Is there a relationship between an electron's energy and its distance from the nucleus?

Question

I have read that in Bohr's model of the atom electrons in the same atomic orbital have the same energy, and that this energy (as a result of the electron existing in a fixed atomic orbital) is related to the distance of the electron from the nucleus. However, using the quantum mechanical model of the atom, in which an orbital is the area of space around an atom where the probability of interacting with an electron is greatest, it is implied that there is no relationship between an electron's energy and its distance from the nucleus. Is there no relationship, or have I misunderstood something else?

Also, it is often mentioned that an atom is 99.99% empty space etc., but since the probability of interacting with an electron is described by its wave function (which suggests it is possible, though unlikely, to interact with an electron very close to the nucleus), this suggests to me that this isn't necessarily true. What am I misunderstanding?

Answer 1

I will start off by addressing Jon's comment above. Yes, the Bohr model is flawed. I think it is still worth learning about it just from a historical standpoint, to see how we discovered the quantum mechanical description of the electron, but when you are studying it you absolutely have to remember that all of what he said was effectively rubbish.

Now on to your questions.

Is there a relationship between an electron's energy and its distance from the nucleus?

The first thing that should be said is that in QM, unlike in the Bohr model, there is no well-defined "distance from the nucleus". Electrons exist everywhere (except at nodes) and the probability density is given by $\lvert\psi\rvert^2$. However, it is possible to calculate the "most probable distance from the nucleus" using the radial distribution function, $r^2 R^2$, where $R(r)$ is the radial part of the wavefunction. Essentially, you take the derivative of this with respect to $r$ and set it to 0 - you can read more here. For a 1s electron in hydrogen, you will find that the most probable radius is approximately $52.9 \text{ pm}$, a quantity called the Bohr radius and denoted $a_0$.

(As an aside: oddly enough, this correlates with the flawed Bohr model, which said that the radius of the 1s electron in hydrogen was $52.9 \text{ pm}$. However, what Bohr said was that the electron was always at this distance, which is wrong. QM says that the electron is most likely to be found at this distance.)

Anyway, you will find that there is some degree of correlation between the most probable radius and the energy of the electron. If we consider the 2s orbital of hydrogen, you can perform exactly the same calculation as you did for the 1s orbital, and you will find that the most probable radius is $4a_0$.

Also, it is often mentioned that an atom is 99.99% empty space etc...

That's just a simplified view presented in popular science (or elementary chemistry classes) that ignores the quantum mechanical model. No serious chemistry book would write something like that. I don't know exactly how it's calculated, but I can make a good guess. You take any atom you like, let's say hydrogen. You look up the atomic radius - Google tells you it's $52.9 \text{ pm}$. (No surprises there.) So you can think of the atom as a sphere with that radius, and calculate the volume of the sphere: $V = \frac{4}{3}\pi r^3$. Then you go and look up the volume of an electron and a proton (a physicist would get a headache from just reading the phrase "volume of an electron"), and add it up, and you find that the "atom" is "99.99% empty space".

As Ivan succinctly said - with the QM description of the atom, there is no longer such a thing as free space since the electron "exists everywhere". Apart from that, just like how the distance from the nucleus is no longer well-defined in QM, the "volume" of an atom is also no longer well-defined. We pretty much threw that view out of the window.