rode_of_the_ruin Posted June 6, 2020 Share Posted June 6, 2020 we know about the E=mc^2 of Einstein recipe, it's not distinguish radioactive or non-radioactive substance, even though 1kg water H(2)O such as power to Uranium 235 constant! But in one react of battery reducting Hydrogen, people consider it very low power, and because of MOL (mole 6,023 x 10^23 in 12gr Carbon) material, so it can add or multiply from lowest level power to a bigger. But if you think opposite, one react reducting an Hydrogen is the biggest, and the reason level of power is medium, this beacuse of quantity of react is being against each other, beacuse i think if two battery is being contrast + with + and - with - is no or very low power output. Also the Uranium 235 is being fission reaction, it decrease the quantity of reaction and the power from very low quantity is bigger out then!! Right? Link to comment Share on other sites More sharing options...
Strange Posted June 6, 2020 Share Posted June 6, 2020 It is not very clear what you are asking. Reduction is a chemical process so there is almost no change in mass, hence very small release of energy. Uranium fission involves a large change in mass and so a greater release of energy. Why have you posted this in Speculations? It looks like a question about Physics. 1 Link to comment Share on other sites More sharing options...
joigus Posted June 6, 2020 Share Posted June 6, 2020 I don't understand it either. Seems like you want to compare the reducing power of hydrogen versus energy released by 235U in fission... [?] I agree with Strange in everything I can think of +1. Another way is seeing it is by scaling/dimensional arguments. Nuclear rearrangements typically are 103 = 1000 times more energetic than electron rearrangements. That's about the ratio mp/me, which is the order of 2000. Link to comment Share on other sites More sharing options...
MigL Posted June 6, 2020 Share Posted June 6, 2020 Whereas a redox reaction involves the binding of an electron from a reducer to an oxidizer, a fusion/fission reaction involves the binding energy of nucleons ( protons and neutrons ) inside the nucleus, which, as Joigus has explained, is several orders of magnitude more energetic, because it involves the strong nuclear force, not the electromagnetic force. Link to comment Share on other sites More sharing options...
swansont Posted June 6, 2020 Share Posted June 6, 2020 2 hours ago, joigus said: Another way is seeing it is by scaling/dimensional arguments. Nuclear rearrangements typically are 103 = 1000 times more energetic than electron rearrangements. That's about the ratio mp/me, which is the order of 2000. Are you saying that this is the result of some interaction, or is it just coincidence? The relative strength of the electromagnetic interaction is tied to the fine structure constant, and has no mass term (the strong nuclear coupling is ~1) Link to comment Share on other sites More sharing options...
joigus Posted June 6, 2020 Share Posted June 6, 2020 (edited) 35 minutes ago, swansont said: Are you saying that this is the result of some interaction, or is it just coincidence? The relative strength of the electromagnetic interaction is tied to the fine structure constant, and has no mass term (the strong nuclear coupling is ~1) Yeah, yeah, yeah. You got me there. I was just thinking about re-editing that bit. Edit1. On the other hand, the average ratio in interactions is of the order 102, so it really is: \[\alpha_{\textrm{QCD}}/\alpha_{\textrm{EM}}\sim100\] Masses don't play a part, AFAI can see. Thank you. Edit2. On second thought. Energy released in fission what really is is stored electrostatic energy from proximity of the protons. So the estimation is different and is not governed by, \[\alpha_{\textrm{QCD}}/\alpha_{\textrm{EM}}\sim100\] 100 times stronger seems to me peculiarly small. What do you think, @swansont? Edited June 6, 2020 by joigus addition Link to comment Share on other sites More sharing options...
swansont Posted June 6, 2020 Share Posted June 6, 2020 55 minutes ago, joigus said: Yeah, yeah, yeah. You got me there. I was just thinking about re-editing that bit. Edit1. On the other hand, the average ratio in interactions is of the order 102, so it really is: αQCD/αEM∼100 Masses don't play a part, AFAI can see. Thank you. Edit2. On second thought. Energy released in fission what really is is stored electrostatic energy from proximity of the protons. So the estimation is different and is not governed by, αQCD/αEM∼100 100 times stronger seems to me peculiarly small. What do you think, @swansont? I agree that it’s largely electrostatic energy, but it’s also true that the protons can only be in such proximity because the nuclear force permits it. So the nuclear coupling plays a role IOW, all else being equal, if the nuclear coupling were weaker, you’d expect the energy released in fission to drop as well, by some factor that was related to the coupling, because the nucleon spacing would increase in the parent and fission fragments 1 Link to comment Share on other sites More sharing options...
