Coulomb ionization: excess charge, energy costs, and outer radii

Source and verification

This article explains family 263 of the October 6, 2026 OpenAI collection. Research conclusions are attributed to its manuscripts. The review covers statements, introductions, and selected proof mechanisms; it is not an independent verification of the complete proofs.

🏷️ Coulomb atoms and the outer-electron problem

An atom’s total energy is dominated by its inner electrons. Ionization asks a finer question: how much energy is needed to remove electrons from its outer region? This family connects uniform bounds on excess charge and first ionization with precise large-ionization energy and radius laws. Its three manuscripts concern the full correlated, nonrelativistic model with two spin states. (OpenAI, 2026) (OpenAI, 2026a) (OpenAI, 2026b)

The prerequisites are Sobolev spaces, quadratic forms and the variational principle, basic probability and conditional expectation, and the Poisson equation. Antisymmetry means that exchanging two electrons, including their spin labels, changes the wave function’s sign. It does not restrict the state to a single Slater determinant.

For a nucleus of charge $Z$ at the origin, the $N$-electron Hamiltonian is

$$\begin{array}{c}{H}_{Z,N}=\sum _{i=1}^N\left(-\frac{1}{2}{\Delta }_i-\frac{Z}{|x_i|}\right)\\ +\sum _{1\le i<j\le N}\frac{1}{|x_i-x_j|}.\end{array}$$

The form domain consists of antisymmetric, spin-valued $H^1$ functions. In three dimensions, Hardy’s inequality and Cauchy–Schwarz make each Coulomb term infinitesimally form bounded relative to the kinetic term. For each fixed system this gives a closed, bounded-below form and an associated self-adjoint operator. Write $E_Z(N)=\inf \sigma (H_{Z,N})$, with $E_Z(0)=0$.

The energies are nonincreasing in $N$: add an electron in a distant, broad orbital, antisymmetrize using disjoint supports, and let its kinetic energy and interaction with the existing state tend to zero. Strict binding in sector $N$ means $E_Z(N)<E_Z(N-1)$. This gap prevents a minimizing sequence from losing an electron at infinity and yields a ground state. Equality of adjacent infima alone does not exclude a ground state at threshold.

For integer $Z>m\ge 1$, the ionization energy is

$$\begin{array}{c}{I}_m(Z)=E_Z(Z-m)-E_Z(Z),\\ I_m(Z)\ge 0.\end{array}$$

If ${\Psi }_Z$ is a normalized neutral ground state, its spin-summed one-electron density ${\rho }_{{\Psi }_Z}$ is characterized by

$${\int }_A{\rho }_{{\Psi }_Z}(x)dx={\mathbb{E}}_{{\Psi }_Z}\#\{i:x_i\in A\}.$$

Thus its total mass is $Z$. Define the expected exterior mass $T_Z(r)={\int }_{|x|>r}{\rho }_{{\Psi }_Z}(x)dx$ and, for $0<m<Z$,

$$\begin{array}{c}{R}_m({\Psi }_Z)\\ =\inf \{r\ge 0:T_Z(r)\le m\}.\end{array}$$

This is a deterministic radius extracted from the density. It is not a sample maximum of electron positions. Moreover, the state still has $Z$ electrons: specifying exterior mass $m$ does not turn it into the positive ion with $Z-m$ electrons.

🏷️ Three scales of conclusions

The foundation manuscript also treats $M$ fixed nuclei at distinct positions, with real charges $z_j\ge 1$ and total charge $Z={\sum }_j{z}_j$. Replace $Z/|x|$ by ${\sum }_j{z}_j/|x-R_j|$. Nuclear repulsion can be omitted because it is the same scalar in every electron sector. (OpenAI, 2026c)

Uniform bounds attributed to the foundation manuscript

A universal finite $C$ satisfies $N\le Z+CM$ whenever a molecular sector binds strictly. For $N>Z+CM$, adjacent sector energies are equal. The constant is independent of all nuclear charges and distinct nuclear positions, including arbitrarily close configurations.

For neutral atoms with integer $Z\ge 1$, there are universal positive lower and upper bounds for both $I_1(Z)$ and $R_{1/2}({\Psi }_Z)$, uniformly over every normalized neutral ground state. The manuscript also supplies ground-state existence in these atomic sectors. No uniqueness or spherical symmetry is assumed.

