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(maybe) simple calculation question

  1. Jan 17, 2006 #1
    Now I'm reading a book QM, scattering problen when an electron is scattered by an atomic nucleus, and stuck at a simple formula; [tex]\int_0^\infty \varrho_p(R) R^4 dR = \langle R^2 \rangle,[/tex] where [tex]\varrho_p(R)[/tex] is the charge density at radius = [tex]R[/tex] In this formula, electric charge unit [tex]e[/tex] is omitted.

    As a matter of course, [tex]4\pi \int_0^\infty \varrho_p (R) R^2 dR = Z[/tex] holds.

    Will anyone tell me why the first equation holds?

    Thanks in advance.
     
    Last edited: Jan 17, 2006
  2. jcsd
  3. Jan 17, 2006 #2

    Meir Achuz

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    The average value of the squared charge radius over a volume is defined as [tex]<r^2>=\int r^2\rho d^3r/\int\rho d^3r=\int r^2\rho d^3r/q.[/tex]
    The volume differential is [tex]d^3r=r^2drd\Omega.[/tex] The angular integral gives a factor of [tex]4\pi[/tex], so the result should be
    [tex]<r^2>=(4\pi/q)\int_0^\infty r^4\rho dr[/tex]. Your book is probably considering unit charge. The absence of the [tex]4\pi[/tex] factor could be due to an unusual use of rationalized notation. Maybe the wave function is normalized to [tex]1/4\pi[/tex].
     
  4. Jan 17, 2006 #3
    I first thought that way, but what annoys me is, in that book, this is written:

    [tex]4\pi \int_0^\infty \rho(R)R^2 dR - \frac {2\pi}{3}s^2 \int_0^\infty \rho(R)R^4 dR \equiv Z - \frac{2\pi}{3}s^2 \langle R^2 \rangle[/tex]

    So [tex]4\pi[/tex] isn't normalized, and that charge number [tex]Z[/tex] isn't present in the second term...??? or is it just a definition?
     
    Last edited: Jan 18, 2006
  5. Jan 18, 2006 #4

    Meir Achuz

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    From what you write, it looks to me like the first term is an effect of the whole atom, while the second term is due to one electron in the atom.
    The factor of s^2/6 in the second term must come from the particular spin-orbital state of that electron. Tell me the name of the book, and
    I'll look at it when I get a chance.
     
  6. Jan 18, 2006 #5
    This appears in a calculation to show electron scattering with small momentum transfer to atomic nuclei will give the mean quadratic radius.
    Original formula is

    [tex]F(s) = 4\pi/s \int \varrho(R) \sin(sR) R dR[/tex]

    and using Taylor expansion

    [tex]\sin(sR) \approx sR - (sR)^3/3! \cdots[/tex]

    Thank you!!!

    edit:
    oops!! My bad! After these formulae, I see

    [tex]F(s) = 4\pi \langle R \rangle - 2\pi/3 s^2 \langle R^2 \rangle \cdots[/tex]

    (I was missing this one line!) So it may be just a typo and this final formula makes sense! This book is "Quantum Mechanics: An introduction" by Greiner, page 320-321. I have found some trivial typos in this book which are easy to spot and correct, but never seen a serious mistake or miscalculation.

    Thanks!
     
    Last edited: Jan 18, 2006
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