MigL Posted June 6, 2020 Share Posted June 6, 2020 It wouldn't simply be the ratio of the colour force to the electromagnetic. The force that binds nucleons is actually 'residual force from that which binds quarks in nucleons. Link to comment Share on other sites More sharing options...
joigus Posted June 6, 2020 Share Posted June 6, 2020 2 minutes ago, swansont said: I agree that it’s largely electrostatic energy, but it’s also true that the protons can only be in such proximity because the nuclear force permits it. So the nuclear coupling plays a role IOW, all else being equal, if the nuclear coupling were weaker, you’d expect the energy released in fission to drop as well, by some factor that was related to the coupling, because the nucleon spacing would increase Aha, right. That totally adds up in my mind. So ignoring the QCD factor, which I wouldn't be able to calculate anyway, for an estimation of electrostatic attraction between proton and electron, we can use the inverse Bohr's radius, which proportional to the mass of the electron. That would give an estimation of chemical energy released in hydrogen reduction. For an estimation of electrostatic repulsion between proton pairs (which is the order of energy released when the nucleus undergoes fission,) you can assume the energy proportional to the mass of the proton. The reason being that the reduced mass of the p-p system is mp/2 while the reduced mass of the p-e system is me. I'm being very schematic here, because p-p is not an EM bound state. So the solutions will not be the same, but they will scale the same. That would give, \[\frac{E_{pp}}{E_{pe}}\sim\frac{r_{pe}}{r_{pp}}\sim\frac{m_{p}}{m_{e}}\sim2000\] A rough estimation does give you order of 1000 times more energy released. And then there would be corrections due to breaking QCD bonds that you're referring to. The rough way to picture it for me would be a rubber sheet that gets very loose when the protons get closer together, but pulls very hard when they pull apart. Until they break. But having to break the gluon bond would subtract from electrostatic energy stored. The higher order term would have no dependence on either alpha_EM or alpha_QCD coupling constants, but the other would carry a QCD-coupling dependence for sure. I think the mp/me does appear as the dominant factor, but the argument was missing. As you suggested, it's not a coincidence. I think it's a consequence of separating the leading term, which is electrostatic in nature. Thanks a lot, @swansont, and corrections most welcome. 1 hour ago, MigL said: It wouldn't simply be the ratio of the colour force to the electromagnetic. The force that binds nucleons is actually 'residual force from that which binds quarks in nucleons. I agree, @MigL. Link to comment Share on other sites More sharing options...
rode_of_the_ruin Posted June 7, 2020 Author Share Posted June 7, 2020 23 hours ago, Strange said: It is not very clear what you are asking. Reduction is a chemical process so there is almost no change in mass, hence very small release of energy. Uranium fission involves a large change in mass and so a greater release of energy. Why have you posted this in Speculations? It looks like a question about Physics. OK, now I have two physic quantity are mass and quantity of react is being proportional to each other, called "ratio together": Hydrogen mass: (1,00794 x 10^-26kg) x 6,023 x 10^23... Oh no! I can not guess this, can you help me? But I think like you, mass changing in chemical react is none, right, mass of Hydrogen doesn't be changed, but quantity? If changed to a smaller, so the power is decrease down, but, but.. If the quantity is "1" value!! Then it magic change into the "divide": 1 / 10^-26 = 10^26 ?? Link to comment Share on other sites More sharing options...
Strange Posted June 7, 2020 Share Posted June 7, 2020 47 minutes ago, rode_of_the_ruin said: Hydrogen mass: (1,00794 x 10^-26kg) x 6,023 x 10^23... Oh no! I can not guess this, can you help me? I don't know what your first number is. Are you trying to calculate the mass of one mole of hydrogen? (mass of hydrogen atom * Avogradro's constant) The mass of a hydrogen atom is: 1.6735575 × 10^-27 kg So the mass of one mole of hydrogen is about 0.001 kg. 50 minutes ago, rode_of_the_ruin said: But I think like you, mass changing in chemical react is none, right, mass of Hydrogen doesn't be changed, but quantity? The quantity of hydrogen (or any other chemical) does not change in a chemical reaction. However, the total mass of the reagents will change: if energy is released, then the total mass will be less at the end. Link to comment Share on other sites More sharing options...
