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Lunar_Lander
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Homework Statement
Determine for the hydrogen atom states 1s and 2p the expectation value of the radius r and the associated mean square error Δr.
Homework Equations
Wave Functions for 1s and 2p from Demtroeder's Experimental Physics Volume 3 (it says "The normalized complete eigenfunctions of an electron in the Coulomb potential [itex]V(r)=-Z\cdot e^2/(4\pi\epsilon_0r)[/itex]", is this what we need?) :
[itex]\psi_{1s}(r,\theta,\phi)=\frac{1}{\sqrt{\pi}}(Z/a_0)^{3/2}\cdot e^{-Zr/a_0}[/itex]
[itex]\psi_{2p_0}(r,\theta,\phi)=\frac{1}{4\sqrt{2\pi}}(Z/a_0)^{3/2}\cdot\frac{Zr}{a_0}\cdot e^{-Zr/2a_0}\cos(\theta)[/itex]
[itex]\Delta r=\sqrt{<(r-\overline{r})^2>}[/itex]
The Attempt at a Solution
Well, as in the neighbour thread (https://www.physicsforums.com/showthread.php?t=562573), the expectation value for r seems to be [itex]<\psi|r|\psi>=\int_0^r\int_0^{\pi}\int_0^{2\pi} r\cdot\psi^2\cdot r^2\sin(\theta)~dr~d\theta~d\phi[/itex], which I would, being naive, simplify to [itex]<r>=\int_0^r\int_0^{\pi}\int_0^{2\pi} \cdot\psi^2\cdot r^3\sin(\theta)~dr~d\theta~d\phi[/itex]. As we are in Hydrogen, Z should be 1.
Then I tried 1s. I get
[itex]\psi^2=\frac{Z^3\exp(-\frac{2rZ}{a_0})}{\pi\cdot a_0^3}=\frac{1}{a_0^3\pi}\exp(-\frac{2r}{a_0})[/itex]
Phi Integration:
[itex]\int_0^r\int_0^{\pi} 2\pi \frac{1}{a_0^3\pi}\exp(-\frac{2r}{a_0}) \cdot r^3\sin(\theta)~dr~d\theta[/itex]
Theta Integration:
[itex]\int_0^r 4\pi \frac{1}{a_0^3\pi}\exp(-\frac{2r}{a_0}) \cdot r^3~dr[/itex]
[itex]\int_0^r \frac{4}{a_0^3}\exp(-\frac{2r}{a_0}) \cdot r^3~dr[/itex]
r Integration:
[itex]\frac{4}{a_0^3} \int_0^r \exp(-\frac{2r}{a_0}) \cdot r^3~dr[/itex]
[itex]<r>_{1s}=\frac{4}{a_0^3} 6a_0^4-a_0\cdot\exp(-r/a_0)(6a_0^3+6a_0^2 r+3a_0r^2+r^3)[/itex]
Is this the right track?
PS: There is a third wave function for 2p, for m=+/-1: [itex]\psi_{2p_{\pm 1}}(r,\theta,\phi)=\frac{1}{8\sqrt{2\pi}}(Z/a_0)^{3/2}\cdot\frac{Zr}{a_0}\cdot e^{-Zr/2a_0}\sin(\theta)\cdot e^{\pm i\varphi}[/itex]
So two more calculations to run?
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