Anisotropic charge screening and the rate of spontaneous fission Fast electrons that pass by a heavy nucleus have a cross section for being deflected from their course, giving rise to bremsstrahlung. The cross section is partially a function of the impact parameter, and a closer approach will yield a larger probability for the electron being deflected through a wide angle.  In the limit, the electron is completely stopped and its kinetic energy converted to a photon.
Spontaneous fission and alpha decay are quantum mechanical tunneling processes that are very sensitive to the width of the Coulomb barrier through which a decay daughter must tunnel.  There are experiments here and there that document small changes in the beta decay rate in response to changes in pressure and the chemical environment, and similar reasoning leads the author of that page to suspect that something might happen in the case of alpha decay and spontaneous fission, although there have been no detected changes in these rates that I am aware of in the latter case.
What amount of screening of the Coulomb barrier of a heavy nucleus occurs as an energetic electron is deflected or stopped upon approach? Is this screening sufficient in theory to modify the rate of spontaneous fission? Presumably any charge screening that occurs is nonuniform; how does anisotropy of the screening factor into any affect on the rate of fission, and how can this problem be modeled?
 A: In answer to a related question by you, I computed a charge of about $-0.5e$ due to $s$-wave electrons in the volume with radius $\rm\sim50\,fm$ probed by tunneling alpha particles in decaying uranium.  You can use a similar method to estimate the charge screening of the nucleus at larger radii.  Your results will be approximate since you generally won't have multi-electron wavefunctions available, but it's a place to start.
I would not expect the charge screening to be anisotropic.  While individual wavefunctions with $\ell\neq0$ have a nontrivial orientation in space, a full shell of electrons with $\ell\neq0$ has a spherically symmetrical distribution of charge.  (There are two ways to see this.  One is a symmetry argument.  For instance with $\ell=1$ and electrons equally likely to be found in $p$-orbitals along the $x$-, $y$-, and $z$-axes, the charge distribution cannot depend on your arbitrary choice of coordinate system.  The second argument, finding the total charge distribution by adding charges in the various orbitals, makes a nice advanced quantum homework problem.)
Any anisotropy in the electron charge near the nucleus would be encoded in, and correlated with, the hyperfine structure of the ground state of the unstable atom.  First off, since hyperfine levels are typically separated by μeV-ish energies, all the hyperfine levels will be equally populated in a macroscopic sample of atoms at room temperature; any anisotropy in a real sample of atoms would average away.
(It gets even messier if your sample of atoms is condensed to a liquid or solid.)
Second, to the extent that the hyperfine structure is an interaction between the nucleus and the valence electrons, the electron charge distribution in the vicinity of the nucleus is negligible.
Unlike in our previous discussion about alpha decay, for fission I think there's not any handwaving argument about tunneling that makes the relevant nuclear volume large; the biggest perturbation to expect is the parts-per-million electron charge that's actually in the nuclear volume.
