# E-Cat Test News Page at PESWiki

We’ve created a page over at PESWiki to track the news as it comes in on this: http://peswiki.com/index.php/News:E-Cat_Fuel_Analysis_and_Validation_Paper_Posted_October_8%2C_2014

We welcome your help in keeping it updated. We try to use GMT time with the date so we can keep things in their proper sequence of arrival.

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We use stars to highlight excellent coverage, and we also have a flag to mark “mainstream” news stories as they trickle in.

As usual, E-CatWorld dominates the coverage, doing an excellent job.

• Alan DeAngelis

Li(7) + Ni(58) > Ni(59) + Li(6) 1.74 MeV

Li(7) + Ni(59) > Ni(60) + Li(6) 4.13 MeV

Li(7) + Ni(60) > Ni(61) + Li(6) 0.569 MeV

Li(7) + Ni(61)) > Ni(62) + Li(6) 3.34 MeV

• Ged

Interesting idea. Where do you get those values, I am curious if there’s supposed to be some partical form to that energy out. Also, what role for the hydrogen? Helping to shuttle neutons by cycling to deuterium and back to protium?

• Alan DeAngelis

Hi Ged and Andreas:
Here is a video that explains how to calculate the energy released in a nuclear reaction.
To do the calculation you just take the deference of the masses of the reactants and the products (I got the masses out of an old handbook but you can find them on the web).
For example for Li(7) + Ni(61) > N(62) + Li(6)

Mass of reactants (amu):
Li(7) is 7.0160041
Ni(61) is 60.931060
Total mass of reactants
67.9470641

Mass of products (amu):
Li(6) is 6.0151223
Ni(62) is 61.928348
Total mass of products
67.9434703

So the mass lost in the reaction is 67.9470641 – 67.9470641 = 0.0035938 amu

Convert to MeV by multiplying by 931 MeV/amu

0.0035938 amu x 931 MeV/amu = 3.35 MeV

I think I have the how but the why is tougher. Yeah,maybe the hydrogen under pressure is acting like the cue ball that starts it off. And/or maybe it’s needed for the following reaction.
Li(7) + H(1) > He(4) + He(4)

And the He(4) is the cue ball.

• Ged

Thank you for the informative reply!

• Andreas Moraitis

That would be LENR without hydrogen… Is the hydrogen primarily utilized to increase the pressure inside the lattice? Maybe one could initiate the same reactions with a hydraulic press, one of the sort that is used for producing diamonds from graphite. These presses reach up to 80000 bar – however, I have no idea if that would be enough.

• Alan DeAngelis

PS
Li(7) + Ni(62) > N(63) + Li(6) -0.413 MeV
It’s endothermic so it stops at Ni(62). So Ni(62) builds up and Li(7) goes to Li(6)

• Andreas Moraitis

If a neutron were stripped off the 7Li and captured by a Ni nucleus, you would get mostly stable isotopes. From the list above, only 59 Ni is unstable and decays by electron capture to 59Co. Since its half-life is relatively long (7.6×10^4 y), one would not expect to see a lot of cobalt in the ash, though. Besides, it could be transmuted to 60Ni before it decays. But at least one could look for X-rays and bremsstrahlung from betas:

http://ie.lbl.gov/toi/nuclide.asp?iZA=280059

• Andreas Moraitis

I assume that the given X-rays are the bremsstrahlung. But perhaps there are additional options if the lattice contains other elements than nickel. I’m not sure.

• Pekka Janhunen

Interesting idea. Three questions. 1) There should anyway be a small amount of Ni(59) produced which is slightly radioactive. But no Ni(59) was seen and no radioactivity was measured. 2) What mechanism destroyed the original Ni-64? I calculated that the reaction with Ni-64 would be endothermic as with Ni-62. 3) The fuel contained about ten times fewer lithium atoms than nickel atoms, but the proposed mechanism consumes several Li-7’s per each Ni-58 transmuted to Ni-62. The reactor must have contained more lithium somewhere, which of course is not impossible.

