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aviv

What makes nuclei disintegrate at a certain point in time as opposed to another? As in, what is the fundamental reason for disintegration finally happening in an unstable nucleus?

jcsd

Gold Member
It's completely random, but the probailty that a nucleus decays can be affected by external factors.

Alpha decay can be reduced to the proabilty of the alpha particle (which can be thought of as reatining it's identity even when it is part of the nucleus) tunelling (as in quantum mechanical tunelling) out of the potential well casued by the strong nuclear force. The most obvious factors which would affect the proabilty of the alpha particle tunelling out of the nucleus is the potential involved.

mormonator_rm

I would agree with jcsd. To rephrase his statement, I would say it is a result of the wavefunction. Both the weak and strong forces have to play a role in it because the strong force will hold the nucleus together very strongly. I would say that the decay must happen through one of at least three modes:

1) quark flavor decay: a quark in one of the constituent nucleons changes flavor, as in the neutron decay n --> p + e- + -v. This mode accounts for beta decay, electron capture, and a small fraction of gamma decay.

2) radiative decay: nucleons moving to lower energy states in the nucleus emit a gamma photon. This mode accounts for the bulk of gamma decay.

3) tunneling decay: the wavefunction of a particular piece or constituent of the nucleus allows it to tunnel out of the nucleus sufficiently far to escape. This mode accounts for some neutron emmission and alpha decay.

I'm sure there are more, and that I haven't covered half of what could go on, but these are some of the basics in nuclear decay. I am not entirely sure how massive nuclear fission occurs exactly, but I would also guess that it results from a wave-function linked to nucleons trying to settle into more suitable energy levels after neutron capture (as in Uranium 235 + n --> Uranium 236 --> fission products).

aviv

So what if we magically figure out the energy levels and all the properties of an unstable nucleus at a certain time (I know this is impossible but just for the sake or argument). Could we then predict the exact time of decay of that nucleus?

jcsd

Gold Member
Originally posted by aviv
So what if we magically figure out the energy levels and all the properties of an unstable nucleus at a certain time (I know this is impossible but just for the sake or argument). Could we then predict the exact time of decay of that nucleus?
The randomness is an intrinsic part of the system, you have a wavefunction which behaves in a perfectly deterministic way which can correspond to a single known energy level and a known potential, but when you go to make a measurment you'll always find that that there is a ceratin probabilty that decay has occured.

jcsd

Gold Member
I'll just add with specific reefrnce to alpha decay:

the wavefunction $u_n(x)$ ($n$ corresponds to the energy level of the alpha particle)as I said behaves in a perfectly detrministic way, but imagine in the case where we have our Alpha particle still inside the nucleus (at least last time we looked) and we haven't yet made a measurement, the wavefunction, though it is centred on the nucleus reaches outside the nucleus through the potential barrier (where it decays exponmentially)and out of the other side into the region where the interplay of the attractive strong force and the repulsive Coulomb force would allow our alpha particle to be at it's present energy level. The wavefunction at a certain point at a certain correspondfs to the probailty of finding our alpha particle at that point at that time via the relationship:

$$|u_n(x)|^2 = P(x)$$

So when we make a measuremnt there is always a probailty that we can find the alpha particle outside of the nucleus.

aviv

Originally posted by jcsd
The randomness is an intrinsic part of the system, you have a wavefunction which behaves in a perfectly deterministic way which can correspond to a single known energy level and a known potential, but when you go to make a measurment you'll always find that that there is a ceratin probabilty that decay has occured.
So what you are saying that the time of decay is not predictible only when you insert measurement. That if I was a being that knew all that was going on at the present, I could predict the exact time a certain nucleus will decay. And that it is only when you bring in us humans trying to measure the situation that you get this probability situation.

Am I right?

jcsd

Gold Member
No, it's deeper than that before you make a measurement of it's postion the alpha particle doesn't have a well-defined postion (you could say it has simultaneously inside and outside of the nucleus, but that's not the conventional interpretation), the act of measutremnt forces the nucleus to decide whether it has decayed or not.

aviv

Originally posted by jcsd
the act of measutremnt forces the nucleus to decide whether it has decayed or not.
How so?

