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Consider a nuclide like $\mathrm{^{232}Th}$, which has a half-life of 1.4e10 years and which decays by $\alpha$ decay to $\mathrm{^{228}Ra}$. Alpha decay is a quantum mechanical tunneling process in which an $\alpha$ particle tunnels through the width of the Coulomb barrier.

Now suppose several of the d- or p-shell electrons of the $\mathrm{^{232}Th}$ atom are excited into higher-level s-shells. The electrons in these new orbitals do not have nodes in the nuclear volume and hence will now spend relatively more of their time there than before. Will the increased electron charge density in the nuclear volume narrow the Coulomb barrier width to $\alpha$ decay and spontaneous fission?

Can a change of decay rate be expected for nuclides unstable against $\beta^-$ decay or electron capture (for somewhat different reasons)? Will there be an effect on the rate of internal conversion?

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What a fun question! The intuitive answer is that it would be very hard to detect any half-life difference because the electrons are smeared out over something like the Bohr radius, while the nucleus is much smaller, so the contribution of the electron charge to the problem is negligible. However if the correct radius to consider is the distance over which the alpha particles "tunnel" to escape from the nucleus, the effect might be quite substantial for highly-charged ions.

Let's consider the tunneling model for alpha decay. We'll assume a nucleus with $Z$ protons and radius $a \approx A^{1/3}\times1.3\,\rm fm$. Inside the nucleus, thanks to the strong interaction, the alpha particle sees some constant potential that we don't really care about. Outside the nucleus, the alpha particle is electrically repelled. So the total potential, in spherical coordinates, is $$ V(r) = \left\{ \begin{array}{cc} V_0 & r < a \\ \displaystyle \alpha \hbar c \frac{2(Z-2)}{r} & r > a \end{array} \right. $$ This potential totally ignores all the atomic electrons.

The electrons will contribute a potential $$ V_\text{e}(r) = -\alpha\hbar c \frac{2q_\text{enc}(r)}r $$ where $q_\text{enc}$ is the charge (in units of the charge quantum $e$) enclosed within the radius $r$. (The shell theorem for electromagnetism says that we can safely ignore charges which are uniformly distributed at large radius.) The spherical harmonics with $\ell\neq0$ vanish at the origin, so we need only consider the $s$-wave electrons. The radial hydrogenic wavefunctions with $\ell=0$ have a uniform probability density near the origin of $$ \left|\psi(r=0)\right|^2 = \frac1{\pi a_e^3} = \frac1\pi \left( \frac{Z}{na_0} \right)^3 $$ where $a_0 \approx 53\,000\,\rm fm$ is the Bohr radius. This gives us, for an electron in the $n$th $s$ shell, a charge enclosed within radius $r$ of $$ q_\text{enc}^n(r) = \frac 43 \left( \frac Zn \frac{r}{a_0} \right)^3. $$ Let's try it for a $1s$ electron in uranium (largest $Z$), and compare the nuclear charge $Z$ to the charge of electrons which are visiting, using the definition of nuclear radius given above: \begin{align} q_\text{enc}^1 (a_{\text{U-238}}) &= \frac43 \left( \frac{92}1 \frac{1.3\,\rm fm}{a_0} \right)^3 A \tag1 \\&\approx 4\times10^{-6} \end{align} However, the nuclear volume isn't quite the right thing to consider here. The alpha particle escapes the nucleus by tunneling through the barrier. The distance where the alpha's wavefunction changes from decaying exponential to sinusoidal is where the final kinetic energy is equal to the electrostatic energy, \begin{align} E_\alpha &= V(r) \\ E_\alpha &= \alpha \hbar c \frac{2(Z-2)}{r} \\ \text{5 MeV from uranium: } r &= \frac{200\rm\,MeV\,fm}{137} \frac{2\cdot90}{5\rm\,MeV} \approx 52\rm\,fm. \end{align} The relation (1) above suggests that this larger volume contains about $0.23e$ from each of the two $1s$ electrons, $0.03e$ from each of the two $2s$ electrons, and the rest don't matter so much. This suggests that the width of Coulomb barrier might be different at the 0.5% level for completely ionized uranium compared to partially ionized uranium (since the $1s$ electrons will be the last to go). The decay rate difference between different ionization states sounds modest but experimentally accessible to me, but your question is the first I've heard of it.


A recent preprint by F. Belloni, with a nice bibliography, studies this phenomenon in much more detail. I neglected to include the change in the $Q$-value of the decay between the bare nucleus and the neutral ion, which also affects the and makes the whole question rather complicated. Belloni computes a lifetime difference of 0.1% between bare Po-210 and hydrogen-like Po-210 and larger changes in α-lifetimes at extremely high electron densities. In general Belloni predicts shorter α-decay lifetimes in the presence of electrons.

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  • $\begingroup$ Why did you drop the discussion of the barrier width, which seems more pertinent to the question? $\endgroup$ – Eric Walker Oct 23 '15 at 1:24
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    $\begingroup$ @EricWalker An excellent point which I was editing in as you commented; it changes the answer quite a bit. $\endgroup$ – rob Oct 23 '15 at 1:38
  • $\begingroup$ Does Figure 3-8 of this discussion alter your assumptions about the radial distribution of the s-orbits for $n>1$? chemistry.mcmaster.ca/esam/Chapter_3/section_2.html $\endgroup$ – Eric Walker Oct 23 '15 at 5:05
  • $\begingroup$ It doesn't. Because of the shell theorem what matters is the electron density near the nucleus; the large-$n$ electrons spend more of their time far from the nucleus. $\endgroup$ – rob Oct 23 '15 at 21:12
  • $\begingroup$ What struck me about the figure was that the nuclear volume and the barrier are a speck at the origin, and that for s-waves at all n there is a local maximum at the origin that overwhelms the nucleus and the barrier. $\endgroup$ – Eric Walker Oct 23 '15 at 21:24
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I think, it may be some misconception on Coulomb barrier. Electrons are too far from the nucleus to really influence nuclear binding. In the other case fully deprived of electrons nuclei would never exists.

So, decay rate should be the same.

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One might call ionization the lifting of electrons to higher levels. No one seems to have noticed the implications of the fact that Alpha emission has the effect of removing both four units of mass and two units of positive charge from an atom. It is very likely that the actual, decaying unit may well be the di-cation form of the Thorium, or any other Alpha emitter.

Looking at the first, simplest "alpha emitter," Be8, there is no obvious reason for this unit to fission, considered as a neutral unit, particularly since symmetric fission is an impossibility. However, there is easily seen why the Be8 di-cation could fission to a Helium 4 atom (tetrahedral) and an Alpha particle, "squarthe implication of the fact e planar."

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    $\begingroup$ Hello, and welcome to Physics SE. Look around, and take the tour. I would contest your assertion that nobody has noticed that an alpha consists of two protons and two neutrons. As for $^{8}$Be, there is a very clear reason for it to fission - it lowers the total energy of the system. Please spend some time on basic nuclear physics. $\endgroup$ – Jon Custer Oct 22 '15 at 18:22
  • $\begingroup$ Squarthe? I find this answer very unclear, but if I could understand it, I'm pretty sure I would think it's wrong. What's the difference between symmetric fission and alpha decay for Be8? $\endgroup$ – ragnar Oct 24 '15 at 17:48

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