Lazarus Posted August 4, 2013 Author Posted August 4, 2013 Strange said: Close, but no banana. Weight of H atom: 1.007825 Weight of Neutron: 1.008664916 Delta: 0.000839916 H / Delta: 1199.9116578323 N / Delta: 1200.9116578323 P / Delta: 1199.2585768099 Using your less accurate numbers gives a result even further from what you claim. And what does "but a better number is .00083985429 AMU" mean? It almost sounds like you have made it up to give the right answer... None of the ratios appear to be exact or have any physical meaning. Reply: That is the mass of a combined electron and positron. 1200 times that number gives the hydrogen mass the exact 7 digits of accuracy to the number you use plus 3 more digits. 1201 times the number gives the neutron mass matching your number to 7 digits of accuracy. The probability of pure coincidence is therefore 1 in 10 million. At least it is unlikely. ******** Strange said: Please show the math to support this. And show that it applies to atoms with large numbers of protons, such as lead and uranium. Reply: For an electron and 2 protons with the protons a distance of 1 from the electron the attraction between the electron and a proton is 1. The proton to proton repulsion is .25. So 1 > .25 For 3 protons the attraction between the electron and a proton is 1. For the repulsion of a proton from another proton the calculation is: Half the p to p distance is d/2 = sqr root of 3 divided by 2. The repulsive force is = 1/3. The force from the other 2 protons is 2/3 times cos(30). Net repulsive force on a proton is 1 / 1.732. So 1 > 1/1.732 The arithmetic for 4 protons is a little work. ******** The 2D models are easy to construct. Models can be constucted in 3D but it is necessary to use the proton consisting of electrons and positrons to build most of them. Greater attractive force is available.. Here are some heavier atoms. Helium 3 Helium 4 alpha particle p p p e e e (or) p p p p p e e p p p Z=2 A=3 Z=2 A=4 Z=2 A=4 Sodium 23 Sodium 24 p e p p p p p e e e p p p p p p e e p e e p p p p p e e epe p p p p p p p p e e e p e e p p p p p p e e p e e p p p p p p p p p p e e e p p p Z=11 A=23 Z=11 A=24 Platinum 198 P P P P E E E E P P P P P P P P P P P P E E E E E E P P P P P P E E E E E E P P P P P P P P P P P P P P P P E E E E E E E E P P P P P P P P P P E E E E E E E E E E P P P P P P P P P P P P P P P P P P P P E E E E E E E E E P P P P P P P P P E E E E E E E E E P P P P P P P P P P P P P P P P P P E E E E E E E E P P P P P P P P E E E E E E E E P P P P P P P P P P P P P P P P P P E E E E E E E E E P P P P P P P P P E E E E E E E E E P P P P P P P P P P P P P P P P P P P P E E E E E E E E E E P P P P P P P P P P E E E E E E E E P P P P P P P P P P P P P P P P E E E E E E P P P P P P E E E E E E P P P P P P P P P P P P E E E E P P P P Z=78 A=198 ******** Strange said: Here's one: you're wrong. Reply: If you can't give a reason just invoke Authority. I really do appreciate your comments.
swansont Posted August 4, 2013 Posted August 4, 2013 Swansont Said: You're jumping the gun (in many ways). A semi-classical model with elliptical orbits would have to be tested first, before you can apply it and claim it to be responsible for a redshift. For starters, explain how you get an S state in hydrogen, which has zero orbital angular momentum, with this model. Reply: Let's rotate the electron enough to zero the rotational momentum. We could name it spin. No, that doesn't work. We know the orbital angular momentum of the S state is zero. In addition to (or separate from) the electron spin. ******** Swansont said: "Heisenberg Uncertainty Principle" Reply: I am somewhat in agreement with Bohr's "Don't tell God what to do" but I agree with Einstein's contention which is something like, "The cause and effect should not be thrown out. The lack of knowledge or ability to measure is behind the need for probability." I take it that this non-response response means you have no justification.
Lazarus Posted August 4, 2013 Author Posted August 4, 2013 Swansont said: No, that doesn't work. We know the orbital angular momentum of the S state is zero. In addition to (or separate from) the electron spin. Reply: I am not sure that applies to this model. If this model generates valid Red Shift values it may not matter. Swansont said:I take it that this non-response response means you have no justification. Reply: There are a great many excellent and very usefull algorithms that incorporate the Heisenberg concept but being able to calculate the probability of something doesn't mean that the real mechanism can't exist. I am still fretting about non radiating electron orbits. That seems to be a solid argument that has to be answered.
