E-cat fraud

CharlesKramer · Post by **CharlesKramer** » Mon Feb 13, 2012 1:18 am

Rossi has a history of making wild claims (his "thermoelectric generator") that turn out not true. And he's done jail time (even though he was later acquitted).

Rossi's record begs suspicion.

To me, the only question is why Rossi's snake oil comedy show got as far as it did.

Possibly those who understand energy just wished his claims could be true. I mean... what a great dream! Who didn't wish it true? Some obscure guy armed with mostly tin foil and household ingredients solves the energy crisis.

A similar explanation may also explain those dreaming of a "hydrogen" economy (coming right up! as soon as someone finds the first hydrogen mine), or the claims that electric cars are 100% clean energy (DUH except for coal-based electricity and the pollution from manufacturing and disposing lithium batteries) or the wildly held view that a secret cabal are keeping secret automobiles that need only water to run.

Needless-to-add, I imply no criticism of THIS (talk-polywell) scienterrific group...

Maui · Post by **Maui** » Mon Feb 13, 2012 3:56 am

You act as though this is the first suggestion of fraud that has come up.

tomclarke · Post by **tomclarke** » Mon Feb 13, 2012 8:18 am

Maui wrote:You act as though this is the first suggestion of fraud that has come up.

This is not a well-written article. It suggests a possible way in which Rossi's results could have been faked (in practice mistake would be unlikely). There are also many other ways in which the results can come from bad experimental technique, with the possibility, but no certainty, of fraud.

So it does not advance things. Not, whatever you think of LENR, that they need much advancement in the case of Rossi.

TallDave · Post by **TallDave** » Mon Feb 13, 2012 11:41 pm

Did they not know about the later tests? Half a MW from bad wiring? Okay then.

My favorite part was when they called Krivit "a journalist who covers cold fusion claims."

Ivy Matt · Post by **Ivy Matt** » Tue Feb 14, 2012 3:06 pm

The announcement on the Australian Skeptics' website can be found here, the full press release can be found here, and a press release covering the technical details can be found here. The full press release contains the following statement:

Some scientists have remained rightly skeptical, such as Professor Peter Ekstrom of Lund University in Sweden. He says that nickel, hydrogen and copper are everyday materials that cannot be involved in a nuclear reaction.

Peter Ekström is a senior lecturer in nuclear physics at Lund University. The full press release also contains a link to a paper by Ekström, translated into English by a member of the Australian Skeptics, but the the paper contains nothing to support the statement contained in the full press release, as far as I can tell. As a matter of fact, it contains the following statement:

Rossi says that the device works by fusion of hydrogen and nickel. [If this was true, it would certainly produce energy], because capture of a proton by a nickel [nucleus] releases 8 MeV, which is the typical binding energy for a nucleus of atomic weight about 60.

Of course, this doesn't prove anything one way or the other about Rossi, but it does make me wonder about the Australian Skeptics, as I have a hard time believing the former statement (which the Australian Skeptics did not enclose in quotation marks) is something Ekström actually said.

ladajo · Post by **ladajo** » Tue Feb 14, 2012 5:50 pm

It now appears that Dick Smith, who brought in the Australian Skeptics to check into an investment request, subsequently denied based on the reporting, has posted a $1 Million offer to Rossi to repeat the March test with the Swedes. He has delineated pretty far terms, with one single caveat, that in addition to the Swedes, they add a mutually agreed thrid body to set up instrumentation and data collect.

If Rossi turns this down or fails to 'produce the body' it is not going to do well for his already tarnished endeavor.

The Rossiworld Turns.

http://blog.newenergytimes.com/2012/02/ ... s-through/

http://blog.newenergytimes.com/2012/02/ ... -cat-demo/

Skipjack · Post by **Skipjack** » Tue Feb 14, 2012 6:13 pm

Is that the James Randi foundations 1 million dollar challenge money, I wonder...
It would be really, really stupid of Rossi to not accept that challenge if he really has something. If he doesnt have anything, then he will of course come up with all sorts of excuses (you just have to see what certain psychics have been saying about Randis challenge and draw the parallels then). In the latter case, I would be completely convinced that he is a fraud.

Ivy Matt · Post by **Ivy Matt** » Tue Feb 14, 2012 6:30 pm

Ah, very good news. I was starting to worry about Rossi's financial situation when I heard the University of Bologna contract fell through, but now it appears Rossi's financial difficulties are finally over....

