Dan,
PS - if you admit the overall reality of the mass balance, I am sure there are a number of people here, including me, who can help you to figure out why your argument does not apply.
E-cat fraud
It amazes me how omniscient folks think themselves to be at times. There are a number of reasons, based on possible factors I (and skip) can know nothing about, that would argue against is doing any such "challennge" other than by the method he already stated, sell a unit under normal contractual safeguards and perform acceptance testing.Skipjack wrote:It would be really, really stupid of Rossi to not accept that challenge if he really has something.
Ignoring your previous incorrect discussions that merely demonstrated your confusion between stability and energy content; you are partially correct here. The p+62Ni = 62Cu reaction does not produce 8MeV of energy. It produces only ~6.1MeV.D Tibbets wrote: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.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.
However, though it is a matter of opinion, but I would not call this "much smaller", at least in this context.
And this yet again demonstrates your confusion. The B/A is a matter of STABILITY, not energy release. It does play into how LIKELY the reaction is to happen under normal circumstances, but not directly into how much energy is released. For that you need to compare TOTAL binding energy.D Tibbets wrote: 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.
(B/A = 8.79 and 8.75 respectively)D Tibbets wrote: For Ni62 this is ~ 8.9 MeV, for Cu63 it is ~8.8 MeV.
Not at all. This is just a misunderstanding on your part.D Tibbets wrote: 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.
Dan, final total state - initial total state = change in state. Final binding energy - initial binding energy = binding energy change. Please learn that and stop making a fool out of yourself!D Tibbets wrote: 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.
Since it has nothing to do with stellar evolution per se, no change is needed, other than your understanding of it.D Tibbets wrote: Anyone that would challenge this point must provide an alternate explanation for stellar evolution.
IN A PLASMA (like a star) where particles are pretty much tiny little balls-o-stuff, when a p/62Ni pair have enough energy to overcome the coulomb barrier, the Cu63 that is produced contains enough "binding energy excitation to rip itself apart again. It would RATHER be a proton and a 62Ni from a stability standpoint. In the absence of any intervening act, that is what it will do a VAST, HUGELY PREDOMINANT proportion of the time. But not ALWAYS. Sometimes, it emits a gamma or ICs a passing electron or whatever and remains a 63Cu, having shed the excess 6.1MeV binding energy difference (plus whatever differential kinetic energy results between the particles). Under stellar conditions, this almost never happens. Billions, trillions, maybe hexillions to one against. But when it does, it releases the energy. Got it?
To others, why do I bother? Because his meme is akin to a destructive virus, and I will continuously try to eradicate that meme. He hasn't learned yet. But I am ever hopeful that he can.
Actually, you have this inversed. He is right, you are wrong.D Tibbets wrote: Both parts of your response is wrong.
Dude. E=Mc². It is the most famous equation in the world. The more mass missing, the more energy released. It is really that simple.D Tibbets wrote: 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.
Because the sum of the masses of the fission products and neutrons is LESS than the sum of the masses of the fissile atom + neutron. Simple.D Tibbets wrote:Otherwise, ehy do heavier elements break down.
If I extract the energy of the volitile combustable gasses from wood, that doesn't mean the charcoal left over has no energy remaining to give.D Tibbets wrote:If you are extracting heat from the reactions, you are going to elements with lower potential energy.
When you combine a neutron with an atom that is larger than 58Ni you release energy, but not ALL that is possible to release. Keep adding neutrons, you release energy from each neutron but not ALL the energy that can be given. After a while, there is a lot of stored up energy potential. That allows fission.
INCORRECT!!!! They have ALL the potential energy in these cases. (potential energy being the opposite of binding energy).D Tibbets wrote:Again (sigh) rember that protons and neutrons have no or almost no aviable potential energy by themselfs.
Using these neat buzz words, it is the gluon binding between the solitary neutron (nucleon) and the rest of the nucleous that overcomes/overwhelms the repulsive coulohm energy and releases all the energy in question.D Tibbets wrote: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.
Except that an alpha is a remarkably stable nucleus. It has converted ALMOST all the mass energy it is possible to convert. Indeed, radioactive decay of large elements STOP when the binding energy/nucleon of the daughter+alpha is no longer greater than that of the parent atom.D Tibbets wrote: 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 .
