# How big is an excited hydrogen atom?

Suppose an empty universe with the exception of a single hydrogen atom (1 proton, 1 electron). The electron may be in its ground state or it may be excited a certain number of levels. Suppose it is at level $n$.

Pick some reasonable measure of size, e.g., the average distance of the electron to the proton or the radius of the sphere surrounding the proton which would, with sufficiently high probability (e.g., $1 - 10^{-20}$), contain the electron. If some other definition of size is easier to compute, I'd be happy to use that instead.

How does this size grow with $n$?

I was inspired by this question about the Bekestein bound, in particular Jerry Schirmer's suggestion to store the information in a single atom. I've seen this idea thrown around before, but never properly analyzed it. In particular, the bound seems to place (weak) limits on just how close the electron could be to the proton in a highly excited state, though perhaps the high escape chance of the electron plays havoc with the bound.

• Highly excited atoms are called Rydberg atoms. They are current hot topic in research but one of the main challenges of reaching arbitrarily high values of $n$ is the fact that the polarizability scales as $n^7$ so the most minute variation in electric field would produce a very large force on the atom. – Asaf Nov 4 '14 at 19:13

## 4 Answers

$r=n^2a_0$ for the Bohr model.

Similarly, in the Schrodinger model, the most probable radius is $r=n^2a_0$, when $n=l+1$

From the Virial Theorem we can say the total energy of the atom is propotional to the potential energy of the atom. The potential energy is given by a coulomb potential and so is it will be roughly proportional to $\frac{1}{\langle r \rangle}$ where $\langle r \rangle$ is the mean radius of the electron's orbital. For a hydrogen atom the energy $E\propto \frac{1}{n^2}$, so we would expect \begin{align}\frac{1}{\langle r \rangle} & \propto \frac{1}{n^2}\\ \langle r \rangle &\propto n^2\end{align}

Unfortunately this does not help you much in storing your infinite amount of information in a single atom. In order to get an estimate of $\langle r \rangle$ you need to make many measurements of the position of the electron (especially if it is in a very spread out distribution such as for a high $n$ state) each of these measurements will collapse the wavefunction and you will have to prepare the atom into its initial state all over again before making the next measurement... but that was exactly what you were trying to ascertain by measuring the electron! To determine the state of your atom you need a set of quantities that can be measured simultaneously and uniquely determine which state your atom was in first time, which for a hydrogen atom essentially means you want the energy and the angular momentum. Now the angular momentum shouldn't give you too much trouble; it is nicely quantised into integer multiples of $\hbar$. The energy however will give you a problem. As $E \propto -n^{-2}$, for large $n$, the separation of energy levels goes like \begin{align} \frac{1}{n^2} - \frac{1}{(n+1)^2} &= \frac{1}{n^2}\left[ 1 - (1+ \frac{1}{n})^{-2}\right]\\ &\approx \frac{2}{n^3}\end{align} That will get very small very fast and how accurately you can measure that separation will limit how much information you can practically store.

• In other words, for large enough $n$, it would be difficult to measure $n$ accurately. – Charles Nov 4 '14 at 19:16

Yes, the mean radius grows as $n^2$ asymptotically, but to have $1-10^{-20}$ for sure you have to add some more ;-). Highly excited atoms are called Rydberg atoms and are dealt with in experiment.

• Thank you! Does the "some more" mean multiplying by a (large) constant, or will it scale up faster as $n$ increases? – Charles Nov 4 '14 at 19:14
• To answer your question, I have to carry out the corresponding research ;-) – Vladimir Kalitvianski Nov 4 '14 at 19:49
• :) No need, just curious. – Charles Nov 4 '14 at 22:48

Here's another, direct method to go with (and agree with) the Virial theorem method given in By Symmetry's answer.

From the Schrödinger equation solution (see the "Mathematical summary of eigenstates of hydrogen atom" heading in the "Hydrogen Atom" Wikipedia page):

$$\psi_{n,\,\ell,\,m}(r,\vartheta,\varphi) = \sqrt {{\left ( \frac{2}{n a_0} \right )}^3\frac{(n-\ell-1)!}{2n(n+\ell)!} } e^{- \rho / 2} \rho^{\ell} L_{n-\ell-1}^{2\ell+1}(\rho) Y_{\ell}^{m}(\vartheta, \varphi )$$

where $n$ is the principal quantum number, $\ell$ and $m$ the angular momentum quantum numbers, $\rho = 2\,r/(n\,a_0$ and $a_0$ is the Bohr radius. Thus:

$$\bar{r}=\frac{\int_{r=0}^\infty\,|\psi(r)|^2\,r^3\,{\rm d} r}{\int_{r=0}^\infty\,|\psi(r)|^2\,r^2\,{\rm d} r}\approx \frac{\int_{r=0}^\infty\,e^{-\frac{2\, r}{a_0\, n}} \left(\frac{r}{a_0\,n}\right)^{2\, n}\,r^3\,{\rm d} r}{\int_{r=0}^\infty\,e^{-\frac{2\, r}{a_0\, n}} \left(\frac{r}{a_0\,n}\right)^{2\, n}\,r^2\,{\rm d} r}= n\,(n+\frac{3}{2})\,a \sim n^2$$

and the Dirac equation solution will give a very like result.

• Everybody gives $a_0 n^2$, but the question is how far one should integrate to get the probability $1-10^{-20}$. – Vladimir Kalitvianski Nov 6 '14 at 7:01
• @VladimirKalitvianski I'll need some more time to mess around with asymptotic series for that one, but I'd be almost certain $r_\epsilon$ will be proportional to $n^2$ for small enough $\epsilon$. – WetSavannaAnimal Nov 6 '14 at 7:22