.
>If you are using a fusion rocket then D would be a very good fuel along
>with 3H or 3He.

Well, for small values of "good". :-)  The problem with D-T or D-D fusion
is the tremendous neutron output, which is very bad for the surroundings
(and creates a nasty cooling problem for superconducting magnets in
particular).  Even D-3He has neutron-producing side reactions.  *Good*
fuels for fusion are things like H-11B... although building a fusion
reactor which will burn them is very tricky.
Microsoft shouldn't be broken up.       |  Henry Spencer
It should be shut down.  -- Phil Agre   |      (aka

From: (Henry Spencer)
Subject: Re: Dueterium as rocket fuel?
Date: Wed, 26 Jul 2000 00:28:13 GMT

In article <8lkpnp$3q9$>,  <> wrote:
>> D-3He has neutron-producing side reactions.  *Good* fuels
>> for fusion are things like H-11B... although building a fusion
>> reactor which will burn them is very tricky.
>H-11B is tricky because it requires high temperatures/pressures?

And the bremsstrahlung losses (nearby electrons move in a curve in the
ion's electric field, and emit radiation as a result) are much worse,
which makes it much more difficult to get positive net energy release.
It's unclear that it's possible to build an H-11B reactor; advanced
cleverness would definitely be needed.

>IIRC, isn't the H-11B reaction:
>H + 11B --> [bang] --> 4 He3? (or 3 He4?)

Three 4He is what you normally get, 4He being far more stable than 3He.
All the products being charged particles has a number of advantages, over
and above avoiding the neutron problem.

>Are there any potentials for neutron-spewing side-reactions
>from the reaction products (especially if they're something
>other than the He3 I'm thinking of)?

Last I heard, it was thought to be relatively clean... but of course,
whatever advanced cleverness is used to build the reactor could change
Microsoft shouldn't be broken up.       |  Henry Spencer
It should be shut down.  -- Phil Agre   |      (aka

From: (Henry Spencer)
Subject: Re: Dueterium as rocket fuel?
Date: Wed, 26 Jul 2000 04:01:48 GMT

In article <>,
Tomorrow Calling  <> wrote:
> H + 11B --> [bang] -->  3 He4
>> Are there any potentials for neutron-spewing side-reactions
>> from the reaction products...
>I believe the He4 + He4 has a very high activation energy so it
>shouldn't be a problem

Besides, 4He + 4He gives you 8Be, which very very very promptly (half-life
of a small fraction of a picosecond) decays into... 4He + 4He.  This is
actually one of the fundamental facts of thermonuclear life, which affects
the way stars evolve etc:  two 4He nuclei do not do anything interesting
together in thermonuclear conditions.  To get higher elements, you need to
smack *three* 4He nuclei together, which is much more difficult to

Microsoft shouldn't be broken up.       |  Henry Spencer
It should be shut down.  -- Phil Agre   |      (aka

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