Phy5645/HydrogenAtomProblem: Difference between revisions

From PhyWiki
Jump to navigation Jump to search
No edit summary
No edit summary
Line 4: Line 4:
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>


<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>


<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>
 
Therefore <math>N^{2}\left(24a^{5}\right) = 1 </math> so <math>N = \sqrt\frac{1}{24a^5}</math>


'''(b)'''
'''(b)'''


<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>
 
<math>\Longrightarrow \left(\frac{1}{24{a_o}^5}\right) {a_o}^2
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>
 
'''(c)'''
 
Average over <math>\phi</math> and <math>\theta</math> at <math>r = 2a_{o}</math>


<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>


<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}
\right)e^{-2} = \frac{0.0013}{a}</math>


'''(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>


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>


<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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N,\!} we simply take the volume integral of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi\psi^\ast.} Note that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Y_1^{-1}\left(\theta, \phi \right) = \sqrt{\frac{3}{8\pi}}\sin(\theta)e^{-i\phi},} and thus the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi\!} dependence in the integral vanishes.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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}e^{-r/a}r^{2}\sin{\theta}\,dr\,d\theta\,d\phi}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\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}

Therefore, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N = \frac{1}{\sqrt{24a^5}}.}

(b)

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi\psi^\ast(r,\theta,\phi) = \frac{1}{24a^5}r^{2}\sin^{2}(\theta)\left(\frac{3}{8\pi}\right)e^{-r/a}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\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}}

(c) We simply integrate Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi\psi^\ast\!} over the spherical shell given by varying Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta\!} with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r = 2a.\!} The spherical harmonics, as we have defined them, are already normalized, so that the probability per unit radial coordinate is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dP}{dr}=\left(\frac{1}{24{a}^5}\right)(2a)^{4}e^{-2} = \frac{2e^{-2}}{3a} = \frac{0.0902}{a}.}

(d) We may read the orbital and magnetic quantum numbers directly off of the spherical harmonic, they are Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l=1\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m=-1.\!}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat L^2 = \hbar^{2} l (l +1) = 2\hbar^{2} }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat L_z = \hbar m = -\hbar }

Back to Hydrogen Atom

Retrieved from "http://wiki.physics.fsu.edu/index.php?title=Phy5645/HydrogenAtomProblem&oldid=15557"