Available energy in β+ and β- nuclear reaction

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 2K views
duchuy
Messages
79
Reaction score
3
Homework Statement
Formula demonstration
Relevant Equations
Ed = [Mn(X) – (Mn(Y) + m(e))] c2
Hi,
I understood that to calculate the available energy in these two reactions could be calculated from Ed = [Mn(X) – (Mn(Y) + m(e))] c^2, but when I have to change use the atoms' mass instead of the nucleons' mass, it gives out two different formulas :
Ed = [M(X) – M (Y)] c2 for β-
Ed = [M(X) – (M(Y) + 2 m(e))].c2 for Ed = [M(X) – (M(Y) + 2 m(e))].c2 for β+
Can someone please explain to me why for β-, the mass of the electron isn't taken into consideration whilst for β+, we'd have to add the mass of two electrons ( when we are using the mass of the atom to calculate ).
Sorry if I have misused any vocabulary, I translated this from french.
Thank you so much for your help!
 
Physics news on Phys.org
kuruman said:
Please be more specific. What does Ed represent? What two reactions is Ed associated with?
I'm so sorry for that.
Ed represents the available energy. The formula Ed = [Mn(X) – (Mn(Y) + m(e))] c2 uses the mass of the entire atom and is used for β+ and - reaction.
But when only the mass of the nuclei are given, we end up with two different formulas :
Ed = [M(X) – M (Y)] c2 for β-
Ed = [M(X) – (M(Y) + 2 m(e))].c2 for Ed = [M(X) – (M(Y) + 2 m(e))].c2 for β+
I just don't understand what happened to the mass of the electrons in these two reactions.
Thank you!
 
duchuy said:
Ed represents the available energy.
The available energy when what happens?
 
kuruman said:
The available energy when what happens?
I think it's the available energy in an atom with either an excess of neutron or proton depending on the β radioactivity. I'm not quite sure though sorry, it's just what's written in the text that we have to learn...
 
I think that the prof doesn't like negative signs, so instead of using the energy released which is going to have a negative value when we consider the system, he just uses the available energy so the energy would be positive. I didn't get the chance to ask him since all classes are uploaded videos...