# How much space does an atom occupy? [closed]

So my stupid question is: we know that in the classical model of a atom there is a nucleus at the middle and electron revolving around it in orbits numbered from 0 to infinity. So according to this an atom must have infinite space to accommodate infinite orbits. How is this possible and where am I going wrong?

## closed as unclear what you're asking by ACuriousMind♦, yuggib, LDC3, Kyle Kanos, JamalSJun 11 '15 at 13:11

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• It would be my GUESS that it is because only the first few are occupied and the size of the atom is determined by it's occupied orbits rather then it's theoretically possible orbits. – Quantum spaghettification Jun 9 '15 at 10:52
• This is unanswerable unless you define what you mean by "space that an atom occupies". Quantum physics, generally, has no precise notion of "occupied space". – ACuriousMind Jun 9 '15 at 15:00
• @ACuriousMind My knowledge about this subject is almost negligible as compared to you people.My question is purely out of curiosity. I think of an atom as a solid object which occupies a definite volume or space as I have often heard my teachers saying "Two atoms collide with each other ......" or "An atom breaks up into two particles ..... . – Sourav Kanta Jun 9 '15 at 15:09
• "Colliding" essentially means "interacting" and "breaking" means an interaction where one system splits into two which don't interact much afterwards. None of these needs notion of "size". Perhaps look also at this question where the notion of "solidity" for an atom is discussed. – ACuriousMind Jun 9 '15 at 15:17

## 1 Answer

Even in the classical model, an infinite amount of levels doesn't necessarily mean that it occupies an infinite amount of space. You can divide any finite distance into infinitely many bits (for instance, $1 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots$). EDIT: I'd forgotten about the $r\sim N^2$ relation that the OP mentions below, so yes, although the above is true, in the classical theory the size of the atom will go to infinity for $N\rightarrow\infty$.

However, the atom is not classical. Rather, the position of its electron is described by a probability density function which is non-zero at all distances even for the ground level, although it quickly reduces to zero-enough for most practical purposes. So your worry is still justified: In a sense, the atom does occupy all of the Universe.

Since in reality the atom is thus infinitely large, any practial definition of its size must be arbitrary, and indeed several definitions exist, which however are of the same order of magnitude, roughly an Ångström or so. From Wikipedia's article on the size of an atom: "Three widely used definitions of atomic radius are Van der Waals radius, ionic radius, and covalent radius."

• I recently completed my 12th standard and we learnt about the Bohr's atomic radius and how it is proportional to $N^2$ where N is the principal quantum number.So as you say if we can visualize that an atom fills the entire universe how can we say which electron belongs to which atom as all atoms will have to overlap. – Sourav Kanta Jun 9 '15 at 12:05
• @SouravKanta: The radius being proportional to $N^2$ gives you the most probable distance at which to find the electron. In reality, the probability decreases so fast that the probability of different level overlapping is extremely small. But in principle, there is a non-zero probability of finding an $N=1$ electron at the distance from the nucleus of $N=1$, at some point in time. At the next instant, it is at some other point. But you won't know until you measure it, so all you can say is "What is the probability of finding the electron at position $x$ when I measure it?" [to be cont'd…] – pela Jun 9 '15 at 12:24
• IMHO you should start with the electron, which has an electromagnetic field that has no outer edge. In QED it's described as an excitation of the electron field. Note that it's quantum field theory, not quantum point-particle theory. An electron isn't some point-particle thing that has a field, instead field is what it is. I think a cyclone isn't a bad analogy for starters, because the electron has a "spinor" nature. The Compton wavelength is like the diameter of the eye, but that isn't the size of the storm. – John Duffield Jun 9 '15 at 12:27
• You can for instance calculate and compare the probabilities of finding an $N=1$ electron and an $N=2$ electron at the same position (not at the same time, though). And if you measure an electron at position $x$, as you suspect you cannot say with 100% certainty which level it belongs to, but you can calculate the probabilities for various level and know it with quite high certainty. – pela Jun 9 '15 at 12:30
• Note also that the probability of being excited to high-$N$ levels decreases quickly with $N$, since the energy required to raise the electron to that level gets increasingly closer to the energy required to ionize the atom, in wcich case the electron is lost. – pela Jun 9 '15 at 12:30