swansont Posted June 7, 2020 Share Posted June 7, 2020 3 hours ago, rode_of_the_ruin said: But I think like you, mass changing in chemical react is none, right, mass of Hydrogen doesn't be changed, but quantity? If changed to a smaller, so the power is decrease down, but, but.. If the quantity is "1" value!! Then it magic change into the "divide": The amount of hydrogen will not change in a chemical reaction. The number of neutrons+protons will not change in either a chemical or nuclear reaction The amount if mass in the system will change; it will typically be much larger in a nuclear reaction, as has already been discussed Link to comment Share on other sites More sharing options...
rode_of_the_ruin Posted June 7, 2020 Author Share Posted June 7, 2020 (edited) I think about reduction of Hydrogen is not simple at the lowest power, different from sodium(Na) reduct potassium chloride(KCl) at high temperature, because Hydrogen only have one proton, proton have 3 quarks hardest connect together, there is no evidence that quark bonds can be destroyed! If reducting from a lot of protons example is Zinc(Zn) on alkaline battery to one proton is Hydrogen, so if Hydrogen "don't want" to be break link of 3 quarks, it must release the big energy to protect the link of 3 quarks, but if have a lot of Hydrogen is being reduct together, big power energy be sure against each other and result give low power! Is this ideal?? Edited June 7, 2020 by rode_of_the_ruin edit Link to comment Share on other sites More sharing options...
joigus Posted June 7, 2020 Share Posted June 7, 2020 5 minutes ago, rode_of_the_ruin said: I think about reduction of Hydrogen is not simple at the lowest power, different from sodium(Na) reduct potassium chloride(KCl) at high temperature, because Hydrogen only have one proton, proton have 3 quarks hardest connect together, there is no evidence that quark bonds can be destroyed! If reducting from a lot of protons example is Zinc(Zn) on alkaline battery to one proton is Hydrogen, so if Hydrogen "don't want" to be break link of 3 quarks, it must release the big energy to protect the link of 3 quarks, but if have a lot of Hydrogen is being reduct together, big power energy be sure against each other and result give low power! Is this ideal?? Please don't do that. I started colouring my posts some days ago (on selected bits,) then I realized how silly I was being, and mended my ways. I recommend you use the standard diacritics. Plus the abuse of colour is against the rules, if I remember correctly. Best luck. Link to comment Share on other sites More sharing options...
rode_of_the_ruin Posted June 7, 2020 Author Share Posted June 7, 2020 2 minutes ago, joigus said: Please don't do that. I started colouring my posts some days ago (on selected bits,) then I realized how silly I was being, and mended my ways. I recommend you use the standard diacritics. Plus the abuse of colour is against the rules, if I remember correctly. Best luck. Yes I have edited Link to comment Share on other sites More sharing options...
swansont Posted June 7, 2020 Share Posted June 7, 2020 11 minutes ago, rode_of_the_ruin said: I think about reduction of Hydrogen is not simple at the lowest power, different from sodium(Na) reduct potassium chloride(KCl) at high temperature, because Hydrogen only have one proton, proton have 3 quarks hardest connect together, there is no evidence that quark bonds can be destroyed! If reducting from a lot of protons example is Zinc(Zn) on alkaline battery to one proton is Hydrogen, so if Hydrogen "don't want" to be break link of 3 quarks, it must release the big energy to protect the link of 3 quarks, but if have a lot of Hydrogen is being reduct together, big power energy be sure against each other and result give low power! Is this ideal?? Perhaps you could present a simpler scenario in order to clear up your confusion. Hydrogen will remain hydrogen in any chemical reaction. Link to comment Share on other sites More sharing options...
joigus Posted June 7, 2020 Share Posted June 7, 2020 1 minute ago, rode_of_the_ruin said: Yes I have edited Thank you. Link to comment Share on other sites More sharing options...
rode_of_the_ruin Posted June 7, 2020 Author Share Posted June 7, 2020 1 minute ago, swansont said: Perhaps you could present a simpler scenario in order to clear up your confusion. Hydrogen will remain hydrogen in any chemical reaction. Yes, hydro is only hydro, but the link of proton and neutron together in Zinc or Uranium 235 is not stronger than link of quarks You say Hydro is only hydro? But I know Uranium the first taking from exploit from mines, and Uranium first is Uranium salt(K(2)UO(2)(SO4)(2)) and when carrying to dump truck, it must have burning and many stages to get the pure metal Uranium, but burning and many stages have Hydrogen from "carbon hydrate" fuel, so I can call "Uranium" is "Hydrogen" with "nickname" is "Hydrogen 235"! Hydrogen taking 75% of materials from universe! Link to comment Share on other sites More sharing options...
