Total Binding energy for all isotopes.

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KitemanSA
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Post by KitemanSA »

WizWom wrote: Furthermore, what the chart misses is the energy required to overcome the coulomb barrier, that is, what energy you need the proton to have to actually get close enough to the nucleus to be considered an event. If you charted that against the energy per nucleon chart, then you find that at about Iron you wind up needing more energy in than you get out... which is the point at which fusion rates would drastically drop.
Actually, the chart includes that "coulomb barrier energy".

I posted this table in another thread. Please note that where the two elements have isotopes that are neutron poor (relatively proton rich) the total binding energy is quite a bit lower at the "proton rich" side (~7.71MeV/A for 50Ni) vice the stable area (8.79 MeV/A for 62Ni). This (in part) represents coulomb energy input but NOT retrieved by gluon linking from the "missing" neutrons.

Code: Select all

p\n    22	    23	    24	    25	     26	    27	    28	    29	    30	    31
28	385451	401170	420457	435254	453152	467347	483988	494235	506454	515453
29				399647	418552	434858	452859	467907	484682	497108	509871	519932
keV   N/A	   -1523	 -1905	  -396	  -293	   560	   695	  2873	  3418	  4479

p\n    32	    33	    34*	    35	     36	    37	    38	    39	    40	    41
28	526842	534662	545259	552097	561755	567853	576830	582615	590430	595393
29	531642	540528	551381	559297	569207	576273	585391	591704	599973	605265		
keV    4800	  5867	  6122	  7201	  7453	  8421	  8560	  9088	  9543	  9872	

p\n    42	    43	    44	    45	     46	    47	    48	    49	    50	    51
28	602572	607048	613908	617527	623894	627251	633124	636073	641377	
29	613140	618510	625677	630293	636967	641042	647288	650832	656603	658517
keV   10568	 11462	 11769	 12766	 13073	 13791	 14164	 14759	 15226	 N/A
* This is the 62Ni+P=63Cu reaction.  It releases ~6.1 MeV.
PS: The numbers may not add up perfectly cuz I rounded some in Excel, and some here.

The binding energy contains all those effects. By the "Semi-Emperical" binding enegry equation, if those effects were not included, each nuucleon would release ~14MeV, not closer to 8.
Seems the prime reason such reactions happen very slowly in stars is 1) there is a derth of H when Fe is present, and 2) the simplest way to shed the excitation energy for such excited nuclei is to eject the proton again. None-the-less, it DOES happen in stars. Indeed, up to half the nuclei of elements <=Bi made by stars are made in main sequence evolution. the other half happens during novae/supernovae.

D Tibbets
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Post by D Tibbets »

I've made the mistake of using the coulomb barrier as part of the binding energy picture. While the coulomb barrier plays a role in the probability of a reaction proceeding, my current understanding is that this barrier does not have anything to do with the binding energy per nucleon. Once the nucleons approach close enough for the strong force to exert it's effect, the barrier is no longer relevant. If you assume you have perfect confinement of particles and energy within the system, then the coulomb barrier only determines the speed of the system, not the energy balance(distribution between potential and kinetic energy) This does not mean that the electromagnetic effect is not still involved. The binding energy per nucleon is determined by the strong force attraction and the electromagnetic repulsion balance. This balance is the key word. Total binding energy per nucleus does not reflect this balance, only the total energy, not the balance of positive and negative energy in relation to the potential energy of the nucleus, that is stored in the nucleus- or rather the energy that has to be poured into the system in order to drag all of the nucleons apart.

KitemanSA's comment about the binding energy per nucleon is revealing. The added neutrons adds to the strong force binding energy, but not to the electromagnetic repulsion of the bound protons within the nucleus. Thus the binding energy per nucleon is relatively larger than in the proton rich isotopes.
If you use the total binding energy graph, then the addition of either a proton or neutron adds ~ the same amount to the mass deficite. The argument that I oppose is that this total binding energy (mass deficite) reflects the exthermic nature of the reaction. This would imply that the nucleus would have the same energy/ mass deficite if it had the same number of nucleons irregardless of whether they were neutrons or protons. Yet the binding energy per nucleon is different. Since they have the same total number of nucleons, the total binding energy / mass deticite should be the same for isotopes that have the same number of nucleons. 56Fe should be the same as 56Ni. 3H (tritium) should have the same total binding energy as 3He, thus the energy balance should be zero. A 4H isotope should be the same total binding energy as a 4He isotope. This assumes your total binding energy (like in chrisMB's efforts) is dericved from the binding energy per nucleon *the number of nucleons. Does not this discrepancy force you to reconsider your viewpoint? you need to have a mechanism that accounts for this. Using the binding energy per nucleon does just this as it accounts for the differences in the binding energy for different combinations- elemental isotopes that have the same number of nucleons, but different distributions of protons and neutrons.

