7
$\begingroup$

On this interactive table of nuclides, there is a region just "north east" of $^{208}\text{Pb}$ and $^{209}\text{Bi}$ with extremely unstable nuclides (the yellow/pink/light green squares). The longest $\text{T}_{1/2}$ between $^{211}\text{Po}$ and $^{224}\text{U}$ is 0.511 seconds, with most being in the millisecond to nanosecond range. Refer to the area circled in red in the below image.

enter image description here

On the nuclide chart, this "gouge" is plainly visible from like a mile away. :-) I'd like to think of it as an "island of instability". From a nuclear physics perspective, What phenomenon best explains this?

$\endgroup$
  • $\begingroup$ arxiv.org/abs/0909.4492 $\endgroup$ – Count Iblis Aug 31 '16 at 0:08
  • $\begingroup$ They're all $\alpha$ emitters, so probably has to do with the Geiger-Nuttall law which states that $\ln (1/\tau)\sim -Z$ $\endgroup$ – Paul Aug 31 '16 at 10:16
3
$\begingroup$

It is somewhat ironic that this "Island of Instability" would occur just after one of the most stable large nuclei, i.e. $Pb^{208}$. $Pb^{208}$ owes its stability to the fact that it is doubly magic (consisting of closed shells of both neutrons and protons). These doubly magic systems are spherical and when they occur near the line of beta stability (as is the case for$Pb^{208}$) their stability is further enhanced. So what could explain the marked instability of the nuclei immediately following?

The answer to this question requires some fine details of the nuclear shell model and specifically the nature of the neutron and proton orbitals that are being filled in this region. If one looks at the ground state spins of both $Pb^{209}$ and $Bi^{209}$, one sees that they both have spin $\frac92$. From calculations that I performed for my PhD, these orbitals are likely to be $0g\frac92$ and $0h\frac92$ respectively for the neutron and proton orbitals. The unique thing about this situation is that both of these orbitals have very high orbital angular momentum (4 and 5 respectively).

When extra neutrons and protons are added to a spherical core, the pairing force acts to render the lowest total nuclear spin possible. For even-even nuclei the ground states are always spin 0. The high angular momentum of the orbitals following $Pb^{208}$ means that these paired entities will form on the nuclear periphery (forced there by the high angular momentum barrier). That means that $\alpha$ decay (for even-even systems) will be more probable in this region than would be the case if either or both of the pairing orbitals had lower angular momentum. As the comments have indicated, these are mostly $\alpha$ emitters, so the enhanced instability is to be expected.

$\endgroup$
  • $\begingroup$ Assuming those horizontal and vertical lines on the above chart represent the so-called "magic" numbers, then isn't it odd (no reverse pun intended) that $^{206}$Pb is doubly magic whereas very few others are as well (e.g. $^{40}$Ca, etc.)? Or mere coincidence? Be aware that I'm teaching myself nuclear physics (mostly here on Physics), and I'm kinda new to all this. Thank you for your answer. $\endgroup$ – pr1268 Sep 3 '16 at 23:05
  • 1
    $\begingroup$ Thanks for a good question. The last time I looked at a chart of the nuclides, it was a folded paper copy (BW) that required a large table. Needless to say, the unstable region you highlighted did not jump out. Surely someone has noticed this before, but I could not find a mention of it in the research lit. As for the sparse number of high mass doubly magic nuclei, the number decreases as N and P increases because the orbital occupancy goes up as $2l+1$. Also, the chance that double shell closures occur along the $/beta $ stability line decreases. Also, nearly degenerate levels mess things up. $\endgroup$ – Lewis Miller Sep 5 '16 at 2:35

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.