You almost had it. Several statements you made here were correct. What you got wrong was equating the binding energy per nucleon graph with potential energy. In this parlance, the potential energy is the amount of mass that CAN be converted to energy.D Tibbets wrote: 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.
- * Free nucleons have lost ZERO mass thus have the most potential energy.
* The nucleons in Ni have lost the MOST mass it is possible to loose in normal matter and have the least potential energy per nucleon.
*If you combine a proton with a Ni, the proton looses a LOT of mass and every nucleon in the Ni gains an itsy bit of mass. But the total mass in the Cu is lower than the sum of the mass of the Ni and proton. Thus the reaction converts mass to energy, EXOthermic.
Oops, I totally forgot about the Talk Polywell p+Ni exothermic/endothermic controversy. My apologies for bringing it up again. 
Regarding Rossi's financial situation, some food for thought:
If I recall correctly, sometime between his first meeting with Focardi and the January 2011 demonstration he sold his factory (the one that been heated for two years by an early E-Cat prototype) and devoted his entire efforts to developing the E-Cat.
In March of 2011 Rossi said he was assuming all the financial risks for the development of the E-Cat.
In May of 2011 the vice president of Ampenergo said they had given Rossi money for the distribution rights to North and South America (I wonder if Rossi still remembers that agreement), and implied that the money had been an important consideration.
On June 21, 2011 Rossi signed a €500,000 research agreement with the University of Bologna. The research was to begin after Rossi had paid the money and delivered a test device.
On July 15, 2011 Rossi agreed to fund a test at NASA Marshall for $50,000 of his own money.
On July 22, 2011 Rossi withdrew his offer and told NASA Marshall they could buy his device for $15 million if they wanted to test it.
On August 2-4, 2011 Rossi met with Quantum Energy Technologies and offered to sell them a device for $150 million. QET agreed to give him $500,000 if the E-Cat passed an initial test, with more money forthcoming.
On August 4, 2011 Rossi announced that he had broken his agreement with Defkalion. The reason for the break was that Defkalion had failed to pay him, apparently because he had failed to provide them with a device that demonstrated the stability they required.
On September 5-6, 2011 QET personnel attempted to test Rossi's device. The results were inconclusive, so they refused payment of the $500,000.
On September 16, 2011 Rossi announced that he had "very big financial problems". He later told PESN he had sold his home.
On October 28, 2011 Rossi held his demonstration of a megawatt E-Cat for an unidentified customer. He later said that the customer had been satisfied and purchased the device.
On November 18, 2011 Rossi told Focus he had an order for a 13 megawatt E-Cat from a military customer.
On November 21, 2011 Rossi announced on his website that he would be taking pre-orders from customers interested in buying a 10 kW E-cat for €400 per thermal kW, and that he would accept orders only if he received 10,000 names.
On December 9, 2011 Ross said he had met his target of 10,000 names.
On January 5, 2012 Rossi told a concerned poster on his website "we are a well fueled warship".
On January 15, 2012 Rossi's research contract with the University of Bologna was terminated, as he had failed to pay the €500,000 by then.
On January 17, 2012 (according to Steven Krivit) Rossi sent Sol Millin an invoice of €100,000 (US $131,000) for the Australasian E-Cat distribution rights. Sol Millin did not have the money because Dick Smith, who had offered him an investment of $200,000 (AUS, presumably) if he could be convinced the E-Cat worked, was not convinced. In response, Millin threatened to sue Smith for $100 million (AUS, no doubt) in damages (due to loss of the distribution rights) for Smith's failure to meet his "obligation". (Millin claimed he had proven sufficiently that the technology worked. One wonders how.)
On February 9, 2012 Rossi said he had received about 100,000 pre-orders.
On February 14, 2012 Rossi rejected Dick Smith's offer of US $1 million to repeat his test of March 29, 2011.
Note: In the interest of full disclosure, I am not one (or more) of the ~100,000 pre-orders.
Regarding Rossi's financial situation, some food for thought:
If I recall correctly, sometime between his first meeting with Focardi and the January 2011 demonstration he sold his factory (the one that been heated for two years by an early E-Cat prototype) and devoted his entire efforts to developing the E-Cat.
