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Homework Help: Hydrogen Wave Function

  1. Sep 24, 2010 #1
    1. The problem statement, all variables and given/known data
    Calculate the probability of finding the electron in a hydrogen within the angle [tex]\pm30\circ[/tex] from the x-y plane.The hydrogen is in the (2,1,1) state.

    2. Relevant equations
    [tex]probability = \int\int\int\left|R_{2,1,1}\right|^{2} \left|Y^{1}_{1}\right|^{2} r^{2} sin(\theta) dr d\phi d\theta[/tex]

    [tex]Y^{1}_{1} = -\frac{1}{2}\sqrt{\frac{3}{2\pi}}sin(\theta)e^{i\phi}[/tex]

    3. The attempt at a solution
    [tex]\int\left|R_{2,1,1}\right|^{2} r^{2} dr = 1[/tex]
    because limits are 0 to infinity.

    limit for [tex]\theta[/tex] is [tex]\frac{\pi}{3}[/tex] to [tex]\frac{2\pi}{3}[/tex]
    limit for [tex]\phi[/tex] is [tex]0[/tex] to [tex]2\pi[/tex]


    [tex]probability = \int\int \left|Y^{1}_{1}\right|^{2} d\phi d\theta[/tex]

    [tex]probability = \int\int \frac{1}{4}\frac{3}{2\pi}sin^{3}\theta e^{\phi} d\phi d\theta[/tex]

    [tex]= \frac{3}{8\pi}\int\int sin^{3}\theta e^{\phi} d\phi d\theta[/tex]
    [tex]= \frac{3}{8\pi}\int sin^{3}\theta \left[e^{\phi}\right]^{2\pi}_{0} d\theta[/tex]

    [tex]= \frac{3}{8\pi} \left(e^{2\pi}-1\right) \int sin^{3}d\theta[/tex]

    and after some algebra..

    [tex]\int^{\frac{2\pi}{3}}_{\frac{\pi}{3}} sin^{3}d\theta = 1-\frac{1}{12}[/tex]

    and so

    [tex]probability = \frac{3}{8\pi} \left(e^{2\pi}-1\right) \left(1-\frac{1}{12}\right)[/tex]

    Now, [tex]e^{2\pi}[/tex] is more than [tex]500[/tex]
    which makes the probability equal to [tex]58.5[/tex]


    Please help and thanks in advance
  2. jcsd
  3. Sep 25, 2010 #2
    ...:smile: Nevermind.

    Found what's wrong...
  4. Sep 25, 2010 #3
    Can you plz tell me whats wrong with it . Cos I have a similar problem . Thanks
  5. Sep 27, 2010 #4
    [tex]probability = \int\int \frac{1}{4}\frac{3}{2\pi}sin^{3}\theta e^{\phi} d\phi d\theta[/tex]

    Should be

    [tex]probability = \int\int \frac{1}{4}\frac{3}{2\pi}sin^{3}\theta d\phi d\theta[/tex]
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