swansont Posted June 7, 2020 Share Posted June 7, 2020 Just now, rode_of_the_ruin said: Yes, hydro is only hydro, but the link of proton and neutron together in Zinc or Uranium 235 is not stronger than link of quarks Not sure why you think this is important Just now, rode_of_the_ruin said: You say Hydro is only hydro? But I know Uranium the first taking from exploit from mines, and Uranium first is Uranium salt(K(2)UO(2)(SO4)(2)) and when carrying to dump truck, it must have burning and many stages to get the pure metal Uranium, but burning and many stages have Hydrogen from "carbon hydrate" fuel, so I can call "Uranium" is "Hydrogen" with "nickname" is "Hydrogen 235"! Hydrogen taking 75% of materials from universe! You can say that, but it’s meaningless. I can’t unpack your misconception(s) from this scenario. Link to comment Share on other sites More sharing options...
Strange Posted June 7, 2020 Share Posted June 7, 2020 36 minutes ago, rode_of_the_ruin said: I think about reduction of Hydrogen is not simple at the lowest power, different from sodium(Na) reduct potassium chloride(KCl) at high temperature, because Hydrogen only have one proton, proton have 3 quarks hardest connect together, there is no evidence that quark bonds can be destroyed! If reducting from a lot of protons example is Zinc(Zn) on alkaline battery to one proton is Hydrogen, so if Hydrogen "don't want" to be break link of 3 quarks, it must release the big energy to protect the link of 3 quarks, but if have a lot of Hydrogen is being reduct together, big power energy be sure against each other and result give low power! Is this ideal?? Chemical reactions only involve the electrons, not protons or quarks. 13 minutes ago, rode_of_the_ruin said: so I can call "Uranium" is "Hydrogen" with "nickname" is "Hydrogen 235"! Uranium has 92 protons and 143 neutrons (in the case of U235) so it is nothing like hydrogen. Link to comment Share on other sites More sharing options...
rode_of_the_ruin Posted June 7, 2020 Author Share Posted June 7, 2020 6 minutes ago, Strange said: Chemical reactions only involve the electrons, not protons or quarks. If follow you, so proton or quarks is only keep or release electron from atom other, but the "strong interaction" researching on the world can not look at electron, and many magnetic field, electric field, "strong interaction" field is not involve the electrons, just only guess what's contact from proton give reason electron to be moved, If looking at electron theory, you can not deploy more can be change life on the world!! Link to comment Share on other sites More sharing options...
Strange Posted June 7, 2020 Share Posted June 7, 2020 9 minutes ago, rode_of_the_ruin said: If follow you, so proton or quarks is only keep or release electron from atom other, but the "strong interaction" researching on the world can not look at electron, and many magnetic field, electric field, "strong interaction" field is not involve the electrons, just only guess what's contact from proton give reason electron to be moved, If looking at electron theory, you can not deploy more can be change life on the world!! Your posts are pretty incomprehensible. Are you using Google translate? If so, stop. If not, try it. Yes, the protons and neutrons are held together by the strong force. This has no effect on the electrons. The electrons are there because of the electric charge of the protons. That is why there are the same number of protons and electrons. Only electrons (and the electromagnetic force) is involved in chemical reactions. Link to comment Share on other sites More sharing options...
rode_of_the_ruin Posted June 7, 2020 Author Share Posted June 7, 2020 6 minutes ago, Strange said: Your posts are pretty incomprehensible. Are you using Google translate? If so, stop. If not, try it. Yes, the protons and neutrons are held together by the strong force. This has no effect on the electrons. The electrons are there because of the electric charge of the protons. That is why there are the same number of protons and electrons. Only electrons (and the electromagnetic force) is involved in chemical reactions. Yes, i used google translate! but in case of new theory about nuclear replace electron theory, you can also calculated the power from low to high to be clearly, my guess can wrong, is my idea make you interest? Link to comment Share on other sites More sharing options...
Strange Posted June 7, 2020 Share Posted June 7, 2020 1 hour ago, rode_of_the_ruin said: but in case of new theory about nuclear replace electron theory, you can also calculated the power from low to high to be clearly, my guess can wrong, is my idea make you interest? As it is very unclear(*) what you are saying, and it is presumably based on your lack of knowledge, then no I am not interested. (*) I mean, impossible to understand Link to comment Share on other sites More sharing options...
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