In March of 2011 Rossi said he was assuming all the financial risks for the development of the E-Cat.
In May of 2011 the vice president of Ampenergo said they had given Rossi money for the distribution rights to North and South America (I wonder if Rossi still remembers that agreement), and implied that the money had been an important consideration.
On June 21, 2011 Rossi signed a €500,000 research agreement with the University of Bologna. The research was to begin after Rossi had paid the money and delivered a test device.
On July 15, 2011 Rossi agreed to fund a test at NASA Marshall for $50,000 of his own money.
On July 22, 2011 Rossi withdrew his offer and told NASA Marshall they could buy his device for $15 million if they wanted to test it.
On August 2-4, 2011 Rossi met with Quantum Energy Technologies and offered to sell them a device for $150 million. QET agreed to give him $500,000 if the E-Cat passed an initial test, with more money forthcoming.
On August 4, 2011 Rossi announced that he had broken his agreement with Defkalion. The reason for the break was that Defkalion had failed to pay him, apparently because he had failed to provide them with a device that demonstrated the stability they required.
On September 5-6, 2011 QET personnel attempted to test Rossi's device. The results were inconclusive, so they refused payment of the $500,000.
On September 16, 2011 Rossi announced that he had "very big financial problems". He later told PESN he had sold his home.
On October 28, 2011 Rossi held his demonstration of a megawatt E-Cat for an unidentified customer. He later said that the customer had been satisfied and purchased the device.
On November 18, 2011 Rossi told Focus he had an order for a 13 megawatt E-Cat from a military customer.
On November 21, 2011 Rossi announced on his website that he would be taking pre-orders from customers interested in buying a 10 kW E-cat for €400 per thermal kW, and that he would accept orders only if he received 10,000 names.
On December 9, 2011 Ross said he had met his target of 10,000 names.
On January 5, 2012 Rossi told a concerned poster on his website "we are a well fueled warship".
On January 15, 2012 Rossi's research contract with the University of Bologna was terminated, as he had failed to pay the €500,000 by then.
On January 17, 2012 (according to Steven Krivit) Rossi sent Sol Millin an invoice of €100,000 (US $131,000) for the Australasian E-Cat distribution rights. Sol Millin did not have the money because Dick Smith, who had offered him an investment of $200,000 (AUS, presumably) if he could be convinced the E-Cat worked, was not convinced. In response, Millin threatened to sue Smith for $100 million (AUS, no doubt) in damages (due to loss of the distribution rights) for Smith's failure to meet his "obligation". (Millin claimed he had proven sufficiently that the technology worked. One wonders how.)
On February 9, 2012 Rossi said he had received about 100,000 pre-orders.
On February 14, 2012 Rossi rejected Dick Smith's offer of US $1 million to repeat his test of March 29, 2011.
Note: In the interest of full disclosure, I am not one (or more) of the ~100,000 pre-orders.
Temperature, density, confinement time: pick any two.
I am not going to repeat myself over and over (I'm sure that is appreciated. 
But please consider reality, how stars evolve, why they go supernova (core collapse supernova), why heavy nuclei do not continue to grow infinitely. Even introductory discussions of stellar fusion points out the progressive decrease in heat output as progressively heavier elements fuse (up to the limit of Ni(or rather iron in stars because of the various likely reactions., and why this plays directly into the aging of the stars- from main sequence hydrogen burning, giant stage helium burning, and super giant carbon, oxygen burning. I defy you (reversing your challenge ) to find a single source that describes fusion past nickel (or iron) as a heat source for powering the star.
Two additional references about the astronomical consideration of endothermic / exothermic nature of nuclear fusion, and a example of the duration of heavier element burning in stars and the relative times that they can provide enough heat to delay further collapse of the star. Admittedly the picture of progressive element burning is complex due to relative densities, and temperatures of the stellar core, and relative cross sections, but the system is in equalibrium. The stars outer envelope may expand and thus radiate away more heat, thus allowing for higher core fusion yields in the core (and adjacent layers) while maintaining equilibrium, but much of this size consideration is present during helium fusion so any further changes do not contribute much. Certainly not for the time differences between ~ 1 million years of helium fusion heat production and a few days of silicon burning. Even when you factor in the decreased number of fusion fuel elements available (eg: 2 heliums are converted into one carbon, etc.) the difference is significant.
