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Technical difficulties in calculating hydrogen probabilities

  1. Mar 6, 2012 #1
    I am trying to prove to myself that the most probable distance for a 1s electron in a H atom is the Bohr radius.

    The probability of finding an electron (for any given state in a hydrogenic atom) in a spherical shell of thickness dr at a distance r from the nucleus is [itex]\left|R_{nl}\right|^{2} r^{2} dr = \left|\psi_{nlm}\right|^{2} 4\pi r^{2} dr [/itex].

    Given this, I need to differentiate this expression wrt r to obtain the radius at which the radial probability is maximum. This is where I face the problem. How do I differentiate either of the above expressions when there's a dr in each of them?
     
  2. jcsd
  3. Mar 7, 2012 #2

    tom.stoer

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    There are several different ways to look at this problem.

    1) Take probability for finding the electron in a thin shell

    [tex]P_{nlm}(R,dR) = \int_\Omega d\Omega \int_R^{R+dR} dr\,r^2\,|\psi_{nlm}|^2[/tex]

    devide by the volume

    [tex]V(R,dR) = \int_\Omega d\Omega \int_R^{R+dR} dr\,r^2[/tex]

    [tex]p_{nlm}(R) = \frac{P_{nlm}(R,dR)}{V(R,dR)}[/tex]

    and determine the maximum via

    [tex]\frac{\partial p_{nlm}(R)}{\partial R} = 0[/tex]

    I think this is what you have in mind. You will find that (when deviding by V) the dR dependency drops out as expected.

    2) Calculate the mean radius (or the more general case of the average value for rn)

    [tex]\langle r^n \rangle = \int_\Omega d\Omega \int_0^\infty dr\,r^{2+n}\,|\psi_{nlm}|^2[/tex]

    Of course in both cases for nlm=n00, i.e. for s-wavefunctions, the dΩ integration is just a constant 4π and need not be considered.
     
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