Dan Tibbets
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WizWom
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Post by WizWom »

Really, Dan, please stop. Wait until you have taken a modern physics class to contribute again. You are showing confusion.

There is a "ballpark estimate" stability formula, or atomic mass formula, which is the same thing. This was worked out in the 1950s, the term is "liquid drop model."
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D Tibbets
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Post by D Tibbets »

WizWom wrote:
D Tibbets wrote: Note that there are errors in this post. See my next post for a more reasoned and fact supported analysis.

Let me try some algebra.

Binding energy/ nucleon chart gives values from ~ 0- 8.9 MeV on the low side of Ni62 and ~ 8.9 MeV to ~ 6 MeV on the high side (these numbers are approximations).
Moron. Using 1 sig fig in a calculation of the change in the 6the and 7th decimal places of mass-energy is going to give you nonsense.
D Tibbets wrote:
f(x) = a/b *c
f(x)= energy yield
a= binding energy
b= nucleon
c= number of nucleons in the nucleus

The values are emperically determined from the graph and I have linked at least one source derived ~ values.

Solving for the equation for Ni62 yields:
energy yield for Ni62 + P ---> Cu63
8.9 MeV / nuclon * 62 nucleons ---> 8.7 MeV/ nucleon * 63 = 551 MeV ---> 548 MeV .
Like I said, Moron. Proper way: c^2*(mass in-mass out) = energy change.
Which is, in MeV and amu: 931.46*(61.9283488+1.0078250321-62.9296011) = 931.46*0.0065727321 = 6.1222 MeV released
D Tibbets wrote: The total binding energy drops by ~ 3 MeV. As this binding energy represents the energy released, it is clear that it took more energy to make the Ni62, than it did to make the Cu63. This is an endothermic reaction.

This is only one simple example, and admittedly the empirical MeV numbers per nucleon is estimates, but I think this shows the trend.
Garbage in, Garbage out. The only trend you show is that you have no clue.
You can complain about my math calculations, and call me what ever names you wish. You should note though that values in the millions does not imply 6 significant figures. You can choose any precision you wish. I chose two significant figures from memory estimates to illustrate my point- of course the effort was negated by errors.

But, your contention that this is inappropriate because there are six significant figures is wrong. Actually the table that this data is from has 2, 4 or mostly 7 significant figures. Many are actual measurements, but some are estimates based on theory. To state that there are 6 significant figures merely because most of the values are in MeV is, well- moronic.

If you are derogatory because someone commits errors you are a hypocrite if you do not apply the same rule to yourself.

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

KitemanSA
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Post by KitemanSA »

D Tibbets wrote:I've made the mistake of using the coulomb barrier as part of the binding energy picture. ...
Dan, I started to go through your missive above to correct you each place you were wrong. But I find I just don't have the time to waste on this anymore. There were just too many such places.

First and formost, learn what mass deficit is! Wow do you have that screwed up in your mind!!!

Most of the rest follows from that, I think.

First you must unlearn what you think you know that is wrong. Good luck with that.

Do the numbers correctly, think Hydrogen.

KitemanSA
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Post by KitemanSA »

D Tibbets wrote:
WizWom wrote: ....
You can complain about my math calculations, and call me what ever names you wish. You should note though that values in the millions does not imply 6 significant figures.
Dan your point is correct here, but you were off in the second significant digit and when you multiple such incorrect real numbers by integers in the mid two digit range, significant errors occur. Do the same calculations again CORRECTLY. PLEASE! You may just learn.







Why do I doubt it?

