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Homework Help: Hydrogen atom (finding electron probability)

  1. Feb 11, 2008 #1
    [SOLVED] Hydrogen atom (finding electron probability)

    1. The problem statement, all variables and given/known data
    For electron in eigensate of Hydrogen we have these expectation values

    <r>= 6a , <r^-1>= 1/4a

    (a is Bohr radius)

    a) find that eigenstate.
    b) find the probability of finding electron in region 0 < phi < Pi/6 , 0 < theta < Pi/8
    and 0 < r < rm
    where rm is the first minimum of radial wavefunction in which electron is.

    2. Relevant equations

    3. The attempt at a solution

    a) Using <nl|r^-1|nl>=1/((n^2)*a) it follows that n=2

    then using <nl|r|nl>=1/2*[3*n^2-l(l+1)]a I got l=0 ,
    so the eigenstate must be Psi(200).

    b) To find the probability, I have to integrate

    [tex]P=\int|\psi_{200}|^{2}r^{2}Sin\theta drd\phi d\theta = \frac{1-Cos(\frac{\pi}{8})}{(2a)^{3}6}\int^{r_{m}}_{0}(1-\frac{r}{2a})^{2}e^{\frac{-r}{a}}r^{2}dr[/tex]

    I can do the integral using tables, but can this integral be simplified. I only have to integrate from 0 to rm so if rm is much less than Bohr radius a, I can make an approximation (series expansion of e^x).
    I'm not shure is rm<<a ? , also how are rm and a (Bohr radius) related ?
    Last edited: Feb 12, 2008
  2. jcsd
  3. Feb 12, 2008 #2


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    How do you determine "r_{m}" ? From what equation ?
  4. Feb 13, 2008 #3
    [tex]\frac{\partial}{\partial r}(R_{20}(r))=0 \Rightarrow \frac{e^{-r/2a}(-4a+r)}{4\sqrt{2}a^{7/2}}=0 \Rightarrow r_{m}=4a[/tex]

    So, rm is four times smaller than a.
    Well, I think this is ok, since I'm just trynig to avoid some integral, not build something that suppose to work :smile: .
  5. Feb 13, 2008 #4
    You can also expand the [tex](1-\frac{r}{2a})^{2}[/tex] term, multiply with [tex]r^2[/tex] and integrate by parts. Quite painful, but exact!
  6. Feb 13, 2008 #5
    and multiply by e^(-r/a), then I have integral of r^4*e^(-r/a)
  7. Feb 13, 2008 #6
    I'll just use [tex]e^{\frac{-r}{a}} \approx 1-\frac{r}{a}[/tex]. [tex]r_{m}[/tex] is small enough for me.
  8. Feb 16, 2008 #7
    I meant that you could integrate by parts the integral

    [tex]\int_0^{r_m} r^4e^{-r/a}dr=r^4\frac{e^{-r/a}}{-1/a}\left|_0^{r_m}-\int_0^{r_m} 4r^3e^{-r/a}\left(-\frac{1}{a}\right)dr=\cdots[/tex]

    and so on...
    Last edited by a moderator: Feb 16, 2008
  9. Feb 17, 2008 #8
    ah, yes , tahnks again.
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