Độ hụt khối $\Delta m_{1}$
=$Z_{1}m_{p}+\left(A_{1}-Z_{1}\right)m_{n}-m_{1}$
=$Z_{1}\left(m_{p}-m_{n}\right)+A_{1}\left(m_{n}-1\right)$ ( vì $m_{1}\approx A_{1}\left(u \right)$)
Vì $m_{p}\approx m_{n}$ nên $\Delta m_{1}$=$A_{1}\left(m_{n}-1 \right)$
Tương tự $\Delta m_{2}=A_{2}\left(m_{n}-1 \right)$
Ta lại có R=1,23.$10^{-15}$.$A^{\dfrac{1}{3}}$
nên khi $\dfrac{r_{1}}{r_{2}}=2$ thì $\dfrac{A_{1}}{A_{2}}=8$