B Radioactive decay, Stability and Halflife

[A M]

TL;DR Summary

I have 3 main questions:

As it's written in the following article, nuclear binding energy is always a positive number; thus it takes energy to disassemble a nucleus into its nucleons.
...The binding energy is always a positive number, as we need to spend energy in moving these nucleons, attracted to each other by the strong nuclear force, away from each other... [Wikipedia]

And according to this diagram, some unstable nuclei like U_235 and U_238 have positive binding energy per nucleon (about 7.5 Mev) even greater than many stable ones'.

So, why should a radioactive nucleus (with BE/N higher than many stable nucleus) decay?

What does the stability of nuclei depend on?

And I also have a question about half life;
If decay energy of the nucleus -or anything else that makes it decay- is greater than nuclear binding energy, the unstable nucleus is expected to suddenly get disassembled into its component NUCLEONS. Or it shouldn't be created at all...
But they are stable for a consistent period of time and then converse to other NUCLEI.
⁉ Would you please explain the reason?

I would be grateful if anyone could elucidate these problems.
A M

A person who never made a mistake, never tried anything new.
-Albert Einstein

 

[mfb]

The decays go to nuclei with an even higher binding energy per nucleon. They never go into "all free protons and neutrons".
A nuclei is stable if there is no possible decay mode that releases energy. Some decay modes are so unlikely that nuclei can appear stable even if they should be able to decay in theory. This happens for many of the heavier nuclei we consider stable, e.g. some lead nuclei: Fission of lead could release energy, but it is so unlikely that it won't happen over any realistic timescale.

If decay energy of the nucleus -or anything else that makes it decay- is greater than nuclear binding energy

It isn't.

 

[A M]

Hmm... I'm still a little confused.
First of all:
If

It isn't.

How can decay energy dominate binding energy?
What exactly makes a radioactive nucleus decay?

- And I've read there are two main parameters on which binding energy depends:
Nuclear strong force & repulsive Coulomb force.
But I guess there should be a great force to overcome binding energy (in particular for heavier elements)
Right?
A M

A person who never made a mistake, never tried anything new.
-Albert Einstein

 

[mfb]

Toy example: An uranium-235 nucleus with 7.5 MeV binding energy per nucleon fissions to two nuclei with 8.5 MeV binding energy per nucleon for both of them. Released energy: 235 MeV.

- And I've read there are two main parameters on which binding energy depends:
Nuclear strong force & repulsive Coulomb force.
But I guess there should be a great force to overcome binding energy (in particular for heavier elements)

For heavier elements the repulsive Coulomb force becomes larger fast, lowering their binding energy per nucleon relative to previous elements.

 

[A M]

Thank you for your quick answer.
But even for heaviest nuclei, Coulomb force is smaller than strong nuclear force. Thus there should be another influencing factor...
A M

 

[Vanadium 50]

But even for heaviest nuclei, Coulomb force is smaller than strong nuclear force. Thus there should be another influencing factor...

That's already included in the binding energy.

 

[A M]

That's already included in the binding energy.

So increasing Coulomb force can't be the reason for instability of heavy nuclei.

 

[A M]

My main question is:
Why are heavy elements mostly unstable?
Lower binding energy per nucleon, Greater Coulomb force,...

 

[Vanadium 50]

So increasing Coulomb force can't be the reason for instability of heavy nuclei.

I said no such thing.

 

[A M]

I said no such thing.

What do you mean?

 

[A M]

Hmm...

 

[Vanadium 50]

I said no such thing.

What do you mean?

Which word didn't you understand?

 

[A M]

Forget about it!

 

[PeterDonis]

Binding energy is always a positive number; so we need energy to disassemble such nuclei.

Radioactive decay is not "disassembling" the nucleus. As has already been said, radioactive decay takes a nucleus with a particular binding energy per nucleon to another nucleus with even higher binding energy per nucleon. So "binding energy is always a positive number" is not a reason for nuclei not to decay.

Binding energy/nucleon of many heavy unstable nuclei is greater than some table ones

That's because there are only two kinds of radioactive decay that take one nucleus to another: alpha decay and beta decay. If there is not another nucleus with higher binding energy per nucleon that is reachable by one of those decays, a nucleus will be stable. The nucleus doesn't "care" that there are other unstable nuclei with higher binding energy per nucleon; those other unstable nuclei just happen to have a nucleus with even higher binding energy per nucleon reachable by alpha or beta decay (for heavy nuclei it's almost always alpha).

Increasing repulsive Coulomb force can't be the reason; because even for the heaviest nuclei, strong nuclear force is greater

The balance of forces is involved in determining what the binding energy per nucleon is for a given nucleus; but since we already know the binding energy per nucleon, we don't care how the particular balance of forces in a nucleus produced it if all we're interested in is stability.

why are heavy nuclei (Z>82) unstable?

Because every nucleus with Z>82 has at least one other nucleus with higher binding energy per nucleon reachable by alpha or beta decay, while at least some nuclei with Z <= 82 don't.

 

[A M]

Extremely helpful!
That makes so much more sense!

 

[A M]

there are only two kinds of radioactive decay that take one nucleus to another: alpha decay and beta decay.

