Conjugate nuclei

Question

I'm trying to find an explanation for the difference in binding energies between conjugate nuclei being governed by the Coulomb interaction. I was looking at the semi-empirical mass formula and trying to draw some conclusion.
We have that:
$$ R = R_0 A^{1/3}$$
and:
$$ Z = A-N$$
so for the Coulomb term I find that it gets a $A^{5/3}$ dependence and of all the terms in the formula that one dominates in terms of $A$ dependence but it's not entirely clear as to why does the Coulomb interaction govern the difference in binding energies for conjugate nuclei. How do we justify this?

Urb · Accepted Answer

In the semi-empirical mass formula
$$E_B(N,Z)=a_VA-a_SA^{2/3}-a_Cfrac{Z(Z-1)}{A^{1/3}}-a_Afrac{(N-Z)^2}{A}+delta(N,Z)$$
the volume and surface terms only depend on $A$, so they give the same value for a nucleus and its mirror nucleus, since the mirror nucleus is obtained by changing $Nleftrightarrow Z$, so $A$ is fixed. The asymmetry term depends on $(N-Z)^2/A$, so it's also the same when $N$ and $Z$ are interchanged. The pairing term $delta(N,Z)$ is either $0$ or only depends on $A$. So the only term that changes between a nucleus and its conjugate is the Coulomb term and the difference in their binding energies $Delta E_B=E_B(N,Z)-E_B(Z,N)$ must come from this term.

Conjugate nuclei

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