Beta-Decay Energy: Li-11 Q=20 MeV & Calculations

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The discussion focuses on the beta-decay processes of Lithium-11 (Li-11) with a decay energy (Q) of 20 MeV. Two potential decay processes are analyzed: (1) Li-11 → Be-9 + 2n and (2) Li-11 → He-3 + He-8. The energy conditions for beta decay are defined using the equations Q={M(Z,A)-M(Z+1,A)-m_e } c^2 and Q={M(Z,A)-M(Z-1,A)-m_e } c^2. The participants clarify that the first process involves a decrease in both neutrons and protons, while the second process involves a decrease in neutrons and an increase in protons, indicating beta-minus decay.

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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!
 
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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?
11=9+2, the total number of nucleons stays the same. As you get one new proton, the number of neutrons decreases by one.

Basically, I don’t understand how the decay process works, and from then how to calculate Q when there are two daughter nuclei.
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.
 

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