Can 41Ca Decay to 40K through Proton Removal in Certain Situations?

[ProjectFringe]

As I understand ⁴¹Ca usually decays to ⁴¹K through electron capture. However, if given a situation like the following:

⁴¹Ca⁺H⁺N=NK:

where ⁴¹Ca⁺ only has one electron, and that electron is bounded to a nitrogen atom (essentially making ⁴¹Ca⁺²), is electron capture still the best option for decay? In this situation electron capture would cause an undesirable number of electrons for the ⁴¹Ca⁺ and unbalance the compound.

So, assuming there are no other atoms to react with apart from what is listed in this compound, what is the most likely method of decay? Is it possible for ⁴¹Ca⁺ to remove a proton instead (decaying to ⁴⁰K and H⁺), creating the compound below?

H⁺KN=NKH⁺

 

[mfb]

Calcium has 20 electrons as neutral atom, a Ca+ ion still has 19. In particular, it still has all the inner s electrons which are most likely to be used for electron capture. The chemistry of the outer electrons is irrelevant for its nuclear decays. The molecule will change or break apart after a radioactive decay but that's a completely separate question.

Proton emission is a very rare decay process in general and it cannot happen with Ca-41.

 

[ProjectFringe]

Calcium has 20 electrons as neutral atom, a Ca+ ion still has 19. In particular, it still has all the inner s electrons which are most likely to be used for electron capture. The chemistry of the outer electrons is irrelevant for its nuclear decays. The molecule will change or break apart after a radioactive decay but that's a completely separate question.

Proton emission is a very rare decay process in general and it cannot happen with Ca-41.

Thanks!

I have two questions regarding your response:

1. Is electron capture still the most likely method of decay for ⁴¹Ca⁺² as well?

2. Can proton emission occur in beta decay from ⁴⁰K to ⁴⁰Ca?

 

[snorkack]

Thanks!

I have two questions regarding your response:

1. Is electron capture still the most likely method of decay for ⁴¹Ca⁺² as well?

Not only most likely but the only.
Found the decay energy of Ca-41. It is quoted as 421 keV
http://www.nucleide.org/DDEP_WG/Nuclides/Ca-41_tables.pdf
This means that there is no alternative to electron capture - positron emission would require 1018 keV.
Ca ions up to ⁴¹Ca⁺¹⁹ still undergo electron capture, just with longer halflives and lower decay energy. ⁴¹Ca⁺²⁰ is stable, having no electron to capture and not enough energy to emit a positron.

2. Can proton emission occur in beta decay from ⁴⁰K to ⁴⁰Ca?

No. For one, it would result in ³⁹Ar, not ⁴⁰Ca. For another, not nearly enough energy. Proton emission requires over 5 MeV energy normally.

 

[ProjectFringe]

Not only most likely but the only.
Found the decay energy of Ca-41. It is quoted as 421 keV
http://www.nucleide.org/DDEP_WG/Nuclides/Ca-41_tables.pdf
This means that there is no alternative to electron capture - positron emission would require 1018 keV.
Ca ions up to ⁴¹Ca⁺¹⁹ still undergo electron capture, just with longer halflives and lower decay energy. ⁴¹Ca⁺²⁰ is stable, having no electron to capture and not enough energy to emit a positron.

No. For one, it would result in ³⁹Ar, not ⁴⁰Ca. For another, not nearly enough energy. Proton emission requires over 5 MeV energy normally.

Thanks for your response! Now I have more questions

When you talk about the energy being released, is it the amount of energy that is being released upon decay of one atom of the element?

If it is, then does the quantity of atoms affect how much energy is available?

And is the energy being constantly released throughout the whole decay process of the atom, or just at the end?

 

[mfb]

When you talk about the energy being released, is it the amount of energy that is being released upon decay of one atom of the element?

Yes.

If it is, then does the quantity of atoms affect how much energy is available?

No, because each nucleus is independent of others.

And is the energy being constantly released throughout the whole decay process of the atom, or just at the end?

