Atoms whose protons have been stripped off can fuse?

[Morbius]

Back on the topic of fusion...

Dr. Greenman,

I'm an undergrad in math, but one of my fascinations is with fusion, which I read as much as I can find on. I have a question regarding radiation losses in the plasma. From what I've read, one of the major losses is bremsstrahlung radiation caused by the braking of the ions by the electrons. Yet, every fusion device I've seen uses a neutral plasma. My question is, why isn't a nonneutral, pure ion plasma used for fusion? Without electrons, bremsstrahlung losses are eliminated. Sorry if the question is stupid or obvious, it's just been nagging me and I had to ask. :)

CrazedMathematician,

As Davorak has already correctly pointed out - a pure ion plasma would
have a very high electric field, and there's the problem of how to make
that pure ion plasma.

Additionally, you think we've got problems now trying to confine an
electrically neutral plasma - it's going to be even tougher when you
have a plasma of particles that are all repelling each other without
any electrons to mitigate the repulsion.

Additionally, you don't eliminate bremstrahlung. You get bremstrahlung
any time you accelerate a charged particle. That is, any time, a charged
particle speeds up, slows down, or changes direction - it is accelerating -
and it will radiate. Unless you have a collisionless plasma - your ions
are going to be colliding.

You are going to get bremstrahlung losses due to ion-ion interactions.
Although in a neutral plasma, the electron-ion interactions are dominant,
the elimination of the electrons doesn't eliminate bremstrahlung - only
the dominant piece of it.

Dr. Gregory Greenman
Physicist

 

[sid_galt]

Does anybody know to what energy (in eVs) must a deuterium nucleus must be accelerated to achieve fusion with
1 - a tritium pellet in a cyclotron?
2 - a tritium nucleus which is not enclosed in a pellet assuming that the deuterium always hits the tritium nucleus so there is no need for high density?

 

[hitssquad]

Does anybody know to what energy (in eVs) must a deuterium nucleus must be accelerated to achieve fusion with [...] tritium [...]?

The resulting energy release is 17.6 MeV. This site says that that is an energy gain of 450 times, so perhaps the deuterium must be accelerated to .039 MeV. However, the Wiki article on fusion says the D-T ignition energy needed is 0.1 MeV:
http://en.wikipedia.org/wiki/Nuclear_fusion

 

[Astronuc]

The deuterium-tritium fusion reaction results in an energy gain of about 450:1!

Given the context of the site, it would seem that the deuteron and triton each have an energy, e.g. 0.039 MeV (~40 keV) as hitssquad mentioned.

However, the question is:

Does anybody know to what energy (in eVs) must a deuterium nucleus must be accelerated to achieve fusion with
1 - a tritium pellet in a cyclotron?
2 - a tritium nucleus which is not enclosed in a pellet assuming that the deuterium always hits the tritium nucleus so there is no need for high density?

Basically, both address the same question, i.e. what is the necessary energy for a deuteron to enable a fusion reaction with a stationary triton?

If one looks at the cross-section of a monoenergetic D beam on a tritiated target, the maximum cross-section occurs 110 keV. However, there is a probability of a fusion reaction at 15 keV, although the cross section is about 2 orders of magnitude lower.

Finally, I was looking at the formula for overcoming the coulomb barrier -

\large E_{CB}(MeV) = \Large \frac{1.44 Z_{1}Z_{2}}{1.16 \left ( A_{1}^{1/3} +<br /> A_{2}^{1/3} + 2 \right ) }

and with Z₁=Z₂=1, A₁=2, A₂=3
the number I get is approximately E_CB=0.264 MeV, which is higher than the energies stated above. I suspect that the equation may not be necessarily valid for light atoms, or due to QM, there is a probability that the reaction occurs at lower energies.

 

[sid_galt]

Thank you for the help.

I suspect that the equation may not be necessarily valid for light atoms, or due to QM, there is a probability that the reaction occurs at lower energies.

You mean barrier tunneling, right?

 

[NEOclassic]

The LANL "zipper" 14 MeV neutron source.

Does anybody know to what energy (in eVs) must a deuterium nucleus must be accelerated to achieve fusion with
1 - a tritium pellet in a cyclotron?
2 - a tritium nucleus which is not enclosed in a pellet assuming that the deuterium always hits the tritium nucleus so there is no need for high density?

