causeeffect.org 
My
Articles

Nuclear Fusion Mass Lost, Crevices between Nucleons, by Carl R. Littmann, 612015 Optional quick Review of some basic Physics: (For those not needing the review, click here to start at ‘Main Introduction’, instead.) (Text may be enlarged by printing it out, or copy & pasting desired section.) (Optional Review): Roughly speaking, the simplest atom, Hydrogen1, is made of one relatively massive, positively charged ‘proton’ at its center or core, and a much lighter negatively charged ‘electron’ orbiting it. That is somewhat like the earth orbiting around the more massive Sun. And there exists atoms more massive than hydrogen, like Helium4, which, roughly speaking, consists of 2 positive protons and 2 neutral neutrons in its fused core, and two negative electrons orbiting around that core. We thus describe the Helium4 atom as having 4 ‘nucleons’, (2 protons + 2 neutrons in its core), and 2 light electrons orbiting that core. Thus, the atom is ‘net neutral’ regarding its net charge. Different books may use different notation styles. But we will use the notation _{2}He^{4} to designate that ‘Helium4’. It is made of 2 positive protons in the nucleus, as denoted by the number ‘2’ at the lower left of the ‘He’ abbreviation. Those 2 protons touch 2 neutron nucleons also in its nucleus (core) – thus giving a total of 4 nucleons in its core, and that ‘4’ is denoted at upper right of that ‘He’ abbreviation. Or we say, Helium4 with an ‘atomic mass number’ of 4 (i.e., 4 nucleons in its fused core). And 2 of those core nucleons are positively charged protons, giving the atom a socalled ‘atomic number’ of 2. (The latter, together with the 2 orbiting electrons, give that atom its important main chemical attributes.) There exists an even lighter form of helium, called ‘Helium3’, but it is rarely found. It is denoted _{2}He^{3} because it has only 1 neutron and 2 protons in its core, instead of 2 neutrons and 2 protons. We say that _{2}He^{3} is the rare ‘Isotope’ of Helium, and _{2}He^{4} is the common ‘Isotope’ of Helium. Or that Helium3 is a rare isotope and Helium4 is a common isotope. Of great importance to this paper is this: When (4) Hydrogen1 atoms fuse together to form (1) Helium4 atom, there is huge energy released. And that is because the mass of the initial ingredients, (4) neutral Hydrogen1 atoms’, at rest, is slightly greater than the final product, (1) neutral Helium4 atom, when it comes to rest. The difference is a small amount of ‘conventional’ mass that is, sotospeak, ‘lost’ or more accurately “emitted with very high energy ‘equivalent of mass’.” (There are some simplifications in this article, like the above model of Helium4, described, “as made from (4) Hydrogen1 atoms”. That simple model was promoted by two prominent early textbook writers, ‘Sears and Zemansky’. They realized, holistically, that that’s how our Sun makes Helium4, in net effect.) And I will make some simplifications too, by just saying this: “The huge ‘energy lost’, (the mass difference between the initial ingredients and final product), is called ‘the Binding Energy’ of the final product – the final product being (1) Helium4 atom, in the above example.” And I will just describe it as follows: “The ‘equivalent energy emitted’ is equal to the ‘mass lost’. And that is equal to the sum of the huge energy radiated at the speed of light, plus the energies transferred to target material as the fastmoving exploded products come to rest. Roughly speaking, we associate all that with an exploding ‘Hydrogen bomb’.” So, summarizing, “if ‘m’ is the ‘mass loss’ (i.e., the ‘mass difference between main products at the start and at the finish’); then the huge Energy emitted is equal to (mc^{2}), where ‘c’ is the speed of light.” That’s Einstein’s famous equation. As previously said, an atom’s ‘mass number’ is given as a ‘whole numbers’, 1 or 2 or 3 or 4, etc., and is the total number of its nucleons, (all its protons plus neutrons added together). But an atom’s exact ‘atomic weight’ generally differs from the atom’s mass number (the whole number) by a very slight amount. For example, science has somewhat arbitrarily set the atomic weight of a special atom, Carbon12, to be exactly ‘12 u’, (‘u’ as in ‘units’). But this results in the atomic weight of Hydrogen1 coming out ‘1.00782503207 u’ (not exactly just ‘1 