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Binding Energy Curve Misunderstanding

by tfortmuller
Tags: binding, curve, energy, misunderstanding
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tfortmuller
#1
Apr22-14, 12:45 AM
P: 1
Here's an example of the graphs I am currently studying.

http://hyperphysics.phy-astr.gsu.edu...gnuk/bcurv.gif

I think I have a fundamental misunderstanding of its meaning that I would love to get past. I understand that Fe-56 is the most stable nucleus (although I'm not sure why) and that nuclei above that mark are more likely to fission and nuclei below Fe in regard to their atomic number are more likely to fuse in order to achieve the thermodynamically favorable, very stable Fe nucleus. I understand that binding energy is the energy required to break apart the nucleus into its constituent nucleons.

Questions:

1) This seems counterintuitive, and not in the good way, 'He' for example has a much lower binding energy than 'U', why is this the case. To me, 'He' having a lower bonding energy means that it requires less energy to break apart and is thus not as strongly held together and would be more likely to fission rather than fuse. Visa versa for 'U', it has a high bonding energy, so it is strongly held together, so it should not readily fission. This is not what is observed, what is the misconception I'm experiencing, please explain and expand.

2) What does it mean to have a binding energy per nucleon in the first place, binding energy is the energy required to break the entire nucleus apart, how can we interpret binding energy based on its value per nucleon.

Please feel free to lend any other insight that might improve my comprehension of this topic!

Thank you.
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The_Duck
#2
Apr22-14, 01:28 AM
P: 860
Quote Quote by tfortmuller View Post
I understand that binding energy is the energy required to break apart the nucleus into its constituent nucleons.
Right (but remember that what is shown in the graph is the binding energy *divided by the number of nucleons*).

Quote Quote by tfortmuller View Post
To me, 'He' having a lower bonding energy means that it requires less energy to break apart and is thus not as strongly held together and would be more likely to fission rather than fuse. Visa versa for 'U', it has a high bonding energy, so it is strongly held together, so it should not readily fission. This is not what is observed, what is the misconception I'm experiencing, please explain and expand.
To assess whether a reaction is likely to happen, you should look at the energy of the initial state and the energy of the final state. If the final state has less energy than the initial state, then the reaction can happen under the right conditions.

Therefore, NO nucleus will ever disintegrate completely into its component nucleons, because all of them have nonzero binding energy, which means that the final state of a bunch of individual nucleons has higher energy than the initial nucleus (unless you supply a bunch of extra energy in the initial state, by, say crashing another particle into the nucleus really hard).

However, a nucleus can fission into two smaller nuclei if the sum of the energies of the final state nuclei is less than the energy of the initial nucleus. This can happen if the final state nuclei have higher binding energy per nucleon (remember than binding energy is actually a negative energy). Then the sum of the energies in the final state will be less than the energy in the initial state (and the extra energy will become kinetic energy of the final state nuclei).

Quote Quote by tfortmuller View Post
2) What does it mean to have a binding energy per nucleon in the first place, binding energy is the energy required to break the entire nucleus apart, how can we interpret binding energy based on its value per nucleon.
Binding energy per nucleon is simply the binding energy divided by the number of nucleons. There's nothing more to it.
snorkack
#3
Apr22-14, 02:53 AM
P: 384
Quote Quote by tfortmuller View Post

Questions:

1) This seems counterintuitive, and not in the good way, 'He' for example has a much lower binding energy than 'U', why is this the case. To me, 'He' having a lower bonding energy means that it requires less energy to break apart and is thus not as strongly held together and would be more likely to fission rather than fuse.
It takes less energy to break apart α into lone protons and neutrons. But still a lot of energy which has to be provided somehow.

And all nuclei smaller than α - there are just three, d, t and He-3 - have hugely smaller bonding energy, so it takes a lot of energy to break α, because little of the binding energy is kept by the remaining nucle/i.

Whereas the nuclei much bigger than α have higher binding energy. You could produce energy by fusing α-s into C-12, or even U.
Quote Quote by tfortmuller View Post
Visa versa for 'U', it has a high bonding energy, so it is strongly held together, so it should not readily fission.
It is strongly held together. But Th is held together even more strongly. So much more strongly that even though α is less strongly held together than U is, there is still energy released when U is fissioned into Th and α.


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