Strange Posted August 4, 2013 Posted August 4, 2013 That is the mass of a combined electron and positron. Is that a reply to "And what does "but a better number is .00083985429 AMU" mean? If so, it is wrong. The sum of the electron and positron masses is 0.0010971598 (based on the Particle Data Group figures). Here are some heavier atoms. Pretty pictures but without the math, I don't believe it works.
Lazarus Posted August 4, 2013 Author Posted August 4, 2013 Is that a reply to "And what does "but a better number is .00083985429 AMU" mean? If so, it is wrong. The sum of the electron and positron masses is 0.0010971598 (based on the Particle Data Group figures). Pretty pictures but without the math, I don't believe it works. Strange said: Is that a reply to "And what does "but a better number is .00083985429 AMU" mean? If so, it is wrong. The sum of the electron and positron masses is 0.0010971598 (based on the Particle Data Group figures). Reply: One possible explaination is that puting a electron and a positon together could result is mass loss. That is rather than annihilating each other. Strange said: Pretty pictures but without the math, I don't believe it works. Reply: It is easy to see that the light elements could work. The heavier elements would not work in this cofiguration. By allowing the protons consisting of electrons and positons share some of them a stronger bond can be made. Also, that makes it easier to constuct 3D models.
Strange Posted August 4, 2013 Posted August 4, 2013 One possible explaination is that puting a electron and a positon together could result is mass loss. That is rather than annihilating each other. Any evidence for that? Or just another ad-hoc guess to shore up a failed idea? It is easy to see that the light elements could work. The heavier elements would not work in this cofiguration. By allowing the protons consisting of electrons and positons share some of them a stronger bond can be made. Also, that makes it easier to constuct 3D models. Without the math, I don't believe it works.
swansont Posted August 4, 2013 Posted August 4, 2013 Swansont said: No, that doesn't work. We know the orbital angular momentum of the S state is zero. In addition to (or separate from) the electron spin. Reply: I am not sure that applies to this model. If this model generates valid Red Shift values it may not matter. It applies to reality; your model has to agree with nature. The ground-state of Hydrogen has no orbital angular momentum. Any model that does not agree is wrong.
Lazarus Posted August 13, 2013 Author Posted August 13, 2013 (edited) Strange said: And what does "but a better number is .00083985429 AMU" mean? Reply: The number .00083985428 divides the mass of both the proton and neutron evenly to 7 significant digits implying a "building block" for both of them. Do you have a better explanation for that or do you just prefer the 1 in 10 million chance that it is pure coincidence. Also, it pure coincidence that the arithmetic of an electron orbiting a nucleus can show the speed of galaxies. In addition, it is pure coincidence that the synchronizing of the movement of electrons in the Bohr atom shows that in elliptical orbits the electrons will take turns approaching the nucleus. ***************************************************** Strange said: Pretty pictures but without the math, I don't believe it works. Reply: I wrote a couple of programs to show the mathematics involved just for you. One does the positioning of the electrons and protons and the other does the force calculations. A picture of the model of the nucleus is followed by the results of the calculations. The programs are at the end of the post. Net force vector (nfx,nfy) on proton number 4 Atomic number is 3 Proton 4 at x= 1.000 y= 1.732 Plus x is repulsive, minus x is attractive Number x y d f nfx nfy q 1 1.000 0.000 1.732 0.333 -0.000 0.333 1 2 -0.500 0.866 1.732 0.333 0.289 