In class I was given a diagram like this: (albeit without the electrostatic force line)

(source: boredofstudies.org)

However, the teacher told us the nucleons are typically separated when the force is zero. So as the strong force crosses the $x$ axis (as does our textbook).

However, initially this did not make sense to me because electrostatic force will still be repelling (in the case of a proton-proton "bond"/interaction) and so surely the separation distance must be on the positive side to counteract the electrostatic force?

  • $\begingroup$ (btw I have a test tomorrow so answers before then will be greatly appreciated, but after is good to know to) $\endgroup$
    – Jonathan.
    Oct 16, 2011 at 20:03
  • $\begingroup$ I think he/she meant that a deuterium's nucleons composed of a proton and a neutron can be separated when the nuclear force is zero, on the right of your diagram. In case of an alpha particle, two protons and two neutrons, the electrostatic repulsion will make the distance a bit shorter, but insignificant to the scale of the strong force. Two protons do not bond because of the electrostatic barrier: they cannot get close enough to see the strong force, unless at high energy scatterings, and then there is too much energy to bond. $\endgroup$
    – anna v
    Oct 17, 2011 at 4:23
  • $\begingroup$ I think the vertical axis of your graph is mislabeled. It should be potential energy. The minimum of the potential energy is where net force will be zero. As stated in the answers, you really need to add the blue and red curves and find the minimum. The force will be the negative of the derivative (or slope) of the total potential energy curve which will be 0 about where the typical nucleon separation line is drawn. $\endgroup$
    – FrankH
    Oct 18, 2011 at 7:50
  • $\begingroup$ @FrankH...no see this page webs.mn.catholic.edu.au/physics/emery/… a pic in it... $\endgroup$ Oct 18, 2011 at 11:54
  • $\begingroup$ @vineet I looked at your link which appears to be identical to the OP graph and yes, that link graph is also wrong. Here is the best reference which clearly shows and says that the minimum of the potential energy is where the force is 0. $\endgroup$
    – FrankH
    Oct 18, 2011 at 19:15

2 Answers 2


The answer is in your graph. You need to ADD the red line you added to the black strong nuclear force line, which will only slightly alter the overall shape. In other words the total sum of the electric and nuclear force at the 'typical nuclear separation' line is say -10 (in some arbitrary units) from the nuclear force, plus +2 from the electric, for a total of -8 at the min point, which does not move the minimum energy point much to the left or right.

  • $\begingroup$ See my other comments... $\endgroup$
    – FrankH
    Oct 18, 2011 at 7:53

Simply draw another graph i.e. the resultant force by adding the blue and red lines; remember red is attractive and blue is repulsive, so effectively you need to subtract the modulus of both.

Its quite clear that at the distance marked as "typical separation" you would find your minima i.e. maximum attractive force or least potential energy.


  • $\begingroup$ But the force is the negative derivative of the potential energy so the force should be zero when the nucleons are in equilibrium and this would correspond to the lowest potential energy where the slope would be 0. See my other comment also. $\endgroup$
    – FrankH
    Oct 18, 2011 at 7:52
  • $\begingroup$ ya...so what's the problem if $U$ is 0 at "typical separation"... $\endgroup$ Oct 18, 2011 at 11:57
  • $\begingroup$ Force should be 0 at typical separation and U, potential energy, should be at a minimum with slope 0 at that point. That is why I think the OP graph is mislabeled. It is not Force, but potential energy on the Y axis. Then the dotted line at typical nuclear separation makes sense. $\endgroup$
    – FrankH
    Oct 18, 2011 at 15:08

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