You mean Isomeric Transition doesn't take one nucleus to another?!
Therefore they have some differences...

 

[PeterDonis]

Isomeric Transition doesn't take one nucleus to another?!

It doesn't change the isotope, no. The atomic number and mass number are the same after the transition as before.

 

[A M]

How about doubleβ decay?

 

[PeterDonis]

How about doubleβ decay?

That would count as beta decay, which I included.

 

[A M]

Oops... yes, yes.
Thanks for your complete explanation!
Good luck!

 

[A M]

The decays go to nuclei with an even higher binding energy per nucleon.

radioactive decay takes a nucleus with a particular binding energy per nucleon to another nucleus with even higher binding energy per nucleon.

unstable nuclei just happen to have a nucleus with even higher binding energy per nucleon reachable by alpha or beta decay

What do you mean "even"?
You mean there are some decays that go to "lower" binding energy?
In that case the final binding energy is higher, therefore energy is "needed"!

 

[PeterDonis]

What do you mean "even"?
You mean there are some decays that go to "lower" binding energy?
In that case the final binding energy is higher, therefore energy is "needed"!

You're making this way harder than it needs to be.

Say we have three nuclei, A, B, and C, with binding energies per nucleon $a$, $b$, and $c$, such that $a < b < c$.

Suppose there is a decay path (alpha or beta) from A to C, but no decay path from B to any nucleus of higher binding energy per nucleon (C or any other).

Then A will be unstable but B will be stable, even though B has higher binding energy per nucleon than A. And B will be stable even though there is at least one nucleus, C, with higher binding energy per nucleon than B, because B has no decay path to any such nucleus.

 

[A M]

I see!

there are only two kinds of radioactive decay that take one nucleus to another: alpha decay and beta decay.

But how about proton emission, neutron emission, spontaneous fission and cluster decay?

 

[PeterDonis]

how about proton emission, neutron emission, spontaneous fission and cluster decay?

These are all rare, but sure, if you want to throw them into the mix, that's fine. Everything I've said would apply to them as well.

 

[A M]

As you said (or I've understood!) if there is not another nucleus with higher binding energy per nucleon that is reachable by α, β, ... decays, a nucleus will be stable. But unstable nuclei have a nucleus with higher binding energy per nucleon reachable by alpha, beta... decays.
Now, we know why a nucleus is stable or unstable.
And we also know that among unstable nuclei, those that have shorter half lives are more radioactive and more unstable.
But why do some unstable nuclei decay faster than the others?
For example:

	Decay mode	Half life
Bismuth_209	alpha	≈2×10^19 years
Beryllium_8	alpha	≈8×10^-17 seconds

 

[PeterDonis]

why do some unstable nuclei decay faster than the others?

Have you looked up the other properties of these nuclei? For example, have you looked up the binding energy per nucleon of Bismuth-209 and Beryllium-8?

 

[A M]

Is it important?

 

[A M]

Well, let me guess...
The more the difference between a nucleus an its decay products, the faster it decays.
Right?!

 

[A M]

Here:

	Binding energy per nucleon (MeV)
Bismuth_209	7.848057507177
Beryllium_8	7.0624385
Thallium_205	7.8784650097561
Helium_4	7.1

 

[PeterDonis]

Here

Where are you getting the value for Beryllium-8 from?

 

[A M]

Wikipedia

 

[PeterDonis]

Wikipedia

Please give a specific link.

 

[A M]

Oh that wasn't Wiki
barwinski

 

[mfb]

Alpha decay tends to have a shorter half life if the reaction releases more energy. It can be a bit more complicated but that is the general trend.

 

[PeterDonis]

The more the difference between a nucleus an its decay products, the faster it decays.

It's not always that simple. Consider a couple of examples.

First, Beryllium-8. Its mass is 8.0053051. It decays into two alpha particles; twice the alpha particle mass is 8.005204. The difference is 0.0001011 amu, which is 94.2 keV; and the Be-8 nucleus is heavier. In other words, a Be-8 nucleus is basically two alpha particles plus 94 keV--the only reason it exists at all is that it's a resonance, i.e., when two alpha particles happen to come close enough together they can briefly form this state before splitting apart again. So the "half-life" of Be-8 is really just the length of time this resonance state exists.

Second, Bismuth-209. (Note that the barwinski.net reference you gave lists this isotope as stable.) Its mass is 208.9803987. It decays into Thallium-205, which has a mass of 204.9744275; that plus the mass of an alpha particle, 4.002602, gives 208.9770295. The difference is 0.0033692 amu, or 3.138 MeV. This is a signficantly bigger difference (decay energy) than Beryllium-8, but Bismuth-209 has a much longer half-life.

However, Beryllium-8 is an extreme outlier, because its "decay product" via alpha decay is an alpha particle itself--i.e., as I said above, it's just two alpha particles close together. Of course no other alpha decay nucleus has this property. For most other alpha emitters, I would expect the general rule @mfb gave to hold, that the larger the decay energy, the shorter the half-life.

 

[mfb]

Here is a graph for half life vs. decay energy of most naturally occurring alpha emitters. Bismuth-209 is off the scale.