Hi sid,
A few decades ago some fission bombs were so designed that they wouldn't suffer predetonation when iniated by a continuous externally sourced neutron stream. The so-called "zipper" had a triton beam that fell upon a deuterated target. The beam was accelerated by a 180,000 volt gradient. I realize that you seem to be concerned with pelletized tritium but the relative rare availability of tritons are better allocated when the deuterons are pelletized. Cheers, Jim

 

[CrazedMathematician]

As Davorak has already correctly pointed out - a pure ion plasma would
have a very high electric field, and there's the problem of how to make
that pure ion plasma.

Additionally, you think we've got problems now trying to confine an
electrically neutral plasma - it's going to be even tougher when you
have a plasma of particles that are all repelling each other without
any electrons to mitigate the repulsion.

Additionally, you don't eliminate bremstrahlung. You get bremstrahlung
any time you accelerate a charged particle. That is, any time, a charged
particle speeds up, slows down, or changes direction - it is accelerating -
and it will radiate. Unless you have a collisionless plasma - your ions
are going to be colliding.

You are going to get bremstrahlung losses due to ion-ion interactions.
Although in a neutral plasma, the electron-ion interactions are dominant,
the elimination of the electrons doesn't eliminate bremstrahlung - only
the dominant piece of it.

Dr. Gregory Greenman
Physicist

This paper I found talks about it, maybe you've read it:

Confinement Of Pure Ion Plasma In A Cylindrical Current Sheet
Plasma Physics Laboratory, Princeton University

http://www.osti.gov/bridge/servlets/purl/15113-w7GWk3/webviewable/15113.pdf

Abstract. A novel method for containing a pure ion plasma at thermonuclear densities and temperatures has been modeled. The method combines the confinement properties of a Penning-Malmberg trap and some aspects of the magnetic field geometry of a pulsed theta-pinch.

Introduction
This paper presents a novel method for achieving a well-known goal: the confinement of non-neutral ion plasmas that are adequately dense for controlled thermonuclear fusion applications.

Now most of the paper goes way over my head, so maybe you can explain the approach they're taking and if you think it's feasible.

 

[Morbius]

Now most of the paper goes way over my head, so maybe you can explain the approach they're taking and if you think it's feasible.

CrazedMathematician,

Interesting idea.

However, I note that this has only been modeled in 1-D.

In 1-D, you don't have the instabilities that break up the plasma. If you
think of a long "rope" of plasma - these instabilities manifest themselves
as "kinks" in the rope. Kinking of the rope is a 2-D or 3-D phenomenon
that isn't considered in their analysis.

Perhaps an actual experiment, as proposed by the paper; will decide if
it is a good idea or not.

Dr. Gregory Greenman
Physicist

 

[CrazedMathematician]

Dr. Greenman,

In your opinion, what is the most promising approach to fusion that will produce net energy? Also, do you think it's really worth it trying to induce fusion in commercially impractical fuels like Deuterium-Tritium or Deuterium-Deuterium, which produce high energy neutrons and, in the case of tritium, require a fission reactor to create the fuel? Shouldn't more effort be put into ways which could allow the fusion of aneuronic fuels like Helium 3-Deuterium, Helium 3-Helium 3, or Hydrogen-Boron 11? The latter of which is not only the most abundant (Hydrogen obviously, and Boron-11 makes up 80% of natural Boron) but also one of the safest and easiest to convert into electricity (directly using the charged particles). So while a device that would allow net energy production in a fuel like Tritium-Deuterium would be an incremental step, a device that could fusion Hydrogen-Boron 11 would be a giant leap, allowing nearly limitless amounts of electrical energy. What are your thoughts on this and the way fusion is being researched at the moment?

 

[hitssquad]

do you think it's really worth it trying to induce fusion in commercially impractical fuels like Deuterium-Tritium or Deuterium-Deuterium, which produce high energy neutrons and, in the case of tritium, require a fission reactor to create the fuel?

Those two issues are related. Fission reactors create tritium by bombarding lithium-6 with neutrons. Fusion reactors can make tritium in the same manner, and shield neutrons at the same time.

