The complete binding energy is (more details in Krane's book, page 68):

$$B = a_v A - a_s A^{2/3} - a_c Z(Z-1)A^{-1/3} - a_{sym} \frac{(A-2Z)^2}{A} + \delta$$

I want to explain the $Z(Z-1)$ term. After reading the book I thought I got the reason (from Krane):

Our binding energy formula must also include the Coulomb repulsion of the protons, which likewise tends to make the nucleus less tightly bound. Since each proton repels all of the others, this term is proportional to $Z(Z-1)$.

Then, in the book, the factor $Z(Z-1)$ is used to calculate the energy needed to charge a homogeneously charged sphere with radius $R$ and charge $Ze$:

$$-\frac{3}{5}\frac{Z^2 e^2}{4 \pi \epsilon R} \propto -\frac{Z^2}{R}$$

Where the negative sign implies a reduction in binding energy.

Then it is clear; $Z(Z-1)$ accounts for the Coulomb's repulsion force between the protons.

However, in the lecture my professor argued that accounts for the subtracted energy used to build up the nucleus (I literally copied this sentence really quick, so I may have copied it wrongly).

Then, how can we argue $Z(Z-1)$ in terms of energy? I think we could argue this repulsion in terms of Coulomb's potential.

  • 1
    $\begingroup$ "build up the proton" Do you mean nucleus? How do you build up a proton? I don't understand what's bothering you because $Z(Z-1) A^{-1/3}$ has all the particle terms and $a_c$ has all the constants buried inside ($e^2,\pi,\epsilon$, shielding factors, etc). $\endgroup$ – Bill N Feb 20 '19 at 19:48
  • $\begingroup$ Ups I meant the nucleus, my bad. What is bothering me is how to explain $Z(Z-1)$ in terms of energy $\endgroup$ – JD_PM Feb 20 '19 at 19:53
  • 5
    $\begingroup$ @JD_PM: When someone points out a mistake in the question, you should just edit the question to correct it. $\endgroup$ – user4552 Feb 20 '19 at 21:55
  • 1
    $\begingroup$ @JD_PM It IS a repulsion factor because it reduces the binding energy. High binding energy means more tightly bound. A negative contribution to BE means something is making the system less bound. Or are you asking how z(z-1) could have energy units? $\endgroup$ – Bill N Feb 21 '19 at 13:10
  • 1
    $\begingroup$ @JD_PM As I implied in my first comment, while $Z(Z-1)$ accounts for the interacting pairs of protons, there is also an $A^{-1/3}$ term which relates to the size of the nucleus, and hence is proportional to 1/R. The $a_c$ is an empirical factor which will absorb all the $e, \pi, \epsilon$ terms as well as account for the non-exactness of the 1/r term, and is the proportionality factor to make it all work out to energy. $\endgroup$ – Bill N Feb 21 '19 at 21:59

The number of combinations of $Z$ things, taken 2 at a time and ignoring order, is $Z(Z-1)/2$. Therefore the number of interacting pairs of protons is proportional to $Z(Z-1)$.

  • 1
    $\begingroup$ And the $A^{-1/3}$ mimics the inverse distance factor. $\endgroup$ – Bill N Feb 21 '19 at 13:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.