# Beta-decay energy

1. Oct 4, 2013

### RESolo

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!

2. Oct 4, 2013

### Staff: Mentor

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.