Tracing the U(235,92) + Neutron Process to Final Stable Nuclei

[Amith2006]

Homework Statement

1) In a sequential process, U(235,92) plus a neutron forms the compound nucleus [U(236,92)]* which then fissions; the fissions then produces decays. If the initial fission fragments are Ba(143,56) and Kr(90,36), illustrate a process leading to the final stable nuclei.

Homework Equations

The Attempt at a Solution

I have the solution, but I am unable to understand it. They say that both Barium and Krypton undergoes negative beta decays 4 times.
The initial process is:
U(235,92) + 1 neutron ---> [U(236,92)]* ---> Ba(143,56) + Kr(90,36) + 3 neutrons

The final reaction is:
U(235,92) + 1 neutron ---> [U(236,92)]* ---> Nd(143,60) + Zr(90,40) + 3 neutrons + 8
electrons + 8 antineutrinos
Why can’t alpha decay take place instead of beta decay? I think it is because the former takes place in isotopes having mass number greater than 200? Also, in such questions how do they say that beta decay takes place only 4 times after which a stable nuclei is obtained?

 

[esalihm]

this is more of a conceptual question I think. and the question is asking why is it beta and not alpha or anything else.
first of all, elements with large mass numbers when compared to their atomic numbers (when Z>2*A) tend to go under beta decay.
this is perhaps the answer they are looking for, but this is not the only reason and it cannot be.
the nature of the element also plays an important role on the type of decay. some rare elements might go under alpha in tihs case.