I've just done an exercise of computing the binding energy of Helium 4, which is around 27,43 MeV. Obviously the binding energy "compensates" for the repulsion between the protons due to their charges. I'd like to compare the repulsion between the protons with the binding energy, but the problem is that using Coulomb's law to compute the "repulsion" between the protons gives a force as a result (in Newtons), whereas the binding energy is in eV or MeV or Joules. My questions are the following. First, is it true that the binding energy's role is mainly to compensate for the repulsion between protons, or is it stronger than that ? Second, how can we compare the repulsive force between the protons, with the binding energy, to prove that hypothesis ? Third, in a more general way, is there a means to compare an energy with a force, if that even makes sense ?


2 Answers 2


I would like to answer your second and third questions first. Although there exist good approximations of the force due to strong nuclear force, a much easier option would be to compare the binding energy with the potential energy created due to electrostatic repulsion.

To make an estimation of the potential energy, I assume the radius of Helium nucleus to be approximately $1\space\mathrm{fm}$$=1\times10^{-15}m.$ $$V=\frac{1}{4\pi\epsilon_0}.\frac{e^2}{r}$$ $$V\approx1.15\times10^{-13}\mathrm{J}=1,439,964\space\mathrm{eV}$$

The electrostatic repulsion potential comes out to about $1.43\space\mathrm{MeV}$ as opposed to about $28\space\mathrm{MeV}$. This is of course a very simplified calculation but accurate enough to be of use (as pointed out by the other answer (+1)).

Hence the overall potential energy is negative (net force is attractive) and the nucleus is in fact very stable.


For some reason I get 28.296 MeV for binding energy of $^4He$.

Coulomb potential energy (for a charged sphere) can be written as $E=\frac{3}{5}(\frac{1}{4\pi\epsilon_0})\frac{Q^2}{R}$. This term is actually plugged into a liquid drop model formula (the Coulomb term $-a_C\frac{Z(Z-1)}{A^{1/3}}$) and from both theory and experiment give you a similar value $a_C \approx 0.691 MeV$. This implies 1.26 MeV for the Coulomb energy of the two protons in $^4He$, quite in agreement with the previous answer (+1).

First - The binding energy does not have a role in this sense, it is just a result of the fact that the strong interaction is stronger than the EM interaction.

Second - as was written, the potential energy can be (and is) compared.

Third - once there is a force in your system, you can define some potential energy (that can be zero at $\infty$) and compare the energies. If you switched off the strong interaction, your protons should end up in $t=\infty$ with a kinetic energy = Coulomb potential energy at $t=0$.


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