0.500 1 3 -0.500 -0.866 3.000 0.111 0.344 0.596 1 5 -2.000 -0.000 3.464 0.083 0.416 0.638 1 6 1.000 -1.732 3.464 0.083 0.416 0.721 1 7 0.500 0.866 1.000 1.000 -0.084 -0.145-1 8 -1.000 -0.000 2.646 0.143 -0.192 -0.238-1 9 0.500 -0.866 2.646 0.143 -0.219 -0.379-1 Attractive force electron to proton -1.000 Total repulsive force on proton -0.437 Net force vector (nfx,nfy) on proton number 7 Atomic number is 6 Proton 7 at x= 1.500 y= 0.866 Plus x is repulsive, minus x is attractive Number x y d f nfx nfy q 1 1.000 0.000 1.000 1.000 0.500 0.866 1 2 0.500 0.866 1.000 1.000 1.500 0.866 1 3 -0.500 0.866 2.000 0.250 1.750 0.866 1 4 -1.000 -0.000 2.646 0.143 1.885 0.913 1 5 -0.500 -0.866 2.646 0.143 1.993 1.006 1 6 0.500 -0.866 2.000 0.250 2.118 1.223 1 8 -0.000 1.732 1.732 0.333 2.407 1.056 1 9 -1.500 0.866 3.000 0.111 2.518 1.056 1 10 -1.500 -0.866 3.464 0.083 2.590 1.098 1 11 0.000 -1.732 3.000 0.111 2.646 1.194 1 12 1.500 -0.866 1.732 0.333 2.646 1.527 1 13 1.000 0.577 0.577 3.000 0.047 0.027-1 14 -0.000 1.155 1.528 0.429 -0.374 0.108-1 15 -1.000 0.577 2.517 0.158 -0.530 0.090-1 16 -1.000 -0.577 2.887 0.120 -0.634 0.030-1 17 0.000 -1.155 2.517 0.158 -0.728 -0.097-1 18 1.000 -0.577 1.527 0.429 -0.869 -0.502-1 Attractive force electron to proton -3.000 Total repulsive force on proton -1.003 Net force vector (nfx,nfy) on proton number 93 Atomic number is 92 Proton 93 at x= 1.058 y= 0.036 Plus x is repulsive, minus x is attractive Number x y d f nfx nfy q 1 1.000 0.000 0.068 214.487 181.980 113.525 1 2 0.998 0.068 0.068 214.487 371.283 12.684 1 3 0.991 0.136 0.121 68.835 409.688 -44.442 1 91 0.991 -0.136 0.185 29.234 489.730 -41.700 1 92 0.998 -0.068 0.121 68.850 524.158 17.925 1 94 1.053 0.108 0.072 191.408 537.220 -173.037 1 274 1.004 -0.173 0.216 21.424 -215.954 166.582-1 275 1.014 -0.104 0.147 46.198 -229.816 122.512-1 276 1.019 -0.035 0.081 151.935 -303.600 -10.304-1 Attractive force electron to proton -643.469 Total repulsive force on proton -303.775 ***************************************************** Programs ' calcbnuc.bas defdbl a-z ' Calc positions of electrons and protons in a nucleus ' Circular models only ' ' Input Atomic number for different atoms ' Formlas ' r=1 Radius of inner circle of atom ' x=r*cos(theta)' Polar to orthogonal location of proton or electron ' y=r*sin(theta) ' Trig functions ' An example of circular models ' beryllium 8 ' p p p ' e e ' p p ' e e ' p p p ' ' Numbered ' ' 6 2 5 ' 10 9 ' 3 1 ' 11 12 ' 7 4 8 ' open "calcnuc.txt" for output as#8 open "junk" for output as#9 print"What is the atomic number of the atom?" input z z=int(z) atomno=z if z<3 then print" The number can't be less than 3" input a$ goto c9999' Outta here end if howmany=atomno' Number of protons in inner circle pi=3.1416' Pi twopi=pi*2 r=1' Set radius of inner circle to 1 anglebig=twopi/howmany' Angle between protons in inner circle angsmall=anglebig/2' Angle between every particle ri=1' ri is radius of inner circle of protons = 1 ' Calculate radiuso, radius of outer circle of protons x1=1' Coordinates of proton number 1 y1=0 x2=r*cos(theta+anglebig)' Coordinates of proton number 2 y2=r*sin(theta+anglebig) xc=r*cos(theta+angsmall)' Coordinates of point on circle to proton 2 yc=r*sin(theta+angsmall) xm=(x1+x2)/2' Mid point between protons 1 and 2 ym=(y1+y2)/2 distp1p2=sqr((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2))' Distance p1 to p2 distmpo=distp1p2*.866'Dist mid to outer.Equilateral triangle rp=distmpo/3' Radius of particles. Diagonal is 2/3 of angle bisector distcl=sqr((xc-xm)*(xc-xm)+(yc-ym)*(yc-ym))'Distance circle vs line ro=1+distmpo-distcl' Radius of outer circle ' Calculate radiusm, radius of middle circle of electrons rm=ro-rp*2' Radius of middle circle theta=0' Angle to particle. A proton is on x axis print#9, print#9," Inner protons" print#9, for i=1 to howmany' Do the inner protons x=ri*cos(theta)' Convert from polar to rectangular cordinates