(Submitted by team 5. This is based on problems 4.13 and 4.14 from Quantum Mechanics by Griffiths)
(a)
ψ = e − r a π a 3 {\displaystyle \psi ={\frac {e^{\frac {-r}{a}}}{\sqrt {\pi a^{3}}}}}
so < r n >= 1 π a 3 ∫ r n e − 2 r a r 2 s i n θ d r d θ d ϕ = 4 π π a 3 ∫ 0 ∞ r n + 2 e − 2 r a d r {\displaystyle <r^{n}>={\frac {1}{\pi a^{3}}}\int \ r^{n}e^{\frac {-2r}{a}}r^{2}sin{\theta }\,dr\,d\theta \,d\phi ={\frac {4\pi }{\pi a^{3}}}\int _{0}^{\infty }\!r^{n+2}e^{\frac {-2r}{a}}\,dr}
< r >= 4 a 3 ∫ 0 ∞ r 3 e − 2 r a d r = 4 a 3 3 ! ( a 2 ) 4 = 3 2 a {\displaystyle <r>={\frac {4}{a^{3}}}\int _{0}^{\infty }\!r^{3}e^{\frac {-2r}{a}}\,dr={\frac {4}{a^{3}}}3!({\frac {a}{2}})^{4}={\frac {3}{2}}a}
< r 2 >= 4 a 3 ∫ 0 ∞ r 4 e − 2 r a d r = 4 a 3 4 ! ( a 2 ) 5 = 3 a 2 {\displaystyle <r^{2}>={\frac {4}{a^{3}}}\int _{0}^{\infty }\!r^{4}e^{\frac {-2r}{a}}\,dr={\frac {4}{a^{3}}}4!({\frac {a}{2}})^{5}=3a^{2}}
(B) What is the most probable value of r, in the ground state of hydrogen?
(b)
ψ = 1 π a 3 e − r a {\displaystyle \psi ={\frac {1}{\sqrt {\pi a^{3}}}}e^{\frac {-r}{a}}}
P =∣ ψ ∣ 2 4 π r 2 d r = 4 a 3 e − 2 r a r 2 d r = p ( r ) d r ; p ( r ) = 4 a 3 r 2 e − 2 r a {\displaystyle P=\mid \psi \mid ^{2}4\pi r^{2}dr={\frac {4}{a^{3}}}e^{\frac {-2r}{a}}r^{2}dr=p(r)dr;p(r)={\frac {4}{a^{3}}}r^{2}e^{\frac {-2r}{a}}}
d p d r = 4 a 3 [ 2 r e − 2 r a + r 2 ( − 2 a e − 2 r a ) ] = 8 r a 3 e − 2 r a ( 1 − r a ) = 0 ⟹ r = a {\displaystyle {\frac {dp}{dr}}={\frac {4}{a^{3}}}[2re^{\frac {-2r}{a}}+r^{2}({\frac {-2}{a}}e^{\frac {-2r}{a}})]={\frac {8r}{a^{3}}}e^{\frac {-2r}{a}}(1-{\frac {r}{a}})=0\implies r=a}
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![{\displaystyle {\frac {dp}{dr}}={\frac {4}{a^{3}}}[2re^{\frac {-2r}{a}}+r^{2}({\frac {-2}{a}}e^{\frac {-2r}{a}})]={\frac {8r}{a^{3}}}e^{\frac {-2r}{a}}(1-{\frac {r}{a}})=0\implies r=a}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3dfeb19589b010add206a106d7ed8ea196e5c90)