A180638 Phan Thành Nam's upper bound on the number of non-relativistic electrons bound to a nucleus of charge n.

4, 6, 8, 9, 11, 12, 14, 15, 17, 18, 20, 21, 23, 24, 25, 27, 28, 30, 31, 32, 34, 35, 36, 38, 39, 40, 42, 43, 44, 46, 47, 48, 50, 51, 52, 54, 55, 56, 58, 59, 60, 62, 63, 64, 65, 67, 68, 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

Offset

1,1

Comments

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).

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."

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

References

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

Links

Table of n, a(n) for n=1..73.

E. H. Lieb, Bound on the maximum negative ionization of atoms and molecules, Phys. Rev. A 29, 3018-3028 (1984).

Phan Thành Nam, New bounds on the maximum ionization of atoms , Communications in Mathematical Physics 312:2 (2012), pp. 427-445. arXiv:1009.2367 [math-ph]

L. A. Seco, I. M. Sigal, and J. P. Solovej, Bound on the ionization energy of large atoms, Communications in Mathematical Physics 131:2 (1990), pp. 307-315.

Formula

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

Example

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

Maple

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

Mathematica

Table[Floor[1.22 n + 3.1 n^(1/3)], {n, 73}] (* Michael De Vlieger, Dec 23 2015 *)

Prog

(PARI) a(n)=floor(1.22*n+3.1*sqrtn(n, 3)) \\ Charles R Greathouse IV, Sep 13 2016

Crossrefs

Cf. A007656.

Sequence in context: A049579 A137242 A098015 * A213320 A024887 A067611

Adjacent sequences: A180635 A180636 A180637 * A180639 A180640 A180641

Keyword

easy,nonn

Author

Jonathan Vos Post, Sep 14 2010

Extensions

Reference converted to link, entries checked - R. J. Mathar, Oct 06 2010

Status

approved