y=ri*sin(theta) print#8,x;",";y;",";1;atomno;rp'x, y, sign and radius of particle print#9,x;",";y;",";1;atomno;rp'x, y, sign and radius of particle theta=theta+anglebig' Increment around circle next i print#9, print#9," Outer protons" print#9, theta=angsmall for i=1 to howmany' Do the outer protons x=ro*cos(theta)' Convert from polar to rectangular cordinates y=ro*sin(theta) print#8,x;",";y;",";1;atomno;rp'x, y, sign and radius of particle print#9,x;",";y;",";1;atomno;rp'x, y, sign and radius of particle theta=theta+anglebig' Increment around circle next i print#9, print#9," Electron" print#9, theta=angsmall for i=1 to howmany' Do the inner electrons x=rm*cos(theta)' Convert from polar to rectangular cordinates y=rm*sin(theta) print#8,x;",";y;",";-1;atomno;rp'x, y, sign and radius of particle print#9,x;",";y;",";-1;atomno;rp'x, y, sign and radius of particle theta=theta+anglebig' Increment around circle next i c9999:' Done ******************************************************************** ' calcrepu.bas ' Calculate the force on a vunerable proton in the nucleus open "calcnuc.txt" for input as#1 open "calcrepu.txt" for output as#8 dim xx(500)' The electrons and protons dim yy(500) dim charge(500) dim theta(500) ct=0' Count of particles c12: if eof(1) goto c19' Read in the electrons and protons ct=ct+1 input#1,xx(ct),yy(ct),charge(ct),atomno,radius goto C12 ' Calculate the net force on a proton c19: particle=ct/3+1' Choose a proton in the outer circle to test atomno=ct/3' Atomic number of atom print#8," Net force vector (nfx,nfy) on proton number"; print#8,particle print#8," Atomic number is";atomno; print#8," Proton"; print#8,particle; print#8,"at x="; print#8,using"##.###";xx(particle); print#8," y="; print#8,using"##.###";yy(particle)' print#8," Plus x is repulsive, minus x is attractive" print#8, Print#8,"Number x y d f nfx nfy q" gosub calcnet' Calculate net force vector. + is out, - is in. repulsef=sqr(nfx*nfx+nfy*nfy)' Calculate repulsive force if nfx<0 then repulsef=-repulsef' Negative if attractive print#8, print#8," Attractive force electron to proton ";using"#####.###";-maxforce print#8," Total repulsive force on proton ";using"#####.###";repulsef goto c8' Done '----------------------------------------------------- ' calcnet subroutine calcnet: x=xx(particle)' Remember which proton to test y=yy(particle) maxforce=0' Remember maximum e to p force nfx=0' Net force vector for the proton nfy=0 for i=1 to ct' Loop through all the particles if i=particle goto c25' Skip ourself if charge(i)=1 then dx=x-xx(i)' Distance vector from a proton dy=y-yy(i) end if if charge(i)<>1 then dx=xx(i)-x' Distance vector from an electron dy=yy(i)-y end if d=sqr(dx*dx + dy*dy)' Distance ux=dx/d' Unit vector uy=dy/d f=1/(d*d)' Force = inverse of distance squared if charge(i)=-1 then if maxforce<f then maxforce=f' Remember maximun e to p force end if fx=ux*f' Unit vector times force fy=uy*f nfx=nfx+fx' Add to net force nfy=nfy+fy print#8,using"###";i; print#8,using"###.###";xx(i);'Print the coordinates, distance and print#8,using"###.###";yy(i);' force of this proton print#8,using"###.###";d; print#8,using"#####.###";f; print#8,using"#####.###";nfx; print#8,using"#####.###";nfy; print#8,charge(i) c25: next i return c8: c9999: Sorry, pictures didn't take. Edited August 13, 2013 by Lazarus
swansont Posted August 13, 2013 Posted August 13, 2013 There is no "positioning of electrons" without a valid model that has electron positions, and you haven't presented one. The Bohr model is demonstrably wrong — anything based on its mechanics is likewise wrong. There is no point in going past that to analyze any model when the premise is flawed. Any conclusion that might be right is purely accidental. Further, the Bohr atom doesn't have elliptical orbits! They are circular (one part of its wrongness, in many ways, for many reasons). Any analysis past this point of not having a valid model is a wasted effort.
