I am trying to gain an intuitive picture of what is referred to by "electron-shell energy". I have read that outer electron shells have higher energy than inner electron shells, and this seemed to make sense to me by analogy to a wheel — a point on the rim of a wheel moves faster than a point on the hub. However, I have also read that electrons in inner shells move faster than those in outer shells, that in particularly heavy atoms, relativistic effects have to be taken into account for the inner shells to determine the chemical behavior of the element. And if inner shells are smaller, doesn't that mean that for an electron to be in such a shell, it would need a shorter wavelength, thus higher energy? Can anybody shed some light on my confusion?

Edit: I posted my question because there is an aspect to this that I feel has not been addressed in that other question about the speed of electrons, but I did not express it clearly enough. I recognize that the billiard-ball picture of electrons is not right, and getting beyond that picture is not my difficulty. It is that for all I read that speed is not a relevant concept in this context, I nevertheless keep running into mention of electron "speed" in technical literature, like in a recent SA article, Cracks in the Periodic Table (Scientific American, June 2013), and in the Wikipedia article Electron Configuration:

For the heavier elements, it is also necessary to take account of the effects of Special Relativity on the energies of the atomic orbitals, as the inner-shell electrons are moving at speeds approaching the speed of light.

So, speed is not a relevant concept in the context of electron shell energies, except when it is? Can somebody help me sort out the conflicting messages here?

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    $\begingroup$ The very first thing you need to do is to stop imagining that bound electrons follow paths and have speeds along them. They have position distributions and momentum distributions. $\endgroup$ Commented Jun 2, 2013 at 19:33

5 Answers 5


Your confusion arises from trying to understand the electron as a classical billiard ball.

The electron is a quantum mechanical entity, when free it displays a dual nature, that of a particle whose location is described by a specific $(x,y,z,t)$ and that of a probability wave.

In particularly an electron caught in a potential well has its energy described by the solutions of the quantum mechanical equations. It is not in an orbit around the nucleus it is in an orbital and best understood as a probability wave:


The shapes of the first five atomic orbitals: $1s, 2s, 2px, 2py$, and $2pz$. The colors show the wave function phase. These are graphs of $\psi(x, y, z)$ functions which depend on the coordinates of one electron.

The square of $\psi(x, y, z)$ gives the probability of finding the electron at that space point.


The five d orbitals in $\psi(x, y, z)**2$ form, with a combination diagram showing how they fit together to fill space around an atomic nucleus.

These are orbitals with higher angular momentum l=2

So velocity of the electron has no meaning here . It is pure quantum mechanical probability distributions. The shell energy defines the amount of energy needed to kick the electron out of its orbital, with a photon or another interaction. The lower the n quantum number the tighter the binding. When free it can exhibit its particle nature and have a velocity assigned to it .


firstly you should not make an analogy with a wheel since electrons are not orbiting the nuclei with equal angular velocity. it is more like satellites orbiting the earth because in both cases the force is proportional to $r^{-2}$. farther the satellite, it has less velocity.
to compare the energies of the shells you should consider electric potential energy too ,which is negative. electrons on outer shells are "on average" farther from the nuclei so they have less negative energy, which make them have higher energies. although you should use quantum mechanics to calculate energies properly, but it should give you an idea about where the higher energies come.
when we say "speed" is irrelevant in quantum context, it doesn’t mean that way. you can measure momentum of the electron in orbit and you will get a value, but the point is if you repeat the experiment probably you will get a different value, there is some subtleties interpreting this difference, but anyway, you can find the average value experimentally and theoretically.
so when you want to consider relativity effects, you look at this average speed value and compare it with the speed of light. if it was comparable, you should consider relativity effects.


