Alpha Decay, Binding Energy and fusion

AI Thread Summary
Alpha particle formation within a nucleus can be viewed as a type of nuclear fusion, as it involves the binding of nucleons. Larger nuclei are more likely to undergo this process due to their greater number of nucleons, which allows for more interactions. The formation of an alpha particle results in a configuration with higher binding energy, indicating a more stable state. Smaller nuclei do not experience this phenomenon as readily because they lack the necessary nucleon density and interactions. Understanding these dynamics is crucial for comprehending nuclear stability and reactions.
esmeralda4
Messages
52
Reaction score
0
Hi there,

When an alpha particle is formed inside a nucleus by binding together is this essentially nuclear fusion occurring?

Why is this more likely to happen in a large nucleus than a small one?

Thanks
 
Physics news on Phys.org
looks like you are completely going off the topic
nuclear fusion is a nuclear process where in which two LIGHTER nuclei of two atoms collide and eventually forms a new nucleus. so it is more likely to happen in smaller nucleus.
 
Thanks for reply.

But considering the alpha particle that is formed. Why do the nucleons within the nucleus get even closer together to for an alpha particle with higher binding energy? And why does this not happen in a small nucleus?

Thanks
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Someone who shows interest in science is initially a welcome development. So are fresh ideas from unexpected quarters. In contrast, there is a scientific community that is meticulously organized down to the last detail, allowing little to no external influence. With the invention of social media and other sites on the internet competing for content, unprecedented opportunities have opened up for...
I am going through this course on collision detection: https://siggraphcontact.github.io/ In this link is a PDF called course notes. Scrolling down to section 1.3, called constraints. In this section it is said that we can write bilateral constraints as ##\phi(\mathbf{x}) = 0## and unilateral constraints as ##\phi(\mathbf{x}) \ge 0##. I understand that, but then it says that these constraints call also be written as: $$\mathbf{J} \mathbf{u} = 0, \mathbf{J} \mathbf{u} \ge 0,$$ where...
Back
Top