And, do you mean that until measured, the alpha particle has no real position and that it only gets a position once measured?

jcsd

Gold Member
Yes that's exactly what I mean, this may seem counterintutitive, but quantum physics isn't intutitive.

What happens when you make a measuremnt is that the wavefunction collapses at random into one of it's eigenvalues.

aviv

Thanks a lot that clears up a few things.

I still don't understand a hell of a lot about this, like how exctly "the act of measutremnt forces the nucleus to decide whether it has decayed or not" but I still have many many many years to figure it out

Thanks again.

jcsd

Gold Member
I wouldn't worry about it as why the wavefunction colapses when a measuremnt is made has baffled people for years; it's called te quantum mechanical measurment problem. The phenoumna of decoherence and alternatiove interpretations of quantum mechanics do try to adress this problem though.

turin

Homework Helper
Originally posted by aviv
I still don't understand a hell of a lot about this, like how exctly "the act of measutremnt forces the nucleus to decide whether it has decayed or not" ...
I wouldn't feel too bad about your level of understanding. If I'm not mistaken, Einstein didn't even ever come to understand (or even accept) it.

russ_watters

Mentor
Originally posted by aviv
So what if we magically figure out the energy levels and all the properties of an unstable nucleus at a certain time (I know this is impossible but just for the sake or argument). Could we then predict the exact time of decay of that nucleus?
Try this analogy on: You have a ball bouncing around in a box with an opening in it. It bounces around completely randomly. You know how big the ball is, how big the opening is, and how fast its moving around. From this you can calculate the average time it should take to find the opening, but you will never be able to caclulate exactly when it will find the opening.

jcsd

Gold Member
I think Einstein understood it, but he certainly never accepted it.

The analogy of ball bouncing around a box (presumably in a non-linear way) is okay, but it suggests that the particle has a postion before it is measured, though there are realist intrepretaions (e.g. Bohm's) where the partcile does have a postion before it's measured, it's not the conventional interpreation of quantum mechanics.

turin

Homework Helper
Originally posted by jcsd
I think Einstein understood it, but he certainly never accepted it.
He understood it more than I do, probably, but, when someone thinks a complete theory is incomplete for the wrong reasons, then can you really say that that person understands it?

Originally posted by jcsd
... there are realist intrepretaions (e.g. Bohm's) where the partcile does have a postion before it's measured, ...
Doesn't the Bohm-DeBroglie interpretation have holes in it, like conservation problems?

aviv

Thanks everyone. I still don't like the idea of a particle having no position until measured because if it has no certain position then how can it be in a certain position once measured, or to be measured? Same with a particle colliding, how can a particle collide in a certain position when it had no certain position before the collision? This is very interesting, I'm gonna read about Bohm's interpretation.

jcsd

Gold Member
Originally posted by turin
He understood it more than I do, probably, but, when someone thinks a complete theory is incomplete for the wrong reasons, then can you really say that that person understands it?
Well I think if he didn't understand it he wouldn't of been able to come up with all his various attempts to disprove it (i.e. EPR paradox, etc.), which actually helped to advance quantum mechanics. I don't think you have to necessarily believe something is true to understand it.

Doesn't the Bohm-DeBroglie interpretation have holes in it, like conservation problems?
It's got problems like the non-locality of the quantum potential, a particle in it desn't seem o have any other real attributes rather tthan postion, etc. So most people reject it. I thikn it just illustrates the problems of trying to create a realist interpretation of quantum mechanics.

jcsd

Gold Member
Originally posted by aviv
Thanks everyone. I still don't like the idea of a particle having no position until measured because if it has no certain position then how can it be in a certain position once measured, or to be measured? Same with a particle colliding, how can a particle collide in a certain position when it had no certain position before the collision? This is very interesting, I'm gonna read about Bohm's interpretation.
Because it has a probailty of having a certain postion once a measuremnt is made.

I wouldn't bother with Bohm's interpretaion, it's interesting, bu it's unlikely to ever be more useful than or to replace the conventional interpretaion of quantum measurement, I were you I'd look at decohernce or Everett's/the many worlds interpretation as they give much more satisfactory answers than Bohm.

aviv

Originally posted by jcsd
I were you I'd look at decohernce or Everett's/the many worlds interpretation as they give much more satisfactory answers than Bohm.
Thanks, will do!

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