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Hydrogen probability problem

  1. Jan 22, 2008 #1
    I am having some trouble starting this one.
    Calculate the probability that a hydrogen 1s electron will be found within a distance 2a0 from the nucleus.

    Any help is appreciated
     
  2. jcsd
  3. Jan 22, 2008 #2

    jtbell

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    Staff: Mentor

    To start with, you need the wave function [itex]\psi[/itex] for the hydrogen 1s state. Do you have that? What do you do with a wave function in order to calculate a probability?
     
  4. Jan 22, 2008 #3
    So, since the wavefunction for 1s has only the r variable, would I square it first and then integrate from 0 to 2a0?
     
  5. Jan 22, 2008 #4
    This is what I have so far. I used the wavefunction (1/sqrt(pi))*(Z/a0)^3/2*exp(-Zr/a0)
    I multiplied this by itself and by r^2 and then integrated from 0 to 2a0.
    I got a long answer so I'm not sure if this is correct. I got
    (exp(-4Z)/8Z^3)*[2(a0)^3*(-8Z^2-5)] + (2a^3/8Z^3)

    If someone can check my work, that would be great
    Thank you
     
  6. Jan 22, 2008 #5

    jtbell

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    Staff: Mentor

    Remember Z = 1 for hydrogen so you can simplify your equations a little bit. You've got the right [itex]\psi[/itex] . Squaring it gives you the probability density [itex]P[/itex]. This is the probability per unit volume so you have to integrate over all three dimensions [itex](r,\theta,\phi)[/itex] in spherical coordinates. The volume element in spherical coordinates is [itex]r^2 \sin \theta dr d\theta d\phi[/itex]. So you still need the [itex]\theta[/itex] and [itex]\phi[/itex] integrals which are pretty easy:

    [tex]P = \int_{r=0}^{r=2a_0} \int_{\phi=0}^{\phi=2\pi} \int_{\theta=0}^{\theta=\pi} P(r,\theta,\phi}) r^2 \sin \theta dr d\theta d\phi[/tex]

    Remember you can check the r integral by taking the derivative of the indefinite integral (before substituting the limits). I haven't had time to work it out myself yet... looks like integration by parts. Actually I normally look it up in a table of integrals. :redface:
     
  7. Jan 23, 2008 #6
    Okay, I got the probability will be .7616. If someone can check if thats correct, I would appreciate it.
     
  8. Jan 26, 2008 #7
    nobody can check for me?
     
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