The other two manuscripts identify asymptotic coefficients. Let $a_{\mathrm{TF}}>0$ denote the Thomas–Fermi ionization constant defined below, and put $b_{\mathrm{TF}}=(81{\pi }^2/2{)}^{1/3}$.

Quantity	Regime	Manuscript conclusion
Ionization energy $I_m(Z)$	Jointly $m\to \infty$, $Z/m\to \infty$	$I_m(Z)/m^{7/3}\to a_{\mathrm{TF}}$
Neutral radius $R_m$	First upper/lower limits in $Z$, then $m\to \infty$	Both scaled limits equal $b_{\mathrm{TF}}$
First ionization and half-electron radius	Every integer $Z\ge 1$	Positive universal lower and upper bounds

For clarity, the radius assertion is

$$\begin{array}{rl}\underset{m\to \infty }{\lim }{m}^{1/3}\underset{Z\to \infty }{\mathrm{limsup}}{R}_m({\Psi }_Z)& =b_{\mathrm{TF}},\\ \underset{m\to \infty }{\lim }{m}^{1/3}\underset{Z\to \infty }{\mathrm{liminf}}{R}_m({\Psi }_Z)& =b_{\mathrm{TF}}.\end{array}$$

Here $Z>m$ are integers, and the assertion holds for every choice of neutral ground states. The energy result also implies its corresponding iterated upper and lower limits. Neither result asserts convergence in $Z$ for fixed $m$. The radius theorem is not replaced here by an unrestricted joint-limit claim. (OpenAI, 2026a) (OpenAI, 2026b)

🏷️ Thomas–Fermi theory and the exponents

For a nonnegative density of prescribed mass, Thomas–Fermi theory minimizes

$$\begin{array}{c}{\mathcal{T}}_Z(\rho )=T\int {\rho }^{5/3}\\ -Z\int \frac{\rho (x)}{|x|}dx+D(\rho ),\\ T=\frac{3}{10}(3{\pi }^2{)}^{2/3},\\ D(\rho )=\frac{1}{2}\iint \frac{\rho (x)\rho (y)}{|x-y|}dxdy.\end{array}$$

The admissible finite-mass densities lie in $L^1\cap L^{5/3}$. The coefficient $T$ incorporates two spin states and kinetic operator $-\Delta /2$. A merely positive constant in a coarse kinetic inequality does not determine this sharp semiclassical coefficient.

Write $E_Z^{\mathrm{TF}}(N)$ for this infimum at mass $N$. The variational characterization of $a_{\mathrm{TF}}$ is

$$\begin{array}{c}\underset{Z\to \infty }{\lim }[E_Z^{\mathrm{TF}}(Z-m)-E_Z^{\mathrm{TF}}(Z)]\\ =a_{\mathrm{TF}}{m}^{7/3}\end{array}$$

for each fixed real $m>0$. This classical-model limit defines the constant; fixed-$m$ convergence is not thereby established for quantum energies. (OpenAI, 2026a)

Let $K(x)=|x{|}^{-1}$ and $F=ZK-K\ast \rho$ be a screened field. The zero-price Euler equation away from the nucleus is

$$\begin{array}{c}\rho =k(F_{+}{)}^{3/2},\Delta F=4\pi \rho ,\\ k=(5T/3{)}^{-3/2}=\frac{2^{3/2}}{3{\pi }^2}.\end{array}$$

Here $F_{+}=\max (F,0)$ and $\Delta K=-4\pi {\delta }_0$. Since $\Delta |x{|}^{-4}=12|x{|}^{-6}$, the positive homogeneous solution is

$$\begin{array}{c}{F}_{\ast }(x)=A_{\ast }|x{|}^{-4},\\ A_{\ast }={\left(\frac{12}{4\pi k}\right)}^2=\frac{81{\pi }^2}{8},\\ {\rho }_{\ast }(x)=k{A}_{\ast }^{3/2}|x{|}^{-6}.\end{array}$$