D Tibbets
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Post by D Tibbets »

KitemanSA wrote:
D Tibbets wrote:
WizWom wrote: ....
You can complain about my math calculations, and call me what ever names you wish. You should note though that values in the millions does not imply 6 significant figures.
Dan your point is correct here, but you were off in the second significant digit and when you multiple such incorrect real numbers by integers in the mid two digit range, significant errors occur. Do the same calculations again CORRECTLY. PLEASE! You may just learn.

Why do I doubt it?
Actually I was not off in my significant digits. I chose two significant digits, eg: 8.9 and 8.8. The significancent digit is not off, the numbers chosen as estimates (I was not trying to quote the data frem a table, but only a recolected estimate that would demonstrate my point. The numbers were wrong, not the significant figures.

As far as repeating the exercise, It is not worth it. I admit the effort was flawed, as I was not addressing the issue that is the bone of contention.

As you admit the strong force and the electromagnetic fore mediated energies both contribute to the mass deficit and the binding energy that is derived from it. What is the real issue is that you do not recognize, or admit, that these energies oppose each other. The strong force is an attractive force and tries to hold the nucleus together. The electromagnetic force tries to tear it apart (provided there are at least two protons within the nucleus). I have struggled to understand and explain this. The uses of the binding energy per nucleon and discussion about it on various physics sites makes it clear that fusion past 62Ni is not exothermic. Trying to understand why and explain it for myself and others is where the confusion derives. Not the basic fact.

As I have posted and linked, an easier way to approach the situation is considering the potential energy of the nucleus- after all it is this lowest potential energy/ most stable end product that determines the energy balance.
Using a negative number for the attractive force and a positive number for the repulsive is perfectly reasonable. It does not imply some strange energy, merely a mathematical convention that makes comparisons more straight forward.
The strong force is negative. When one or more nucleons is added to a nucleus, it always increases- becomes more negative (the potential energy goes down). The effect decreases with each additional nucleon (roughly) , but never reaches zero. If this was the only consideration, then nucleosynthesis through adding nucleons would always release energy, and there would be no limit on the nuclear growth. Elements of ziptillions of nucleons would be permissible, at least from an energy perspective. Each succeeding nuclei would be more stable, which means it would have less potential energy (a more negative number by the above convention). The opposite- removing nucleons- fission would always be endotherimic- raising the potential energy- decreasing the stability of the nucleus.

The second component is also important though. This is the electromagnetic repulsion between protons. You can consider the coulomb barrier that needs to be overcome as a loading of a spring, that wants to push the protons apart. Once within the range of the strong force, this spring cannot be easily unload. The Coulomb repulsion can be considered as loading a cross bow. The tension is maintained by the trigger block (strong force) that prevents the release of the tension. You have stored this kinetic energy (drawing the bow) as potential energy. By the above convention this is a positive number. Considering quantum tunneling, or magical catalysts that modifies this Coulomb barrier complicates things, but basically, you can ignore how the proton got into the nucleus so long as you recognize that once there it will be like a loaded spring pushing against all of the other protons. It represents a per nucleon (actually per proton) positive force or energy that is stored in the nucleus. It doesn't disappear or convert into some homogenous force.
The electromagnetic force is generally considered to be ~ 1,000,000 times weaker than the strong force- but this comparison is dependent on some assumptions about distance and number of particles involved. Remember the discussion a few years ago where the impressive repulsive magnitudes that could be obtained by a few Coulombs of charge/ cubic meter, such that many millions of volts (equivalent to eV) are required to contain them. Admittedly this is very many more charges than in a nucleus, but the nucleus also represents a very much smaller containment volume than a cubic meter. And with the inverse square law....
This is to illustrate that with the right conditions, the electromagnetic repulsion could actually exceed the strong force adhesion and stable nuclei become impossible. This is actually reported as becoming relatively dominate above atomic weights beyond lead (~206 AMU) . Beyond a few islands of relative stability, elements above plutonium are so unstable, that they can only be found by making them in a lab, and observing them very quickly before they fall apart- due to the electromagnetic repulsion dominating over the strong force attraction/ adhesion. These single or clumps of nucleons falling out of the nucleus have kinetic energy. This kinetic energy comes from the electromagnetic energy stored in the nucleus (the cross bow firing). The strong force is actually also decreasing, but this is an endothermic process and in this instance it is of lower magnitude compared to the electromagnetic energy released. It is just that there is less strong force mediated energy absorbed (increasing strong force mediated potential energy- the number becomes less negative) than there is electromagnetic force mediated energy being released. The net result is decreased potential energy of the product nuclei (s) compared to the starting bound nuclei. The opposite of this of course drives the release of energy from light element fusion.

