- #1
duchuy
- 79
- 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!
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!