## Total Binding energy for all isotopes.

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chrismb
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### Total Binding energy for all isotopes.

(At great personal effort to prepare!!.. behold!)

(Actually, I plotted it as nucleon number rather than atomic mass, but it's close enough for the purpose.)

KitemanSA
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Marvelous!

D Tibbets
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Well, this is perhaps an accurate depiction, but only of part of the story. Your graph is essentially the second in this (fig 2.5) illustration depicting the repulsive force, the attractive force and the net force (the binding energy curve).
Taken from the textbook with the front cover included.

There are several preceding pages that describe the derivation of the graphs, but for that you will need the book.

http://www.cambridge.org/gb/knowledge/i ... cale=en_GB

Dan Tibbets
To error is human... and I'm very human.

chrismb
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No, my graph above is bimodal, just very slightly so.

Initially the curve is concave, then goes convex. If you put a grazing line from the origin, it would touch the curve at 62Ni.

D Tibbets
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Concave and convex? Your eyes must be better than mine. From the X axis it curves concave all the way. And you could draw a straight line to any nuclei assuming the graph is continuously concave.
And if my eyes deceive me and if your graph is indead S shaped, then why? At some point does each added nuclei added start magically adding ever increasingly greater binding energy.?

And, if so, how do you address the multiple sources of the binding energy curve that accept the reality that it represents. Are they all wrong?

PS: granted there is some variation when the range of isotopes of each element is included, and these are obvious (after all Ni62 is the most stable, not Ni64, or Ni58). but theses variations due to neutron numbers, etc. are only locally significant when compared to near neighbors on the elemental sequence. The most obvious point where a too tightly focused view could be confusing is at Helium 4. The general curve is concave throughout.

Dan Tibbets
To error is human... and I'm very human.

KitemanSA
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chrismb wrote:No, my graph above is bimodal, just very slightly so.

Initially the curve is concave, then goes convex. If you put a grazing line from the origin, it would touch the curve at 62Ni.
Sorry Chris, I think you just confused him more.

The curve is concave down all the way but with changing curvature. If you plot a line that has the slope of ~8.7MeV/A and adjust it so that it skims the curve it will touch at about A=62.

The plot is the BE vs A curve. Not the more typical BE/A vs A curve I showed earlier. But it does show that if you add more nucleons, you release more energy.

Dan, this is not the "first" part of anything. It IS the BE vs A curve.

D Tibbets
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I am appalled!.
I quote and show the specifics from a plasma fusion text and you ignore it.
I am not the one who is confused and in denial. Other than criticizing my intelligence, I have yet to see any strong defense against these expert sources.
Chrismb's curve is of the attractive nuclear force. It is clearly so when you compare it to the second of three graphs, and read the text. This has also been presented from several other sources (in another thread). Why he ignores the coulomb force which has been spelled out repeatedly by multiple authorities escapes me. The graph Chrismb presented is not the Nuclear Binding Energy, unless you reject and explain that rejection of the Total Binding Energy graph which has been posted and referenced in multiple links. The Binding Energy is clearly shown to peak at Ni62 and to fall off on either side. Please read the text included with the illustration. Better yet get the book and read the pertinent ~ 8 pages.

Dan Tibbets
To error is human... and I'm very human.

hanelyp
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Can we get a graph of binding energy/nucleon? Or the spreadsheet so we can do such graphs ourselves?

chrismb
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I re-plotted the total binding enery graph on new axes. I transformed the plane of the graph to show more of the detail.

This is too cool! Just look at the structure within the data!

Shall I call this the 'Bradley diagram of binding energies'!

