Final answer:
The question relates to the conservation of charge, angular momentum, and linear momentum in beta decay, like that of Co-60 to Ni-60, and the normalization of the angular distribution of decay products into a probability distribution. Specific distribution details require the examination of spins and momenta of decay products.
Step-by-step explanation:
The question given pertains to the angular distribution of decay products in beta decay and involves utilizing conservation laws to determine how energy is distributed between decay products. In beta decay such as that of Cobalt-60 (Co-60) to Nickel-60 (Ni-60), both charge conservation and angular momentum conservation play critical roles. For the proper normalization of the angular distribution into a probability distribution, one must account for all possible directions in which the decay products can be emitted and the associated probabilities.
In this particular case of Co-60, the decay is a beta decay process where the initial nucleus at rest decays into a daughter nucleus (Ni-60), an electron (beta particle), and an antineutrino. The total charge before and after decay is conserved at 27. The lighter decay products, namely the electron and the antineutrino, will carry away most of the energy due to their smaller masses, as linear momentum is also conserved. Without detailed information on the specific spins and angular momenta of the decay products, only a general description of the conservation principles can be given, rather than a precise formula for the angular distribution. The decay energy absorbed can be estimated using given data such as the fraction of Carbon-14 and the decay energy average.