$A$ nuclide $1$ is said to be the mirror isobar of nuclide $2$ if $Z_1 = N_2$ and $Z_2 = N_1$. $(a)$ What nuclide is a mirror isobar of $_{11}^{23}Na$? $(b)$ Which nuclide out of the two mirror isobars has greater binding energy and why?

(A) For the given nuclide $_{11}^{23}Na$,the atomic number $Z_1 = 11$ and the mass number $A = 23$. The number of neutrons is $N_1 = A - Z_1 = 23 - 11 = 12$.
For a mirror isobar,the new atomic number $Z_2 = N_1 = 12$ and the new number of neutrons $N_2 = Z_1 = 11$. The mass number remains $A = Z_2 + N_2 = 12 + 11 = 23$. The element with atomic number $12$ is Magnesium $(Mg)$. Thus,the mirror isobar is $_{12}^{23}Mg$.
$(b)$ The binding energy of a nucleus is influenced by the symmetry of protons and neutrons. In $_{11}^{23}Na$,there are $11$ protons and $12$ neutrons,while in $_{12}^{23}Mg$,there are $12$ protons and $11$ neutrons. The nucleus with more neutrons relative to protons generally experiences a slightly stronger net nuclear attraction due to the absence of additional Coulomb repulsion between protons. Therefore,$_{11}^{23}Na$ has a greater binding energy than $_{12}^{23}Mg$.