Also regarding fusion neutrons, some ideas for fusion reactors involve hybridization with fission such that fission fuel surrounds the fusion portion of the hybrid reactor reactor and uses the fusion-generated neutrons to produce controlled fission reactions. Similarly as above, this use of the fusion neutrons would simultaneously act as a shield.
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/fusion.html

--
The most promising source of tritium seems to be the breeding of tritium from lithium-6 by neutron bombardment with the reaction

{_3^6}Li + {_0^1}n \rightarrow {_2^4}He + {_1^3}H + 4.8 MeV

which can be achieved by slow neutrons. This would occur if lithium were used as the coolant and heat transfer medium around the reaction chamber of a fusion reactor.
--

 

[Astronuc]

You mean barrier tunneling, right?

Yes. When the minimum energy is less than the Coulomb 'barrier', it implies tunneling through the barrier.

 

[Morbius]

Dr. Greenman,

In your opinion, what is the most promising approach to fusion that will produce net energy?

CrazedMathematician,

Right now - I would say inertial confinement is the most promising.

Of course, since I work in the field - I'm an inertial confinement partisan.

Also, do you think it's really worth it trying to induce fusion in commercially impractical fuels like Deuterium-Tritium or Deuterium-Deuterium, which produce high energy neutrons

I see nothing commercially impractical about deuterium - especially
when compared to something like Helium-3. You are going to have to
go to the Moon to get large quantities of Helium-3.

Deuterium can be extracted from ocean water quite readily. About
1 in 140 atoms of Hydrogen in the ocean is Deuterium. Heavy water,
D20; is made on an industrial scale all the time.

The fast neutrons aren't a problem - in fact - that's where the ENERGY
is. Of the 17.6 MeV, that you get from D-T fusion; 14.1 MeV is in that
neutron.

Shielding for the neutrons and capturing their energy is NOT a big
problem. For example, LLNL proposed a concept in which the "first wall"
of an inertial confinement fusion reactor was a "shower" of liquid
lithium. The lithium shields the fast neutrons, captures their energy,
and breeds the needed Tritium.

and, in the case of tritium, require a fission reactor to create the fuel?

You don't need a fission reactor to create the Tritium - you can create
it from the neutrons from the D-T reaction.

Shouldn't more effort be put into ways which could allow the fusion of aneuronic fuels like Helium 3-Deuterium, Helium 3-Helium 3, or Hydrogen-Boron 11? The latter of which is not only the most abundant (Hydrogen obviously, and Boron-11 makes up 80% of natural Boron) but also one of the safest and easiest to convert into electricity (directly using the charged particles). So while a device that would allow net energy production in a fuel like Tritium-Deuterium would be an incremental step, a device that could fusion Hydrogen-Boron 11 would be a giant leap, allowing nearly limitless amounts of electrical energy. What are your thoughts on this and the way fusion is being researched at the moment?

For a whole host of reasons; I would put Helium-3 and Boron-11 fusion
on the absolute BOTTOM of my priority list. They look good if you
take a superficial look at them - but when you get into the real physics;
they're not all that promising and should be relagated to last place.

Dr. Gregory Greenman
Physicist

 

[sid_galt]

I have another question regarding d-t fusion in cyclotrons using external electric and magnetic fields.

Since around .264 MeV of energy per deuterium particle is required for d-t fusion, it means a particle has to be accelerated in around .264MV of voltage. So what inhibits the actual energy delivered to a particle to make such fusion costly?

 

[Morbius]

I have another question regarding d-t fusion in cyclotrons using external electric and magnetic fields.

Since around .264 MeV of energy per deuterium particle is required for d-t fusion, it means a particle has to be accelerated in around .264MV of voltage. So what inhibits the actual energy delivered to a particle to make such fusion costly?

sid_galt,

The energy you get out of the fusion doesn't pay for running the cyclotron.

If you want to get net energy out of fusion reactions - think of a fire.

An ordinary fire is self perpetuating - the energy needed to initiate the
next oxidation reaction comes from the oxidation reaction that has
just happened.

That's what you have to do with fusion - get the fusion reaction itself to
supply the energy needed to trigger the next fusion reaction.

In a cyclotron, you have to supply a lot of energy to the electromagnets
that keep the particles in their spiral orbits - energy which just goes
away as heat in the coils of the magnets.

You have to supply energy to the big RF oscillators that provide the
accelerating voltage on the cyclotron's "Dees". Again, that's energy
down the drain.

A cyclotron or any particle accelerator only accelerates a rather
small number of particles. You don't get moles and moles of particles
out of a cyclotron.

