Nuclear Models — Question 5

Question 5

Pairing creates an even-odd staggering beyond the smooth mass curve

A mass model adds $\delta=+a_pA^{-1/2}$ for even-even nuclei, zero for odd A, and $-a_pA^{-1/2}$ for odd-odd nuclei, with $a_p=12.0\,\mathrm{MeV}$. Compare only this pairing contribution, holding other terms aside.

1. Find the pairing terms for even-even and odd-odd isobars at A=100 and their difference in binding.

2. If the smooth model places an even-even isobar 1.00 MeV higher in mass energy than an odd-odd isobar, determine the ordering after the pairing correction.

Question 5 - Solution

Pairing creates an even-odd staggering beyond the smooth mass curve

(a) Method and calculation

$$\boxed{\delta_{ee}=+1.20\,\mathrm{MeV},\quad\delta_{oo}=-1.20\,\mathrm{MeV}}.$$$$\boxed{B_{ee}-B_{oo}\big|_{\mathrm{pairing}}=2.40\,\mathrm{MeV}}.$$

(b) Interpretation and further calculation

$$Mc^2=M_{\mathrm{smooth}}c^2-\delta\Rightarrow\Delta(Mc^2)=1.00-(1.20-(-1.20)).$$$$\boxed{(M_{ee}-M_{oo})c^2=-1.40\,\mathrm{MeV}}\quad\text{even-even is now lower}.$$

Physical check

Extra binding lowers mass. The example isolates pairing; it does not imply that every even-even isobar is lighter than every odd-odd one.