http://aether.lbl.gov/www/tour/elements ... lar_a.html
http://casswww.ucsd.edu/archive/public/tutorial/SN.html
But please consider reality, how stars evolve, why they go supernova (core collapse supernova), why heavy nuclei do not continue to grow infinitely. Even introductory discussions of stellar fusion points out the progressive decrease in heat output as progressively heavier elements fuse (up to the limit of Ni(or rather iron in stars because of the various likely reactions., and why this plays directly into the aging of the stars- from main sequence hydrogen burning, giant stage helium burning, and super giant carbon, oxygen burning. I defy you (reversing your challenge ) to find a single source that describes fusion past nickel (or iron) as a heat source for powering the star.
Two additional references about the astronomical consideration of endothermic / exothermic nature of nuclear fusion, and a example of the duration of heavier element burning in stars and the relative times that they can provide enough heat to delay further collapse of the star. Admittedly the picture of progressive element burning is complex due to relative densities, and temperatures of the stellar core, and relative cross sections, but the system is in equalibrium. The stars outer envelope may expand and thus radiate away more heat, thus allowing for higher core fusion yields in the core (and adjacent layers) while maintaining equilibrium, but much of this size consideration is present during helium fusion so any further changes do not contribute much. Certainly not for the time differences between ~ 1 million years of helium fusion heat production and a few days of silicon burning. Even when you factor in the decreased number of fusion fuel elements available (eg: 2 heliums are converted into one carbon, etc.) the difference is significant.
http://aether.lbl.gov/www/tour/elements ... lar_a.html
As the fusion process continues, the concentration of Fe increases in the core of the star, the core contracts, and the temperature increases again. When the temperature reaches a point where Fe can undergo nuclear reactions, the resulting reactions are different than the ones that have previously taken place. Fe nuclei are the most stable of all atomic nuclei. Because of this, when they undergo nuclear reactions, they don't release energy, but absorb it.
http://casswww.ucsd.edu/archive/public/tutorial/SN.html
Dan TibbetsSupernovae
Following the Helium Burning Main Sequence in massive stars, a series of nuclear burning stages transforms the star into an onion-like shell structure, until Silicon and Sulfur burning create a core of iron (and other iron-peak elements. During this phase the star will criss-cross the upper regions of the H-R diagram from Red Supergiant to Blue Supergiant and back as different core and shell burning stages ignite. Each successive nuclear burning stage releases less energy than the previous stage, so the lifetime in each stage becomes progressively shorter. For a 20 M star:
Main sequence lifetime ~ 10 million years
Helium burning (3-) ~ 1 million years
Carbon burning ~ 300 years
Oxygen burning ~ 2/3 year
Silicon burning ~ 2 days
To error is human... and I'm very human.
Dan, these reactions are with He4, not with ULM neutrons.D Tibbets wrote:I am not going to repeat myself over and over (I'm sure that is appreciated.
But please consider reality, how stars evolve, why they go supernova (core collapse supernova), why heavy nuclei do not continue to grow infinitely. Even introductory discussions of stellar fusion points out the progressive decrease in heat output as progressively heavier elements fuse (up to the limit of Ni(or rather iron in stars because of the various likely reactions., and why this plays directly into the aging of the stars- from main sequence hydrogen burning, giant stage helium burning, and super giant carbon, oxygen burning. I defy you (reversing your challenge ) to find a single source that describes fusion past nickel (or iron) as a heat source for powering the star.
Two additional references about the astronomical consideration of endothermic / exothermic nature of nuclear fusion, and a example of the duration of heavier element burning in stars and the relative times that they can provide enough heat to delay further collapse of the star. Admittedly the picture of progressive element burning is complex due to relative densities, and temperatures of the stellar core, and relative cross sections, but the system is in equalibrium. The stars outer envelope may expand and thus radiate away more heat, thus allowing for higher core fusion yields in the core (and adjacent layers) while maintaining equilibrium, but much of this size consideration is present during helium fusion so any further changes do not contribute much. Certainly not for the time differences between ~ 1 million years of helium fusion heat production and a few days of silicon burning. Even when you factor in the decreased number of fusion fuel elements available (eg: 2 heliums are converted into one carbon, etc.) the difference is significant.