The decay process doesn't have a duration for all practical purposes. It's instantaneous. The energy is released at the time of the decay.

 

[ProjectFringe]

1. So where does the energy go when it is released during decay? Is it confined to the nucleus of the atom?

2. Also if the nucleus of an atom is going through multiple decays simultaneously, is there a possibility that the energy produced could be compounded in some way?

3. Lastly, is there any way possible then, that instead of ⁴¹Ca⁺ decaying to ⁴⁰K and H⁺, that ⁴¹Ca⁺ can decay to ⁴¹K through electron capture, and then ⁴¹K can decay to ⁴⁰K and a neutron through neutron emission.

⁴¹Ca⁺ > ⁴¹K > ⁴⁰K + Neutron

And I really mean is there ANY way for ⁴¹K to decay to ⁴⁰K through neutron emission? I tired to look it up, but it was quite confusing

 

[mfb]

So where does the energy go when it is released during decay?

Kinetic energy of the decay products, sometimes photons, sometimes excited states of the nucleus, sometimes some energy is transferred to electrons in the atom.

And if the nucleus of an atom is going through multiple decays simultaneously

It's not.

None of the nuclei you discuss here can decay via neutron or proton emissions. They are too close to the stable nuclei.

 

[ProjectFringe]

The decay process doesn't have a duration for all practical purposes. It's instantaneous. The energy is released at the time of the decay.

How does an atom of ⁴¹Ca, for example, go through decay (electron capture)? I was under the impression that energy was required for the decay to occur, but it sounds like energy is just being produced as a result of the decay. Is that correct?

 

[ProjectFringe]

Also if the nucleus of an atom is going through multiple decays simultaneously, is there a possibility that the energy produced could be compounded in some way?

Sorry not simultaneously. I meant in succession, like 116Pd > 116Ag > 116Cd > 116Sn.

 

[romsofia]

I'll just leave this here for you, it seems like it would be worth checking out if you haven't:
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map:_A_Molecular_Approach_(Tro)/20:_Radioactivity_and_Nuclear_Chemistry/20.04:_The_Valley_of_Stability-_Predicting_the_Type_of_Radioactivity

This concept is called "valley of stability" in nuclear physics if you have a textbook.

EDIT: The page before should be read as well.

 

[mfb]

Electron capture of Ca-41 releases energy. It doesn't need energy. If a process would need energy then it cannot happen as decay, as energy is conserved.

Sorry not simultaneously. I meant in succession, like 116Pd > 116Ag > 116Cd > 116Sn.

These are independent decays and you can study each one individually.

 

[Drakkith]

How does an atom of ⁴¹Ca, for example, go through decay (electron capture)? I was under the impression that energy was required for the decay to occur, but it sounds like energy is just being produced as a result of the decay. Is that correct?

That's right. The binding energy of ⁴¹Ca is smaller than that of ⁴¹K (takes more energy to pull ⁴¹K apart into its constituent particles), meaning that it is energetically favorable for ⁴¹Ca to undergo decay into ⁴¹K. This process releases energy.

You can see this in the difference in the mass of the two:
⁴¹Ca : 40.962278 Da
⁴¹K : 40.961825 Da

This is a difference of about 0.000453 Da, or 421,967 eV, which is given off as energy during the decay process. Note that rounding errors abound in my calculation, which is why my number doesn't exactly match the one published.

 

[ProjectFringe]

Got it! Thanks everyone for your help and patience!

 

[snorkack]

Neutron emission with beta decay is a common and very important process. This is why nuclear reactors are ever possible.
However, K-40 is nowhere near decaying this way
The beta active isotopes of K start:

1. K-40 1250 million years
2. (K-41 stable)
3. K-42 12 h
4. K-43 22 h
5. K-44 22 min
6. K-45 17 min
7. K-46 2 min
8. K-47 17 s
9. K-48 7 s beta and n

I could not look up the exact decay energyes, but K-40 has nowhere near enough energy to emit a neutron. Even K-47 does not have quite enough.