Integrating over the exterior of a ball gives

$$\begin{array}{rl}{\int }_{|x|>r}{\rho }_{\ast }(x)dx& =\frac{4\pi k{A}_{\ast }^{3/2}}{3r^3}\\ & =4A_{\ast }{r}^{-3}=b_{\mathrm{TF}}^3{r}^{-3}.\end{array}$$

Thus exterior mass $m$ corresponds to radius $b_{\mathrm{TF}}{m}^{-1/3}$. The field at that radius has size $m^{4/3}$, suggesting a removal cost of order $m\cdot m^{4/3}=m^{7/3}$. These calculations explain the exponents and the radius coefficient. The difficult step is proving that correlated quantum states approach the required screened profiles. (OpenAI, 2026b)

Bulk Thomas–Fermi convergence does not settle this question. The bulk length is $Z^{-1/3}$, while fixed exterior mass represents a fraction $m/Z\to 0$ of the electrons. Similarly, subtracting two total-energy asymptotics with errors larger than $m^{7/3}$ does not determine $I_m(Z)$.

🏷️ Conditional screening and uniform bounds

The common mechanism is a local comparison that survives arbitrary correlations. A spatial square partition records removed positions and spins and leaves an antisymmetric conditional core. A trial electron cloud is inserted into an empty patch. Coherent-state trial filling gives an upper energy estimate; counting Neumann eigenvalues gives a lower estimate with the same Thomas–Fermi coefficient. The comparison controls a local density and its screened field. These matched estimates are substantial source inputs. (OpenAI, 2026c)

Allowing particle number to change is essential. Initially work relative to $E={\inf }_{N\ge 0}{E}_Z(N)$, so every actual integer-sector trial has energy at least $E$. A state with energy at most $E+\delta$ retains an explicit offset $\delta$. Subsequent conditioning or particle deletion changes this offset, which must remain in the estimates.

The manuscript observes positions with small independent noise and forgets their labels. For an event $A$ of probability $p>0$, its likelihood $p_A(x)$ is symmetric, so the tilted state $\sqrt{p_A/p}\Psi$ stays antisymmetric. For the prescribed observation law and a sector eigenstate of energy $e_{\Psi }$, its form energy obeys

$$\mathfrak{q}[{\Psi }_A]\le e_{\Psi }+C{\ell }^{-2}{\log }^5(e/p).$$

The constant has no electron-number factor; multiple observation scales require the source’s summability condition on inverse squared widths. This bound lets a hypothetical failure of a local field comparison be tested by a legitimate conditional quantum state. The original observation posterior and a fresh spatial-cut posterior are different laws; the argument compares them through separate conditional-expectation identities. (OpenAI, 2026c)

Positive nonlinear subsolutions then propagate through successive scales. Before forgetting one observation array, the proof replaces the model reaction by actual conditional density. Exceptional comparisons contribute an error density whose expected integral is summed. For molecules, the geometric counting of empty nuclear annuli must cost $O(M)$ in total rather than once per scale. Together with the exterior electron count, this produces the excess-charge bound.

The atomic argument first obtains $E_Z(Z-1)\le E+C$ by deletion. It then inserts an orbital into the singly ionized ground state. Average its center over a distant annulus of radius $S$ and choose orbital width $b=S^{3/5}$. Newton’s spherical-average identity gives a remaining attraction at least $1/(2S)$ from the nucleus minus $Z-1$ electrons, without assuming a radial state. The kinetic cost is $O(b^{-2})=O(S^{-6/5})$.

Making a hole for that orbital also loses attraction. The source controls this loss using a complementary localized state and offset-sensitive screening; its averaged loss is $o(S^{-1})$. Fixing a sufficiently large universal $S$ gives a positive ionization gap. The previously obtained offset supplies the upper bound. A separate shell estimate, weighted by revealed shell mass, then gives $T_Z(R)\le C/R$ for large $R$. The positive interior deficit gives the lower half-electron radius. These deletion, insertion, and tail steps are distinct; the radius does not follow just from strict binding. (OpenAI, 2026c)

🏷️ Recovering ionization energies from a price

Introduce a positive price per electron,

$$P_Z(\lambda )=\underset{N\ge 0}{\min }(E_Z(N)+\lambda N).$$

As the minimum of affine functions, $P_Z$ is concave. Several sectors may minimize at the same price, and a prescribed sector need not minimize at any price. The proof does not assume convexity of $N\mapsto E_Z(N)$. (OpenAI, 2026a)