D Tibbets · Post by **D Tibbets** » Tue Feb 14, 2012 7:58 pm

Ivy Matt wrote:....
The full press release also contains a link to a paper by Ekström, translated into English by a member of the Australian Skeptics, but the the paper contains nothing to support the statement contained in the full press release, as far as I can tell. As a matter of fact, it contains the following statement:

Quote:
Rossi says that the device works by fusion of hydrogen and nickel. [If this was true, it would certainly produce energy], because capture of a proton by a nickel [nucleus] releases 8 MeV, which is the typical binding energy for a nucleus of atomic weight about 60.

If that quote is accurate, it is another example of misinformation, confusion. Fusing a single nucleon to nickel (any isotope) does not release ~ 8MeV of energy. Even if you ignore my previous discussion about Ni 62 having the least potential energy of any nucleus, and thus the turnaround point for exothermic vs endothermic fusion (and conversely fission). The energy yields either way is much smaller. Without going through the derivations again, a simple consideration demonstrates this point. A star will burn hydrogen to helium for billions of years. Helium will burn for hundreds of millions of years, carbon, oxygen, etc will burn for millions of years. If the star is hot enough the light nucleons can fuse up to iron (due to pathways involved). This is very close to Ni and fusions (or fissions) in this neighborhood yield little difference in free energy (heat). This is why the last stages of fusion in a star can produce heat to resist gravitational collapse for only a few thousand years. This is well understood and accepted and explains the evolution of stars. Each step of fusion to heavier elements up to Fe or Ni releases progressively less energy, thus each stage of the stars evolution lasts a shorter period of time.

The binding energy per nucleon of an isotope needs to be compared to the isotope that is the product (s) of the reaction. Keep in mind that protons and neutrons have ~ 0 binding energy per nucleon so arguments that they are contributing to the missing energy/ net binding energy is inappropriate. Only nucleons (protons and neutrons) contained within a parent nucleus with a Z>1 has any missing mass. On their own they are irrele4vent. Of course hydrogen (a proton) does not have any missing mass because it is the standard by which everything else is measured against.

If you take Ni62 and Cu63, the energy released (or absorbed- depending on which side of the argument you are on) will be the difference in the binding energy per nucleon of the respective isotopes. For Ni62 this is ~ 8.9 MeV, for Cu63 it is ~8.8 MeV. Converting from one to the other would thus involve a potential energy- kinetic energy difference of ~ 0.1 MeV (actually if carried out to several significant decimal places, the difference is ~ 0.04 MeV). This explains and is compatible with the stellar evolution time frames.

Again, some argue that the total binding energy is pertinent, which is wrong. We are not talking about the disassembly of every nucleon from a nucleus, but only removing (or adding) one nucleon between these close neighbors. It also ignores the importance of the two main opposing forces that are involved in the missing mass, but that is an issue of the energy direction of the reaction and I am not arguing that point here, only the magnitude of the difference between reactant and product.
Anyone that would challenge this point must provide an alternate explanation for stellar evolution.

Dan Tibbets

dkfenger · Post by **dkfenger** » Tue Feb 14, 2012 9:20 pm

Solar fusion reaction, from <http://library.thinkquest.org/3471/fusion.html>
4H1 + 2e => He4 + (gammas+neutrinos)
delta-E = [(4)(1.007825u) - 4.002603u][931 MeV/u] = 26.7 MeV

Apply the same math to the Hydrogen-Nickel reaction: (masses from Wiki, feel free to provide alternate numbers if you have a better source.)
H1 + Ni62 => Cu63
delta-E = [1.007825u + 61.9283451u - 62.9295975u][931 MeV/u]
delta-E = [0.0065726][931 MeV/u]
delta-E = 6.11 MeV (positive, same as above.)

Stellar evolution - when a star runs out of Hydrogen, it has to fuse heavier stuff. Iron, Nickel and higher do not produce energy when fused to other nuclei of similar weight (ie, what you have left once the Hydrogen, Helium, etc have run out). But as long as there is Hydrogen left, and things to fuse it to, energy production is still possible. It is not the presence of Iron that dooms stars, it's the lack of anything lighter.