http://aether.lbl.gov/www/tour/elements ... lar_a.html
As the fusion process continues, the concentration of Fe increases in the core of the star, the core contracts, and the temperature increases again. When the temperature reaches a point where Fe can undergo nuclear reactions, the resulting reactions are different than the ones that have previously taken place. Fe nuclei are the most stable of all atomic nuclei. Because of this, when they undergo nuclear reactions, they don't release energy, but absorb it.
http://casswww.ucsd.edu/archive/public/tutorial/SN.html
Dan TibbetsSupernovae
Following the Helium Burning Main Sequence in massive stars, a series of nuclear burning stages transforms the star into an onion-like shell structure, until Silicon and Sulfur burning create a core of iron (and other iron-peak elements. During this phase the star will criss-cross the upper regions of the H-R diagram from Red Supergiant to Blue Supergiant and back as different core and shell burning stages ignite. Each successive nuclear burning stage releases less energy than the previous stage, so the lifetime in each stage becomes progressively shorter. For a 20 M star:
Main sequence lifetime ~ 10 million years
Helium burning (3-) ~ 1 million years
Carbon burning ~ 300 years
Oxygen burning ~ 2/3 year
Silicon burning ~ 2 days
ULM neutrons have a mass of 1.0086u per nucleon
He4 has a mass of 4.0026/4 = 1.0007 u/nucleon.
Notice the difference? He4 nucleans have much more binding energy than n!
so a reaction He4+X will be much less likley to be exothermic than n + X
slow neutron capture is significant in stellar evolution:
http://en.wikipedia.org/wiki/S-process
This is not a complete picture. When silicon burning starts, there is no hydrogen left in the core of the star. The reactions are between He4 and Si28 (and heavier products). All the Hydrogen is above the oxygen, neon, carbon and helium-burning layers, a long way away from the Si or Ni core.D Tibbets wrote: http://aether.lbl.gov/www/tour/elements ... lar_a.html
As the fusion process continues, the concentration of Fe increases in the core of the star, the core contracts, and the temperature increases again. When the temperature reaches a point where Fe can undergo nuclear reactions, the resulting reactions are different than the ones that have previously taken place. Fe nuclei are the most stable of all atomic nuclei. Because of this, when they undergo nuclear reactions, they don't release energy, but absorb it.
Wikipedia's page on silicon burning ( http://en.wikipedia.org/wiki/Silicon-burning_process ) shows it as a simple chain of X+He4 fusions, Si28 to S32 to Ar36 to Ca40 to Ti44 to Cr48 to Fe52 to Ni56. Where it ends, because Ni56 + He4 => Zn60 is endothermic.
The reality is more complex than that, according to Stan Woosley's Astronomy 12 lecture notes. http://www.ucolick.org/~woosley/ay220-1 ... e12.4x.pdf
Still, the broad point is that the process ends because adding He4 to Ni56 or anything heavier is endothermic. Note, He4, not H1.
If the proposed reaction was He + Ni, the stellar evolution argument would have weight. But since we're talking about H + Ni, it doesn't.
And yet you do. Sigh.D Tibbets wrote:I am not going to repeat myself over and over (I'm sure that is appreciated.
Dan,D Tibbets wrote:As the fusion process continues, the concentration of Fe increases in the core of the star, the core contracts, and the temperature increases again. When the temperature reaches a point where Fe can undergo nuclear reactions, the resulting reactions are different than the ones that have previously taken place. Fe nuclei are the most stable of all atomic nuclei. Because of this, when they undergo nuclear reactions, they don't release energy, but absorb it.
ASSUMING A p+Fe reaction, yes, the IRON absorbs energy, but the proton releases MUCH more, so the net is EXOthermic.
ASSUMING a Fe+Fe reaction, both Fes absorb energy so the reaction is ENDOthermic.
