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.