In-Reply-To: <[EMAIL PROTECTED]>
OK, I think I've found answers to some of my questions, but
not others, and I've got some new ones, too.
Choosing a Mills paper at random -- "Formation of a Hydrogen
Plasma from an Incandescently Heated Hydrogen-Catalyst Gas
Mixture with an Anomalous Afterglow Duration" - H_Plasma1.pdf
-- he writes first of all that n = 1 for the ground state,
not 0, so I was wrong about that, and also:
hydrino atom binding energy = 13.6 ev / n^2
where n = 1/2, 1/3, 1/4, ... 1/p
Thus binding energy = 13.6 ev * p^2
So the difference in binding energy between n = 1/p and n = 1/(p+1)
would be:
13.6 ev * ((p + 1)^2 - p^2) = 13.6 (2p + 1)
Mills writes that the transition from n = 1 to n = 1/2 gives 40.8 ev.
This would be p = 1 in the formula, 13.6 * 3, which does equal 40.8 .
The change from n = 1/10 to 1/11 plus 1/11 to 1/12 would give:
13.6 * (2 * 10 + 1 + 2 * 11 + 1) = 598.4 ev
and from 1/120 to 1/122:
13.6 * (2 * 120 + 1 + 2 * 121 + 1) = 6582.4 ev
exactly as Robin wrote.
I guess this also explains Jones Beene's recent remark:
BTW, if one wished to maximize hydrino "manufacture" then it
would seem that a combination of both Rb, K and Sr
electrolytes would be an improvement as they cover different
IP ranges. Since you need to get to the first stage quickly,
I would suggest that half or more of the mole% be Rb hydroxide.
since various hydrino level increases require catalysts that absorb
different amounts of energy.
This still leaves the questions of:
1. How are deeper hydrino level transitions catalyzed, since
chemical catalysts can't absorb hundreds or thousands of ev,
and many-body collisions are too improbable?
If it's hydrinos catalyzing other hydrinos, does this release
any net energy?
2. What determines the partition of the liberated energy
between the catalyst and the hydrino or UV photon?
This would be very important, because if the catalyst only absorbs
part of the released energy, the fraction it absorbs determines
the size of the electron-energy transition that it has to possess,
and thus which elements or molecules are eligible as catalysts.
Or does the catalyst absorb all of the energy and then give some
back to the hydrino and/or UV photon?
3. And on another subject, the HSG FAQ says:
"Being extremely light, [hydrinos] rapidly float up into the
atmosphere and diffuse into space."
But since a hydrino is much smaller than a normal H atom, and
still weighs 1 amu, wouldn't it be very dense and (since it is
so tiny) tend to fall toward the center of the Earth?
4. Also, are hydrinos toxic? Deuterium is (mildly). If we can
create a practical hydrino power generator, will it be necessary
to trap and store the hydrinos to keep them from contaminating
the ground water?
5. If so, will the hydrino storage tanks blow up some day from
cross-hydrino reactions, or even turn into fusion bombs if the
hydrinos get small enough to approach the nuclei of other atoms
or each other within reach of the strong nuclear force? Which
is my favorite theory of cold fusion and transmutation, because
it's simple enough that I can understand it! 8^)
Mark
On Sun, Jan 30, 2005 at 10:35:59AM -0800, Mark S Bilk wrote:
>In-Reply-To: <[EMAIL PROTECTED]>
>On Sun, Jan 30, 2005 at 04:29:47PM +1100, Robin van Spaandonk wrote:
>
>>When H[n=1/3 (or more)] is formed from H, a total of 108.8 eV
>>is liberated. Of this, 54.4 eV goes to the catalyst, leaving
>>54.4 eV either in the form of UV, or as kinetic energy of the
>>hydrino.
>
>This change is from n = 0 to n = 1/2, and n = 1/2 to n = 1/3.
>So each incremental change in hydrino "level" -- i.e., change
>of n from 1/k to 1/(k+1) -- liberates 54 ev.
>
>I thought 27.2 ev is liberated for each hydrino level increase
>from n = 0 to n = 1/2, or n = 1/k to n = 1/(k+1), where k > 1.
>
>But you're saying that the catalyst gets 27.2 ev per level
>increase and the hydrino, or a UV photon, also gets 27.2 ev.
>
>>by the time n gets to e.g. n=1/10, a drop of 2 levels,
>>such as would be catalyzed by O++, to n=1/12, results in an
>>energy release of 598 eV, ...
>
>>(For n=1/120 -> n=1/122 this is 6582 eV according to Mills).
>
>You're saying that the drop from n = 1/10 to n = 1/12 produces
>598 ev, or 299 ev per level increment, and n=1/120 to n=1/122
>produces 3291 ev per level increment!
>
>I thought the energy released for each increase in hydrino level
>was the same, 27.2 ev -- at least this is the amount that the
>catalyst has to absorb -- for any change of n = 1/k to n = 1/(k+1).
>
>But if the energy released per level increment increases at
>greater levels, then the original catalysts would only work for
>the first few levels. What kind of catalyst would absorb 299 ev
>or 3291 ev?
>
>What is the formula for energy released for an incremental
>hydrino level increase, at a given level?
>
>
