# Electron capture, electrons binding energy

## 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?