At 62Ni there is a balancing point where the negative strong force mediated energy change per nucleon is balanced by the electromagnetic mediated positive energy change per nucleon. The sum of the two is zero [this sentance needs some clearification]. This means that the potential energy is relatively zero at this point - a relative baseline that means the potential energy of the nucleus is as low as it can get with any permissible combination of the competing forces. And, the lowest possible potential energy means that the maximum of available kinetic energy has been released. Moving away from 62Ni in either direction means you are adding kinetic energy in order to transform it into a nucleus with greater potential energy. It is endothermic. Of course moving towards 62Ni is thus exothermic- ie: fission above 62Ni and fusion below 62Ni, just as multiple instances of the binding energy per nucleon charts are labeled in various links.

At 62Ni, the strong force binding energy is still greater than the electromagnetic repulsive energy , but the electromagnetic repulsive energy is now increasing at a rate faster then the strong force attractive energy. As a result the nucleus is less tightly packed, less stable, has a higher potential energy as you add or subtract nucleons from this point. That a hydrogen nucleus/ proton has no binding energy is not relevant. The energy release or absorption depends on the balance between the strong force and the electromagnetic force within a bound nucleus. The total absolute energy made up of the opposing forces without regard to their competitive natures yields the total binding energy per nucleus curve. But as this ignores the competition between the forces it only tells you the total energy, not the very important interactions and their effects on the potential energy of the nucleus within this system of pertinent nuclear reactions.

The complexity with neutral neutrons, spin pairing preferences and other reactions complicates the picture, but the end result- is the binding energy per nucleon chart which shows the special relationships to 62Ni.

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

KitemanSA
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Post by KitemanSA »

D Tibbets wrote:
KitemanSA wrote:
D Tibbets wrote: You can complain about my math calculations, and call me what ever names you wish. You should note though that values in the millions does not imply 6 significant figures.
Dan your point is correct here, but you were off in the second significant digit and when you multiple such incorrect real numbers by integers in the mid two digit range, significant errors occur. Do the same calculations again CORRECTLY. PLEASE! You may just learn.

Why do I doubt it?
Actually I was not off in my significant digits. I chose two significant digits, eg: 8.9 and 8.8. The significancent digit is not off, the numbers chosen as estimates (I was not trying to quote the data frem a table, but only a recolected estimate that would demonstrate my point. The numbers were wrong, not the significant figures.
Actually you were. You had 8.9 and 8.8 but they are both 8.8 to two significant figures. You seem to have take 8.78 and inverted it to 8.87 and rounded to 8.9. Your last significant digit on Ni was wrong.
As I pointed out before...
62Ni: 8.78*62 ~ 544
63Cu: 8.76*63 ~ 552
63Cu has a larger binding energy, EXOthermic.

KitemanSA
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Post by KitemanSA »

D Tibbets wrote: As you admit the strong force and the electromagnetic fore mediated energies both contribute to the mass deficit and the binding energy that is derived from it. What is the real issue is that you do not recognize, or admit, that these energies oppose each other. The strong force is an attractive force and tries to hold the nucleus together. The electromagnetic force tries to tear it apart (provided there are at least two protons within the nucleus). I have struggled to understand and explain this. The uses of the binding energy per nucleon and discussion about it on various physics sites makes it clear that fusion past 62Ni is not exothermic. Trying to understand why and explain it for myself and others is where the confusion derives. Not the basic fact.
Right back at you guy.

Review the semi-emperical binding energy equation section in Wikipedia. The strong force is the POSITIVE "saturation" value that leads off the equation. It has a value of 14MeV per nucleon. All the other factors are negative (except the last one). The coulomb repulsion value is negative. The "surface tension" factor is negative. The symmetry factor is negative. All those factors subtract from the 14MeV per nucleon and produce the final value of binding energy per nucleon. None of which changes the fact that when ALL FACTORS are taken into account, the calculation using the actual binding energy per nucleon results in 62Ni + p = 63Cu being EXOthermic.

