(adsbygoogle = window.adsbygoogle || []).push({}); 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]

so...

[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

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# Homework Help: Hydrogen Wave Function

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