# Au-196 decay: help reading tables

1. Oct 22, 2009

### JustinLevy

I'm sorry, this is probably a pretty stupid question as I probably am just not understanding something basic about nuclear reactions. Can someone please help?

I am trying to understand what happens when Au-196 goes to Hg-196.
I am assuming it is something like:
$Au^{196}_{79} \rightarrow Hg^{196}_{80} + e^- + \bar{\nu}_e + 0.686 MeV$
according to this table:
http://atom.kaeri.re.kr/cgi-bin/nuclide?nuc=Au196

What is confusing me is that it says the Au isotope is spin 2, and the table for Hg says the isotope is spin 0.
http://atom.kaeri.re.kr/cgi-bin/nuclide?nuc=Hg196

How is this possible?
Somehow the electron and neutrino must carry away some specific orbital angular momentum?

Also, what is the +/- they use in front of the spin numbers. How would a +2 be different than a -2?

If I was to look at an electron capture:
$Hg^{196}_{80} + e^- + energy \rightarrow Au^{196}_{79} + \nu_e$
Are there certain configurations/polarizations I can rule out due to the spin?
Does the negative beta decay energy it lists in the table mean this capture would require the electron have have at least ~ 4.4MeV of energy?

Any help understanding these tables would be great.

2. Oct 24, 2009

### arivero

Ah, you are looking for the Stone :tongue:. Ok, trick is first to do alpha from Hg to Pt, then beta from Pt to Au stable. In this way the energy balance is favorable, but as you say you could still have some issue with angular momentum, plus a very very slow rate (so you need the Stone, of course... and it is not easy to find in the market Joachimsthal black, nowadays. Of course you could do a "particular".).

About reading the tables, I have always some doubt about the electron binding energy. I believe that the tables are always for neutral atoms, so it is accounted. But no sure.

3. Oct 24, 2009

### JustinLevy

What do they mean by spin in those tables?
I'm starting to realize it clearly doesn't mean what I thought (I was taking it to be the total angular momentum of the nucleus. But if that was the case, it wouldn't mean anything to give a sign in front of it ... yet there are some nuclei in the full table it lists as spin=+2 and some as spin=-2).
http://atom.kaeri.re.kr/cgi-bin/nuclide?nuc=Au184 ... shows a state with spin = 2+
http://atom.kaeri.re.kr/cgi-bin/nuclide?nuc=Au196 ... shows a state with spin = 2-

Just understanding what they mean by spin here, and what the sign means, would be quite useful.

4. Oct 24, 2009

### OmCheeto

The http://www.nndc.bnl.gov/chart/reColor.jsp?newColor=dm" refers to your charts "spin" numbers as "Jπ"

Last edited by a moderator: Apr 24, 2017
5. Nov 9, 2009

### bcrowell

Staff Emeritus
This looks correct to me.

Right. The maximum angular momentum they could take away in the form of spin would be 1/2+1/2=1. But they also carry orbital angular momentum, which brings the total up to 2 in this case.

This is a notation for the parity of the state. The usual notation in nuclear physics is to write the parity to the right, as a superscript.

Ordinarily when we talk about electron capture, we're talking about the capture of an electron in an inner shell of the atom. Such an electron would have an energy in the low keV range, so there's no way you'd get one with 4.4 MeV of energy. The reaction you've written is still possible in theory, but it's not obvious to me that there's any situation where it would actually happen. In an accelerator, I think the cross-section would be way too low to detect, since it's a weak-force process. Conceivably this could happen in a white dwarf or something, assuming you had some heavy nuclei around. I don't think there's any easy way to rule out any possibilities according to spin. The minimum angular momentum for the final state is 3/2, so the incoming electron has to have at least 1 hbar of orbital angular momentum, which is perfectly possible. I don't think the parity of the incoming electron is constrained, since the outgoing neutrino's parity could be + or -.