For the same principal quantum number, on increasing the value of the azimuthal quantum number does the average radius of the subshell increase or decrease?

In other words, which out of, say, 3s, 3p, 3d would have the largest average radius and why?

  • 1
    $\begingroup$ This looks like a homework question. To get answers on this site for homework questions, you need to show the work you've done first. $\endgroup$
    – theorist
    Jul 12, 2020 at 2:48
  • $\begingroup$ @theorist no, it's not. It is just something I thought about while reading my text and found nowhere on the internet. It may feel so due to the way I have worded it, but I did that only to be crystal clear. And it's a theoretical question, what work should I show? $\endgroup$
    – Amadeus
    Jul 12, 2020 at 3:18
  • $\begingroup$ Well, you'd want to clarify if you're thinking about a single-electron atom (typically this would be hydrogen), or a multi-electron atom. Probably the former would be what you'd want to consider, since then you'd be looking at just the inherent nature of the orbitals themselves. And you could then do some searching on the internet for effect of azimuthal quantum number on average electron distance in hydrogen and, based on that, say something to the effect of: this is what I've found/understand thus far, but I'm still confused about this, etc etc.. $\endgroup$
    – theorist
    Jul 12, 2020 at 4:21

2 Answers 2


For a hydrogen atom, according to Hertel and Schulz [1]:

$$\langle r\rangle_{n,\,l}=\frac{a_0 n^2 \left(\frac{1}{2} \left(1-\frac{l (l+1)}{n^2}\right)+1\right)}{Z}$$


$$\langle r^2\rangle_{n,\,l}=\frac{a_0^2 n^4 \left(\frac{3}{2} \left(1-\frac{l (l+1)-\frac{1}{3}}{n^2}\right)+1\right)}{Z^2},$$

where $\langle r\rangle_{n,\,l}$ is the average distance of the electron from the nucleus as a function of n and l, and $\langle r^2\rangle_{n,\,l}$ is the average distance squared of the electron from the nucleus as a function of n and l, $a_0$ is the Bohr radius, and Z is the nuclear charge (Z = 1 for hydrogen). These are known as expectation values, because these are the expected values based on the probabilistic distribution of electron density.

Using these, we can compare $\langle r\rangle_{n,\,l}$ and $\langle r^2\rangle_{n,\,l}$ for the $3s \,(n = 3, l = 0), 3p \,(n = 3, l = 1) \text{ and } 3d\, (n = 3, l = 2)$ subshells:

$$\langle r\rangle_{3,\,0}= \frac{27 a_0}{2 Z}$$ $$\langle r\rangle_{3,\,1}= \frac{25 a_0}{2 Z}$$ $$\langle r\rangle_{3,\,2}= \frac{21 a_0}{2 Z}$$

$$\langle r^2\rangle_{3,\,0}= \frac{207 a_0^2}{Z^2}$$ $$\langle r^2\rangle_{3,\,1}= \frac{180 a_0^2}{Z^2}$$ $$\langle r^2\rangle_{3,\,2}= \frac{126 a_0^2}{Z^2}$$

You can see that, as the azimuthal quantum number (l) increases, the electron's average distance from the nucleus decreases. Note, however, that in hydrogen (or any single-electron atom, such as $\ce{He^+}$ or $\ce{Li^{2+}}$, etc.), all orbitals in the same shell have the same energy. Thus the energies of the 3s, 3p, and 3d orbitals are identical. As a consequence, a hydrogen electron excited to the third energy level is actually in a weighted linear combination of all of these orbitals, and its actual average distance would likewise be a weighted average of these.

But still, why does the average distance decrease with increasing azimuthal quantum number? Well, quantum is not my field, but I believe this is what's going on (perhaps a quantum expert can let us know if this is correct):

As I noted, in any single-electron atom, the 3s, 3p, and 3d orbitals have the same energy. The electron's energy is a combination of potential energy due to its distance from the nucleus, and the energy due to its angular momentum. As the azimuthal quantum number increases, the electron's angular momentum increases (it has no angular momentum in 3s, some in 3p, and more in 3d), and thus the energy associated with this angular momentum increases. But since the total energy is constant, as the energy due to angular momentum increases, the potential energy associated with distance from the nucleus must decrease, and hence the average distance from the nucleus gets lower.