Commenting on just this interpretation, you are wrong. Consider chemical reactions which from a Kinetic energy vs potential energy consideration are the same as nuclear reactions. All reactions have three basic components. The potential energy of the reactants, the products, and the threshold energy needed to permit the reaction to occur. In chemistry this is referred to Gibbs free energy. This is what catalysts effect. The reaction is more likely to proceed if this energy threshold is reduced, therefor the speed of the reaction is increased, but the energy balance does not change. There is no change in the potential energy of the reactants and products. The energy in or energy out is directly linked to the potential energy difference between the materials on both sides of the reaction. If there are is heat lost from the system due to radiation, etc, it still comes from this energy difference. You cannot produce KE without extracting it from the potential energy of the involved materials. If the end potential energy is less in the product(s) then energy is released as KE.KitemanSA wrote:And this yet again demonstrates your confusion. The B/A is a matter of STABILITY, not energy release. It does play into how LIKELY the reaction is to happen under normal circumstances, but not directly into how much energy is released. For that you need to compare TOTAL binding energy.D Tibbets wrote: 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.
Stability in this context means the material with the lowest potential energy is statistically unlikely to reverse to a higher potential energy material, even if the excess KE (Gibbs free energy) is supplied. The reaction has a preferred direction because the lower potential energy product is more stable. Heating oxygen and hydrogen to high enough temperatures is very likely to produce water. The reverse is unlikely (not impossible). Of course you can play tricks, by removing one or more of the elements from the system before an equilibrium is reached. You can drive a reaction to higher potential energy by this method (electrophoresis would be an example of this) but you always have to spend energy to do it.
Dan Tibbets
To error is human... and I'm very human.
This is a bone of contention that is unlikely to be resolved. But, look at the graph of binding energy per nucleon ( or use the table referenced multiple times). The MeV per nucleon is a specific property of a nucleus. Ni62 has a value as does Cu63. If you add or remove nucleons the nucleus is different and you have to look up the value for that nucleus. Hopefully we agree that the missing mass, binding energy per nucleon, or for the sake of this arguement only , the total binding energy is directly proportional to the potential energy of the nucleus which can be added to or subtracted from in a reaction. Look at the graph or table. What is the value for hydrogen? Is it ZERO? You are claiming that something with zero potential energy (zero binding energy) somehow is providing potential energy to a reaction. It is not and this causes considerably confusion. How can you get energy by burning hydrogen to helium if the hydrogen does not have any potential energy? This could lead you to think that hydrogen somehow is the source for all of the energy.KitemanSA wrote:And yet you do. Sigh.D Tibbets wrote:I am not going to repeat myself over and over (I'm sure that is appreciated.Dan,D Tibbets wrote:As the fusion process continues, the concentration of Fe increases in the core of the star, the core contracts, and the temperature increases again. When the temperature reaches a point where Fe can undergo nuclear reactions, the resulting reactions are different than the ones that have previously taken place. Fe nuclei are the most stable of all atomic nuclei. Because of this, when they undergo nuclear reactions, they don't release energy, but absorb it.
ASSUMING A p+Fe reaction, yes, the IRON absorbs energy, but the proton releases MUCH more, so the net is EXOthermic.
ASSUMING a Fe+Fe reaction, both Fes absorb energy so the reaction is ENDOthermic.
If you accept that hydrogen (a proton) has potential energy/ binding energy =0, you have to reconize that zero is not an endpoint but just another value in a range from negative to positive numbers. Recall that potential energy is assigned a negative value by convention. This allows for more convient book keeping as relationships are compared.
Consider the reaction Ni62 + P -> Cu63. Each element has a binding energy per nucleon. If you ignore the energy flow direction, the absolute difference in the missing mass/ binding energy of the reaction is the difference between the two sides. ~ 8.9 MeV + 0 MeV -> ~ 8.8 MeV. The hydrogen contributes nothing to the energy balance. Another way of stating it is that the numbers assigned to the nickel and copper already incorporates the contribution of the proton which is what is different between them.
The energy difference between the Ni and Cu is the energy difference is a consequence of their makeup. The binding energy per nucleon is the energy nessisary to tear off a nucleon. This fits the reverse reaction better. Cu63 -> Ni62 + P. Again looking up the values for thes, the energy is ~ 8.8 MeV -> 8.9 MeV + 0 MeV. The energy required to convert a Cu63 to a Ni62 is ~ 8.8 MeV. The energy needed to convert a Ni62 into a Cu63 is ~ -(8.9 MeV). It takes more energy to convert to Cu than the reverse. IE: Fusion of Ni62 to Cu63 requires the difference or ~ 0.1 MeV of input energy, thus endothermic. The Fission of Cu63 into Ni62 plus a proton releases ~ 0.1 MeV of energy, thus exothermic.