D Tibbets
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Post by D Tibbets »

KitemanSA wrote:
D Tibbets wrote:
KitemanSA wrote: Dan your point is correct here, but you were off in the second significant digit and when you multiple such incorrect real numbers by integers in the mid two digit range, significant errors occur. Do the same calculations again CORRECTLY. PLEASE! You may just learn.

Why do I doubt it?
Actually I was not off in my significant digits. I chose two significant digits, eg: 8.9 and 8.8. The significancent digit is not off, the numbers chosen as estimates (I was not trying to quote the data frem a table, but only a recolected estimate that would demonstrate my point. The numbers were wrong, not the significant figures.
Actually you were. You had 8.9 and 8.8 but they are both 8.8 to two significant figures. You seem to have take 8.78 and inverted it to 8.87 and rounded to 8.9. Your last significant digit on Ni was wrong.
As I pointed out before...
62Ni: 8.78*62 ~ 544
63Cu: 8.76*63 ~ 552
63Cu has a larger binding energy, EXOthermic.
You are picking hairs, and irrelevent hairs at that. You are wrong about the significant figures, I did not round from three figures, I simply stated erroneous values. End of story.

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

KitemanSA
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Post by KitemanSA »

D Tibbets wrote:
KitemanSA wrote: Actually you were. You had 8.9 and 8.8 but they are both 8.8 to two significant figures. You seem to have take 8.78 and inverted it to 8.87 and rounded to 8.9. Your last significant digit on Ni was wrong.
As I pointed out before...
62Ni: 8.78*62 ~ 544
63Cu: 8.76*63 ~ 552
63Cu has a larger binding energy, EXOthermic.
You are picking hairs, and irrelevent hairs at that. You are wrong about the significant figures, I did not round from three figures, I simply stated erroneous values. End of story.
Well, we agree that HOW you got to the wrong value is irrelevant. The FACT that you got the wrong values has been corrected and the results demonstrated. The FACT that the reaction in question is EXOthermic has been demonstrated. I will be ecstatic if this can indeed finally be "End of story"

D Tibbets
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Post by D Tibbets »

Using a negative number to represent the attractive force is reasonable and is stated in the same Wikipedia article where the empirical mass formula is discussed. Like gravity, the attraction is considered a negative force
The energy of the nucleus is negative with regard to the energy of the particles pulled apart to infinite distance (just like the energy of planets of the solar system),

That this attractive force is represented in the formula without this convention, is not really important, so long as the additive or subtractive contributions of the various interactions are treated consistantly within the formula.
I have looked at the liquid drop model and the emperical mass formula multiple times. I don't claim to follow the derivations as they involve complex math, but the implications of opposing forces and the net effect of same is obvious. It is also obvious that these forces are all involved in the mass deficit total.

Also, in the second link below (the Wikipedia article), the quoted second paragraph states the relationship. Can it be any more clearly stated?. If you are calling me a liar and/or fool, you are also calling this author (and any reviewers) the same.- not to mention all of the links that say the same.

http://hyperphysics.phy-astr.gsu.edu/hb ... qdrop.html


http://en.wikipedia.org/wiki/Nuclear_binding_energy
Combining nuclei

Other small nuclei can similarly combine into bigger ones and release energy, but in combining such nuclei, the amount of energy released is much smaller. The reason is that while the process gains energy from letting the nuclear attraction do its work, it has to invest energy to force together positively charged protons, which also repel each other with their electric charge.[5]

Once iron is reached—a nucleus with 26 protons—this process no longer gains energy. In even heavier nuclei, we find energy is lost, not gained by adding protons. Overcoming the electric repulsion (which affects all protons in the nucleus) requires more energy than what is released by the nuclear attraction (effective mainly between close neighbors). Energy could actually by gained, however, by breaking apart nuclei heavier than iron.[5]

The energy of the nucleus is negative with regard to the energy of the particles pulled apart to infinite distance (just like the energy of planets of the solar system),

The net binding energy of a nucleus is that of the nuclear attraction, minus the disruptive energy of the electric force. As nuclei get heavier than helium, their net binding energy per nucleon (deduced from the difference in mass between the nucleus and the sum of masses of component nucleons) grows more and more slowly, reaching its peak at iron. As nucleons are added, the total nuclear binding energy always increases—but the total disruptive energy of electric forces (positive protons repelling other protons) also increases, and past iron, the second increase outweighs the first. One may say 56Fe is the most efficiently bound nucleus.[6]