If one allows for the end product to be something else than Li(6), the most exothermic reactions seem to be the following (although I did not make quite exhaustive search):

Li(7) + Ni(58) > Ni(60) + He(4) + H(1) 9.44 MeV
Li(7) + Ni(60) > Ni(62) + He(4) + H(1) 7.47 MeV
Li(7) + Ni(61) > Ni(60) + He(4) + He(4) 10.3 MeV
Li(7) + Ni(62) > Ni(61) + He(4) + He(4) 7.53 MeV
Li(7) + Ni(64) > Ni(63) + He(4) + He(4) 8.47 MeV
Li(7) + Ni(64) > Fe(59) + C(12) 8.47 MeV
So this would explain why there is no Ni(59). But Ni(64) would produce Ni(63) and Fe(59) which are radioactive. Ni(62) would also be consumed, and the end products would be isotopes 60, 61 and 62. It’s a promising avenue of thought, anyway.

• Pekka Janhunen

Sorry I made an elementary error… all He(4)+He(4) branches are wrong… I’ll fix it

• Andreas Moraitis

The 64Ni question seems to be the decisive one. I’m not quite sure about the Li. If the fuel is a mixture of Ni powder and LiAlH4, the tiny amounts they have analyzed might not have been representative for the overall composition. 59Ni could be transmuted quickly into 60Ni before is has the chance to decay, see my comment below.

• Alan DeAngelis

I’ve been thinking about IR stretching of the Nickel hydride bonds enabling Ni(64)H2absorb its protons and form Ni(62) by the following
reactions.
H(1) + Ni(64) > Cu(65)* Step1

H(1) + Cu(65)* > Ni(62) + He(4) Step 2
________________________
Over all
2 H(1) + Ni(64) > Ni(62) + He(4) 11.8 MeV

Or if it’s symmetrical stretching.

H(1) + Ni(64) > Zn(66)* Step1

Zn(66)* > Ni(62) + He(4) Step 2

Over all
2 H(1) + Ni(64) > Ni(62) + He(4) 11.8 MeV

• Alan DeAngelis

Yeah, in the excitement I didn’t think of the stoichiometry. Not enough lithium. Coming back down to earth.

• Alan DeAngelis

PS
On second thought, what if he’s adding just plain old lithium hydride LiH (melting point 688 C). He could packing a lot of that in mole wise without adding too much weight wise.

• Alan DeAngelis

PPS
Pardon me “ten times fewer lithium atoms than nickel atoms”.
Coming down to earth again.

• Pekka Janhunen

But the amount of lithium is something which is not completely set in stone. It’s not impossible that there might be some additional lithium source inside the reactor. The amount of lithium in the ash is not very reliable measure since lithium is mostly in gaseous form at 1400 C so it condenses to all surface when the reactor cools. It doesn’t necessarily stay in the nickel particles.

• tlp

This has been discussed also in vortex-l, where mixent first presented those same as Alan DeAngelis:

https://www.mail-archive.com/[email protected]/msg98050.html

• Pekka Janhunen

(to tip, to Andreas M) Thanks, seems a very good comment by mixent (and written on the same day that the report was released!). If it’s neutron tunnelling from lithium to nickel as he says, then my helium etc producing branches are irrelevant. Maybe Ni(64) was not actually depleted. Its count rate in Table 1 of Appendix 3 (page 42) is given as “approximately zero”. Maybe the measurement accuracy just wasn’t enough to detect it, in case of ash analysis.

• Alan DeAngelis

• Alan DeAngelis

I like yours better. You have an answer for the carbon. Did they mention carbon (stopped reading after awhile to give my eyes a rest)?

• Alan DeAngelis

PS

Maybe the high MeV protons go onto make this reaction:

2H(2) + Li(7) > 2He(4) 17.4 MeV

With your route there’s a chain reaction. You don’t have to increase the Li(6) just decrease the Li(7).
Much better. But I’m forgetting that we need more lithium.