D Tibbets · Post by **D Tibbets** » Tue Feb 14, 2012 11:20 pm

dkfenger wrote:Solar fusion reaction, from <http://library.thinkquest.org/3471/fusion.html>
4H1 + 2e => He4 + (gammas+neutrinos)
delta-E = [(4)(1.007825u) - 4.002603u][931 MeV/u] = 26.7 MeV

Apply the same math to the Hydrogen-Nickel reaction: (masses from Wiki, feel free to provide alternate numbers if you have a better source.)
H1 + Ni62 => Cu63
delta-E = [1.007825u + 61.9283451u - 62.9295975u][931 MeV/u]
delta-E = [0.0065726][931 MeV/u]
delta-E = 6.11 MeV (positive, same as above.)

Stellar evolution - when a star runs out of Hydrogen, it has to fuse heavier stuff. Iron, Nickel and higher do not produce energy when fused to other nuclei of similar weight (ie, what you have left once the Hydrogen, Helium, etc have run out). But as long as there is Hydrogen left, and things to fuse it to, energy production is still possible. It is not the presence of Iron that dooms stars, it's the lack of anything lighter.

Both parts of your response is wrong. You are talking about missing mass. That always goes up as you go to hevier ellements. But to equate that to the energy balance is a fallicy. Otherwise, ehy do heavier elements break down. If you are extracting heat from the reactions, you are going to elements with lower potential energy. You are not creating energy, you can only transfer the energy between the aviable total of kinetic energy and potential energy in the system. If heavier elements (>Ni62) are relaesing KE then they have lost potential energy. Thus they are more stable, and they will not breakdown- decay. There would be no limit on the mass of an element. Uranium would be one of the lighter elements possible. Atomic weights of 23 trillion would be possible and preferred from an energy viewpoint. The missing mass goes up, of course. But this missing mass is primarily in the form of the Strong force, and the electromagnetic force. They have opposite effects on the stability (potential energy) of the nucleus, they grow at different rates and there is a crossover point (there has to be , the slopes of the graphs cross) At this cross over point the attractive Strong force balances against the electromagnetic force. There is almost zero energy that can be converted into KE. the nuclei on either side have more stored energy, thus moving towards Ni62 releases KE, moving away absorbs KE.

Again (sigh) rember that protons and neutrons have no or almost no aviable potential energy by themselfs. It is the incorperation of these nucleons into heavier nuclei that changes the energy picture- through the gluon of photon mediators of the Strong and Electromagnetic effect.
You seem to assume that adding a alpha particle or heavier nuclei is a completely different process than adding protons or neutrons. It isn't . What is important is the potential energy of the resultant product/ nuclei. If the product has a lesser potential energy , then KE is released. If the potential energy is greater (less negative by convention) then it is thermodynamically impossible for KE to be released, it is an endothermic process. The binding energy per nucleon graphs, and tabular data shows this.

Who ever developed this nomenclature did a disservice. The issue that is paramount is the KE of the product. If it is less (more negative) than the sum of the potential energy of the reactants, then KE is released. Think of the binding energy per nucleon as the measure of the potential energy of the nucleus. Here you have to reconize the the negative sign that is used by convention for potential energy. The graph makes more sise if it was inverted. Also the baseline (zero ) should be set at the proton. A Ni62 with a potential energy of ~ minus 8.9 MeV this is a value that reflects the balance between the opposing forces, not the absolute sum of the opposing forceswhich would be the missing mass). Add a proton with 0 potential energy yields a Cu63 with ~ minus 8.8 MeV. The proton with zero potential energy is neutral to the energy balance., but the Cu 63 has a higher potential energy (less negative). That this nucleus hass more energy tied up in the nucleus means that KE has to be absorbed for the reaction to occur, it is endothermic.

Your stellar arguement is a catch 22. You say heavier nuclei fuse together and perhaps but protons do not contribute to the process because they have all been consumed in earlier steps. Of course this means that Cu63 could not be produced in nature through proton absorption. That means it could only exist due to heavy nuclei fusioning together. Neucleosynthesis is a complicated process, but even working very hard with various convolutions you would have a very difficult time getting to all of the elements/ isotopes.