u’), relative to the nice ‘12 u’ of the atom ‘Carbon12’. Most other atomic weights also do not come out exactly as whole numbers relative to the Carbon12 atomic weight, 12 u. But that ‘sets the stage’ for a slight amount of mass lost and huge emission of energy – either when small nucleons fuse together,  or when a large nucleus undergoes ‘fission’, which is the breaking up of its large core nucleus into several smaller pieces). The First major topic or question that we attempt to address in this paper is this: Why do so many atoms (or atomic isotopes) have nucleons that ‘bind’ so tightly together to form a stable nucleus, instead of coming apart? In other words, why is a nucleus, in so many cases, so stable? And, generally, a huge amount of energy emitted when such core nucleus is formed from the fusing nucleons? Click to Fig. 1 for our proposed answer: Near the top of that sketch, we note that under certain conditions, a structural ‘narrowing of flow’ can lead to greatly reduced pressure, even suction, and thus even lead to a relative ‘attractionlike’ force. We note that occurs in the region of a gradual narrowing of pipe crosssection shown, before it fully widens again. And near the bottom of the sketch, a similar ‘narrowing’ is illustrated, but using three touching spheres, or in that case, nucleons, instead of a narrowing pipe crosssection  to bring about the narrowing. (Optionally, one may click Fig. 2 to glance at the 3 nucleons from other directions, but then promptly return here to our first topic and to Fig.1.) The nucleons are thus assisted in ‘fusing tightly’ by the narrowing passageway or crevice near the center of the pattern created by three touching nucleons, which form the atom’s nucleus. A ‘Venturilike effect’ arises there. And that effect can be calculated by applying “Bernoulli’s Equation” to such narrowing flow regions. ((Bernoulli’s equation, in effect, imposes ‘the conservation of energy’ to the case of a constant amount of flow; but where such flow must speed up as it goes through a constricted region. That ‘speedup’ in the constricted region is required if the same amount of flow (in ‘volume per second’) occurs there  as also occurs in the wider regions upstream from there, and wider regions downstream. Here, we also assume that the fluid is nearly ‘incompressible’ and that the flow is ‘orderly’ and that there is very low friction.)) The Second major topic or question that we try to address is this: Why is the ‘binding energy’ (the energy released when Helium4 is formed) nearly 4 times as much energy as when Helium3 is formed? This seems strange at first, since when 3 sticky balls or nucleons touch in a triangular pattern, that seems like 3 ‘sticking points’, and when 4 sticky nucleons touch in a tetrahedral pattern, that seems like 6 ‘sticking points’. Thus, the simplest view (a ‘sticky’ model approach) would seem like merely the doubling of the binding energy should be expected, but not the quadrupling of it (but, in fact, the quadrupling does occurs)! Click to Fig. 3  for our proposed answer: There we will note that a tetrahedral formation (using 4 nucleons), in effect, ‘defines’ 4 triangular planes, which contrasts with the use of only 3 nucleons that merely defines 1 triangular plane, not 4. So it looks like the 4 planes (as generated by 4 fused nucleons) might reasonably be associated with the release of about 4 times the energy as just 1 plane (as generated by 3 fused nucleon). We also importantly note the 4 ‘donutlike’ holes or narrow passageways that are associated with the 4 touching nucleons, but only 1 ‘donutlike’ hole is associated with just 3 touching nucleons. That also ‘goes along’ with the 4fold jump described in the underlined sentence above. (The effect of a narrowed passageway was addressed previously in our discussion of Fig. 1.) There is barely enough room for 4 imaginary small spheres, tetrahedrally aligned, to fit inside the cavern space between 4 large touching spheres, also tetrahedrally aligned. That is – with each of the imaginary small spheres barely fitting through the ‘donut holes’, one at a time, into the interior. And yet, when forming the small tetrahedral array inside the ‘cavern’ – with no part of those small interior spheres quite touching any of four imaginary triangular planes that touch the centers of the four larger spheres. (More