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.


1 Answer 1


You are thinking of it as an attractive potential energy, but it is all protons in both nuclei . If you let them go they would fly away, nothing to do with binding energy.

The binding energy is an interplay between the repulsive electromagnetic and the attractive nuclear force. That is why you are asked to go beyond the range of the nuclear force.

The nuclear force (or nucleon–nucleon interaction or residual strong force) is a force that acts between the protons and neutrons of atoms. Neutrons and protons, both nucleons, are affected by the nuclear force almost identically. Since protons have charge +1 e, they experience an electric force that tends to push them apart, but at short range the attractive nuclear force is strong enough to overcome the electromagnetic force. The nuclear force binds nucleons into atomic nuclei.


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