- #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

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!