Phy5645/HydrogenAtomProblem

(a) To find $N,\!$ we simply take the volume integral of $\psi \psi ^{\ast }.$ Note that $Y_{1}^{-1}\left(\theta ,\phi \right)={\sqrt {\frac {3}{8\pi }}}\sin(\theta )e^{-i\phi },$ and thus the $\phi \!$ dependence in the integral vanishes.

$1={\frac {3}{8\pi }}\int _{\phi =0}^{2\pi }\int _{\theta =0}^{\pi }\int _{r=0}^{\infty }N^{2}r^{2}\sin ^{2}{\theta }e^{-r/a}r^{2}\sin {\theta }\,dr\,d\theta \,d\phi$

$={\tfrac {3}{4}}N^{2}\int _{0}^{\pi }\sin ^{3}{\theta }\,d\theta \int _{0}^{\infty }r^{4}e^{-r/a}\,dr=24a^{5}N^{2}$

Therefore, $N={\frac {1}{\sqrt {24a^{5}}}}.$

(b)

$\psi \psi ^{\ast }(r,\theta ,\phi )={\frac {1}{24a^{5}}}r^{2}\sin ^{2}(\theta )\left({\frac {3}{8\pi }}\right)e^{-r/a}$

$=\left({\frac {1}{24a^{5}}}\right)a^{2}\sin ^{2}\left({\frac {\pi }{4}}\right)\left({\frac {3}{8\pi }}\right)e^{-1}={\frac {\pi e^{-1}}{128a^{3}}}={\frac {0.009}{a^{3}}}$

(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.\!} 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 \langle\hat{\mathbf{L}}^2\rangle=2\hbar^{2}}

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 \langle\hat{L}_z\rangle=-\hbar.}

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