Strange Posted August 13, 2013 Posted August 13, 2013 (edited) The number .00083985428 divides the mass of both the proton and neutron evenly to 7 significant digits implying a "building block" for both of them. Except it doesn't. Dividing the neutron and proton masses by this gives 1200.999897256 and 1199.3466947701 respectively. Also, it pure coincidence that the arithmetic of an electron orbiting a nucleus can show the speed of galaxies. As you have a fake model and arbitrary numbers, no, not really. You can "prove" anything with numerology. I wrote a couple of programs to show the mathematics involved just for you. Well, if you can't present the math ... I am not going to try and reverse engineer your code. Edited August 13, 2013 by Strange
Lazarus Posted August 14, 2013 Author Posted August 14, 2013 Strange said: Well, if you can't present the math ... I am not going to try and reverse engineer your code The test proton has the greatest x value and is at coordinates (a,b). q is the charge of a particle. (1 or -1) d is the distance from the test particle. d = sqr((a-x)^2 + (b-y)^2) The force vector on the test particle from another particle is: q(a-x)i q(b-y)j F = ------- + ------- d^3 d^3 The sum of all the force vectors is: gi + hj The net force, f is; f = sqr(g^2 + h^2) The attraction or repulsion is dependent on the sign of g. Swansont said: Further, the Bohr atom doesn't have elliptical orbits! They are circular (one part of its wrongness, in many ways, for many reasons). Any analysis past this point of not having a valid model is a wasted effort. Reply: This model is not the Bohr Model, the Bohr-Summerfield Model, the Bohr-Kramers-Slater Model nor the Rydenberg Model. The biggest similarity is that attempts are being made to describe the actual reality that underlies the probability waves. These people and others made a lot of progress accounting for split lines, fine lines and other relationships. If they had known what we now know that it is possible for nuclei to have rotating magnetic fields, which allows distinct orbits, things might have gone in a different direction. Elliptical orbits were considered long ago. The only substantial objection to this model is your point that an electron on a curved path should radiate. Not only is the momentum changing but the kinetic energy of the electron would vary in the orbit. One possible explanation is that the photon is so much bigger than a atom that an electron could make one or more complete circuits while a photon was coming or going. That could allow the kinetic energy the photon leaving in one part of the orbit to be reacquired on the opposite side. As for the momentum, the photon should leave in opposite directions on opposite sides of the ellipse, which might cause them to cancel each other.
Mellinia Posted August 14, 2013 Posted August 14, 2013 The test proton has the greatest x value and is at coordinates (a,b). what is x?
Lazarus Posted August 15, 2013 Author Posted August 15, 2013 what is x? x is the distance from the origin of a 2 dimentional coordiate system. I really don't understand why I am getting so many complaints about the arithmetic. It is just high school math. I am not complaining, I appreciate any comment.
Klaynos Posted August 15, 2013 Posted August 15, 2013 I really don't understand why I am getting so many complaints about the arithmetic. It's because of unconventional formatting I , look at any text book they all do equation followed by where x is the position in metres etc... Also the maths is where the predictions are where you can tell if an idea agrees with reality our not.
swansont Posted August 15, 2013 Posted August 15, 2013 This model is not the Bohr Model, the Bohr-Summerfield Model, the Bohr-Kramers-Slater Model nor the Rydenberg Model. You referred to it as an "elliptical Bohr atom". If it's not a Bohr atom, maybe you should stop doing that. The biggest similarity is that attempts are being made to describe the actual reality that underlies the probability waves. These people and others made a lot of progress accounting for split lines, fine lines and other relationships. If they had known what we now know that it is possible for nuclei to have rotating magnetic fields, which allows distinct orbits, things might have gone in a different direction. Elliptical orbits were considered long ago. Considered? Sure. People considered phlogiston for heat, too. Were these actually adopted as a working model? Why is that? The only substantial objection to this model is your point that an electron on a curved path should radiate. Not only is the momentum changing but the kinetic energy of the electron would vary in the orbit. That's one objection. Angular momentum is another. The larger issue is that you haven't presented an actual model, with comparison to experiment. There may be many other objections beyond these. It's hard to be specific when all there is is hand-waving.
Mellinia Posted August 15, 2013 Posted August 15, 2013 x is the distance from the origin of a 2 dimentional coordiate system. I really don't understand why I am getting so many complaints about the arithmetic. It is just high school math. I am not complaining, I appreciate any comment. Then why is there a maximum x for a test photon? You have not explained your maths with physic concepts.
swansont Posted August 15, 2013 Posted August 15, 2013 The net force, f is; f = sqr(g^2 + h^2) The attraction or repulsion is dependent on the sign of g. You're squaring g, so how does the sign matter?
