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| '''(a)''' Take the volume integral of <math>\psi\psi*</math>. <math>Y_{1,-1}\left(\theta, \phi | | '''(a)''' To find <math>N,\!</math> we simply take the volume integral of <math>\psi\psi^\ast.</math> Note that <math>Y_1^{-1}\left(\theta, \phi \right) = \sqrt{\frac{3}{8\pi}}\sin(\theta)e^{-i\phi},</math> and thus the <math>\phi\!</math> dependence in the integral vanishes. |
| \right) = \sqrt{\frac{3}{8\pi}}sin(\theta)e^{-i\phi} </math> and as such the | |
| phi dependence in the integral vanishes : | |
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| <math>\int _{\phi= 0}^{2\pi} \int_{\theta = 0}^{\pi} \int_{r=0}^{\infty} | | <math>1=\frac{3}{8\pi}\int_{\phi= 0}^{2\pi} \int_{\theta = 0}^{\pi} \int_{r=0}^{\infty} |
| N^{2}r^{2}sin^{2}(\theta) \frac{3}{8\pi}e^{-\frac{r} | | N^{2}r^{2}\sin^{2}{\theta}e^{-r/a}r^{2}\sin{\theta}\,dr\,d\theta\,d\phi</math> |
| {a_o}}r^{2}sin(\theta)drd\theta d\phi</math>
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| <math>\Longrightarrow \frac{3N^2}{8\pi} \int_{0}^{\pi} sin^{3}(\theta)d | | <math>=\tfrac{3}{4}N^2 \int_{0}^{\pi} \sin^{3}{\theta}\,d\theta \int_{0}^{\infty}r^{4}e^{-r/a}\,dr=24a^5N^2</math> |
| \theta \int_{0}^{2\pi} e^{-2i\phi}d\phi \int_{ 0}^{\infty}r^{4}e^{- | |
| \frac{r}{a_o}}dr = 1 </math>
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| <math>\Longrightarrow \frac{3N^2}{8\pi} \frac{4}{3}(2\pi)(24a^5) = 1</math> | | Therefore, <math>N = \frac{1}{\sqrt{24a^5}}.</math> |
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| Therefore <math>N^{2}\left(24a^{5}\right) = 1 </math> so <math>N = \sqrt\frac{1}{24a^5}</math>
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| '''(b)''' | | '''(b)''' |
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| <math>\psi\psi* = N^{2}r^{2}sin^{2}(\theta)\left(\frac{3}{8\pi}\right)e^{- | | <math>\psi\psi^\ast(r,\theta,\phi) = \frac{1}{24a^5}r^{2}\sin^{2}(\theta)\left(\frac{3}{8\pi}\right)e^{-r/a}</math> |
| \frac{r}{a_o}}</math>
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| <math>\Longrightarrow \left(\frac{1}{24{a_o}^5}\right) {a_o}^2
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| sin^{2}\left(\frac{\pi}{4}\right)\left(\frac{3}{8\pi}\right)e^{-1} = | |
| \left(\frac{\pi e^{-1}}{128{a_o}^3}\right) = \frac{0.009}{{a_o}^3}</math>
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| '''(c)'''
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| Average over <math>\phi</math> and <math>\theta</math> at <math>r = 2a_{o}</math>
| | <math>=\left(\frac{1}{24a^5}\right)a^2\sin^{2}\left(\frac{\pi}{4}\right)\left(\frac{3}{8\pi}\right)e^{-1} = |
| | \frac{\pi e^{-1}}{128a^3} = \frac{0.009}{a^3}</math> |
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| <math>\left| Y_{1,-1}\right|^2 = \left(\frac{3}{8\pi}\right)sin^{2}(\theta) | | '''(c)''' We simply integrate <math>\psi\psi^\ast\!</math> over the spherical shell given by varying <math>\phi\!</math> and <math>\theta\!</math> with <math>r = 2a.\!</math> The spherical harmonics, as we have defined them, are already normalized, so that the probability per unit radial coordinate is |
| = \left(\frac{3}{16\pi}\right)</math>
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| <math>\psi\psi* = N^{2}r^{2}e^{-\frac{r}{a}}s \left| Y_{1,-1}\right|^{2} </math> | |
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| <math>\Longrightarrow \left(\frac{1}{24{a}^5}\right)(2a)^{2}e^{-\frac{2a} | | <math>\frac{dP}{dr}=\left(\frac{1}{24{a}^5}\right)(2a)^{4}e^{-2} = \frac{2e^{-2}}{3a} = \frac{0.0902}{a}.</math> |
| {a}}\left(\frac{3}{16\pi}\right) = \left(\frac{1}{32\pi a}
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| \right)e^{-2} = \frac{0.0013}{a}</math>
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| '''(d)''' | | '''(d)''' We may read the orbital and magnetic quantum numbers directly off of the spherical harmonic, they are <math>l=1\!</math> and <math>m=-1.\!</math> Therefore, |
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| l=1, m = -1 are the l and m of the eigenstate <math>Y_{1,-1}(\theta, \phi)</math>
| | <math>\langle\hat{\mathbf{L}}^2\rangle=2\hbar^{2}</math> |
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| <math>\hat L^2 = \hbar^{2} l (l +1) = 2\hbar^{2} </math>
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| <math>\hat L_z = \hbar m = -\hbar </math> | | <math>\langle\hat{L}_z\rangle=-\hbar.</math> |
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| Back to [[Hydrogen Atom]] | | Back to [[Hydrogen Atom#Problems|Hydrogen Atom]] |
(a) To find 
we simply take the volume integral of 
Note that 
and thus the 
dependence in the integral vanishes.
Therefore, 
(b)
(c) We simply integrate 
over the spherical shell given by varying and 
with 
The spherical harmonics, as we have defined them, are already normalized, so that the probability per unit radial coordinate is
(d) We may read the orbital and magnetic quantum numbers directly off of the spherical harmonic, they are 
and 
Therefore,
and
Back to Hydrogen Atom