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Homework Help: The mean value of (1/r^2) given a normalised wavefunction

  1. Apr 28, 2010 #1
    The normalised wavefunction for the 1s electron in the hydrogen atom is

    ψ=(1/((PI^1/2).a^3/2)).exp(-r/a)

    where a is the bohr radius.

    What is the mean value of (1/r2) in terms of the Bohr radius a0?



    Answer: 2/(a^2)
     
  2. jcsd
  3. Apr 28, 2010 #2

    kuruman

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    How do you think you should proceed? How does one calculate mean values?
     
  4. Apr 28, 2010 #3
    well basically i tried using the method of finding the expectation value of (1/r^2) although i was not ble to get the final answer, i thought about doing the integration using spherical polars, becuase there is a factor of pi that will need to be cancelled out, but i still couldn't get the correct answer.
     
  5. Apr 28, 2010 #4

    kuruman

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    Yes you need to find the expectation value of 1/r2 and spherical coordinates is the way to go. Can you show exactly how you set up the integral?
     
  6. Apr 29, 2010 #5
    i set it up as

    integral of (1/(pi)a^3).(exp(-2r/a)/r^2)dr

    but i do not no how to convert this into spherical polar i know that the volume element of spherical polar is (r^2)sin(theta)drd(theta)d(phi) but i don't no how to convert the dr line elemnt into spherical polars
     
  7. Apr 29, 2010 #6

    kuruman

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    This is a triple integral. There is nothing to convert. You should just do the integral over r. Now the angular part of the integral is

    [tex]\int_0^{\pi} sin\theta d\theta\int_{0}^{2\pi} d \phi[/tex]

    You need to do these two integrals separately and multiply the result with the radial integral.
     
  8. Apr 29, 2010 #7
    i have done that and the integral of them is equal to 2pi

    but once i do the integral of (exp(-2r/a)dr from 0<r<inifinity

    i get a/2

    this gives me a final answer of 1/(a^2) and the correct answer is 2/(a^2)

    do u know where i am missing the extra factor of 2?
     
  9. Apr 29, 2010 #8
    sorry i went over the angular part and got an answer of 4pi, so i have the right answer now

    thanks for your help
     
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