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%I #15 Aug 29 2020 02:54:03
%S 1,3,3,7,0,3,0,1,5,6,8,6,1,7,8,0,1,5,8,9,5,6,7,2,8,6,3,2,2,9,5,8,4,7,
%T 4,9,4,8,1,1,3,9,3,8,9,7,5,1,3,9,1,1,1,3,9,0,4,3,8,5,9,4,0,9,0,6,8,7,
%U 8,2,7,7,4,5,8,4,9,5,9,2,8,8,2,2,2,6,0
%N Mean radius of the orbit of the hydrogen atom's electron, in Bohr radii.
%C The wave function of the 1s orbital of the hydrogen atom is psi(r)=exp(-r/a0)/(sqrt(Pi*a0^3)) where a0 is the Bohr radius A003671. The radial distribution function is psi^2(r)*4*Pi*r^2 which then simplifies to 4*r^2*exp(-2r/a0)/a0^3. Integrating from 0 to infinity and finding the value R where the value inside is the same as outside yields a0 times the root of 2 + 4R + 4R^2 = e^(2R).
%H NIST, <a href="https://physics.nist.gov/cgi-bin/cuu/Value?bohrrada0">Fundamental Physical Constants: Bohr radius</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Classical_electron_radius">Classical electron radius</a>
%e 1.33703015686178015895672863229584749481139389751391113904385940906878277458....
%t RealDigits[r /. FindRoot[2 + 4*r + 4*r^2 - Exp[2*r] == 0, {r, 3/2}, WorkingPrecision -> 100], 10, 87][[1]] (* _Amiram Eldar_, Aug 28 2020 *)
%o (PARI) solve(r=1,2,2+4*r+4*r^2-exp(2*r))
%Y Cf. A335727, A003671.
%K nonn,cons
%O 1,2
%A _Charles R Greathouse IV_, Jun 19 2020
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