KitemanSA
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D Tibbets wrote: I am appalled!.
I quote and show the specifics from a plasma fusion text and you ignore it.
I am not the one who is confused and in denial. Other than criticizing my intelligence, I have yet to see any strong defense against these expert sources.
And I am disappointed. You have shown a nifty set of graphs, one that shows the "nuclear" forces (graph b) and the "coulomb" forces" (graph a) and the combination of the two that results in the BE/A vs A graph that is labeled c. Nifty, neat, huzaw! And it seems infinitely more confusing to you than to me.
Then you wrote: Chrismb's curve is of the attractive nuclear force.
No, it is not. It is the FINAL curve of your trio above (graph c) except that it has been INTEGRATED over A. Graph c above is a differently labeled version of the curve I provided way back when. It is the "BE/A vs A" curve. Chrismb's curve is the BE vs A curve. A simple integration of graph c above; the SAME DATA plotted differently.
Then you wrote:It is clearly so when you compare it to the second of three graphs, and read the text. This has also been presented from several other sources (in another thread).
Graph b is in fact the curve you mention. What you should be taking from this text that you show is that the BE/A vs A curve we have been discussing HAS THE COULOMB EFFECT build in. It is the END product curve. And you can use it without worrying its subcomponent parts.
Then you wrote:Why he ignores the coulomb force which has been spelled out repeatedly by multiple authorities escapes me. The graph Chrismb presented is not the Nuclear Binding Energy, unless you reject and explain that rejection of the Total Binding Energy graph which has been posted and referenced in multiple links. The Binding Energy is clearly shown to peak at Ni62 and to fall off on either side. Please read the text included with the illustration. Better yet get the book and read the pertinent ~ 8 pages.
He hasn't. He simply integrated the last curve, graph c. Again. the BE/A vs A curve has the coulomb issues (AND ALL OTHERS) built in. It is the Final compilation of ALL the issues. You use IT, the FINAL compilation curve to do your calcs. But it is on a PER NUCLEON basis. Chris just plotted it on a "per nucleus" basis, i.e., integrated it (multiplied each term) by the sum of the nucleons (A) in each nucleus.

KitemanSA
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chrismb wrote:I re-plotted the total binding enery graph on new axes. I transformed the plane of the graph to show more of the detail.

This is too cool! Just look at the structure within the data!

Shall I call this the 'Bradley diagram of binding energies'!
Chris, I stand corrected. With this plot it become appearant that in the REALLY SMALL "A" regime there is in fact a convex section. WELL PLOTTED!!

So now we have the "Chris" part and the "b= Bradley" part. What is the "m"? The inquiring mindless want to know

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KitemanSA wrote:
chrismb wrote:I re-plotted the total binding enery graph on new axes. I transformed the plane of the graph to show more of the detail.

This is too cool! Just look at the structure within the data!

Shall I call this the 'Bradley diagram of binding energies'!
Chris, I stand corrected. With this plot it become appearant that in the REALLY SMALL "A" regime there is in fact a convex section. WELL PLOTTED!!

So now we have the "Chris" part and the "b= Bradley" part. What is the "m"? The inquiring mindless want to know
It has made me curious. I am wondering two things. First is what is the driving issue for the boundary. Second is what is the magical isotope that resides at about 500MEV & 100(ish) Nucleons & 7.4(ish)Mev/Nucleon?

KitemanSA
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I think it is a spot on your screen cuz there is no dot at that location on MY screen!

Indeed, that LOCATION doesn't exist on my screen. In order for BE=500MeV AND A=100 to cross, the MeV/A would have to be 5, not 7.4.

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KitemanSA wrote:I think it is a spot on your screen cuz there is no dot at that location on MY screen!

Indeed, that LOCATION doesn't exist on my screen. In order for BE=500MeV AND A=100 to cross, the MeV/A would have to be 5, not 7.4.
Yes, I made an eyeball of the trending lines back to a source point. Originally I wrote it as "Magical Isotope" but then changed it.

It is curious to me that away from the boundary, there seems to be a convergence point.

KitemanSA
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ladajo wrote: Yes, I made an eyeball of the trending lines back to a source point. Originally I wrote it as "Magical Isotope" but then changed it.

It is curious to me that away from the boundary, there seems to be a convergence point.
I am curious. Which point?