Simple pictures showing orbital shapes are intended to describe the angular forms of regions in space where the electrons occupying the orbital are likely to be found. The diagrams cannot, however, show the entire region where an electron can be found, since according to quantum mechanics there is a non-zero probability of finding the electron anywhere in space. Instead the diagrams are approximate representations of boundary or contour surfaces where the probability density | ψ(r, θ, φ) |2 has a constant value, chosen so that there is a certain probability (for example 90%) of finding the electron within the contour. Although | ψ |2 as the square of an absolute value is everywhere non-negative, the sign of the wave function ψ(r, θ, φ) is often indicated in each subregion of the orbital picture.
enter image description here
[Cross-section of computed hydrogen atom orbital (ψ(r, θ, φ)2) for the 6s (n = 6, ℓ = 0, m = 0) orbital.]

I have a question here, if electrons can be found anywhere in the space with non-zero probability, can we give a definite boundary for the atom? i.e can we determine the radius of atom?

My sir has said me that, radius of atom is around $10^{-10}$m (from the $X$-ray experiments), but as we can have non-zero probability of finding the electron even beyond $10^{-10}$m, how can we say specific radius of an atom?


1 Answer 1


You are quite correct that atoms don't have a precise size. When defining the size of atoms we tend to use either bond lengths, if the atoms are reactive, or interatomic potentials for atoms that aren't reactive.

For example take Argon atoms, which are unreactive. The force between two argon atoms is well described by the London dispersion force, which in the case of Argon looks like:


Typically we get a minimum in the energy (at about 380pm in this case) then a hard core repulsion i.e. the energy rises steeply as you push the atoms past the minimum. We can take this as a measure of the size of the atom.

Where atoms react we can use the spacing in the molecules as a guide. For example the O to O distance in the O$_2$ molecule is 121pm giving us a radius for the Oxygen atom of a bit over 60pm.

However we will get different values for the atomic radii depending on how exactly we define it. For example Carbon forms single, double and triple C-C bonds, and they all have different lengths giving us different radii for the Carbon atom. This means the figures for the sizes of atoms are a guide rather than a precise value.

Wikipedia has a list of atomic sizes here that shows the different values for the radii obtained using different measures.

  • $\begingroup$ Thank you for the answer Sir. If we are able to determine the radius of an atom either by bond length or interatomic potential, won't it mean we are saying indirectly that there is zero probability of finding electron beyond that particular said value of radius. It mean we are violating the non-zero probability claim of quantum mechanics. If I am wrong any where, pardon me. $\endgroup$
    – Sensebe
    Jan 6, 2014 at 15:07
  • $\begingroup$ No, we're just saying that you can define an effective size for a somewhat fuzzy object. $\endgroup$ Jan 6, 2014 at 15:08
  • $\begingroup$ I agree with you Sir that, we are just defining an effective size for an atom. Quantum mechanics says that "there is non-zero probability of finding electron anywhere in the space (I meant to say for any distance from the nuclei)". Orbitals what we speak about, describe the region where there is 90% chances of finding an electron, so as we go far and far from the nuclei, the percentage should come down to 0 (no chances of finding electron). So, isn't there a need to redefine the statement as "there is non-zero probability of finding an electron with in a finite distance from the nuclei" $\endgroup$
    – Sensebe
    Jan 6, 2014 at 15:32
  • $\begingroup$ @VINAY: I'm not sure what you're asking. If you take for example a hydrogen 1s orbital the probability of finding the electron at a distance between $r$ and $r + dr$ from the nucleus is: $$\frac{4r^2}{a_0^3}e^{-2r/a_0}dr$$ $\endgroup$ Jan 6, 2014 at 16:54
  • $\begingroup$ @JohnRennie : This question has been bugging me. Say the last electron of the element goes into a d sub-shell. How do we define the radius of such an atom? $\endgroup$
    – ShankRam
    Dec 12, 2015 at 3:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.