(a) Take the volume integral of ψ ψ ∗ {\displaystyle \psi \psi *} . Y 1 , − 1 ( θ , ϕ ) = 3 8 π s i n ( θ ) e − i ϕ {\displaystyle 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 :
∫ ϕ = 0 2 π ∫ θ = 0 π ∫ r = 0 ∞ N 2 r 2 s i n 2 ( θ ) 3 8 π e − r a o r 2 s i n ( θ ) d r d θ d ϕ {\displaystyle \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 }
⟹ 3 N 2 8 π ∫ 0 π s i n 3 ( θ ) d θ ∫ 0 2 π e − 2 i ϕ d ϕ ∫ 0 ∞ r 4 e − r a o d r = 1 {\displaystyle \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}
⟹ 3 N 2 8 π 4 3 ( 2 π ) ( 24 a 5 ) = 1 {\displaystyle \Longrightarrow {\frac {3N^{2}}{8\pi }}{\frac {4}{3}}(2\pi )(24a^{5})=1}
Therefore N 2 ( 24 a 5 ) = 1 {\displaystyle N^{2}\left(24a^{5}\right)=1} so N = 1 24 a 5 {\displaystyle N={\sqrt {\frac {1}{24a^{5}}}}}
(b)
ψ ψ ∗ = N 2 r 2 s i n 2 ( θ ) ( 3 8 π ) e − r a o {\displaystyle \psi \psi *=N^{2}r^{2}sin^{2}(\theta )\left({\frac {3}{8\pi }}\right)e^{-{\frac {r}{a_{o}}}}}
⟹ ( 1 24 a o 5 ) a o 2 s i n 2 ( π 4 ) ( 3 8 π ) e − 1 = ( π e − 1 128 a o 3 ) = 0.009 a o 3 {\displaystyle \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)
Average over ϕ {\displaystyle \phi } and θ {\displaystyle \theta } at r = 2 a o {\displaystyle r=2a_{o}}
| Y 1 , − 1 | 2 = ( 3 8 π ) s i n 2 ( θ ) = ( 3 16 π ) {\displaystyle \left|Y_{1,-1}\right|^{2}=\left({\frac {3}{8\pi }}\right)sin^{2}(\theta )=\left({\frac {3}{16\pi }}\right)} ψ ψ ∗ = N 2 r 2 e − r a s | Y 1 , − 1 | 2 {\displaystyle \psi \psi *=N^{2}r^{2}e^{-{\frac {r}{a}}}s\left|Y_{1,-1}\right|^{2}}
⟹ ( 1 24 a 5 ) ( 2 a ) 2 e − 2 a a ( 3 16 π ) = ( 1 32 π a ) e − 2 = 0.0013 a {\displaystyle \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)
l=1, m = -1 are the l and m of the eigenstate Y 1 , − 1 ( θ , ϕ ) {\displaystyle Y_{1,-1}(\theta ,\phi )}
L ^ 2 = ℏ 2 l ( l + 1 ) = 2 ℏ 2 {\displaystyle {\hat {L}}^{2}=\hbar ^{2}l(l+1)=2\hbar ^{2}}
L ^ z = ℏ m = − ℏ {\displaystyle {\hat {L}}_{z}=\hbar m=-\hbar }
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and as such the
phi dependence in the integral vanishes :
so 
and 
at 
