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Normalizing a radial wave function

  1. Nov 2, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that the radial function R[tex]_{31}[/tex] is normalized.


    2. Relevant equations
    [tex]\frac{1}{a_{0}^{3/2}}\frac{4}{81\sqrt{6}}\left(6-\frac{r}{a_{0}}\right)\frac{r}{a_{0}}e^{-r/3a_{0}}[/tex]

    [tex]\int^{\infty}_{0}r^{2}R_{31}*R_{31}dr=1[/tex]

    3. The attempt at a solution
    So I plugged that radial function in and got [tex]\int^{\infty}_{0}a_{0}^{2}u^{2}\left(6u-u^{2}\right)^{2}e^{-2u/3}du=1[/tex] all multiplied by some constant and [tex]u=\frac{r}{a_{0}}[/tex]

    I'm getting [tex]\frac{243}{4}[/tex] times the constant, and that does not equal one. So I feel like I'm not using the right equation for this one.
     
    Last edited: Nov 2, 2009
  2. jcsd
  3. Nov 2, 2009 #2
    Remember what coordinate system you are working in.
     
  4. Nov 2, 2009 #3
    Spherical coordinates. So there's an r^2 thrown in there. Did I not account for everything in my integral?
     
  5. Nov 3, 2009 #4
    Well, it looks like you didn't in your equation with the u's.
     
  6. Nov 3, 2009 #5
    I'm seeing an extra [itex]a_o^2[/itex], since there's an [itex](\frac{1}{a_0^{3/2}})^2 = \frac{1}{a_0^3}[/itex] term, an [itex]r^2 = a_0^2u^2[/itex] term, and a [itex] dr = a_0 du[/itex] term:

    [tex]\frac{1}{a_0^3}a_0^2u^2a_0 du = u^2 du[/tex]

    I'm curious as to how you're going about integrating

    [tex] C\int_0^\infty (u^6 - 12u^5 + 36u^4)e^{-2u/3} du [/tex] ?

    Does this have to be done by repeated parts (after you split it up of course), or is there a trick?
     
  7. Nov 4, 2009 #6

    kuruman

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    A very useful definite integral that was given to me when I learned about the hydrogen atom is

    [tex]\int^{\infty}_{0}x^n \; e^{-ax}dx=\frac{n!}{a^{n+1}}; \; (a>0;\; n\; integer \:>0)[/tex]
     
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