Question

Demonstrate that the binding energy of a nucleus with mass number `A` and charge `Z` can be found from Eq.(6.6b).

Answer 1

From the basic formula
`E_(b)=Z_(mH)+(A-Z)m_(n)-M`
We define `Delta_(H)=m_(H)-1 am u`
`Delta_(n)=m_(n)-1 am u`
`Delta=M-A am u`
Then clearly `E_(b)=Zdelta_(H)+(A-Z)Delta_(n)-Delta`