And actually, the hydrogen in the star is only consumed to a small percentage. The core may be depleted, but the higher layers have plenty of hydrogne that becomes aviable in the final stages when the mixture and temperatures change rapidly. Nucleosynthesis progresses through three or four major processes. Proton absorption, an mostly neutron absorption through a slow or rapid process. then there is the alpha particle absorption which is another major pathway. During supernovas the neutron absorption and to a lesser degree proton absorption are more significant because of the mixing and the profuse fusions that are occurring in the superheated plasma of the exploding star. with the elevating temperatures and the zoo of various neutron producing fusions and rapid decays. These reactions are ongoing not because they are leading to lower potential energy states, thus releasing KE, they are occuring (the endothermic ones) because there is plenty of excess KE available, not from exothermic fusion processes, but due to runaway gravitational energy release. It is this gravitational collapse that provides the energy that drives the nucleosynthesis past Nickel/ iron. All elements heavier than Ni62 are batteries. Potential energy (specifically EXCESS electromagnetic energy) has been stored in them- an endothermic process, and this energy can be released through fissions, and decays of various types. But only if the product/s have a net lower potential energy. While the binding energy per nucleon graph is confusing for the way it is setup and defined, this aspect of the Ni62 nuclei being the most stable (thus the least potential energy) nucleus and on the peak of the graph illustrates the direction of energy flow without ambiguity. This is not my conclusion. Look at any graph. If energy direction is indicated, it always points towards Ni62 (or iron in some older graphs) for exothermic reactions. Note this is a generalization- there is some jaggedness to the graph for various reasons, but Ni62 is always the endpoint with the greatest stability. Also note that the slope of the graph (plus or negative) is very flat in this region. This goes with the relatively little energy change as you progress from one element to the next in this region. This fits with what I said earlier about stellar evolution. Read the graph. Everyone accepts that hydrogen to Helium4 is a very energetic reaction and provides a lot of heat for the star. This is reflected by the position of the elements on the chart. The same applies to uranium fission to it's daughter products. The differences in the height of the graph reflects this. Again note that the height of the graph changes very little between Ni and Cu. This reflects the significantly smaller energy yields when converting between them. If there was a KE difference of ~ 8 MeV between Ni62 and Cu63, they would reflect this on the graph. Basic graph interpretation immediately revels when when you are off*.

*Again. The energy difference is the sums of the binding energy per nucleon for the reactants and products. For the questioned reaction it is Ni62 with ~8.9 MeV + proton with 0 MeV yields Cu63 with ~8.8 MeV. The proton with 0 binding energy does not effect the energy balance, so the difference is ~ 0.1 MeV. Binding energy per nucleon is defined as the energy input needed to tear off one nucleon. Of course this definition results in zero for a proton. You cannot rip off a nucleon, so the proton has zero potential energy in this system ( things change if you are talking about Quarks, but that is an unneeded complication. If you do the same comparison with Uranium fusion, the two daughter products sum together as they have non zero binding energies per nucleon, but the ~ 2 neutrons produced are ~ 0 MeV binding energy per nucleon, so they are not contributing to the energy balance comparison. They do have considerable KE imparted to them by the reaction, but the driving energy came from the energy difference between the other reactants. It does not imply that the neutrons contributed to the energy, but are merely a passive carrier of the resultant energy release.

Dan Tibbets

Giorgio · Post by **Giorgio** » Wed Feb 15, 2012 12:23 am

Skipjack wrote:It would be really, really stupid of Rossi to not accept that challenge if he really has something. If he doesnt have anything, then he will of course come up with all sorts of excuses (you just have to see what certain psychics have been saying about Randis challenge and draw the parallels then). In the latter case, I would be completely convinced that he is a fraud.

He just refused the offer:

----------------------------------------------------------
Andrea Rossi

February 14th, 2012 at 6:23 PM <http://www.journal-of-nuclear-physics.c ... ent-185047>

Dear Archibald Fields:

This is a Clownerie. If this guy wants to test a 1 MW plant and has 1 million to spend he can buy a 1 MW plant, with a regular contract, that gives him all the necessary guarantees and to us the logic financial guarantees. Our plants Are tested by Our Customers and the Consultants they choose. I have not time at all for this clownery. Besides: when Our E-CATS will be in the market, this “millionaire” will have the chance to buy for few hundred dollars an E-Cat and test it as he wants, so why waste money? I do not need his money.

Warm Regards,
A.R.
----------------------------------------------------------

Skipjack · Post by **Skipjack** » Wed Feb 15, 2012 1:53 am

Ok, I am so done with Rossi now. I mean really?
He allegedly has his reactors running all the time. He obviously does not have money to pay the UoB. Yet he says that he does not want 1 million USD for simply doing the thing he is allegedly doing every day, but under controlled conditions?
I am pretty certain now that Rossi is a fraud.

ladajo · Post by **ladajo** » Wed Feb 15, 2012 3:41 am

He even twists the offer. That is classic...

"If he wants to test a 1MW..." blah blah blah.