Lazarus Posted August 16, 2013 Author Posted August 16, 2013 Then why is there a maximum x for a test photon? Reply: The most vunerable protons for expulsion are at the outer edge of the nucleus. ***************************************************************************************** You have not explained your maths with physic concepts Reply: Place the center of the nucleus at (0,0) of the coordinate system, Then rotate the nucleus until one of the protons furthest from the origin is on the positive x axis. That proton is the test proton. If the net force vector is positive the proton will be expelled, if it is negative the proton will be held in place. The calculation of the force vector between two particles is straight forward. Then you have to add up all the force vectors to find the net force and the direction of the force. . You're squaring g, so how does the sign matter? Reply: g is the component of the net force vector along the x axis. If g is positive the proton will be expelled. if negative ithe proton will stay in place. Correction to the reply to Mellinia Reply: Place the center of the nucleus at (0,0) of the coordinate system, --------------------------------------------------------------------------------------------------------- Wrong Then rotate the nucleus until one of the protons furthest from the origin is on the positive x axis. That proton is the test proton. Should read: Then rotate the nucleus until one of the protons closest to the origin is on the positive x axis. The proton with the smallest angle from it in the counter clockwise direction is the test proton. --------------------------------------------------------------------------------------------------------------- If the net force vector on the test proton is positive the proton will be expelled, if it is negative the proton will be held in place. The calculation of the force vector between two particles is straight forward. Then you have to add up all the force vectors to find the net force and the direction of the force.
imatfaal Posted August 16, 2013 Posted August 16, 2013 ... c12pict.JPG Carbon - with 12 protons and 6 electrons? Perhaps check your very basic chemistry - very positively charged and very not Carbon
Lazarus Posted August 16, 2013 Author Posted August 16, 2013 Carbon - with 12 protons and 6 electrons? Perhaps check your very basic chemistry - very positively charged and very not Carbon That is the nucleus, not the atom.
swansont Posted August 16, 2013 Posted August 16, 2013 Surely a nucleus is three-dimensional, and the lowest energy state would not be a flat ring structure.
imatfaal Posted August 16, 2013 Posted August 16, 2013 That is the nucleus, not the atom. You are trying to re-write physics and you seriously think carbon has 12, XII, twelve, TWELVE protons!!! http://www.webelements.com/carbon/ This is first year school chemistry/physics H, He, Li, Be, B, C...
swansont Posted August 16, 2013 Posted August 16, 2013 You are trying to re-write physics and you seriously think carbon has 12, XII, twelve, TWELVE protons!!! http://www.webelements.com/carbon/ This is first year school chemistry/physics H, He, Li, Be, B, C... The claim appears to be that the 6 neutrons are really 6 protons + 6 electrons, but now we have a problem with e.g. C-14 which would have 14 protons and 8 electrons, and undergoes beta decay. You have the electron emitted from the nucleus, but now have to explain where the antineutrino came from. Worse, you have neutron-deficient nuclei which undergo beta-plus decay. Like most of the "model" we have seen, it looks only at the first layer of the issue and ignore the complex layers below them which mainstream science has hashed out through the years; investigations and experiments which led to similar proposals being discarded when failed to predict or disagreed with observation.
Lazarus Posted August 16, 2013 Author Posted August 16, 2013 Surely a nucleus is three-dimensional, and the lowest energy state would not be a flat ring structure. Agreed. The ring structure just demonstrates that electrons can bind multiple protons and it is easier to visualize. There is something else that had to be resolved to make more satisfactory models. Three protons attached to an electron is insufficient to build good 3D models and 4 protons attached to an electron is too weak of a bond. The solution is indicated by the loss of nominal mass as we go up the atomic numbers and the fact that the difference between the mass of the neutron and the hydrogen atom is evenly divisable to 7 digits of accuracy into both indicating a building block which could consist of an electron and a positron.. That would imply that the joining of an eledtron and a positron could resust in some mass loss as apposed to annialating each other. The reduced mass of bigger nuclei can result form sharing the building blocks resulting in a stronger bond between protons. That allows compact models to be constructed. Another problem that had to be solved. The radius of an electron used to be considered more than twice the radius of the proton. Now the electron is thought bo be smaller than that. That helps to let the protons overlap.
swansont Posted August 16, 2013 Posted August 16, 2013 Agreed. The ring structure just demonstrates that electrons can bind multiple protons and it is easier to visualize. A real structure would be important, because from that you can calculate if the nucleus has any multipole moments and compare that to experiment. But the bigger issue is how this fits in with decay, as I have mentioned, or the complete contradiction with the Heisenberg Uncertainty Principle I think I brought up before.
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