(a) To find N , {\displaystyle N,\!} we simply take the volume integral of ψ ψ ∗ . {\displaystyle \psi \psi ^{\ast }.} Note that Y 1 − 1 ( θ , ϕ ) = 3 8 π sin ( θ ) e − i ϕ , {\displaystyle Y_{1}^{-1}\left(\theta ,\phi \right)={\sqrt {\frac {3}{8\pi }}}\sin(\theta )e^{-i\phi },} and thus the ϕ {\displaystyle \phi \!} dependence in the integral vanishes.
1 = 3 8 π ∫ ϕ = 0 2 π ∫ θ = 0 π ∫ r = 0 ∞ N 2 r 2 sin 2 θ e − r / a r 2 sin θ d r d θ d ϕ {\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 }
= 3 4 N 2 ∫ 0 π s i n 3 θ d θ ∫ 0 ∞ r 4 e − r / a d r {\displaystyle ={\tfrac {3}{4}}N^{2}\int _{0}^{\pi }sin^{3}{\theta }\,d\theta \int _{0}^{\infty }r^{4}e^{-r/a}\,dr}
⟹ 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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we simply take the volume integral of 
Note that 
and thus the 
dependence in the integral vanishes.
so 
and 
at 