Does he honestly think folks won't follow that Smith simply offered him to run HIS test AGAIN, and with HIS testers for $1 Million? And he even goes so far as to make out that Smith is a fraud... "this “millionaire” will have the chance to buy for few hundred dollars"

Rossi, you come across as a complete boob at this point. How deep can you dig?

As posted in the other thread, I am all ears for Rossibots to dismiss this as trivial. I see more empty rhetoric coming. Sigh.

tomclarke · Post by **tomclarke** » Wed Feb 15, 2012 10:05 am

D Tibbets wrote:
dkfenger wrote:Solar fusion reaction, from <http://library.thinkquest.org/3471/fusion.html>
4H1 + 2e => He4 + (gammas+neutrinos)
delta-E = [(4)(1.007825u) - 4.002603u][931 MeV/u] = 26.7 MeV

Apply the same math to the Hydrogen-Nickel reaction: (masses from Wiki, feel free to provide alternate numbers if you have a better source.)
H1 + Ni62 => Cu63
delta-E = [1.007825u + 61.9283451u - 62.9295975u][931 MeV/u]
delta-E = [0.0065726][931 MeV/u]
delta-E = 6.11 MeV (positive, same as above.)

Stellar evolution - when a star runs out of Hydrogen, it has to fuse heavier stuff. Iron, Nickel and higher do not produce energy when fused to other nuclei of similar weight (ie, what you have left once the Hydrogen, Helium, etc have run out). But as long as there is Hydrogen left, and things to fuse it to, energy production is still possible. It is not the presence of Iron that dooms stars, it's the lack of anything lighter.
Both parts of your response is wrong. You are talking about missing mass. That always goes up as you go to hevier ellements. But to equate that to the energy balance is a fallicy. Otherwise, ehy do heavier elements break down. If you are extracting heat from the reactions, you are going to elements with lower potential energy. You are not creating energy, you can only transfer the energy between the aviable total of kinetic energy and potential energy in the system. If heavier elements (>Ni62) are relaesing KE then they have lost potential energy. Thus they are more stable, and they will not breakdown- decay. There would be no limit on the mass of an element. Uranium would be one of the lighter elements possible. Atomic weights of 23 trillion would be possible and preferred from an energy viewpoint. The missing mass goes up, of course. But this missing mass is primarily in the form of the Strong force, and the electromagnetic force. They have opposite effects on the stability (potential energy) of the nucleus, they grow at different rates and there is a crossover point (there has to be , the slopes of the graphs cross) At this cross over point the attractive Strong force balances against the electromagnetic force. There is almost zero energy that can be converted into KE. the nuclei on either side have more stored energy, thus moving towards Ni62 releases KE, moving away absorbs KE.

Again (sigh) rember that protons and neutrons have no or almost no aviable potential energy by themselfs. It is the incorperation of these nucleons into heavier nuclei that changes the energy picture- through the gluon of photon mediators of the Strong and Electromagnetic effect.
You seem to assume that adding a alpha particle or heavier nuclei is a completely different process than adding protons or neutrons. It isn't . What is important is the potential energy of the resultant product/ nuclei. If the product has a lesser potential energy , then KE is released. If the potential energy is greater (less negative by convention) then it is thermodynamically impossible for KE to be released, it is an endothermic process. The binding energy per nucleon graphs, and tabular data shows this.

Who ever developed this nomenclature did a disservice. The issue that is paramount is the KE of the product. If it is less (more negative) than the sum of the potential energy of the reactants, then KE is released. Think of the binding energy per nucleon as the measure of the potential energy of the nucleus. Here you have to reconize the the negative sign that is used by convention for potential energy. The graph makes more sise if it was inverted. Also the baseline (zero ) should be set at the proton. A Ni62 with a potential energy of ~ minus 8.9 MeV this is a value that reflects the balance between the opposing forces, not the absolute sum of the opposing forceswhich would be the missing mass). Add a proton with 0 potential energy yields a Cu63 with ~ minus 8.8 MeV. The proton with zero potential energy is neutral to the energy balance., but the Cu 63 has a higher potential energy (less negative). That this nucleus hass more energy tied up in the nucleus means that KE has to be absorbed for the reaction to occur, it is endothermic.

Your stellar arguement is a catch 22. You say heavier nuclei fuse together and perhaps but protons do not contribute to the process because they have all been consumed in earlier steps. Of course this means that Cu63 could not be produced in nature through proton absorption. That means it could only exist due to heavy nuclei fusioning together. Neucleosynthesis is a complicated process, but even working very hard with various convolutions you would have a very difficult time getting to all of the elements/ isotopes.

