1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook