Binding Energy of Helium-4's Nucleus

[Laen]

Edit: I believe this might be in the wrong forum, if it is I am sorry I don't know how to delete or move :(

Hey guys quick question!

I am studying basic radioactivity just for fun and ran into something that my text doesn't elaborate on. Could someone please clarify?

The mass of Helium-4's nucleus is 4.0026 amu, but the mass of a proton is 1.00783 and a neutron is 1.00866.

Do the math and we are missing 0.03038 amu from the theoretical and the actual mass. The text atributes this missing mass to the 'Binding Energy' of the nucleus.

Converting that mass into energy 28 MeV... 7MeV per Nucleon.

Okay so basicly what I am asking is how is this mass that is lost taken from the nucleon, or the neutron to be more exact? The text states that their is a relationship between the neutron and the binding energy. Is just a little bit of mass 'chipped' off each neutron and used as the binding energy?

I just keep thinking of how this mass could be lost, but I can always shoot down any possible solution I am able to come up with. I am sure tho it has to do with the residual strong force and the gluons being exchanged. Maybe there is a special property of the up,down,down configuration of a neutron? I have also read about something called 'rest mass' but am not to familar with it. Could it be that gluons have no mass while in transit, but mass while 'resting'? My text says gluons have mass ... but if they don't have mass while being exchanged one can assume that there will always be exchanging done and there will always be some missing mass, aka the binding energy.

And well if my hypothesis is correct then I guess you could assume the replusion of the protons override the attraction of the gluons and therefore neutral agents, neutrons are needed to supply the extra gluons/residual force to bind the nucleus together.

And a little off topic:
A quark loses some mass when releasing a gluon, while the accepting quark gains?

Wew okay I have rambled on enough, I must say I have deepened my understanding of it just by the introspection I had to do by just asking this question. I started writing this without any idea, and then came out with a hypothesis while writing and even if it is completely wrong I still have learned so much and have developed a better picture of physics. Who would have thought asking a question would facultate learning :rofl: Well thank you for any input given!

 

[mathman]

The process of getting He4 from H1 goes through several steps. At each stage energy is released, mostly as photons, some in the recoil of the particles involved.

 

[Laen]

Umm I am sorry I made my main question a little vague.

I am really wanting to know:
-How is this missing mass(stated above) taken from the nucleus and used as the binding energy.-

-Also what is this binding energy.- The residual strong force?


This question really has nothing to do with decay, I was just attempting to give the overall goal of my studying.
Thanks again!

 

[cesiumfrog]

The thing to remember is that energy and mass are the same thing. Here's a simpler example:

Say you have an atom, and you want to remove one of the electrons (leaving a positively charged ion). Since the electron is tightly attracted to the nucleus of the atom, you will need to supply more energy to pull it away (this can be supplied in the form of a particularly energetic photon, or just by "heating" the atom). Consequently, if you measure the weight of the ion and the electron, they will add up to give something greater than the initial weight of the atom (and will actually equal the mass of the atom plus the relativistic mass of the photon).

So in helium, the missing mass leaves the nucleus as the heat of fusion. The binding energy is the amount of work it takes to pull the nucleons apart.

 

[Laen]

Ah I see what you two are getting at, I have been making it way too complex. But I still have a little misunderstanding, if you could please look at my logic...

We can use He-4 for example
4 lone nucleons 4.0329 amu

We would like to bind these 4 nucleons together to make He-4, and from what I sense you need to introduce a large amount of energy to bind them(huge particle accelerators etc) yet even with adding this energy you end up with He-4 at 4.00260 amu.

You mentioned energy is lost as heat of fusion. Are you saying with all the energy that it took to bind the nucleus together even more is lost as heat of fusion? Is this right??

Now going from being a bound nucleus to a broken makes complete sense.
He-4 + .0303amu energy = 2P + 2N

Alright and then for a not so important question, just curiosity. So the radius of 1 lone proton of Hydrogen is greater than the radius of of say Germanium. Or is it more dense or? Just wondering how the protons and neutrons pack on that extra mass! I imagine growing and shrinking balls in my head :tongue2: ... alright thanks guys I really appreciate your responses!

 

[quetzalcoatl9]

Umm I am sorry I made my main question a little vague.

I am really wanting to know:
-How is this missing mass(stated above) taken from the nucleus and used as the binding energy.-

-Also what is this binding energy.- The residual strong force?


This question really has nothing to do with decay, I was just attempting to give the overall goal of my studying.
Thanks again!

i don't think you are being vague.

the deal is this - it is not known how to write the hamiltonian for a nuclear system. the strong nuclear potential is only empirically known. but, Heisenberg noticed that the mass of the proton and the mass of the neutron were almost (but not quite) the same. he postulated that this was due to some charge-independent force (i.e. the strong force) that was resulting in the mass difference. the result of this postulate is that the proton and neutron are actually just different "iso"-spin states of the _same_ particle.

this is where my knowledge ends - see Arfken, 3rd Ed., pg. 267 for more.

 

[cesiumfrog]

You mentioned energy is lost as heat of fusion. Are you saying with all the energy that it took to bind the nucleus together even more is lost as heat of fusion? Is this right??

Yes. That is why people are building a fusion power reactor now in France.

Think of it like burning gunpowder.. or better still, hydrogen and oxygen: it just sits there and nothing happens until you push it with a burning match. Even if a reaction is going to release plenty of energy, reactions normally won't happen unless you provide a little activation energy to trigger it. Of course, the match is neglegible compared to the explosive heat/pressure/"mass-energy" released when you let the hydrogen and oxygen fuse together chemically.