[ExI] [ACT] cold fusion

Tue May 8 04:04:26 UTC 2007

According to the abstract of:

Further evidence of nuclear reactions in the Pd/D lattice: emission of
charged particles

http://www.springerlink.com/content/75p4572645025112/

"A model based upon electron capture is proposed to explain the
reaction products observed in the Pd/D-D2O system."

 I don't think this is the same as deuteron to deuteron fusion.

-- 
Best, Jeff Davis

               "Everything's hard till you
                     know how to do it."
                              Ray Charles

On 5/7/07, Eugen Leitl <eugen at leitl.org> wrote:
> ----- Forwarded message from "Perry E. Metzger" <perry at piermont.com> -----
>
> From: "Perry E. Metzger" <perry at piermont.com>
> Date: Mon, 07 May 2007 15:45:24 -0400
> To: act at crackmuppet.org
> Subject: [ACT] cold fusion
>
>
> [People should feel free to forward this if they like.]
>
> Some people got a brand new "cold fusion in Palladium" paper published
> in the last couple of weeks, and it has made several news sources,
> including the Kurzweil blog.
>
> See, for example:
>  http://www.dailytech.com/article.aspx?newsid=7168
>
> A friend asked me what I thought, so I did a quick Fermi estimate.
>
> The answer is, I have a lot of trouble believing it. See below for
> detailed calculations.
>
> The general hypothesis given is that, if the phenomenon is real, what
> is happening in such cases is that the deuterons are close enough
> together to tunnel over the coulomb barrier into each other -- it
> would have to be that because the number of deuterons with enough
> thermal energy to fuse with another is essentially zero.
>
> So, what sort of tunneling probabilities are we talking about?
>
> I don't now enough to do the calculation "right", but, as I said, lets
> do a "Fermi estimate". Lets consider the coulombic repulsion of two
> protons held 5fm apart. A helium nucleus is order 1fm.
>
> An elementary charge is 1.6E-19 coulombs. The potential energy between
> two particles of that charge is
>
> U = \frac{e^2}{4\pi\epsilon_0 r}
>
> So we're talking (1.6E-19 ^ 2) / (4 * 3.14159 * 8.854E-12 * 5E-15)
>
> Plug that in to calc and I get about 4.60E-14 Joules.
>
> That doesn't seem like a lot, but an electron-volt is just 1.60E-19
> Joules -- so that's a barrier of 287,500 eV, which is pretty high for
> wee particles like this, and we're still probably several times too far
> away.
>
> Unfortunately, at this point I have a bit of a problem because the
> textbook solutions for tunneling through a potential well assume that
> we're talking about a square well (a step function) and the potential
> function here is not really like that. However, I've gone this far
> with the back-of-the-envelope, so lets go all the way. I doubt I'm
> *that* far off.
>
> The distance between nucleii in typical bonds is not less than about
> an angstrom, 1E-10m. I presume the way that Pd solvation would
> catalyze fusion would be by getting the deuterons closer together. So,
> lets do the calculation for .1 angstroms -- impossibly close given the
> energies that would require.
>
> We also need to estimate the thermal energy of the deuterons. Normally
> we'd say at room temperature (where these things are being run) that
> it would be about 3kT/2 with T=300K and k=1.38E-23J/K  or so, but
> we'll double it just to be kind and say 3kT for a value of about
> 1.25E-20J. Given the height of the barrier this isn't going to make
> much of a difference anyway, even if I multiplied it by ten.
>
> The mass is about 3.32e-27kg. I'll assume that the potential energy
> barrier is pretty "smooth" and pick a value for the square well
> barrier of 2.3E-14J -- an "average" of zero and something order of
> magnitude of the maximum.
>
> I'll follow the formula in one of my texts that says that the
> probability of getting through is very roughly:
>
>  prob = e^[-(2a/hbar)*sqrt(2m(U-E))]
>
> where "a" is the distance, U is the potential of the barrier and E is
> the kinetic energy of the particle attempting to tunnel.
>
> At .1 angstrom, the probability of tunneling across the square
> potential is somewhere in the range of 1 in 7.8E1021 events. Already
> we can see this isn't pretty.
>
> Now, how many "events" can we expect? Lets say we have a mole of
> deuterons (6.023E23) and that we have (probably far more than we can
> realistically expect, based on usual vibrational spectra, but it won't
> matter even if we're off by thousands) somewhere around around 1E10
> collisions a second. Lets consider the events over 1 hour.
>
> 6.023E23*1E10*3600 is about 2.2E37. Divide.
>
> That means we can expect to wait something like 3.5E986 hours between
> seeing fusion events -- something like 3.9E970 times the life of the
> universe.
>
> I might be off by ten or twenty orders of magnitude. Who cares, though?
> This mechanism doesn't look like a realistic possibility.
>
> Perry
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> ----- End forwarded message -----
> --
> Eugen* Leitl <a href="http://leitl.org">leitl</a> http://leitl.org
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