A180638 - OEIS

%I #25 Sep 15 2016 11:57:48

%S 4,6,8,9,11,12,14,15,17,18,20,21,23,24,25,27,28,30,31,32,34,35,36,38,

%T 39,40,42,43,44,46,47,48,50,51,52,54,55,56,58,59,60,62,63,64,65,67,68,

%U 69,71,72,73,75,76,77,78,80,81,82,84,85,86,87,89,90,91,93,94,95,96,98,99,100,102

%N Phan Thành Nam's upper bound on the number of non-relativistic electrons bound to a nucleus of charge n.

%C From the abstract: "We prove that the maximal number N_c of non-relativistic electrons that a nucleus of charge Z can bind is less than 1.22Z + 3.1Z^(1/3).

%C This improves Lieb's (1984) upper bound N_c < 2Z+1 when Z <= 6. Our method also applies to non-relativistic atoms in magnetic field and to pseudo-relativistic atoms. We show that in these cases, under appropriate conditions, limsup_{Z->infty} N_c/Z <= 1.22."

%C Note that the published version of the paper, as revisions v2 and v3 on the arXiv, use the stronger bound 1.22Z + 3Z^(1/3) instead, where the constant 1.22 can be improved to 1.21684.... In particular this improves the bound for lithium, aluminium, scandium, chromium, cobalt, zinc, arsenic, krypton, yttrium, zirconium, etc. Asymptotically Seco, Sigal, & Solovej show that the maximum ionization is n + O(n^(5/7)) and it is conjectured that the maximum ionization is n + O(1). - _Charles R Greathouse IV_, Sep 13 2016

%D E. H. Lieb and R. Seiringer, The stability of matter in quantum mechanics, Cambridge University Press, 2009.

%H E. H. Lieb, <a href="http://dx.doi.org/10.1103/PhysRevA.29.3018">Bound on the maximum negative ionization of atoms and molecules</a>, Phys. Rev. A 29, 3018-3028 (1984).

%H Phan Thành Nam, <a href="http://arxiv.org/abs/1009.2367">New bounds on the maximum ionization of atoms </a>, Communications in Mathematical Physics 312:2 (2012), pp. 427-445. arXiv:1009.2367 [math-ph]

%H L. A. Seco, I. M. Sigal, and J. P. Solovej, <a href="http://projecteuclid.org/euclid.cmp/1104200838">Bound on the ionization energy of large atoms</a>, Communications in Mathematical Physics 131:2 (1990), pp. 307-315.

%F a(n) = floor(1.22*n + 3.1*n^(1/3)).

%e a(5) = floor(1.22*5 + 3.1*5^(1/3)) = floor(11.40...) = 11, which means that boron (the element with atomic number 5, i.e., with 5 protons) can have no more than 11 bound electrons, which would give it a -6 charge. B^5- has been observed (in Al_3BC) so this bound is reasonably tight. - _Charles R Greathouse IV_, Sep 14 2016

%p A180638 := proc(n) 1.22*n+3.1*n^(1/3) ; floor(%) ; end proc: seq(A180638(n),n=1..100) ; # _R. J. Mathar_, Oct 06 2010

%t Table[Floor[1.22 n + 3.1 n^(1/3)], {n, 73}] (* _Michael De Vlieger_, Dec 23 2015 *)

%o (PARI) a(n)=floor(1.22*n+3.1*sqrtn(n,3)) \\ _Charles R Greathouse IV_, Sep 13 2016

%Y Cf. A007656.

%K easy,nonn

%O 1,1

%A _Jonathan Vos Post_, Sep 14 2010

%E Reference converted to link, entries checked - _R. J. Mathar_, Oct 06 2010