For the energy that you get out in the beam - the cyclotron, or any
other particle accelerator is very, very, inefficient. You can't throw
energy away like that - and get to breakeven - let alone a net power
production.

Cyclotrons and particle accelerators are research tools. Their basic
operating principles were not conceived to be efficient. They were
designed to get a small number of particles to study.

In order to have a fusion reactor that produces power; you need to fuse
macroscopic quantities of fuel. Cyclotrons are many, many, many,
orders of magnitude down in scale from what you need. Don't bother
with trying to base anything on cyclotron-induced fusion - that's a
dead end.

Dr. Gregory Greenman
Physicist

 

[CrazedMathematician]

Shielding for the neutrons and capturing their energy is NOT a big
problem. For example, LLNL proposed a concept in which the "first wall"
of an inertial confinement fusion reactor was a "shower" of liquid
lithium. The lithium shields the fast neutrons, captures their energy,
and breeds the needed Tritium.

But not ALL of the neutrons react with the lithium, some will smash into the reactor walls and make them radioactive. Obviously it doesn't produce nearly as much radioactive waste as fission but it still produces some.

For a whole host of reasons; I would put Helium-3 and Boron-11 fusion
on the absolute BOTTOM of my priority list. They look good if you
take a superficial look at them - but when you get into the real physics;
they're not all that promising and should be relagated to last place.

That's disappointing. Another major application of fusion is space travel. With fuels like Helium-3 and Boron-11 you can use the charged particles as thrust directly, allowing for extremely high specific impulses (over 1,000,000 seconds compared to 450 seconds for liquid oxygen and liquid hydrogen). With deuterium-tritium, since you can't use the neutrons directly for thrust, you would have to heat a working fluid which would significantly decrease your efficiency.

What do you think of this website:
http://www.focusfusion.org/
Is it for real? It talks about using a "plasma focus device for hydrogen-boron nuclear fusion".

 

[Morbius]

But not ALL of the neutrons react with the lithium, some will smash into the reactor walls and make them radioactive. Obviously it doesn't produce nearly as much radioactive waste as fission but it still produces some.

CrazedMathematician,

Is that a problem?

That's disappointing. Another major application of fusion is space travel. With fuels like Helium-3 and Boron-11 you can use the charged particles as thrust directly, allowing for extremely high specific impulses (over 1,000,000 seconds compared to 450 seconds for liquid oxygen and liquid hydrogen). With deuterium-tritium, since you can't use the neutrons directly for thrust, you would have to heat a working fluid which would significantly decrease your efficiency.

When I was in graduate school at MIT, we had one student that was
extolling the virtues of using tokamaks as space vehicle power plants.

So the professor put a problem on the final exam. He assumed a
2 Gigawatt tokamak; and made a lot of very optimistic assumptions -
and had us compute the thrust from diverting plasma out of the 2Gw
tokamak.

It turned out to be something like 2 Newtons of force.

"Plasma drive" turned out to be inefficient in the extreme.

Dr. Gregory Greenman
Physicist

 

[CrazedMathematician]

Is that a problem?

Are you trying to say that radioactive waste isn't a problem? If that was so then why bother with fusion at all, just use fission. We do have enough fuel for millions of years.

When I was in graduate school at MIT, we had one student that was
extolling the virtues of using tokamaks as space vehicle power plants.

So the professor put a problem on the final exam. He assumed a
2 Gigawatt tokamak; and made a lot of very optimistic assumptions -
and had us compute the thrust from diverting plasma out of the 2Gw
tokamak.

It turned out to be something like 2 Newtons of force.

"Plasma drive" turned out to be inefficient in the extreme.

Ummmm, I don't think you understand space propulsion. Efficiency is measured in specific impulse not thrust. So a thruster that has 2 GW output and 2 N of thrust is actually extremely efficient, with a specific impulse of about 200,000,000 seconds (which is not only about 2 orders of magnititude better than the best fusion reactions, it would also mean the exhaust velocity is 2 billion m/s, or faster than the speed of light which is obviously impossible, so your calculations are wrong). That is, one kg of fuel can provide one kg of thrust for 200 million seconds, or in this case, one kg of fuel can provide 2 N of force for 1 billion seconds. Yeah the thruster might only put out 2 N, but it can stay on for over 18,000 years on just 1 kg of fuel.