• Alan DeAngelis

Maybe not a chain reaction but a two for one.

• Alan DeAngelis

Li(7) + Ni(58) > Ni(59) + Li(6) 1.74 MeV

Li(7) + Ni(59) > Ni(60) + Li(6) 4.13 MeV

Li(7) + Ni(60) > Ni(61) + Li(6) 0.569 MeV

Li(7) + Ni(61)) > Ni(62) + Li(6) 3.34 MeV

• Ged

Interesting idea. Where do you get those values, I am curious if there’s supposed to be some particle form to that energy out. Also, what role for the hydrogen? Helping to shuttle neutons by cycling to deuterium and back to protium?

• Alan DeAngelis

Hi Ged and Andreas:
Here is a video that explains how to calculate the energy released in a nuclear reaction.
To do the calculation you just take the deference of the masses of the reactants and the products (I got the masses out of an old handbook but you can find them on the web).
For example for Li(7) + Ni(61) > N(62) + Li(6)

Mass of reactants (amu):
Li(7) is 7.0160041
Ni(61) is 60.931060
Total mass of reactants
67.9470641

Mass of products (amu):
Li(6) is 6.0151223
Ni(62) is 61.928348
Total mass of products
67.9434703

So the mass lost in the reaction is 67.9470641 – 67.9470641 = 0.0035938 amu

Convert to MeV by multiplying by 931 MeV/amu

0.0035938 amu x 931 MeV/amu = 3.35 MeV

I think I have the how but the why is tougher. Yeah,maybe the hydrogen under pressure is acting like the cue ball that starts it off. And/or maybe it’s needed for the following reaction.
Li(7) + H(1) > He(4) + He(4)

And the He(4) is the cue ball.

• Ged

Thank you for the informative reply!

• Andreas Moraitis

That would be LENR without hydrogen… Is the hydrogen primarily utilized to increase the pressure inside the lattice? Maybe one could initiate the same reactions with a hydraulic press, one of the sort that is used for producing diamonds from graphite. These presses reach up to 80000 bar – however, I have no idea if that would be enough.

• Alan DeAngelis

PS
Li(7) + Ni(62) > N(63) + Li(6) -0.413 MeV
It’s endothermic so it stops at Ni(62). So Ni(62) builds up and Li(7) goes to Li(6)

• Andreas Moraitis

If a neutron were stripped off the 7Li and captured by a Ni nucleus, you would get mostly stable isotopes. From the list above, only 59 Ni is unstable and decays by electron capture to 59Co. Since its half-life is relatively long (7.6×10^4 y), one would not expect to see a lot of cobalt in the ash, though. Besides, it could be transmuted to 60Ni before it decays. But at least one could look for X-rays and bremsstrahlung from betas:

http://ie.lbl.gov/toi/nuclide.asp?iZA=280059

• Andreas Moraitis

I assume that the given X-rays are the bremsstrahlung. But perhaps there are additional options if the lattice contains other elements than nickel. I’m not sure.

• Pekka Janhunen

Interesting idea. Three questions. 1) There should anyway be a small amount of Ni(59) produced which is slightly radioactive. But no Ni(59) was seen and no radioactivity was measured. 2) What mechanism destroyed the original Ni-64? I calculated that the reaction with Ni-64 would be endothermic as with Ni-62. 3) The fuel contained about ten times fewer lithium atoms than nickel atoms, but the proposed mechanism consumes several Li-7’s per each Ni-58 transmuted to Ni-62. The reactor must have contained more lithium somewhere, which of course is not impossible.

