Probable distance in the 2p state (H atom)

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SUMMARY

The discussion centers on determining the most likely distance from the origin for an electron in the 2p state of the hydrogen atom, concluding that the maximum probability occurs at r = 4a0. The probability density function is derived as |R21|2·r2 = (r4 / 24a05)·e-r/a0. Additionally, the normalization of the spherical harmonic Y1,1 is confirmed through integration, yielding a result of 1 without the necessity of a table of integrals.

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[SOLVED] Probable distance in the 2p state (H atom)

Homework Statement



1. Determine the most likely distance from the origin for an electron in the 2p state of hydrogen.

[tex]R_{21} = \frac{1}{\sqrt{24a_{0}^{3}}}\frac{r}{a_{0}}e^{-\frac{r}{2a_{0}}}[/tex]

2. Show explicitly by integration that the spherical harmonic [tex]Y_{1,1} = -\sqrt{\frac{3}{8\pi}}sin{\theta}e^{i\phi}[/tex] is normalised. You may use a table of integrals.

The Attempt at a Solution



1. It seems to me the simplest approach is to obtain the probability density function [tex]|R_{21}^2|.r^2 = \frac{r^4}{24a_{0}^5}.e^{-\frac{r}{a_{0}}}[/tex], and find the maxima. I find then that differentiating with respect to r and setting to 0 gives solutions [tex]r = 0, r = 4a_{0}[/tex]. The maxima is then [tex]r = 4a_{0}[/tex]. Does that seem sensible? I can't seem to find any textbook values out there to check it against.

2. My query about the second one is that I find no need to use a table of integrals. The [tex]e^{i\phi}[/tex] happily disappears on taking the square modulus, and we are left (are we not?) with the integral

[tex]\int^{\pi}_{0}d\theta . \int^{2\pi}_{0}d\phi . sin^{3}(\theta) \frac{3}{8\pi}[/tex]

A simple trig identity dissolves the [tex]sin^3[/tex] into a couple of sin functions. And the outcome is indeed 1. Why should we need to use a table of integrals...?

Cheers!
 
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Correct on both counts.

You can look up integrals for powers of trig functions if you want to avoid the trouble of doing a little bit of trig. IMO, it'd probably be quicker in this case to just do the trig than to hunt down the integral in a handbook.
 
Gokul43201 said:
Correct on both counts.

You can look up integrals for powers of trig functions if you want to avoid the trouble of doing a little bit of trig. IMO, it'd probably be quicker in this case to just do the trig than to hunt down the integral in a handbook.

Ok. Thank you :-)
 

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