Mass defect is called the difference between the mass of a nucleus and the sum of the masses of the nucleons of which it is composed. The mass defect is determined by calculating three ascertained quantities.[2]

These are: the actual mass of the nucleus, the composition of the nucleus (number of protons and of neutrons), and the masses of a proton and of a neutron. This is then followed by converting the mass defect into energy. This quantity is the nuclear binding energy, however it must be expressed as energy per mole of atoms or as energy per nucleon.[2]

The amount of energy required to break the nucleus of an atom into its isolated nucleons is called nuclear binding energy. The measured mass deficits of isotopes are always listed as mass deficits of the neutral atoms of that isotope, and mostly in MeV. As a consequence, the listed mass deficits are not a measure for the stability or binding energy of isolated nuclei, but for the whole atoms. This has very practical reasons, because it is very hard to totally ionize heavy elements, i.e. strip them of all of their electrons.
This last link has been used before (or one very much like it). It shows the derivation of the binding energy from the mass deficit in a measured way (not theoretical). It also goes to the trouble of determining the binding energy per nucleon (or per mole).
Using the term "mole and Avogadro's Number set me to thinking. Why do this if the total binding energy per nucleus is, as you contend, the pertinant information.I will try to expound on this in the next posting.

http://www.chem.purdue.edu/gchelp/howto ... gy.htm#Top

D Tibbets
Last edited by D Tibbets on Sat Jul 30, 2011 2:57 am, edited 1 time in total.
To error is human... and I'm very human.

D Tibbets
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Post by D Tibbets »

Take 63Cu and as per the definition of binding energy per nucleon it will cost you 8.752 MeV to rip off one nucleon. This is an energy expenditure, it is endothermic.

Now consider the binding energy per nucleon for 62Ni. It is 8.794 MeV per nucleon. This is the end product of ripping a nucleon off of 63Cu all of the other nucleons in the corresponding nuclei are irreverent beyond this specific reaction. The ripped off nucleon is in this case, irrelevant as it has no binding energy associated with it. The weight of the corresponding nuclei is irrelevant, the weight of the nuclei contained in a mole of the isotope is irrelevant (perhaps). If you do consider them complications enter in- see my edit below.The only thing that is relevant is the energy expended to rip off this single nucleon. or conversely to add this nucleon to the selected nuclei or moles worth of the isotope.
63Cu at 8.752 MeV/ nucleon (which is directly proportional to the energy in a mole of 63Cu) Multiplying the binding energy per nucleon by the number of nucleons is not the same as considering a mole of the material(more on this later). This vs 62Ni at 8.794 MeV per nucleon..

So, one mole of 63Cu would require 8.752 MeV * (6.02 *10^23 nuclei / mole ) = x amount of energy to remove this Avagodros number of nucleons from one mole of 63Cu.
One mole of 62Ni would require 8.794 MeV * (6.03 *10^23 nuclei/ mole)= Y
The moles cancel out and the proportions do not change so using the binding energy per nucleon is the same proportionatly of using energy per mole. By using the total energy per nucleon * the number of nucleons in the nucleus, you are getting a measurement involving weight, not moles.

[edit-I think it is legitimate to say that 63 grams of 63Cu has more binding energy than 62 grams of 62Ni, but this does not mean that an equivalent wt of 62 Ni does not have more energy. Still not sure this is what I'm trying to get at. Another way of stating things may be that one mole of 63Cu (~63 grams) has 551.3849 MeV * Avagadro's Number of binding energy. but ~ 63 grams (1.0161 moles) of 62Ni has a total binding energy of 554.0401 MeV * Avagadros Number of binding energy. The 62 Cu is short 2.6539 MeV on a weight to weight comparison. This raises more questions but it does illustrate that you need to consider everything if you are using the total binding energy per nucleus data. The binding energy per nucleon avoids this complication (?). Remember the binding energy per nucleon = the binding energy per mole. Things are still unsettled though. The larger difference compared to the difference due to using the single nucleon reaction above may reflect the appropriate total binding energy difference, but not the competing energy difference. The delta strong force vs delta electromagnetic force change may be reflected in the difference in the two numbers . The Electromagnetic mediated force may have increased by ~ 0.042 MeV more than the strong force contribution, thus this amount of binding energy per nucleon change. The total binding energy per nucleus went up by6.1227 MeV This would fit with the strong force going up by ~ 3.0403 MeV, while the opposing electromagnetic force increased by 3.0823 MeV. This assumes that the delta energy change of the two forces were balanced at 62Ni, and ignores other contributing interactions, but it does (perhaps conviently) shows that the destabilizing electromagnetic mediated effects have become relatively bigger. Remember though, that the strong force has built up a big lead and it will take another ~ 150 nucleon additions before it catches up fully.. ]