detailed discussion of this topic 2 is given later in this article.) The Third major topic or question is this: It is empirically known that the fusing together of at least 3 nucleons causes a relatively great energy released and significant amount of mass lost. But the fusing together of only 2 nucleons does not lead to nearly such great effect. Therefore, is there some way we can ‘glean’ out a rough quantitative prediction (or conception) of the expected amount of mass lost and energy resulting when 3 nucleons fuse together? That is, might our proposed geometric model of 3 touching nucleons aid our prediction? Click to Fig. 2 again, but this time read the small print that discusses the volume of a very small (imaginary) sphere that would just barely fit through the small ‘donut hole’ – the empty space at or near the center of the that 3nucleon pattern. But imagine 2 such very small spheres instead of 1. (The relative volume of 2 such very small spheres gives nearly the exact amount of mass lost, relative to the much bigger 3nucleon mass forming the patterns. It might be more appealing if the answer was simply (1) such very small spherical mass fitting through the donut hole, instead of double that, ((i.e., (2).)) And we will not, here, try to argue as to why ‘double the 1’, (or 2 very small spheres) results, although I have some thoughts about why. ((Optionally, the reader might recall or later see my earlier website article, ‘Book of Large and Small Spheres …. Etc.’. There we previously showed how small spheres just barely fit into the grooves between large spheres in abstract patterns, and how that volume ratio (big to small sphere) corresponds, proportionally, to actual particles in the real world  the ratio of large mass particles to small mass particles!)) But our article’s treatment of the fusion of big stable nuclei is a little more complicated than treating unstable particles or individual particles. That is because (after calculating the volume of a sphere fitting through the donut hole), we multiply that by 2 to get our ‘mass loss’ prediction. So we proceed as described in the case of Fig. 2  as follows: We imagine that each ‘average’ nucleon (sphere) has the mass of onethird the total mass of the three nucleons forming the stable Helium3 isotope. That means we must subtract the mass of the 2 orbiting electrons from the atomic weight of the entire Helium3 atom to accurately get the actual mass of the atom’s 3nucleon core (although that core constitutes almost all of the atom’s mass). As the calculations in Fig. 2 show, we compare our predicted mass loss (based on our ‘geometric pattern approach’) to the ‘mass loss’ (based on an empirical related ‘electronproton based’ Binding Energy result). Our ‘predicted’ mass loss has an error of only about 0.06% relative to the empiricalrelated ‘binding energy’. Other rather similar geometric approaches, not detailed here, give a very slightly greater or smaller error than that, but are still quite close! Because of that very small error, it looks like that ‘mass loss’, in the above basic fundamental ‘3nucleon’ case, may involve a ‘quantumlike action’ based on the triangular pattern and one donut hole. And that seems possible also in our later treatment of the 4nucleon case. But that ‘quantum’ hypothesis involves some speculation. (More detailed discussion given later.) Also, it is important to note this: By far, most of the mass in the universe seems due to so many Hydrogen1 atoms (or so many of its nuclei) existing. And to a much lesser extent, the Helium4 atoms (or many of its nuclei) existing. If all the other atoms (or their nuclei) were added together, that would not contribute more than a few percent to the total. That is one reason why this article mainly addresses the 1 to 4 nucleon cases, rather than heavier nuclei. (And the great preponderance of neutral Hydrogen1 atoms is a major reason why we prefer to imagine it as the ‘standard starting mass’ to compare the final fused neutral atoms to – to calculate the latters’ binding energies. That is what Sears and Zemansky did for the case of Helium4.) The Fourth topic or question is this: Can we model here or address well here, the ‘Hydrogen2’ atom (or isotope), also known as ‘the Deuteron’? My not very helpful answer, here, is NO! I do NOT present a very detailed model of ‘Hydrogen2’, but reader may click Fig. 4. There, I do make some, hopefully, somewhat helpful comments, as follows: As previously noted, the Binding Energy of the (2nucleon) Deuteron is much less than for the Helium3 (3nucleon case) – the latter having the appreciablesized ‘donuthole’. The true (net) deuteron’s binding energy is only about an (mc^{2}) amount, where the total small mass ‘m’ in that (mc^{2}) term is only a few electron masses. (Or, say, only roughly the small mass difference between a proton and a neutron  the proton and the neutron, in their ‘free’ unbound state, being almost equal in mass.) I therefore believe that the deuteron’s bond is only a socalled ‘electrostatic’ or ‘electrodynamic’ related bond, and not a true strong ‘nuclear’ bond, in nature. That being alleged, readers should also note that various people have various models for ‘charged particles’ or the socalled ‘electric’ bond. And I have some thoughts about that too, but all that is beyond this article. The Fifth major topic or question is this: This article earlier inferred that the old, commonlyused ‘Binding Energy method’ is only ‘pretty useful’, compare to a newer, better method. So, yes, we will show an Alternate ‘Binding Energy method’ that is simpler and often more useful! And see how that is so and why! Now, first, the example of the old binding energy method. ((It is based on imagining Hydrogen3 as fundamentally made from these alleged basic starting ingredients: 1 proton orbited by 1 electron (i.e., 1 Bohr atom) and 2 free neutrons. And that old method regards Helium3 as made of the following basic starting ingredients: 2 protons each orbited by their electron (i.e., 2 Bohr atoms) and 1 free neutron. Below are the ‘starting ingredients’ and ‘final fused products’ and the results: Old method for Hydrogen3: (One ref. free neutron has mass = 1.008664916 u.) Old method for Helium3: (1 free Bohr Hydrogen atom has mass = 1.007825032 u.) More about that and related matters later, but now lets see how our Newer (proton and electron based) Binding Energy method is successful at predicting what is stable vs. unstable at such a basic, simple level: New method for Hydrogen3: (1 Bohr Hydrogen atoms has mass = 1.007825032 u.) New method for Helium3: (Again, 1 Bohr Hydrogen atom has mass = 1.007825032 u.) Important: The New method of determining Binding Energy nicely and correctly gives the binding energy of Helium3 as just slightly exceeding that of Hydrogen3, implying that Helium3 is a bit more stable than Hydrogen3. And that nicely seems to be the case, because Hydrogen3 is unstable and decays, although very slowly; contrasting with Helium3, which is stable, but not nearly as commonly occurring as Helium4, (which has much more ‘binding energy per nucleon’). Although most of this article’s themes are somewhat discernable using the old, often used binding energy method, the themes are more strongly apparent using our newer binding energy method. For example, when going from 3 stable nucleons to 4, the new method gives a jump in ‘binding energy’ as from 0.0074458 u to 0.028698 u, an almost full ‘4fold’ increase. That is only about 3.64% short of that 4fold approximation. But using the old method, the jump in binding energy is from 0.0082857 u to 0.030378 u. We note that it then seems to be only very roughly a 4fold increase. That is, it falls short of a full 4fold increase by about 8.34%. So comparing that with our newer method’s miss of only 3.64%, we note that our newer method comes closer to the ‘4fold’ approximation. Although our improved ‘binding energy’ method was discussed last, it is just as important as the previous topics! Since the topic has been compressed and oversimplified in the above, the reader can, optionally, ‘Read More about the Older and Newer Binding Energy Methods’ by clicking here. To optionally view or printout ‘all illustrations scrolled together’, click Fig.1Fig.2Fig.3Fig.4 —Scrolled. This article has mainly focused on the simple atoms or nuclei with only 1, 2, 3 or 4 nucleons, but to find a few (optional) ‘Comments about Larger Atoms or nuclei’, click here. For a list of ‘References and their relevance to this article’, click here.  End of Main Article  Click to return to Beginning of Main Article Click to return to Home Page 

Causeeffect.org
Carl R. Littmann (Readers’ comments
always welcome)
