# What is the expected distance of the electron from the nucleus in the hydrogen atom?

Specifically, I would like to know the general formula, in terms of $n$ and $l$, assuming the electron is in an orbital (i.e. simultaneous eigenstate of $H$, $L^2$, and $L_z$).

I understand that it involves integrating an associated Laguerre polynomial, but I wasn't able to find the formula for the integral. I looked on Wikipedia and in Abramowitz & Stegun, but no luck.

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You mean. 0.53 angstroms ? Bohr radius –  Ali Gajani Feb 25 '14 at 6:12
That's only the case for $n=1$ –  Brian Bi Feb 25 '14 at 6:16

$\langle r\rangle_{n,l,m}=\frac{a_0n^2}{Z}[1+\frac{1}{2}(1-\frac{l(l+1)}{n^2})]$

Source:McQuarrie, Quantum Chemistry.

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Is there a derivation? –  Brian Bi Feb 25 '14 at 6:22
Sadly the book does not mention any :( But in principle does not seem too hard, assuming we know the general form of the Laguerre Polynomial (since the angular parts will just drop out) –  user41298 Feb 25 '14 at 6:25

Recall that the expectation value for any quantity $Q = Q(\mathbf x)$ in the (normalized) state $\psi$ is $$\langle Q\rangle_\psi = \int_{\mathbb R^3}d^3\mathbf x\, |\psi(\mathbf x)|^2Q(\mathbf x).$$

Let $(r,\theta, \phi)$ be the usual spherical coordinates. If we choose $Q(\mathbf x) = r$ and an energy eigenstate $\psi_{n, \ell, m}(\mathbf x) = R_{n,\ell}(r)Y_{\ell}^m(\theta, \phi)$, then we find the following integral expression in spherical coordinates for the expectation value of the radial distance between the electron and the nucleus: $$\langle r\rangle_{n,\ell,m} = \int_0^\infty dr \,r^3\, |R_{n,\ell}(r)|^2\int_0^{2\pi}d\phi\int_0^\pi d\theta\sin\theta\, |Y_\ell^m(\theta, \phi)|^2$$ Now you just have to lookup the functions $R_{n,\ell}$ and $Y_\ell^m$ in any quantum textbook.

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I wanted a closed form expression. I would do that integral, if I could. –  Brian Bi Feb 25 '14 at 6:22
@BrianBi Oh ok well I'll try to include more detail as soon as I have time. –  joshphysics Feb 25 '14 at 6:25
@BrianBi It turns out the required integrals are a lot trickier than I thought; I've worked on computing them for a while, but I've failed. I'm going to continue to try though; I'll let you know if I make progress. Nice question! –  joshphysics Feb 26 '14 at 3:47