Phy5645/HydrogenAtomProblem3

(a) The ground state wave function for hydrogen is

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

so that

$\langle r^{n}\rangle ={\frac {1}{\pi a^{3}}}\int \ r^{n}e^{-2r/a}r^{2}\sin {\theta }\,dr\,d\theta \,d\phi ={\frac {4}{a^{3}}}\int _{0}^{\infty }\!r^{n+2}e^{-2r/a}\,dr={\frac {4}{a^{3}}}\left({\frac {a}{2}}\right)^{n+3}\int _{0}^{\infty }\!x^{n+2}e^{-x}\,dx={\frac {(n+2)!}{2^{n+1}}}a^{n}.$

In particular, $\langle r\rangle ={\tfrac {3}{2}}a$ and $\langle r^{2}\rangle =3a^{2}.$

(b)

The probability of finding the electron within a spherical shell of thickness $dr\!$ is

$P=|\psi |^{2}\cdot 4\pi r^{2}\,dr={\frac {4}{a^{3}}}r^{2}e^{-2r/a}\,dr=p(r)\,dr.$

The probability per unit length $p(r)$ is then

$p(r)={\frac {4}{a^{3}}}r^{2}e^{-2r/a}.$

The most probable value of $r\!$ is then found by maximizing $p(r):\!$

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

The only non-trivial solution is $r=a.\!$ We know that the probability per unit length goes to zero at $r=0\!$ and as $r\to \infty ,\!$ so that $r=a\!$ must be a maximum.

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