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I Demerits radial distribution functions

  1. Apr 6, 2016 #1
    i have a question, why is the plot of r2(Ψ2p)2 not a good representation of the probability of finding an electron at a distance r from the nucleus in a 2p orbital
     
  2. jcsd
  3. Apr 7, 2016 #2

    vanhees71

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    Well, in spherical coordinates you have
    $$\mathrm{d}^3 \vec{x} = \mathrm{d} r \mathrm{d} \vartheta \mathrm{d} \varphi r^2 \sin \vartheta.$$
    The wave function squared is the probability distribution. So the probability for finding a particle described by the wave function at a distance ##r## is
    $$P(r)=\int_0^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi \, r^2 \sin \vartheta |\psi(r,\vartheta,\varphi)|^2.$$
    Note that sometimes one writes
    $$\psi(r, \vartheta,\varphi)=\frac{1}{r} u(r,\vartheta,\varphi),$$
    because this ansatz simplifies the solution of the Schrödinger equation in spherical coordinates. In terms of ##u## the factor ##r^2## cancels in the above integral.
    $$P(r)=\int_0^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi \, \sin \vartheta |u(r,\vartheta,\varphi)|^2.$$
     
  4. Apr 7, 2016 #3
    What do you mean by cancels?
    I apologize, I am relatively new at quantum mechanics
     
  5. Apr 7, 2016 #4
    and thanks so much.... you have really helped me alot
     
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