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Wave functions and probabilities

  1. May 15, 2009 #1
    1. The problem statement, all variables and given/known data

    http://img200.imageshack.us/img200/9268/29360438.jpg [Broken]


    2. Relevant equations

    [tex]P=\left|\Psi \right| ^{2}dV[/tex]

    3. The attempt at a solution

    Okay, so [tex]r^{2} = x^{2}+y^{2}+z^{2}[/tex] and [tex]\left|\Psi \right| ^{2} = A^{2}e^{-2\alpha r^{2}}[/tex] .

    The volume of the of the bit we're interested in should be

    [tex]dV = 4\pi(r+dr)^{3} - 4\pi r^{3} \approx 12\pi r^{2}dr[/tex] if we ignore the [tex](dr)^{2}[/tex] and [tex](dr)^{3}[/tex] terms. Have I done something wrong here, as I was expecting to just end up with [tex]4\pi r^{2}dr[/tex]?

    Assuming it's correct, the probability is then

    [tex]P = A^{2}e^{-2 \alpha r^{2}}12\pi r^{2}dr[/tex] .

    To find where this has a maximum value, would I set [tex]\frac{dP}{dr}=0[/tex] and then find the corresponding r values? But how do I take the derivative when there's a dr term in there? :confused:

    Thanks for any help!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 15, 2009 #2

    tiny-tim

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    Hi kidsmoker! :smile:
    Yup! :biggrin:

    4 is for areas

    try 4/3 ! :wink:

    (or just multiply the area by dr)
     
  4. May 15, 2009 #3
    Oh yeah, hahaha, i always get that wrong!

    So we have [tex]P = A^{2}e^{-2 \alpha r^{2}}4\pi r^{2}dr[/tex] then what happens to the dr when i take the derivative? :s

    Thanks.
     
  5. May 15, 2009 #4

    tiny-tim

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    Just ignore it :wink: … it's a constant

    think of it as called something other than dr! :rolleyes:
     
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