# Why does the potential energy between two daughter nuclei not give the binding energy per nucleon?

I have attempted a nuclear physics question, given below:

Calculate the electrostatic potential energy of two fragments, a Cesium and a Zirconium nucleus, when separated by a distance $$2D$$, twice the range of the strong nuclear force. $$D = 5.6$$ fm

Using $$U = \frac{kQ_1Q_2}{r}$$, where $$k = 9 \cdot 10^9 \frac{\text{Nm}^2}{\text{C}^2}$$. For Cesium $$Z = 55$$ and for Zirconium $$Z = 40$$. gives $$U = 283 \text{ MeV}$$

However, as a sanity check I wanted to confirm this value gives the $$~7$$ MeV binding energy per nucleon given in the classic BE per nucleon curve. Isn't this $$U$$ (kinetic energy of products) equal to the binding energy released?

$$\frac{283}{2 \cdot 55+2 \cdot 40} = 1.3 \text{ MeV}$$ per nucleon... Can someone explain the flaw in my reasoning please? Many thanks in advance.