If one allows for the end product to be something else than Li(6), the most exothermic reactions seem to be the following (although I did not make quite exhaustive search):

Li(7) + Ni(58) > Ni(60) + He(4) + H(1) 9.44 MeV
Li(7) + Ni(60) > Ni(62) + He(4) + H(1) 7.47 MeV
Li(7) + Ni(61) > Ni(60) + He(4) + He(4) 10.3 MeV
Li(7) + Ni(62) > Ni(61) + He(4) + He(4) 7.53 MeV
Li(7) + Ni(64) > Ni(63) + He(4) + He(4) 8.47 MeV
Li(7) + Ni(64) > Fe(59) + C(12) 8.47 MeV
So this would explain why there is no Ni(59). But Ni(64) would produce Ni(63) and Fe(59) which are radioactive. Ni(62) would also be consumed, and the end products would be isotopes 60, 61 and 62. It’s a promising avenue of thought, anyway.

• Pekka Janhunen

Sorry I made an elementary error… all He(4)+He(4) branches are wrong… I’ll fix it

• Andreas Moraitis

The 64Ni question seems to be the decisive one. I’m not quite sure about the Li. If the fuel is a mixture of Ni powder and LiAlH4, the tiny amounts they have analyzed might not have been representative for the overall composition. 59Ni could be transmuted quickly into 60Ni before is has the chance to decay, see my comment below.

• Alan DeAngelis

I’ve been thinking about IR stretching of the Nickel hydride bonds enabling Ni(64)H2absorb its protons and form Ni(62) by the following
reactions.
H(1) + Ni(64) > Cu(65)* Step1

H(1) + Cu(65)* > Ni(62) + He(4) Step 2
________________________
Over all
2 H(1) + Ni(64) > Ni(62) + He(4) 11.8 MeV

Or if it’s symmetrical stretching.

H(1) + Ni(64) > Zn(66)* Step1

Zn(66)* > Ni(62) + He(4) Step 2

Over all
2 H(1) + Ni(64) > Ni(62) + He(4) 11.8 MeV

• Alan DeAngelis

Yeah, in the excitement I didn’t think of the stoichiometry. Not enough lithium. Coming back down to earth.

• Alan DeAngelis

PS
On second thought, what if he’s adding just plain old lithium hydride LiH (melting point 688 C). He could packing a lot of that in mole wise without adding too much weight wise.

• Alan DeAngelis

PPS
Pardon me “ten times fewer lithium atoms than nickel atoms”.
Coming down to earth again.

• Pekka Janhunen

But the amount of lithium is something which is not completely set in stone. It’s not impossible that there might be some additional lithium source inside the reactor. The amount of lithium in the ash is not very reliable measure since lithium is mostly in gaseous form at 1400 C so it condenses to all surface when the reactor cools. It doesn’t necessarily stay in the nickel particles.

• tlp

This has been discussed also in vortex-l, where mixent first presented those same as Alan DeAngelis:

https://www.mail-archive.com/[email protected]/msg98050.html

• Pekka Janhunen

(to tip, to Andreas M) Thanks, seems a very good comment by mixent (and written on the same day that the report was released!). If it’s neutron tunnelling from lithium to nickel as he says, then my helium etc producing branches are irrelevant. Maybe Ni(64) was not actually depleted. Its count rate in Table 1 of Appendix 3 (page 42) is given as “approximately zero”. Maybe the measurement accuracy just wasn’t enough to detect it, in case of ash analysis.

• Alan DeAngelis

• Alan DeAngelis

I like yours better. You have an answer for the carbon. Did they mention carbon (stopped reading after awhile to give my eyes a rest)?

• Alan DeAngelis

PS

Maybe the high MeV protons go onto make this reaction:

2H(2) + Li(7) > 2He(4) 17.4 MeV

With your route there’s a chain reaction. You don’t have to increase the Li(6) just decrease the Li(7).
Much better. But I’m forgetting that we need more lithium.

• Alan DeAngelis

Maybe not a chain reaction but a two for one.

• kc

Could you add me to the wiki? I’m definitely not a spam bot. I’m a 3d printing nut and a very interested follower of this technology.

• sterlingda

To sign you up, we need your name and email. You can send that to [email protected]

• kc

Thanks Stuart. I’ve tried to email that address however it keeps coming back failed.