Consider the common fission example of Uranium235 into two daughter products plus ~ 2 neutrons. Again, each of the elemental isotopes have a binding energy per nucleon. Comparing these gives the energy difference and direction. Note that the neutrons, which are ~ the same as protons in binding energy, are insignificant for calculating the yield. The neutrons have other very important consequences, but for the energy balance calculation of this specific reaction they are irrelevant, just as is the proton in the Ni62- Cu63 reaction.
Dan Tibbets
To error is human... and I'm very human.
The hydrogen p is a nucleon.Dan wrote: Consider the reaction Ni62 + P -> Cu63. Each element has a binding energy per nucleon. If you ignore the energy flow direction, the absolute difference in the missing mass/ binding energy of the reaction is the difference between the two sides. ~ 8.9 MeV + 0 MeV -> ~ 8.8 MeV. The hydrogen contributes nothing to the energy balance. Another way of stating it is that the numbers assigned to the nickel and copper already incorporates the contribution of the proton which is what is different between them.
Although it has zero binding energy as p. When incorporated into Cu63 it contributes to the overall binding energy.
As you know, around Ni, the binding energy per nucleon is roughly constant. But of course adding a nucleon adds that constant to the result.
All of this complexity is unnecessary if you simply look at nuclear masses, and balance. You get the ~8MeV immediately. But if you look at binding energy per nucleon you can get the same answer, you just have to realise that around Ni/Cu, adding a nucleon will generally increase overall binding energy (because binding energy per nucleon stays roughly constant).
When you get up to U things are rather different. Binding energy per nucleon decreases significantly with increasing Atomic number. So much that fission can be energetically preferred.
This confuses reaction rates with reaction equilibrium positions.Dan wrote: Stability in this context means the material with the lowest potential energy is statistically unlikely to reverse to a higher potential energy material, even if the excess KE (Gibbs free energy) is supplied. The reaction has a preferred direction because the lower potential energy product is more stable.
Stable systems are so because the reaction rates (to anything elese) are very low.
Given infinite time any reaction will have an equilibrium position at which the reaction rates in both directions are equal.
You are saying that higher potential energies have equilibria in which they are favoured over lower potential energies. That is often true, because the collision of suitable energy reactants to form the hogh PE product is unlikely, even given high temperatures. But it is not intrinsically so. Thus with H2O you can find temp and pressure regimes where the equilibrium moves strongly towards dissociated H2 and O2 (although probably dissociated H+ & O6+, at temperatures high enough for this).
Dan,D Tibbets wrote: This is a bone of contention that is unlikely to be resolved. But, look at the graph of binding energy per nucleon ( or use the table referenced multiple times). The MeV per nucleon is a specific property of a nucleus. Ni62 has a value as does Cu63. If you add or remove nucleons the nucleus is different and you have to look up the value for that nucleus. Hopefully we agree that the missing mass, binding energy per nucleon, or for the sake of this arguement only , the total binding energy is directly proportional to the potential energy of the nucleus which can be added to or subtracted from in a reaction. Look at the graph or table. What is the value for hydrogen? Is it ZERO? You are claiming that something with zero potential energy (zero binding energy) somehow is providing potential energy to a reaction.
This demonstrates your fundamental confusion. Mass (potential energy) is not binding (kinetic) energy. They are opposite quantities.
QUESTION: How much gravitational potential energy does a 1kg mass have when infinitely far from any gravitational body?
ANSWER: ZERO.
As it comes closer it LOSES potential energy and said potential goes NEGATIVE.
It is analogous for the nucleaon "infinitely" far from the strong force. it's potential energy (mass change) is ZERO. That "potential energy" goes DOWN (yes, negative) when it releases energy in the form of BINDING energy.
Mass change NEGATIVE from "ZERO",
Binding energy POSITIVE from "ZERO".
The free nucleon has ZERO binding energy because it has lost no mass. thus it has the greatest potential energy.... ZERO!
Delta potential + delta kinetic = ZERO! Jeez, Dan, even a high schooler should know that.