 

[Morbius]

Are you trying to say that radioactive waste isn't a problem? If that was so then why bother with fusion at all, just use fission. We do have enough fuel for millions of years.

Ummmm, I don't think you understand space propulsion.

WRONG! I know propulsion theory VERY WELL.

Efficiency is measured in specific impulse not thrust.

WRONG AGAIN - efficiency is efficiency and specific thrust is specific
thrust.

You can have a high specific thrust - however, in this case you
have an extremely low efficiency.

That is, an exceedingly small amount of the energy produced by
the tokamak went into propelling the craft.

Specific thrust is NOT efficiency - you can tell that right off the
bat by their dimensions.

Specific thrust has units of time [ seconds ] while
efficiency is dimensionless.

Yeah the thruster might only put out 2 N, but it can stay on for over 18,000 years on just 1 kg of fuel.

WRONG again - you didn't do the arithmetic.

1 kg of fusion fuel, at 2 GW of power will last you not quite 2 days.

Don't tell me that I don't know propulsion when you are just pulling
numbers out of the air.

Dr. Gregory Greenman
Physicist

 

[CrazedMathematician]

Power = 1/2 * Exhaust Velocity * Force

2000000000 watts = 1/2 * Ve * 2 N = Ve * 1 N
2000000000 watts / 1 N = Ve = 2000000000 m/s

Specific Impulse = Exhaust Velocity / Acceleration of Gravity
Isp = 2000000000 m/s / 9.8 m/s^2 = 204081632.6531 seconds

Now, if you're talking about efficiency in terms of Fusion Power Toward Thrust / Total Fusion Output Power (as I think you are) I'm wondering how you can truly speculate on the engineering details of a system in such an early stage. Are you saying that with all the future advances we won't improve efficiency? Is there some insurmountable barrier preventing us from getting higher efficiency like 50% or 75% or 99%? Seems very speculative.

 

[Morbius]

Power = 1/2 * Exhaust Velocity * Force

NO - the power that goes into driving the vehicle is equal to the
product of the force or thrust and the speed of the vehicle.

Your formula above is a limiting case.

2000000000 watts = 1/2 * Ve * 2 N = Ve * 1 N
2000000000 watts / 1 N = Ve = 2000000000 m/s

You've made the assumption above that ALL the energy of the 2 GW
tokamak goes into kinetic energy of the vehicle. This is not the case!

You are spewing some very hot plasma out the back of this vehicle - and
a miniscule portion of that is going into kinetic energy of the vehicle.

Specific Impulse = Exhaust Velocity / Acceleration of Gravity
Isp = 2000000000 m/s / 9.8 m/s^2 = 204081632.6531 seconds

Now, if you're talking about efficiency in terms of Fusion Power Toward Thrust / Total Fusion Output Power (as I think you are) I'm wondering how you can truly speculate on the engineering details of a system in such an early stage. Are you saying that with all the future advances we won't improve efficiency? Is there some insurmountable barrier preventing us from getting higher efficiency like 50% or 75% or 99%? Seems very speculative.

The way the MIT professor who wrote the exam problem posed it - he
made some very optimisitic assumptions about the details - what
fraction of the plasma could be diverted by diverters that were 100%
efficient at their function...

It's been a few decades since - so I don't recall all the specifics. However,
I do recall that the assumptions ranged from reasonable to optimistic.

This is a little like those problems you get in thermodynamics - the
specification of a power plant running on a Rankine cycle - with
14 stages of reheat, and extremely efficient turbines ... and with all
that - the efficiency is limited by the 2nd Law of Thermodynamics to
the limiting case of the Carnot efficiency. So you know - it doesn't
matter how advanced the heat engine - the Laws of Physics limit it to
the Carnot efficiency - and that's not speculative at all.

Dr. Gregory Greenman
Physicist

 

[CrazedMathematician]

You didn't comment on http://www.focusfusion.org Do you think it's for real? I searched on "plasma focus" and it seems like a real device. Also, do you know anything about Colliding Beam Fusion? http://fusion.ps.uci.edu/beam/introb.html It also looks promising.

 

[Morbius]

You didn't comment on http://www.focusfusion.org Do you think it's for real? I searched on "plasma focus" and it seems like a real device.

CrazedMathematician,

I haven't heard of this before - and one would think someone in the
scientific community would have.

