I have to work out which if these two beta-decay processes are energetically possible, given that the beta-decay energy of Li-11 is Q = 20 MeV. The two processes are: (1) 〖Li〗_11→〖Be〗_9+2n (2) 〖Li〗_11→〖He〗_3+〖He〗_8 I am supposed to use the energy conditions for beta-decay equations, i.e.: Q={M(Z,A)-M(Z+1,A)-m_e } c^2 Q={M(Z,A)-M(Z-1,A)-m_e } c^2 But I don’t know which one; in the first process the numbers of neutrons and protons both decrease by one – how is this possible? (In the second process, the number of neutrons decreases by one and the number of protons increases by one, so I guess that is beta-minus). Basically, I don’t understand how the decay process works, and from then how to calculate Q when there are two daughter nuclei. So an explanation would be much appreciated. Thanks!
11=9+2, the total number of nucleons stays the same. As you get one new proton, the number of neutrons decreases by one. Just add both nuclei. This is energy conservation - the total available energy is given by the mass of ##{}^11Li##, and you have to check if that is sufficient to produce all daughter particles.