Alpha decay question

[Rade]

Unlike other types of decay, alpha decay (i.e., release of a [NNPP] particle) as a process must have a minimum-size atomic nucleus which can support it. The smallest nuclei which have to date been found to be capable of spontaneous alpha emission are the lightest nuclides of tellurium (element 52), with isotope mass numbers between 106 and 110.

 

I have a question. Given that the alpha cluster model is well known to explain many aspects of nuclei lighter than element # 52, tellurium, (for example, the structure of C-12), why does spontaneous radioactive emission of the alpha not occur at some element lighter than tellurium. WHY TELLURIUM (with Z = 52) is the proton size needed ? Is this understood or a topic for future research ? Thanks for help with understanding.

 

edit: I now see that I need to correct my facts---the "first" example of alpha decay is for berellium-8, this isotope has a structure of two alpha structures--very well documented experimentally. So, the first example of alpha decay is for isotope Be-8, then we never see this again until many, many elements higher in mass, Te-106 (note: it is unknown at this time if Te-105 shows alpha decay).

 

For anyone with interest, you can study decays of ALL known isotopes here, last column is type of decay mode if not a stable isotope:

 

http://httP://amdc.in2p3.fr/nubase/nubtab03.asc