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Electron charge distribution

  1. Feb 14, 2008 #1
    [SOLVED] Electron charge distribution

    Edit: [strike] meant to post this in coursework help. Forums won't let me delete thread. Please move:redface: . [/strike]


    "A quantum mechanical treatment of the hydrogen atom shows that the electron can be treated as a smeared out distribution of charge, which has the form:
    [tex]\rho[/tex](r) = [tex]\rho[/tex](0)e[tex]^{-2r/a}[/tex], where r is the radius and a is the Bohr radius.
    (a) Calculate [tex]\rho_{0}[/tex] using the fact that the atom is uncharged."

    I'm not looking for answers on how this is solved necessarily, I am just not really sure what everything is :/ .

    My interpretation is, rho(r) is the charge distribution of the electron's charge as a function of the distance from the center of the atom. with rho(0) being the charge distribution at r = 0.
    I also assume "e" is the electron charge and not the constant 2.71....

    Is this the correct interpretation? If so how is rho(0) not zero? Seems like the total distribution would be outside that point.
    Note that any actual quantum mechanics should not be needed, as this is a basic physics course.

    Thanks in advance for any replies.
    Last edited: Feb 15, 2008
  2. jcsd
  3. Feb 14, 2008 #2
    No, assume e is Exp, the 2.71, because its a decay as you get farther from the center.
  4. Feb 15, 2008 #3
    Thanks, K.J. , that makes sense seeing as the term would increase with r for e(in coulombs anyway).
    I found my prof. ripped this from the text, the answer in the back ([tex]\rho[/tex][tex]_{0}[/tex] = -e/[tex]\pi[/tex]a[tex]^{a'}[/tex]) at least tells me the units, C/m[tex]^{3}[/tex]. Which tells me it's charge density.
    The mention of quantum-mechanics and distribution had me thinking the equation was for probability's.
    Still am not really seeing what is going on here.

    *Thanks for the move.*

    Edit: it appears to me that what I am missing here is the correct picture of the charge distribution at r = 0.
    Even thinking about a distribution over a point of zero dimension doesn't make any sense to me.
    I think it should relate to the net charge (zero) of the atom using Gauss's law. But how......
    Last edited: Feb 15, 2008
  5. Feb 15, 2008 #4
    Double posting because I think I got it.

    From Gauss's law I know that, [tex]\Phi[/tex] = 0 = Q[tex]_{inside}[/tex]/[tex]\epsilon_{0}[/tex]. Q[tex]_{inside}[/tex] then has to equal zero. So, e[tex]^{+}[/tex] = [tex]\int_{0}^{R}[/tex][tex]\rho(0)[/tex]e[tex]^{-2r/a}[/tex]dr
    Now I solve for [tex]\rho(0)[/tex].
    I haven't done the calculation yet, will when I have the time.
    But I'm hoping someone could confirm/deny my method.

    Edit: Think I left out the volume term needed to multiply with the integral.

    Had some more time to look at this, I believe the equation I need to use to solve for [tex]\rho_{0}[/tex] is,

    [tex]e = -\rho_{0}4\pi \int_{0}^{R}r^{2}e^{-2r/a}dr[/tex]

    Using integration by parts twice I can get to,

    [tex]-e/\rho_{0}4\pi = e^{-2R/a}(-R^{2}a/2 - Ra^{2}/2 - a^{3}/4) + a^{3}/4[/tex]

    I can see from this that if this "[tex]e^{-2R/a}(-R^{2}a/2 - RA^{2}/2 - a^{3}/4)[/tex]" term equals zero my answer will equal the books given answer. Problem is, that term never equals zero, that I can see.......

    ... Fuc...........

    The way the problem is worded and according to my prof (who I asked) r is the radius of the hydrogen atom. But if I take r to just be the distance from the nucleus of the atom and let it go to infinity (which is more in line with my very basic knowledge of QM's) the lim of that term is zero.
    So I think I solved this myself now, but I would still be more content with confirmation if it can be provided.

    Now playing: By the End of Tonight - Ready? Aim. Fire!
    via FoxyTunes
    Last edited: Feb 16, 2008
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