I don't see many research labs like LLNL with a website that is hooked
to PayPal to accept contributions.

I'm dubious.

Also, do you know anything about Colliding Beam Fusion? http://fusion.ps.uci.edu/beam/introb.html It also looks promising.

Looks a little like the Mirror Fusion Test Facility that LLNL had in
the 1970s - except no yin-yang magnets:

http://www.specktech.com/MFTF.html

That project didn't get anywhere - that part of the Lab is still blocked
off for the dismantlement effort.

Dr. Gregory Greenman
Physicist

 

[Astronuc]

Focus Fusion claims are indeed questionable.

I addressed them in this thread - Plasma focus fusion
https://www.physicsforums.com/showthread.php?t=57321

Specific impulse is in some sense a measure of 'effectiveness' (and not really efficiency) of the utilization of propellant mass, or more specifically, mass flow rate to achieve thrust. Furthermore, it is not a measure of the overall 'system efficiency', which includes energy production/conversion efficiencies.

Yes, a plasma has high temperature/kinetic energy per unit mass, however the densities and mass flow rates are extremely small.

A plasma density is about 10¹⁴ particles/cm³, and that is about one-millionth of the density at STP (~3x10¹⁹ molecules/cm³), so a high Isp does not buy much if the mass flow rates are extremely small.

2N thrust is very small, if a spacecraft has a mass of 1000 MT.

One must be very careful in applying basic stoichiometric equations to complex propulsion systems. Fusion reactors are massive structures.

 

[Morbius]

Specific impulse is in some sense a measure of 'effectiveness' (and not really efficiency) of the utilization of propellant mass, or more specifically, mass flow rate to achieve thrust. Furthermore, it is not a measure of the overall 'system efficiency', which includes energy production/conversion efficiencies.

Astronuc,

Exactly correct.

Specific impulse tells you how well you are using a given amount of
reaction mass.

But it doesn't tell you how well you are using your energy.

For high specific impulse, you want a low mass propellant.

As a thought experiment; image we use a cyclotron to accelerate
deuterons, which we then blast out the back of the vehicle.

You may have a high specific thrust owing to the low mass of the
propellant; however - a cyclotron is terribly inefficient from an
energy economy standpoint. You are going to be depositing a lot
of energy in your magnetic field windings...

If you had a source of electricity - you sure wouldn't want to waste it
by heating up the windings and pole pieces of the cyclotron.

Using a cyclotron as a "rocket" is an obvious loser.

Too bad the tokamak as a "rocket" doesn't fair much better.

Dr. Gregory Greenman
Physicist

 

[sid_galt]

If a normal hydrogen nucleus is hit with a neutron, does it combine to form deuterium?

 

[Morbius]

If a normal hydrogen nucleus is hit with a neutron, does it combine to form deuterium?

Sid,

That is ONE of the things that can happen.

If it does - it is called "radiative capture" because the incident neutron is,
of course, not bound to the proton. When the neutron and proton combine
to form deuterium - a bound state - there will be excess energy. That
excess energy will be radiated away as a gamma - hence radiative capture.

The notation is:

H(n,\gamma)D

But that's not the only thing that can happen. The neutron can scatter
off the proton. That's why water is used as a moderator in nuclear
reactors. One wants to slow the neutron down, since the fission
cross-section is greater at low energy than at high. The way one does
that is to let the neutron scatter off the proton - thereby losing energy.

Dr. Gregory Greenman
Physicist

 

[sid_galt]

Although muon catalyzed fusion is not of much practical use, how is it affected by scattering?
If the problem of scattering is significantly lower for muons, is it because of its charge, its mass or both?

 

[Morbius]

Although muon catalyzed fusion is not of much practical use, how is it affected by scattering?
If the problem of scattering is significantly lower for muons, is it because of its charge, its mass or both?

sid,

Muons don't do anything to combat the nuclear scattering problem.

It's not the muons that are scattering - it's the nuclei that are going to fuse.

The only thing muons do is that they "orbit" closer to the nucleus than does
an electron - therefore they reduce the radius at which the fusing atom
is electrically neutral.

Muons only help out on Coulomb repulsion - and, of course, Coulomb
scattering.

Muons don't do anything to help out with nuclear processes - like nuclear
scattering.

Dr. Gregory Greenman
Physicist

 

[sid_galt]

Thanks for the Info