Finally, you can see from the formula that $\langle r\rangle$ is much more sensitive to changes in n than l. For instance, if we compare the sizes of the s, p, and d subshells in the 3rd shell (n = 1, l = 0 to 2), $\langle r\rangle$ ranges from $\frac{27 a_0}{2 Z}$ to $\frac{25 a_0}{2 Z}$ to $\frac{21 a_0}{2 Z}$. By contrast, if we compare the sizes of the s-subshells in the first three shells (l = 0, n = 1 to 3), $\langle r\rangle$ ranges from $\frac{3 a_0}{2 Z}$ to $\frac{12 a_0}{2 Z}$ to $\frac{27 a_0}{2 Z}$.

  1. Hertel, Ingolf V., and Schulz, Claus-Peter. Atoms, molecules and optical physics. Berlin: Springer, 2015
  • $\begingroup$ Sorry if this was too much of a bother, I tried searching for what you suggested but wasn't able to find something useful to me. Well, I think I would, but the thing is, I am not yet familiar with the maths in quantum. Actually, I had thought of posting the question again after I had gotten better at it and done my research. But thanks, your answer was very comprehensive and helped me understand. $\endgroup$
    – Amadeus
    Jul 14, 2020 at 13:36
  • $\begingroup$ Just one question- if energy is a function of distance from nucleus and the distance is a function of both the principal and azimuthal quantum number, then how is energy of 3s,3p,3d equal? chemistry.stackexchange.com/questions/76111/…) I previously saw this answer and in it they said that the radius was only a function of the principal quantum number (in hydrogen-like atoms) so I understood then but now I am confused because the formula you used is a function of both $\endgroup$
    – Amadeus
    Jul 14, 2020 at 13:38
  • $\begingroup$ @l1mbo For your first question, that's what I explained in my answer: If one contributor to the energy (e.g., the angular momentum) goes up, the other (poential energy due to distance from nucleus) has to go down, to keep the total energy the same. I.e., you start with the fact that the total energy depends only on n, and then understand the effect of changing l from that. $\endgroup$
    – theorist
    Jul 14, 2020 at 18:07
  • $\begingroup$ And as to the second, clearly the average radius is not a function of n alone, so whoever said that either didn't know this, or meant to say it depends principally on n, with only a smaller added effect from l (which is correct) (i.e., changing n has much more effect on <r> than changing l). When I get a chance I'll add some examples illustrating the latter to my answer. $\endgroup$
    – theorist
    Jul 14, 2020 at 18:08

Before we start answering the question, we should first understand what the radius of electron is. According to quantum mechanics, the electron density is smeared across the space, and strictly speaking, the density becomes radially only if you have radial nodes, and at $r=0$ (at the nucleus) and at $r=\infty$. This means that you cannot specifically say that this is the radius of my shell. In quantum mechanics, the average radial distance of the wave function is defined as an "average radius", and the formula is given as $$\langle r\rangle=\frac{\langle\psi|r|\psi\rangle}{\langle\psi|\psi\rangle}$$ For hydrogen atom, the average radius of the $1s$ orbital is known as Bohr radius ($0.529 Å$).

Considering this idea, you can take the wave function of the system you like and calculate the average radius. However, you have to keep the following in mind:

  1. The notion of orbitals ($1s,2s$ etc.) exists for hydrogen or hydrogenic atoms (one electron systems) only. Beyond that, the analytical form of the wave function (for multi-electron systems) is impossible to obtain. In fact, it is one of the major challenges in the field of quantum mechanics. In many cases, people use approximate wave function to estimate the orbitals and their energies, and in such cases, it does not make sense to speak about the hydrogenic orbitals at all.
  2. Does it really make sense to talk about average radius in an non-radially symmetric system? This is because, if you change the value of $\theta$ and $\phi$ for a non-radially symmetric system, you might get different values for average radius.

I know I have not answered your question directly, but in case you are interested in orbitals of hydrogen (or hydrogenic systems) only, you can get the expression on internet. Combine that with a bit of knowledge of mathematics and/or coding, you will get the average radius of any (or any property) of any orbital you wish to know.


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