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| N^{2}r^{2}\sin^{2}{\theta}e^{-r/a}r^{2}\sin{\theta}\,dr\,d\theta\,d\phi</math> | | N^{2}r^{2}\sin^{2}{\theta}e^{-r/a}r^{2}\sin{\theta}\,dr\,d\theta\,d\phi</math> |
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| <math>=\tfrac{3}{4}N^2 \int_{0}^{\pi} sin^{3}{\theta}\,d\theta \int_{0}^{\infty}r^{4}e^{-r/a}\,dr</math> | | <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> |
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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>
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| <math>\left| Y_{1,-1}\right|^2 = \left(\frac{3}{8\pi}\right)sin^{2}(\theta) | | <math>=\left(\frac{1}{24a^5}\right)a^2\sin^{2}\left(\frac{\pi}{4}\right)\left(\frac{3}{8\pi}\right)e^{-1} = |
| = \left(\frac{3}{16\pi}\right)</math>
| | \frac{\pi e^{-1}}{128a^3} = \frac{0.009}{a^3}</math> |
| <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} | | '''(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 |
| {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)'''
| | <math>\frac{dP}{dr}=\left(\frac{1}{24{a}^5}\right)(2a)^{4}e^{-2} = \frac{2e^{-2}}{3a} = \frac{0.0902}{a}.</math> |
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| l=1, m = -1 are the l and m of the eigenstate <math>Y_{1,-1}(\theta, \phi)</math>
| | '''(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> |
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| <math>\hat L^2 = \hbar^{2} l (l +1) = 2\hbar^{2} </math> | | <math>\hat L^2 = \hbar^{2} l (l +1) = 2\hbar^{2} </math> |
Revision as of 23:52, 1 September 2013
(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 
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