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To find atomic polarizability

  1. Nov 11, 2005 #1
    The Charge density of an electron cloud for a Hydrogen atom is given by:
    [tex]\rho (r) = \left(\frac{q}{\pi a^3}\right)e^{\frac{-2r}{a}}[/tex]
    Find its polarizability([itex]\alpha [/itex]).

    My work:

    Dipole moment p is:
    [tex]\vec p = \alpha \vec E[/tex]

    I need to calculate the electric field first.
    The electric field is given by Gauss's law:
    [tex]\vec E = \left(\frac{1}{4\pi \epsilon_0}\right)\frac{Q_{total}}{r^2}\hat r[/tex]

    [tex]Q_{total} = \int_{0}^{r} \rho (r)d\tau[/tex]

    [tex]Q_{total} = \int_{0}^{r} \left(\frac{q}{\pi a^3}\right) e^{\frac{-2r}{a}} 4\pi r^2 dr[/tex]

    [tex]Q_{total} = \frac{4q}{a^3} \int^{r}_{0} e^{\frac{-2r}{a}} r^2 dr[/tex]

    How is this integral evaluated?
    Last edited: Nov 11, 2005
  2. jcsd
  3. Nov 11, 2005 #2


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    Integrate by parts. (*groan*)
  4. Nov 12, 2005 #3
    :yuck: Thought so, but is there an appropriate substitution for the e^() term?
  5. Nov 12, 2005 #4
    This is a very easy one. You will need two steps in the integration by parts. Just start like this

    [tex] \frac{4q}{a^3} \int^{r}_{0} e^{\frac{-2r}{a}} r^2 dr = \frac{-a}{2} \frac{4q}{a^3} \int^{r}_{0} r^2 de^{\frac{-2r}{a}} [/tex]

  6. Nov 12, 2005 #5
    You don't really need one but I suppose you could go with r'=2r/a to simplify the algebra a little bit.
  7. Nov 12, 2005 #6
    Thanks, marlon and inha, I'll try it.
  8. Nov 15, 2005 #7
    OK, the integration part was pretty lengthy and I found the magnitude of the electric field of the electron cloud.
    [tex]E_e = \left(\frac{1}{4\pi \epsilon_0}\right)\frac{q}{r^2}\left(1 - e^{\frac{-2r}{a}} \left(1 + \frac{2r}{a} + \frac{2r^2}{a^2}\right)\right)[/tex]

    This is the field of the electron cloud. The proton will be shifted from r = 0 to a point 'd' where the applied field E equals field of the electron cloud.

    [tex]E = \left(\frac{1}{4\pi \epsilon_0}\right)\frac{q}{d^2}\left(1 - e^{\frac{-2d}{a}} \left(1 + \frac{2d}{a} + \frac{2d^2}{a^2}\right)\right)[/tex]

    How do I find the dipole moment term from this equation and hence the atomic polarizability?
    Last edited: Nov 15, 2005
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