Frank Gordon's "Lattice Energy Converter (LEC)"...replicators workshop

Hi guys! I always followed this forum with great interest, but I never posted before, though. However, this topic sparked in my mind some ideas that I would like to share with you.

A first thought is that the unknown radiation emitted by metals while adsorbing hydrogen was indeed observed and described by a number of researchers. Apart from Srinivasan, also E. Storms published an interesting paper on this topic: Storms, E. and B. Scanlan, Nature of Energetic Radiation Emitted from a Metal Exposed to H2. J. Condensed Matter Nucl. Sci., 2013. 11: p. 142-156.

Secondarily, I would suggest a very simple way to detect and potentially exploit this radiation. This method does not involve at all bi-metallic configurations and is completely immune to elettro-chemical effects. If a semiconductive p-n junction is introduced into the cell, the impinging radiation will generate electrons and holes in the p-n junction, and this will produce a measurable potential and current (the current will be proportional to the radiation flux). This is indeed the same principle of betavoiltaic batteries. The p-n junction can be a small solar cell or even a photodiode. If a voltage and/or current can be obtained this way, the presence of the radiation will be confirmed in the cell, behond any doubt realted to hypotetical electrochemical effects. Pheraps even an usefull level of power may be obtained (probably by using a planar configuration, with surface larger area, not a coaxial one).

From the basic reaction H-H --> H*-H* (or UDH) we know that we get about 495eV net energy.

Further the math and experimenst do show that the electron orbit binding force for the H*-H* is only 2/3 of H-H. Thus in average one elctron can be removed by 4.3 eV. This has also been shown by Holmlid over a verys larger temperature range. So for one H*-H* condensation you get about 415 free electrons. This explains the current. The voltage is as usual the delta in electro negativity of the involved partners.

Quote from Curbina

Can you point us to a reference of Holmlid about this, it would be appreciated Wyttenbach .

Fig. 11: ref. 28 in my paper.

99.9% D2O 104 euros. 100 mls.anything cheaper?
https://www.eurisotop.com/deuterium-oxide-5

Quote from Alan Smith

That's about par for the course for small. I have normally bought 1 kilo from Cambridge Isotopes UK for around $1000.

Here a link : https://shop.isotope.com/productdetails.aspx?itemno=DLM-4-PK

But it is only 99.9% D₂O so you have to buy a molecular sieve for further purification after the electrolyser! This is net 200g Deuterium.

Quote from Alan Smith

I see it's gone up quite a bit since this time last year.

The $ (dollar) went down too and the pound may be a tick up....Explains half of it.

As far as I can make out from Frank Gordon's slide presentation, there is no step where D2O is actually required required for producing this effect.

Quote from Alan Smith

Try looking at 3o minutes in.

I don't see anything in Gordon's slide presentation saying that D2O is needed to achieve the results he sees. In particular, I don't see anything saying this about 30 minutes in. He is pretty explicit throughout that either H2O or D2O work. And he doesn't seem to indicate that one works better than the other.

Deuterium is not necessary, I used only hydrogen. At this point, I don't know if deuterium is better or not than hydrogen, but this has to be checked

Pam's recipe specifies D20.... probably the silver /Cu wire control is unnecessary.

but that would be

a possible joie de vivre for some experimenter..

D2O is a bit expensive ... but less expensive than printer ink in $/ml..

As a potential experimenter I would do D2O./Pd/Li

then variations..on the theme

...H2O

...codeposition with Ni/ Ag/Er/Sm....

Maybe its possible to get into the milliwatt power range?

Note: Pam's recipe did do H2O but showed less tracks than with D20..

1. Chemicals and Materials
1.1 The following chemicals were obtained from Sigma-Aldrich: palladium chloride,
copper chloride, lithium chloride, sodium hydroxide, deuterated water, 0.25 mm diameter
silver wire, and 0.25 mm diameter platinum wire. These chemicals and metal wires are
readily available at any university.
1.2 Prepare plating solutions that is 0.03 M PdCl2 (or 0.03 M CuCl2 for the control
experiment) and 0.3 M LiCl. In a 50 mL volumetric flask, weigh out 0.27 g PdCl2
(0.2017 g CuCl2) and 0.636 g LiCl. Add about 20 mL D2O to the volumetric flask. The
PdCl2 will slowly go into solution (it forms a complex with the Clions from the LiCl, PdCl42-
). Once in solution, add the rest of the D2O (solution will be red-brown in color).

http://article.sapub.org/Supplementary%20Material.pdf

I wonder if much of the apparatus that Gordon and Whitehouse describe in their slide presentation is really necessary for seeing an effect. They hypothesize that the LEC current they measure is due to energetic charged particles emanating from a Pd-H or Pd-D co-deposited on the working electrode. They also mention in passing that their LEC generates current even if the interstitial space between the working and counter-electrodes is air. If all this is true, then shouldn't one be able to measure radiation simply from a co-deposited surface held in air? Looked at in this way, the rest of the LEC is simply just a way to measure energetic particles radiating from the prepared lattice and so could be replaced by commercial sensors with known characteristics.

Gordon and Whitehouse calculate that at 180 degC the current from their LEC corresponds to ~ 6 x10^13 Bq (assuming singly charged particles). Given the layout of their device, this implies that the co-deposited surface is emitting at something like 10^12 Bq/cm^2. Even at 80 degC, the LEC current they report corresponds to something on the order of 10^10 Bq/cm^2 .... a signal that could be measurable without the need to construct the whole LEC apparatus.