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Probable distance in the 2p state (H atom)

  1. Dec 2, 2007 #1


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

    1. The problem statement, all variables and given/known data

    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.

    3. 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...?

    Last edited: Dec 2, 2007
  2. jcsd
  3. Dec 2, 2007 #2


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    Staff Emeritus
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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.
  4. Dec 3, 2007 #3


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    Ok. Thank you :-)
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