Phy5645/HydrogenAtomProblem

(a) Take the volume integral of $\psi \psi *$. $Y_{1,-1}\left(\theta ,\phi \right)={\sqrt {\frac {3}{8\pi }}}sin(\theta )e^{-i\phi }$ and as such the phi dependence in the integral vanishes :

$\int _{\phi =0}^{2\pi }\int _{\theta =0}^{\pi }\int _{r=0}^{\infty }N^{2}r^{2}sin^{2}(\theta ){\frac {3}{8\pi }}e^{-{\frac {r}{a_{o}}}}r^{2}sin(\theta )drd\theta d\phi$

$\Longrightarrow {\frac {3N^{2}}{8\pi }}\int _{0}^{\pi }sin^{3}(\theta )d\theta \int _{0}^{2\pi }e^{-2i\phi }d\phi \int _{0}^{\infty }r^{4}e^{-{\frac {r}{a_{o}}}}dr=1$

$\Longrightarrow {\frac {3N^{2}}{8\pi }}{\frac {4}{3}}(2\pi )(24a^{5})=1$

Therefore $N^{2}\left(24a^{5}\right)=1$ so $N={\sqrt {\frac {1}{24a^{5}}}}$

(b)

$\psi \psi *=N^{2}r^{2}sin^{2}(\theta )\left({\frac {3}{8\pi }}\right)e^{-{\frac {r}{a_{o}}}}$

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

(c) What is the probability per unit radial interval (dr) of finding the electron at $a=a_{o}?$

Average over $\phi$ and $\theta$ at $r=2a_{o}$

$\left|Y_{1,-1}\right|^{2}=\left({\frac {3}{8\pi }}\right)sin^{2}(\theta )=\left({\frac {3}{16\pi }}\right)$ $\psi \psi *=N^{2}r^{2}e^{-{\frac {r}{a}}}s\left|Y_{1,-1}\right|^{2}$

$\Longrightarrow \left({\frac {1}{24{a}^{5}}}\right)(2a)^{2}e^{-{\frac {2a}{a}}}\left({\frac {3}{16\pi }}\right)=\left({\frac {1}{32\pi a}}\right)e^{-2}={\frac {0.0013}{a}}$

(d) If $\mathbf {{\hat {L}}^{2}}$ and $\mathbf {{\hat {L}}_{z}}$ are made, what will the results be?

l=1, m = -1 are the l and m of the eigenstate $Y_{1,-1}(\theta ,\phi )$

${\hat {L}}^{2}=\hbar ^{2}l(l+1)=2\hbar ^{2}$

${\hat {L}}_{z}=\hbar m=-\hbar$

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