Phy5645/HydrogenAtomProblem3

(Submitted by team 5. This is based on problems 4.13 and 4.14 from Quantum Mechanics by Griffiths)

Solution

$\psi ={\frac {e^{\frac {-r}{a}}}{\sqrt {\pi a^{3}}}}$

so $<r^{n}>={\frac {1}{\pi a^{3}}}\int \ r^{n}e^{\frac {-2r}{a}}r^{2}sin{\theta }\,dr\,d\theta \,d\phi ={\frac {4\pi }{\pi a^{3}}}\int _{0}^{\infty }\!r^{n+2}e^{\frac {-2r}{a}}\,dr$

$<r>={\frac {4}{a^{3}}}\int _{0}^{\infty }\!r^{3}e^{\frac {-2r}{a}}\,dr={\frac {4}{a^{3}}}3!({\frac {a}{2}})^{4}={\frac {3}{2}}a$

$<r^{2}>={\frac {4}{a^{3}}}\int _{0}^{\infty }\!r^{4}e^{\frac {-2r}{a}}\,dr={\frac {4}{a^{3}}}4!({\frac {a}{2}})^{5}=3a^{2}$

(B) What is the most probable value of r, in the ground state of hydrogen?

Solution

$\psi ={\frac {1}{\sqrt {\pi a^{3}}}}e^{\frac {-r}{a}}$

$P=\mid \psi \mid ^{2}4\pi r^{2}dr={\frac {4}{a^{3}}}e^{\frac {-2r}{a}}r^{2}dr=p(r)dr;p(r)={\frac {4}{a^{3}}}r^{2}e^{\frac {-2r}{a}}$

${\frac {dp}{dr}}={\frac {4}{a^{3}}}[2re^{\frac {-2r}{a}}+r^{2}({\frac {-2}{a}}e^{\frac {-2r}{a}})]={\frac {8r}{a^{3}}}e^{\frac {-2r}{a}}(1-{\frac {r}{a}})=0\implies r=a$

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