Edit: Here with Bi-209 drawn in:

 

[A M]

Sorry for the extremely late reply.
I don't know the reason, but the more new things I learn from Physics, the more questions are formed in my mind.
First of all, why is alpha decay much more common than other types of fission?

Beryllium-8 is an extreme outlier, because its "decay product" via alpha decay is an alpha particle itself

Why is alpha particle that special?
For example there are some nuclei with higher binding energy per nucleon. Why α?

Here is a graph for half life vs. decay energy of most naturally occurring alpha emitters.

What is wave function?
(When I want to find my answers, there are always some unfamiliar words like wave function, spin, angular momentum,... that are related to QM -about which I don't know anything!
Actually I've just entered high school and my main subject is Biology; so the only reason why I learn Physics is my great interest.)

 

[PeterDonis]

why is alpha decay much more common than other types of fission?

Alpha decay is not normally considered a type of fission. It's much more common than spontaneous fission because it's much more likely that there will be some nucleus with higher binding energy per nucleon reachable by alpha decay, than that there will be some pair of nuclei that a heavy nucleus can exactly split into by fission. Also it's much harder to split a heavy nucleus nearly in half than for just an alpha particle to come out, because many more nucleons have to se

Why is alpha particle that special?

It's not that the alpha particle itself is special in the case of Beryllium-8; it's that thinking of Beryllium-8 as "emitting an alpha particle" is a misnomer. What is really happening is that two alpha particles came together for a very short time into a resonance state called "Beryllium-8" and are now separating again. Beryllium-8 is the only nucleus for which this is the case, because it's the only nucleus that is exactly two alpha particles bound together.

That said, the alpha particle--the nucleus of Helium-4--actually is special as far as nuclei are concerned, in that its binding energy per nucleon is very high for such a light nucleus. The detailed reasons behind this are beyond the scope of this thread, but it is part of the reason that alpha particle emission is a fairly common form of radioactive decay.

What is wave function?

If you label a thread as "I" level in the quantum forum, you are expected to already know the answer to this. See further comments below.

When I want to find my answers, there are always some unfamiliar words like wave function, spin, angular momentum,... that are related to QM -about which I don't know anything!

Yes, and that means you will need to spend the time learning about it in order to have the background necessary to understand the answers. You can't expect to just ask individual questions and understand the answers without that background.

I've just entered high school

That means you have plenty of time to learn more. In particular, you have plenty of time to learn more about the basics of quantum mechanics, such as what a wave function is. It is beyond the scope of PF to give detailed explanations of basic concepts like that. The basic answer is that the wave function is how the state of a quantum system is represented mathematically in quantum mechanics; but for that answer to make sense you need to take the time to study the material on your own.

 

[A M]

it's much harder to split a heavy nucleus nearly in half than for just an alpha particle to come out, because many more nucleons have to se

That makes more sense!

Yes, and that means you will need to spend the time learning about it in order to have the background necessary to understand the answers. You can't expect to just ask individual questions and understand the answers without that background.

OK, I'll do my best to learn the background.

It is beyond the scope of PF to give detailed explanations of basic concepts like that.

As a matter of fact even much more basic explanations (like binding energy) are beyond my school lessons!

 

[PeterDonis]

As a matter of fact even much more basic explanations (like binding energy) are beyond my school lessons!

Binding energy is actually fairly easy, so I'll give you a basic pointer.

Consider a simple bound system like a hydrogen atom in its ground state. It has a binding energy of 13.6 eV. What does this mean? It means that, if you wanted to separate the proton and the electron and make them both free particles, not bound to each other at all, you would have to add 13.6 eV to the atom to do it.

In other words, the binding energy of a bound system is the energy you would have to add to the system to separate all of its constituents and make each of them a free system, not bound to any of the others at all.

 

[A M]

In other words, the binding energy of a bound system is the energy you would have to add to the system to separate all of its constituents and make each of them a free system

What is a bound state?!

 

[PeterDonis]

What is a bound state?

A state where constituents, like the proton and electron in a hydrogen atom or all of the nucleons in a nucleus, are confined to a small region of space and can't escape. This is another of those basic terms that you really need to spend some time studying quantum mechanics on your own to learn.

 

[A M]

This is another of those basic terms that you really need to spend some time studying quantum mechanics on your own to learn

A state where constituents, like the proton and electron in a hydrogen atom or all of the nucleons in a nucleus, are confined to a small region of space and can't escape.

So, I've read that according to Pauli exclusion principle, bound state of identical fermions is forbidden. (Identical fermions can't occupy the same quantum state.)
I know what fermions are, but I want to know what "the same quantum state" is.
Thank you!

 

[PeterDonis]

I've read that according to Pauli exclusion principle, bound state of identical fermions is forbidden.

No, you didn't. What you read was this:

Identical fermions can't occupy the same quantum state.

That's not the same as "bound states of identical fermions is forbidden".

I want to know what "the same quantum state" is.
Thank you!

This is one of those basic aspects of QM that you really need to take the time to study for yourself.

 

[PeterDonis]

The OP question has been answered. Thread closed.