I think the important thing to keep in mind is that the electron releases the energy it absorbed to fall back down into its more stable energy state based on its original "velocity". So you're not exactly wrong to suggest it was traveling at a faster "velocity" than it should be because maintaining this velocity would be unstable. That said, physics at this scale makes any intuition difficult, if even completely accurate. We assume the best intuition based on science that works, while simultaneously knowing it is a lie. For example: the plum pudding model was entirely reasonable and useful, until it wasn't.

I think it is important to consider how properties of waves intervene when the billiard ball analogy breaks down. I know it is popular and practical to reference its probability distribution because quantum mechanics, except for the fact that anything can be reduced to statistics. It's helpful for solving a physical property we want to solve for, but it wasn't statistics that ascertained the fundamental property. It's a tool that leverages hindsight, and sometimes it's conflated with physical property itself and so understanding suffers. Imo.

Consider De Broglie and Noether. Noether convincingly demonstrated that for a given field, you have an associated particle. You can think of orbitals as "localized" fields of the same type that manifest differently based on the geometry of 3 dimensions, and by constructive/destructive interference. in my mind, this field is born from the vibration/"a kind of orbit with no center"/"a particle somehow maintaining an occupancy in space" of the nucleus.

The nuclear vibration isn't 100% symmetrical all the time (because environmental influences are unavoidable), and so it gives rise to perturbations (I.e. electrons) in these orbitals/fields in like manner to itself being a purturbation in an non-localized field. This explains why electrons "spin" instead of "vibrate", and also suggests that these particles don't travel up or down between orbitals, but trade orbitals without travelling to maintain symmetry and stable energy states all dictated by the nucleus.

This also suggests something very interesting: a stable nucleus with field orbitals and no electrons is possible in ideal conditions.

But to actually answer your question with regard to relativity: electrons have mass, and if you can associate mass with an energy state, you can solve for a "velocity" property for its kinetic energy. Quantum mechanics wouldn't be able to say much about the mass dilation (I don't think?), In the same intuitive sense. It would just say that for the probability of the electron's location, the electron increased in mass based on its energy state. I see it as a situation where QM and relativity aren't exactly compatible, but they don't exactly contradict each other.


Kinetic energy $K$ depends upon speed $v$ as $K=\frac12mv^2$. The energy of the electron is the sum of its kinetic energy $K$ (which is positive, $K>0$) and its potential energy $U$ (which is negative, $U<0$): $$E=K+U$$ Inner, core electrons have a larger (positive) kinetic energy and also a larger (negative) potential energy than outer, valence electrons. The total energy of inner electrons is lower than that of outer electrons.


An electron cloud, is a zone of high probability. If you have an electron orbiting a proton, the cloud is an area where it is extremely likely to find an electron. You can't stick your finger in and pull out an electron. The cloud is just an area where the electron is likely to be found. More on that here.

The electron(- charge) wants to be as close as possible to the proton(+ charge). Therefore, it takes more energy to have a larger shell, than a smaller one. In fact, when an electron shifts to a lower shell, it releases the energy that was pent-up keeping away from the proton, as light. Photon emission

I like to think of it like a baseball on earth. If carry the baseball up a hill 10 meters above sea-level, if has more potential energy than a baseball that you are carrying at sea-level. If you drop the baseball of a cliff from 10m, it changes some of the potential energy into kinetic energy, and moves closer to sea-level.

Just like a baseball, a photon shifting shells converts potential energy into light, as it "falls" a distance toward the proton.

If someone were to catch your hypothetical baseball, and throw it back, it would gain potential energy as it traveled upwards. Likewise, if an atom absorbs energy in the form of heat, or light, the electrons in the cloud will gain potential energy, and move "upward" away from the atom.

Baseball exampleenter image description here

I hope I was able to help.

  • $\begingroup$ I should point out, that anna V is right. You need to think about the orbitals as they are depicted, and not as the billiard ball model. My graphic is just as an example. In nature, the electron shells are much more complex than the graphic makes them to be. $\endgroup$
    – CoilKid
    Commented Aug 29, 2014 at 20:01

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