Calculating avg distance from nucleus

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SUMMARY

The discussion focuses on calculating the average distance from the Hydrogen nucleus for an electron in the 2p orbital. The correct formula for the average distance is = ∫[r^3 |R|^2]dr from 0 to infinity, where R is the radial wavefunction. The participant initially calculated an average distance of 6a, which is incorrect as the expected value for 2p is 5a, indicating a misunderstanding of the wavefunction's properties. The discussion emphasizes the need for proper integration over spherical coordinates when dealing with non-spherically symmetric wavefunctions.

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Homework Statement



Calculate the avg. distance from the Hydrogen nucleus for an electron in 2p.

Homework Equations



<r> = int[r^3 |R|^2]dr from 0->infinity

For Hydrogen 2p, R = (1/a)^3/2 (1/(2*sqrt(sigma))*sigma*exp(-sigma/2)

where sigma = r/a

The Attempt at a Solution



I get 1/4a^4 (24 a^5) = 6a but the answer's supposed to be 5a. (5a for 2p is supposed to be less than 6a, which is the avg distance for 2s.) What am I doing wrong?
 
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jaejoon89 said:
<r> = int[r^3 |R|^2]dr from 0->infinity

That's only true when the wavefunction is spherically symmetric. In general,

[tex]\langle r\rangle=\int_{\text{all space}}r|\psi(\textbf{r})|^2d^3\textbf{r}=\int_0^{\infty}\int_0^{\pi}\int_0^{2\pi}|\psi(\textbf{r})|^2r^3\sin\theta dr d\theta d\phi[/tex]

For Hydrogen 2p, R = (1/a)^3/2 (1/(2*sqrt(sigma))*sigma*exp(-sigma/2)

where sigma = r/a

That doesn't look quite right. Where did you get this from?
 
Last edited:

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