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

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In summary, the conversation discusses the process of beta-decay and determining its energetic possibility. The two processes being considered involve the decay of Li-11 into either Be-9 and 2 neutrons or He-3 and He-8. The conversation also mentions using energy conditions for beta-decay equations and the confusion around how the decay process works and calculating Q for two daughter nuclei. It is explained that in the first process, the total number of nucleons remains the same and in the second process, the number of neutrons decreases by one and the number of protons increases by one. The summary concludes by emphasizing the importance of energy conservation in this process.
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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.
 

What is beta-decay energy?

Beta-decay energy is the energy released during the radioactive decay of a nucleus. This energy is released in the form of beta particles, which are high-energy electrons or positrons.

What is Li-11 Q=20 MeV?

Li-11 Q=20 MeV refers to the beta-decay of the lithium-11 nucleus, where the Q-value (or energy released) is 20 MeV. This is the amount of energy released during the decay process.

How is beta-decay energy calculated?

Beta-decay energy is calculated using the formula E = (m_initial - m_final)c^2, where E is the energy released, m_initial is the initial mass of the nucleus, m_final is the final mass of the nucleus, and c is the speed of light. This formula takes into account the difference in mass between the initial and final nucleus.

What is the significance of beta-decay energy in nuclear reactions?

Beta-decay energy is important in nuclear reactions as it is a fundamental aspect of radioactive decay and plays a crucial role in understanding the stability and properties of different nuclei. It is also used in various applications such as nuclear power and medical imaging.

Are there any factors that can affect the beta-decay energy of a nucleus?

Yes, there are several factors that can affect the beta-decay energy of a nucleus. These include the type of nucleus, the amount of energy released, and the surrounding environment. Additionally, the decay process can be influenced by external factors such as temperature and pressure.

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