Krdytkyu

In this youtube video it is claimed that electrons orbit their atom's nucleus not in well-known fixed orbits, but within "clouds of probability", i.e., spaces around the nucleus where they can lie with a probability of 95%, called "orbitals".

It is also claimed that the further away one looks for the electron from the nucleus, the more this probability decreases, yet it never reaches 0. The authors of the video conclude that there is a non-zero probability for an atom to have its electron "on the other side of the Universe".

If this is true, then there must be a portion of all atoms on Earth whose electron lies outside the Milky Way. Which portion of atoms has this property?

The quantity you should consider first is the Bohr radius, this tells you an idea of the relevant atomic scales,

$$
a_0 = 5.29times 10^{-11} ~{rm m}
$$

For hydrogen (the most abundant element), in its ground state, the probability of finding an electron beyond a distance $r$ from the center looks something like (for $r gg a_0$)

$$
P(r) approx e^{-2r/a_0}
$$

Now let's plug in some numbers. The virial radius of the Milky Way is around $200 ~{rm kpc} approx 6times 10^{21}~{rm m}$, so the probability of finding an electron outside the MW from an atom on Earch is around

$$
P sim e^{-10^{32}}
$$

that's ... pretty low. But you don't need to go that far to show this effect, the probability that an electron of an atom in your foot is found in your hand is $sim 10^{-10^{10}}$

Given a single electron, what is the probability that it is found outside the Milky Way?
We can estimate it using the ground state wave function of the Hydrogen atom,
$$ psi_{100} = frac{1}{sqrt{pi a_0^3}} e^{-r/a_0} , $$
where $a_0 approx 5*10^{-11}, m$ is the Bohr radius.
$|psi|^2$ is the probability density, integrating gives
$$ p_1 = int_R^infty |psi_{100}|^2 4pi r^2, dr = frac{e^{-2R/a_0}(a_0^2 + 2a_0 R + 2R^2)}{a_0^2} . $$
Plugging in $R approx 5*10^{20}, m$ the radius of the Milky Way, we get
$$ p_1 approx exp(-2*10^{31}) approx 10^{-10^{31}} . $$

This number is so small, it is hardly possible to grasp really how small it is.
There are a lot of electrons in the Earth - about $N = 10^{51}$ - but the number of electrons is utterly tiny compared to these odds. The chance that any electron is found outside the milky way is
$$ p = 1 - (1 - p_1)^N approx N p_1 = 10^{51} , cdot , 10^{-10^{31}} $$
which doesn't even make any dent.

What is said in the video is true, but... remember that the atomic theory is just that: a theory. The theory itself predicts that perturbations will have a really big influence on the results.

Take into account that the models are based on hypothesis, which are easily violated. For example, spherical symmetry, which allows finding the solution in the hydrogen atom (or more accurately, the Coulomb's potential in QM). Reality is never like that, but we can say that "it is close enough" if the atom is far enough from other objects.

Nevertheless, from here to outside of the milky way there are so many perturbations that the model would just fail. You can say that there's a level $n=1324791$, but there are so many particles out there that the effect of your atom is absolutely beaten by ANY other.

So, does it really make sense to calculate such probability if anything can capture that electron much more easily? I don't think so.