And actually, the hydrogen in the star is only consumed to a small percentage. The core may be depleted, but the higher layers have plenty of hydrogne that becomes aviable in the final stages when the mixture and temperatures change rapidly. Nucleosynthesis progresses through three or four major processes. Proton absorption, an mostly neutron absorption through a slow or rapid process. then there is the alpha particle absorption which is another major pathway. During supernovas the neutron absorption and to a lesser degree proton absorption are more significant because of the mixing and the profuse fusions that are occurring in the superheated plasma of the exploding star. with the elevating temperatures and the zoo of various neutron producing fusions and rapid decays. These reactions are ongoing not because they are leading to lower potential energy states, thus releasing KE, they are occuring (the endothermic ones) because there is plenty of excess KE available, not from exothermic fusion processes, but due to runaway gravitational energy release. It is this gravitational collapse that provides the energy that drives the nucleosynthesis past Nickel/ iron. All elements heavier than Ni62 are batteries. Potential energy (specifically EXCESS electromagnetic energy) has been stored in them- an endothermic process, and this energy can be released through fissions, and decays of various types. But only if the product/s have a net lower potential energy. While the binding energy per nucleon graph is confusing for the way it is setup and defined, this aspect of the Ni62 nuclei being the most stable (thus the least potential energy) nucleus and on the peak of the graph illustrates the direction of energy flow without ambiguity. This is not my conclusion. Look at any graph. If energy direction is indicated, it always points towards Ni62 (or iron in some older graphs) for exothermic reactions. Note this is a generalization- there is some jaggedness to the graph for various reasons, but Ni62 is always the endpoint with the greatest stability. Also note that the slope of the graph (plus or negative) is very flat in this region. This goes with the relatively little energy change as you progress from one element to the next in this region. This fits with what I said earlier about stellar evolution. Read the graph. Everyone accepts that hydrogen to Helium4 is a very energetic reaction and provides a lot of heat for the star. This is reflected by the position of the elements on the chart. The same applies to uranium fission to it's daughter products. The differences in the height of the graph reflects this. Again note that the height of the graph changes very little between Ni and Cu. This reflects the significantly smaller energy yields when converting between them. If there was a KE difference of ~ 8 MeV between Ni62 and Cu63, they would reflect this on the graph. Basic graph interpretation immediately revels when when you are off*.

*Again. The energy difference is the sums of the binding energy per nucleon for the reactants and products. For the questioned reaction it is Ni62 with ~8.9 MeV + proton with 0 MeV yields Cu63 with ~8.8 MeV. The proton with 0 binding energy does not effect the energy balance, so the difference is ~ 0.1 MeV. Binding energy per nucleon is defined as the energy input needed to tear off one nucleon. Of course this definition results in zero for a proton. You cannot rip off a nucleon, so the proton has zero potential energy in this system ( things change if you are talking about Quarks, but that is an unneeded complication. If you do the same comparison with Uranium fusion, the two daughter products sum together as they have non zero binding energies per nucleon, but the ~ 2 neutrons produced are ~ 0 MeV binding energy per nucleon, so they are not contributing to the energy balance comparison. They do have considerable KE imparted to them by the reaction, but the driving energy came from the energy difference between the other reactants. It does not imply that the neutrons contributed to the energy, but are merely a passive carrier of the resultant energy release.

Dan Tibbets

dkfenger - I almost posted the same.

Dan,

I don't see how you can get out of this one. If you have some more complex argument that disagrees, then I suggest you look for holes in it.

The logic is simple.

Take 1 X Ni62 atom.
Take one slow neutron.

the total mass of the reactants is what goes in, no energy needed.

What comes out is:
1 X Cu63+ ion +
one beta particle +
excess energy (as K.E. of the beta).

The beta rest mass matches that of the extra electron counted in the Cu63 isotopic mass when compared with Ni62 mass but missing in the Cu63+ ion that is generated. (OK, there is a tiny correction for the electronic binding energies, which we ignore).

Otherwise, the product mass is that of the Cu63.

Now, unless you are way, way out on a limb, the energy produced must be equivalent to the mass lost, which is real. You could in principle transform a large block of Ni62 into Cu63 this way, with a number of nuetrons. The total masses must be as measured, which means a measurable mass decrease.