# Electron capture, electrons binding energy

• rayman123

## Homework Statement

Hello! I am doing some calculations with electron capture for the reaction
$$\stackrel{196}{79}Au\stackrel{EC}{\rightarrow}\stackrel{196}{78}Pt$$
The Q-value for the reaction is
$$Q=[m(\stackrel{196}{79}Au-\stackrel{196}{78}Pt]c^2-B_{n}$$ where $$B_{n}$$ stands for electron's binding energy.
We calculated this value and got
$$Q=1.429 MeV$$ but I do not know how to find the value of $$B_{n}$$ the teacher just wrote value for $$B_{n}=0.08MeV$$
Can someone please explain to get this value?

## The Attempt at a Solution

I would say it's the binding energy of the atomic electron. Usually you can find this tabulated somewhere. I don't know what you can use as aid but normally you can consult a table of listed binding energies for electrons in different shells in different elements. If this is a course in nuclear physics I don't think you're expected to calculate it anyway. What you have to do is probably to do some assumptions about where the electron comes from, and then perhaps use more than one value for the binding energy (depending on which atomic shell the electron comes from).

yes you are right! I found the value in the tables. Thank you!