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

There exists something called Kramers's recursion rule and I think it is what are you looking for.

$\frac{k+1}{n^2} \left\langle r^k \right\rangle - \frac{a_0}{Z} \left(2k+1\right)\left\langle r^{k-1} \right\rangle + \frac{k a_0^2}{4Z^2} \left( \left(2l+1\right)^2 - k^2 \right) \left\langle r^{k-2} \right\rangle$ ,

where $k$ is integer and $a_0$ Bohr radius. For deriving $\left\langle r \right\rangle$ you have to calculate $\left\langle r^{-1} \right\rangle$ at first by setting $k=0$ and then you can set $k=1$ and calculate $\left\langle r \right\rangle$. And of course you know $\left\langle r^{0} \right\rangle = 1$.

The result is

$\left\langle r \right\rangle = \frac{a_0}{2Z}\left(3n^2-l\left(l+1\right)\right)$

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

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