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Radial probability density

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Calculate the probability that the electron in the ground state of a hydrogen atom is in the region 0 < r < 3.75a0.

    2. Relevant equations

    a0=.0529 nm

    P(r)=4(Z/a0)^3*r^2*e^(-2Zr/a0)

    3. The attempt at a solution

    I am confused because I am not sure if I am supposed to use 3.75a0 as my radius. I guess I am not supposed to since I did that and got it wrong. I cannot figure out what else to do. However, I have also seen the probability density written as:

    P(r)[tex]\Delta[/tex]r=[4(Z/a0)^3*r^2*e^(-2Zr/a0)][tex]\Delta[/tex]r

    If this is the correct formula, I am not sure what to use for delta r.

    Any help would be much appreciated.
     
  2. jcsd
  3. Oct 13, 2009 #2

    kuruman

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    You don't use dr. This is a volume integral in radial symmetry. You use dV = 4πr2dr. You carry the integral from lower limit r = 0 to upper limit r = 3.75a0.
     
  4. Oct 13, 2009 #3
    Okay I do that and I get a volume of 9.81e-29m^3. How do I get this into a probability function?
     
  5. Oct 13, 2009 #4

    kuruman

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    You misunderstood what I said. The integral includes the probability density so that

    [tex]P=\int \psi^*\psi\; 4\pi r^2 dr[/tex]
     
  6. Oct 13, 2009 #5
    Okay. Gotcha. Thanks for the help.
     
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