The product of ripping off one nucleon from Cu63 is 62Ni. The difference is one free proton which has zero binding energy . The net is that the binding energy of the product nucleus +proton went up compared to the reactant nucleus by the difference of product - reactant.= (8.794 +0) - 8.752. =0.042 MeV . That the binding energy of 62Ni per nucleon (or per mole) went up, means it released energy (attractive binding energy is considered positive in this evaluation- the more common graphical presentation where the 62Ni forms a positive peak on the binding energy per nucleon graph. Reverse the reaction and you will see that the fusion (adding one nucleon) is endothermic- requires energy input) The proton can be ignored, but I included it to keep the book keeping intact. This also illustrates that considering that the proton is somehow contributing energy is faulty. It is essentially along for the ride. What is important is the binding energy of the BOUND nucleons within a specific nucleus. You can not start with the binding energy / nucleon for a certain nucleus, then tear off all of the nucleons and say - see here is the energy released/ or consumed. It ignores the energy balance of all of the steps it took o get there. You have to sequentially proceed one nucleus at a time in the chain and recalculate (or look up ) the relevant binding energy per nucleon for that nucleus, etc. etc. Using the total binding energy would only be a valid predictor of the energy balance if the binding energy per nucleon in different nuclei stayed constant or changed in a linear manner, or intermediates were totally ignored. It does not work because of the competing forces. The relationships are non linear and even reverses- as is obvious when looking at the binding energy per nucleon (or binding energy per mole) of a nucleus/ isotope graph or data tables.

The total binding energy does describe the energy difference between a bound nucleus and the energy of the nucleons that make it up, but only in this absolute sense. It cannot be applied to intermediate steps as the results vary for each step. If nature was all or nothing, either you have completely free nucleons, of you have them bound into selected final nucleus without any intermediate steps, then it is meaningful. But except for H-H fusion or H+Neutron fusion, it occurs only vary rarely or not at all.

Oh no, another analogy :roll:

Consider a trip from San Francisco, Ca. to Reno, Nev. You can say you started at Sea Level and ended up at ~ 4,000 ft (?). So the energy expenditure (or gain on the return trip ) was 4,000 ft * a constant.
But, there are mountains in the way. You climb to perhaps ~ 10,000 feet then cost down to ~ 4,000 feet. If you look at the end points alone you have a very distorted view of what is actually occurring.

Considering power output per weight vs molar output can be miss leading. One mole of Uranium 235 fission releases much more energy than one mole of hydrogen (D-T) fusion, but one gram of hydrogen fusion yields much more energy than one gram of uranium. I once saw a formula for energy balance expressed as binding energy per nucleon *A /A. The first A would give you the total binding energy, but they divide by A again! It seemed silly. But, it does keep the bookkeeping in order (I think) while reflecting the true molar relationship of the binding energy per nucleon graph.

And another example on the other side. Rip off a nucleon from 62Ni to leave 61Fe. Binding energy difference between product and reactant is
8.703 MeV - 8.794 MeV = -0.091 MeV. This is negative, implying a endothermic fission, or the reverse - an exothermic fusion. This is reasonable and consistent with the above example and multiple references that unequivocally state that 62Ni is the turning point, the most tightly bound and stable nucleus. Also note that the energy yields or expenditures are closer to what I would expect for reactions close to Ni. This is also widely supported by the literature. You just do not get much energy out of these reactions. That is why a star that reaches this stage is on it's last leg. Using the total binding energy per nucleus to get results of multiple MeV differences bothered me and should have been a red flag to everyone that something was wrong. My two previous attempts to do this comparison was faulty due to the use of the inappropriate data (the total binding energy per nucleus data obtained by BE/nucleon *A) which I was arguing against. Dumb, I know. It didn't help that I screwed the math up also,


http://www.nndc.bnl.gov/amdc/masstables ... mass.mas03

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

KitemanSA
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Post by KitemanSA »

