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Longer answer: Quantum mechanics permits only a couple of pathways to fuse hydrogen into helium. They all produce characteristic fast neutrons. No neutrons, no fusion.

If it actually worked, you would Not Want one in your car. Hot neutrons are useful in materials science laboratory beam lines, stars and fast breeder reactors. Everywhere else they are radioactive pollution.

- Jake

If you only spend 20 minutes of the rest of your life on economics, go spend them here.

by JakeS (JangoSierra 'at' gmail 'dot' com) on Sat Jan 12th, 2013 at 09:34:11 AM EST
[ Parent ]
Iwamura is not talking about fusing hydrogen into helium, but apparently about catalysing other reactions in which metals are turned into other metals by absorbing a deuteron (followed by beta decay).

This is not cold fusion, and it could be an energy source (it must be exothermic or it wouldn't happen). Whether it can be done safely from the point of view of both chemical and radioactive contamination is less obvious:



I distribute. You re-distribute. He gives your hard-earned money to lazy scroungers. -- JakeS

by Migeru (migeru at eurotrib dot com) on Sat Jan 12th, 2013 at 10:20:53 AM EST
[ Parent ]
With the caveat that I never did proper nuclear physics, I have a very hard time seeing how the chemical bonds in a palladium matrix can overcome the sort of potentials you would need to overcome to get a hydrogen isotope nucleus close enough to fuse with a tungsten nucleus. The deuterium should be facing an unshielded 10+ elemental charge Coulomb potential long before it gets close enough to have a non-trivial probability of tunneling through it. The sort of energy you need to overcome that potential barrier is sufficient to tear apart the crystal lattice a thousand times over.

For that matter, I have a very hard time seeing how you can do any of that without getting at the very least hard x-rays when the electron structure re-settles.

So even though I haven't solved the quantum mechanic equations to conclusively prove that it's impossible, it is not something I would bet money on.

- Jake

If you only spend 20 minutes of the rest of your life on economics, go spend them here.

by JakeS (JangoSierra 'at' gmail 'dot' com) on Sat Jan 12th, 2013 at 11:19:58 AM EST
[ Parent ]
Haven't checked this particular research, but afaik previous attempts with using transsubstitution have been based on sending neutrons to change the target into an isotope that degradres faster.

Not anything anyone should have in a car, no.

A vote for PES is a vote for EPP! A vote for EPP is a vote for PES! Support the coalition, vote EPP-PES in 2009!

by A swedish kind of death on Sat Jan 12th, 2013 at 11:33:52 AM EST
[ Parent ]
I suspect this is not a matter of nuclear physics but of solid state physics. The point is that hydrogen has an unusually high affinity for adsorption into a palladium matrix. Maybe that means deuterium can easily propagate within the palladium matrix. Quantum mechanics tells us that a particle can propagate through a pure enough crystal as if it were in empty space, so maybe once the deuterium is adsorbed into the palladium the single electron is stripped into the metal's electron gas and the resulting free deuteron forms a deuterium gas. This might increase the interaction cross-section of the deuterons with the impurity metallic nuclei, increasing the rate of the interaction (for instance)

20Ca + 2D -> 22Sc* -> 22Ti + e- - νe

(I cannot write a bar over the nu for an antineutrino so I write 'minus one neutrino' :)

I distribute. You re-distribute. He gives your hard-earned money to lazy scroungers. -- JakeS

by Migeru (migeru at eurotrib dot com) on Sat Jan 12th, 2013 at 12:30:27 PM EST
[ Parent ]
Wikipedia: Chemical structure and properties of palladium hydride
Neutron diffraction studies have shown that hydrogen atoms randomly occupy the octahedral interstices in the metal lattice (in an fcc lattice there is one octahedral hole per metal atom). The limit of absorption at normal pressures is PdH0.7, indicating that approximately 70% of the octahedral holes are occupied. The absorption of hydrogen is reversible, and hydrogen rapidly diffuses through the metal lattice. Metallic conductivity reduces as hydrogen is absorbed, until at around PdH0.5 the solid becomes a semiconductor.
The bulk adsorption and rapid diffusion mean that the hydrogen atoms can tunnel from the centre of one octahedral site to the next with a relatively high amplitude. If metal impurities (say, a Calcium atom) also like to occupy the same insterstices that hydrogen propagates in, then the interaction cross-section of the hydrogen with the impurity atoms would be enhanced with respect to the gas phase.

I distribute. You re-distribute. He gives your hard-earned money to lazy scroungers. -- JakeS
by Migeru (migeru at eurotrib dot com) on Sat Jan 12th, 2013 at 12:47:53 PM EST
[ Parent ]
You should still get a gamma signature from the nuclear decay.

There are off-the-shelf detectors that can split a gamma signal into energy channels, so unless that signature is smack, dab in the middle of the potassium decay channel, you should be able to tell it from background easily enough.

- Jake

If you only spend 20 minutes of the rest of your life on economics, go spend them here.

by JakeS (JangoSierra 'at' gmail 'dot' com) on Sat Jan 12th, 2013 at 01:05:11 PM EST
[ Parent ]
You don't need a gamma ray because the entire energy could be carried away as kinetic energy by the electron and antineutrino. It's beta decay, not gamma decay!

