# How do we decide whether an electron orbital has a non-zero or zero probability of lying inside the nucleus of an hydrogen atom?

How do we decide whether an electron orbital has a non-zero or zero probability of lying inside the nucleus of an hydrogen atom? It is mostly from the radial function, as to what I think but how exactly is this determined?

In this link you will see the radial hydrogen wavefunctions. It is only the l=0 states, S states, that have a value different than zero at r=0.

The other angular momentum states get a very small contribution to the probabilities from r>0 to r=1 fermi ( the charged radius of the proton) as 1 fermi is of order 10^-15meters, and the probability is the square of the wave function.

Nothing happens when they do overlap, because of energy considerations:there is no energy level within the proton to allow for a transition to a different state and the neutron is much heavier than the proton. Anyway, to get there the weak interaction will be needed which is the reason that electron capture in nuclei where energy allows it, is very small.

• I think this is the link, you wanted? hyperphysics.phy-astr.gsu.edu/hbase/hydwf.html#c1 Scroll down for 2p.
– Gert
Nov 13, 2015 at 21:01
• I'm sure if we take the interaction of e and p into account at subatomic length scales, the I=0 state will also not take kindly to having an electron in the nucleus. Nov 13, 2015 at 23:03
• @LLlAMnYP It is energy conservation and the quantum numbers that just let the electron be. With an appropriate nucleus for energy one gets electron capture en.wikipedia.org/wiki/Electron_capture Nov 14, 2015 at 4:54
• @Gert I wanted the math. One can see that for l=0 the function is a constant, not zero. The others have a r mutliplying the function. Nov 14, 2015 at 4:57

One calculates the probability that the electron is inside the nucleus by integrating $\psi^*\psi$ over the volume of the nucleus.

For example, the radial part of the hydrogen ground state wavefunction is $\psi=\frac{e^{-\frac{r}{a_0}}}{\sqrt{\pi a_0^3}}$,

so the integral is $\frac{1}{\pi a_0^3}\int_0^b e^{-2r/a_0} 4\pi r^2 dr$.

In the above, $a_0$ is the Bohr radius and $b$ is the proton radius. The integral is easy enough, but there is a trick that you may like. Since the proton radius is so small (~1 fm) compared to the Bohr radius (~53000 fm), the exponential term is practically a constant ($\approx 1$) over the entire nucleus, so just ignore it! In this approximation, the integral evaluates to $\frac{4b^3}{3a_0^3}$, which is a pretty small probability, but is it really non-zero?

One way to look at a probability is how often the event occurs. If we could perform an experiment to locate the electron once every second, we would find it inside the nucleus once every 5 million years or so. So, it is safe to say you or I will never see it happen, so the probability is zero for all practical purposes.