D Tibbets wrote:Take 63Cu and as per the definition of binding energy per nucleon it will cost you 8.752 MeV to rip off one nucleon.
And Dan, this is where you continuously make your mistake. The 8.752MeV is the AVERAGE energy needed per nucleaon to rip the nucleus totally apart. It does NOT tell you how much energy for any nucleon since the amount will change depending on what TYPE of nucleon you remove.

In order to determine the energy needed to remove ONE nucleon, you must also know the FINAL binding energy of the resultant nucleus. To remove ONE proton requires a different amount of energy that ONE neutron.... usually. The difference is determine by the changes in coulomb repulsion, symmetry and pairity. Study and understand the Semi-empirical Binding Energy Equation.

The binding energy PER NUCLEON is simply an averaged value. It gives ONLY some info on the stability of the nucleus. It CANNOT be used alone to tell you anything other than an APPROXIMATION of the binding energy released or absorbed by a reaction. You need to know the start state and end state of TOTL binding energy to determine the energy needed to remove a specific nucleon..

Since the rest of your missive was predicated on this precedant error, (as far as I read) I will ignore it as useless and not respond.

KitemanSA
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Post by KitemanSA »

D Tibbets wrote: Using a negative number to represent the attractive force is reasonable and is stated in the same Wikipedia article where the empirical mass formula is discussed. Like gravity, the attraction is considered a negative force
The energy of the nucleus is negative with regard to the energy of the particles pulled apart to infinite distance (just like the energy of planets of the solar system),
That this attractive force is represented in the formula without this convention, is not really important, so long as the additive or subtractive contributions of the various interactions are treated consistantly within the formula.
Concur. The convention used is immaterial if applied properly.
However, please note that the energy of the nucleous being negative wrt the starting particles means that a positive energy must be released to meet the deltaE = 0 law. As I understand the convention, "binding energy" is the inverse of the negative potential energy (mass) of the nucleous relative to its particles. The total mass goes DOWN (is negative), the "Binding Energy" goes UP (is positive).
D Tibbets wrote: I have looked at the liquid drop model and the emperical mass formula multiple times. I don't claim to follow the derivations as they involve complex math, but the implications of opposing forces and the net effect of same is obvious. It is also obvious that these forces are all involved in the mass deficit total.
The "liquid drop model" seems to have most of the same terms as the Semi-empirical model but with slightly different constants. I may one day do a side-by-side comparison of the two.
None-the-less, they show what the FINAL binding energy is for a specific nuclear configuration. I've seen formulations where that final binding energy is divided by the number of nucleons and some where it isn't.
In EVERY case that I have done (~400) if you start with a stable nucleous and add a proton, the reaction is EXOthermic (except He4 and H1 wherein the product p-p, (aka He2) doesn't exist).
D Tibbets wrote: Also, in the second link below (the Wikipedia article), the quoted second paragraph states the relationship. Can it be any more clearly stated?. If you are calling me a liar and/or fool, you are also calling this author (and any reviewers) the same.- not to mention all of the links that say the same.

http://hyperphysics.phy-astr.gsu.edu/hb ... qdrop.html

http://en.wikipedia.org/wiki/Nuclear_binding_energy
Sorry Dan, I don't see what you are getting huffy about. Nor do I recall having called you a liar. And all this info supports my side, not yours. I am truly baffled why you don't see that.
D Tibbets wrote:
Combining nuclei
...
Once iron is reached—a nucleus with 26 protons—this process no longer gains energy. In even heavier nuclei, we find energy is lost, not gained by adding protons. Overcoming the electric repulsion (which affects all protons in the nucleus) requires more energy than what is released by the nuclear attraction (effective mainly between close neighbors). Energy could actually by gained, however, by breaking apart nuclei heavier than iron.[5]
Thank you for pointing out this error. It has been corrected! :)

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