# What effects would an electron in the nucleus have?

Under the understanding of quantum chemistry:

There is a very miniscule chance that an electron could appear inside the nucleus of an atom.

What could possibly happen if this "phenomenon" were to occur? Would it wreak havoc upon the atom and tear it apart? Or will it bind with the protons and create a temporary vacuum?

Under my current research, the electron would jump out of the nucleus as soon as possible because this phenomenon should not occur for more than a fraction of a second. Though it shouldn't leave without some adverse effects.

• In electron capture electron does get to nucleus. Dec 17, 2014 at 1:54
• I'm not referring to electron capture, but rather quantum mechanics. The possibility that an electron can appear inside the nucleus. Dec 17, 2014 at 2:36
• In quantum chemistry, having two things simultaneously at the same place is not such a big deal. Jan 31, 2016 at 21:55

An electron being at/in the nucleus results in the Fermi Contact Interaction.

This interaction can be observed experimentally through NMR, EPR, and electron capture.

In NMR of paramagnetic compounds, chemical shifts can be far outside the usual diamagentic range. The contribution to chemical shift of the electron density at the nucleus is referred to as the "contact shift".

For more information on Fermi contact, see "Origin and meaning of the Fermi contact interaction" Theor Chim Acta (1988) 73:173-200.

If you do a (simple) calculation of a quantummechanical rotator in a spherical potential — i.e. calculate where an electron would be in the vicinity of an nucleus — you will find that the eigenfunction $\left | \Psi_1 \right \rangle$ corresponding to the lowest energy $E_1$ is dependent only on the distance of the electron from the nucleus ($\left | \Psi_1 \right \rangle = f(r)$ — use spherical coordinates because it makes your job a million times easier!) — which gives us a sphere like the s-orbitals are often drawn.

If you carry on, you realise that $\left | \Psi_1 \right \rangle$ is more or less a decaying exponential function, so its value is highest for $r = 0$. The probability of an electron being at a specific point is typically given as $\left \langle \Psi_1 \middle | \Psi_1 \right \rangle$ and evaluates as $\Psi^2_1$ in this case. Since the function has only one phase, the highest $\Psi_1^2$ value is — in the nucleus.

So the nucleus already carries the highest single probability of having the electron right inside it. It’s not even ‘special’ for the electron to be in the nucleus.

However, check Dave’s answer for the physical implications, which I didn’t even know about!