Jump to content

Recommended Posts

Posted

I have been doing some analysis work on isotope data, and got some results which dont seem to appear in the literature.

I attach an extract from my findings

Any comments please?

Posted

I don't see anything surprising in there, other than not really believing a graph having data associated with Z=285 and the vicinity. Any specific questions?

Posted
They all look to be results of binding energy that can be found from the semi-empirical mass formula... :|

The only binding energy graphs are 6.5b; 6.6b; and 6.6e

 

I don't see anything surprising in there, other than not really believing a graph having data associated with Z=285 and the vicinity. Any specific questions?

Ah! Graph 6.1 should have read total nucleon A number. Thank you, I will amend

 

Any specific questions?

Some have queried the finding

• There is only one fully stable “Z” isotope for a given atomic Nucleon number “A” and there are only 194 of these.

 

Have you any problem with this?

Posted

Some have queried the finding

• There is only one fully stable “Z” isotope for a given atomic Nucleon number “A” and there are only 194 of these.

 

Have you any problem with this?

 

Since it's not true, then yes. I didn't realize this was being claimed.

Posted

The main data came from the tables of Isotopes from Wikipedia

http://en.wikipedia.org/wiki/Isotopes_of_hydrogen etc

 

To this I have identified the decay mode, from

http://periodictable.com/Isotopes/001.3/index.p.full.dm.html etc.

 

The upload module would not let me attach the working spreadsheet, or an extract in doc format, so I hope you can decipher this. If not send me a message with a contact e-mail, and I will send you the original file.

Graham

Posted

You haven't made it clear what you're trying to prove.

 

As far as the "one stable Z for a given A" this is trivially falsified. Fe-58 and Ni-58 are stable. Se-80 and Kr-80 are stable. Sr-86 and Kr-86 are stable. And there are more.

Posted
As far as the "one stable Z for a given A" this is trivially falsified. Fe-58 and Ni-58 are stable. Se-80 and Kr-80 are stable. Sr-86 and Kr-86 are stable. And there are more.

 

I agree these pairs are listed as stable in most general sources but as I wanted the decay modes I consulted periodictable.com published by Wolfram research, which I presume is a reputable source, and found additional data, for instance the page

http://periodictable.com/Isotopes/028.58/index.p.full.dm.html

shows 58Ni has a 2B+ decay of -118.68keV to 58 Fe

and

http://periodictable.com/Isotopes/034.80/index.p.full.dm.html

shows 80 Se is 2B- decay 132.56keV to 80Kr

and

http://periodictable.com/Isotopes/036.86/index.p.full.dm.html

86Kr is 2B- decay 1258.01keV to 86 Sr.

And so on.

Having crawled through all 3177 listed isotopes, I found that these additional decays accounted for all the duplicates in the A numbers, leaving just one Z number for each which is fully stable (according to Wolfram). In 14 cases between 1 and 208. there is no stable Z number.

Which is why I raised the question (in several forums). It appears that no one has noticed this before.

Posted

The decays are possible because of the energy difference — in any list, only one value can be smallest. But are these decays actually observed?

Posted
But are these decays actually observed?

I think they must be. Although the isotope tables in Wikipedia show these as “stable” the remark is qualified by a very long half-life figure, I cannot imagine that these would be included without some evidence. However it looks like I’m going to have to do more research for supporting evidence.

Posted
In most cases, the double-beta decay is so rare as to be nearly impossible to observe against the background of other radiation.[/i]

And of course you have answered your question. It may be rare, but for these isotopes the preobability of decay is greater than zero.

All I have done is eliminate these from the list so that only the isotopes with zero decay probability remain.(which is where I came in!).

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.