I distribute. You re-distribute. He gives your hard-earned money to lazy scroungers. -- JakeS
by Migeru (migeru at eurotrib dot com) on Sat Jan 12th, 2013 at 02:33:05 PM EST
[ Parent ]
From the beta decay, yes. But after the beta decay the nucleus will be in an excited state because something that used to be a neutron is now a proton.

It would be quite extraordinary for you to find that replacing a neutron with a proton would not move the minimum energy configuration of the nucleus. And the transition to that new minimum energy configuration would most likely be accompanied by a gamma fingerprint, courtesy of the strong nuclear force.

- Jake

If you only spend 20 minutes of the rest of your life on economics, go spend them here.

by JakeS (JangoSierra 'at' gmail 'dot' com) on Sat Jan 12th, 2013 at 04:03:00 PM EST
[ Parent ]
Gah, I mixed up atomic mass and atomic number in that equation. Anyway...

Looking at the isotopes of Ca, Sc ad Ti, the observed transmutation looks definitely _special_.

Let's see... 40Ca is by far the most common Calcium isotope, but its atomic weight is way too low to support the transmutation to Titanium, as

Naturally occurring titanium (Ti) is composed of 5 stable isotopes; 46Ti, 47Ti, 48Ti, 49Ti and 50Ti with 48Ti being the most abundant (73.8% natural abundance).
(wikipedia)

Now, for Calcium

There are five observationally stable isotopes (40Ca, 42Ca, 43Ca, 44Ca and 46Ca), plus one isotope (48Ca) with such a long half-life that for all practical purposes it can be considered stable.
Although 40Ca is the most abundant isotope by far, only 44Ca, 46Ca and 48Ca can result in stable Titanium isotopes by Deuterium absorption followed by beta decay. 44Ca, with a natural abundance of a couple percent, would seem like the only suitable candidate.

The intermediate states would have to be 46Sc, 48Sc or 50Sc respectively, and given that

The isotopes of scandium range in atomic weight from 36 u (36Sc) to 60 u (60Sc). The primary decay mode at masses lower than the only stable isotope, 45Sc, is Beta-plus or electron capture, and the primary mode at masses above it is beta-minus. The primary decay products at atomic weights below 45Sc are calcium isotopes and the primary products from higher atomic weights are titanium isotopes.
it would seem to fit.

Now let's look at the energy balance. The isotopic masses of the Calcium and Titanium isotopes are (in atomic units):

Calcium    Titanium   Difference
43.9554818 45.9526316 1.9971498 
45.9536926 47.9479463 1.9942537
47.952534  49.9447912 1.9922572

while the mass of Deuterium is 2.01410178. The smallest energy release would be from the
44Ca -> 46Ti
reaction. The energy difference is 0.01695198u. The electron rest mass is 0.0005486u, so almost the entire energy difference would be released: 0.0164034u or 15MeV. That's a lot of energy...

I distribute. You re-distribute. He gives your hard-earned money to lazy scroungers. -- JakeS

by Migeru (migeru at eurotrib dot com) on Sat Jan 12th, 2013 at 02:30:17 PM EST
[ Parent ]
Watching the youtube, Iwamura talks about

44Ca + 22H -> 48Ti

"analogous to alpha absorption". There is no beta decay, therefore just gamma emission from nuclear rearrangement. And the goal is not energy but treatment of nuclear waste. If you can make it so that nuclear waste treatment doesn't cost energy, that's obviously a plus. But this isn't promising a hydrogen car.

I distribute. You re-distribute. He gives your hard-earned money to lazy scroungers. -- JakeS

by Migeru (migeru at eurotrib dot com) on Sun Jan 13th, 2013 at 05:22:12 AM EST
[ Parent ]
They are not talking about impurities in the Palladium matrix. They make a palladium "membrane". through which Deuterium flows due to a pressure difference of 1 Atmosphere. On the high pressure side of the membrane they deposit the metals they want to transmute, and they collect them after the experiment.

I distribute. You re-distribute. He gives your hard-earned money to lazy scroungers. -- JakeS
by Migeru (migeru at eurotrib dot com) on Sun Jan 13th, 2013 at 05:24:11 AM EST
[ Parent ]
My suggestion was simply that an unintended LENR of a type other than expected might have occurred in the infamous cold fusion experiments. Were it due to some uncontrolled contamination any products might not have been produced in a manner that their instruments would have detected. Others, or even they themselves, could have tried the intended experiment but have missed the contamination that allowed something to happen.

As the Dutch said while fighting the Spanish: "It is not necessary to have hope in order to persevere."
by ARGeezer (ARGeezer a in a circle eurotrib daught com) on Sun Jan 13th, 2013 at 05:28:12 PM EST
[ Parent ]
Iwamura is not talking about fusing hydrogen into helium, but apparently about catalysing other reactions in which metals are turned into other metals by absorbing a deuteron (followed by beta decay).
Apparently not: two deuterons are absorbed, without beta decay.

I distribute. You re-distribute. He gives your hard-earned money to lazy scroungers. -- JakeS
by Migeru (migeru at eurotrib dot com) on Sun Jan 13th, 2013 at 05:25:09 AM EST
[ Parent ]

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