What Determines the Allowed Values of lα in Alpha-Decay?

[Kulkid]

Hi, I am reading about alpha-decay. I read that the decay 2^- --> 2^x requires l_α = 1, 2, 3 or 4. Why does it have to be one of these numbers?

 

[AndyW]

Hello, thank you for your question about alpha-decay. To answer your question, let's first define what lα represents. In nuclear physics, lα is the orbital angular momentum quantum number for the alpha particle. This value is determined by the number of nodes in the radial wave function of the alpha particle as it tunnels through the nuclear potential barrier during the decay process.

Now, to understand why lα can only be 1, 2, 3, or 4 in the decay 2- --> 2x, we need to look at the conservation laws in nuclear physics. One of these laws is the conservation of angular momentum, which states that the total angular momentum of a system must remain constant. In the case of alpha-decay, the initial state (2-) has a total angular momentum of 0, since it is a spherically symmetric state. Therefore, the final state (2x) must also have a total angular momentum of 0.

Since the alpha particle has a spin of 0, the only way for the final state to have a total angular momentum of 0 is for the orbital angular momentum lα to be equal to 0 as well. However, according to the quantum mechanical rules, the value of lα must be greater than or equal to 0. This means that the only possible values for lα in this decay process are 0, 1, 2, 3, or 4.

In conclusion, the reason why lα can only be 1, 2, 3, or 4 in the decay 2- --> 2x is due to the conservation of angular momentum and the quantum mechanical restrictions on the value of lα. I hope this helps clarify your understanding of alpha-decay. If you have any further questions, please don't hesitate to ask.