The main analytic input is the priced-deficit law. With $\lambda =s^{-4}$, as $s\to 0$ and $Z{s}^3\to \infty$, every minimizing sector satisfies $s^3(Z-N)\to q>0$. Here $q$ is the exterior charge of the unique decaying singular solution

$$\begin{array}{c}\Delta F=4\pi k((F-1{)}_{+}{)}^{3/2}\\ \text{on }{\mathbb{R}}^3\setminus \{0\}.\end{array}$$

Its field is comparable to $|x{|}^{-4}$ near zero and equals $q/|x|$ outside its density support. Rescaling the priced quantum problem makes the price equal to one. Conditional screening yields the equation; a positive barrier preserves its singularity; an exterior cutoff supplies the boundary condition. In this argument the unpriced energy offset can grow like $s^{-7}$, so a fixed-offset neutral-atom estimate alone is insufficient.

The remaining energy deduction is an instructive elementary argument. Along $m\to \infty$, $Z/m\to \infty$, let $N_t$ minimize at price $t{m}^{4/3}$ and define

$$\begin{array}{c}{d}_m(t)=\frac{Z-N_t}{m},\\ G_m(t)=m^{-7/3}[P_Z(t{m}^{4/3})\\ -t{m}^{4/3}Z-E_Z(Z)].\end{array}$$

The priced law says $d_m(t)\to q{t}^{3/4}$ for every fixed $t>0$. Testing the two endpoint minimizing sectors against each other’s prices gives, for $0<v<t$,

$$\begin{array}{c}-d_m(t)\le \frac{G_m(t)-G_m(v)}{t-v}\\ \le -d_m(v).\end{array}$$

Sum these inequalities on a finite partition, pass to the limit, and refine the partition. This integrates the limiting deficit. To determine the integration constant, use the neutral competitor and $E\ge E_Z(Z)-C$:

$$-C{m}^{-7/3}-v{d}_m(v)\le G_m(v)\le 0.$$

Letting $v\downarrow 0$ after the large-system limit gives

$$\begin{array}{c}{G}_m(t)⟶-\frac{4}{7}q{t}^{7/4},\\ \frac{E_Z(N_t)-E_Z(Z)}{m^{7/3}}⟶\frac{3}{7}q{t}^{7/4}.\end{array}$$

Choose prices $t_{-}<q^{-4/3}<t_{+}$. Their deficits eventually bracket $m$, so their particle numbers bracket $Z-m$. Monotonicity of sector energies therefore brackets the desired ionization energy. Let both prices approach $q^{-4/3}$ to obtain

$$\frac{I_m(Z)}{m^{7/3}}⟶\frac{3}{7}{q}^{-4/3}=a_{\mathrm{TF}}.$$

The analogous priced Thomas–Fermi calculation identifies this coefficient with its variational characterization. This deduction tolerates changes of minimizing sector and never differentiates through a price discontinuity. (OpenAI, 2026a)

🏷️ Neutral profiles and expected radii

For radii, take an intermediate physical scale $s$ and rescale a conditional smoothed density $\mu$ and field $H$ by $\sigma (x)=s^6\mu (sx)$ and $F_s(x)=s^{4}H(sx)$. Exact neutrality gives $\int \sigma =Z{s}^3$, which can diverge. Nevertheless, Newton’s identity gives

$$\begin{array}{c}{\overline{F}}_s(d)\\ ={\int }_{|z|>d}\left(\frac{1}{d}-\frac{1}{|z|}\right)\sigma (z)dz\ge 0,\end{array}$$

where the bar denotes spherical average. A pointwise upper bound on $F_s$, combined with this nonnegative mean and $|F_s|=2(F_s{)}_{+}-F_s$, bounds its local $L^1$ norm. Adding back a local Coulomb potential produces a harmonic function, so interior estimates give compactness of these possibly signed screened fields. Positive barriers then force every limit to satisfy $c|x{|}^{-4}\le F(x)\le C|x{|}^{-4}$. (OpenAI, 2026b)

The source’s profile rigidity has a short mechanism. Let $b=\sup |x{|}^{4}F(x)$ and $h=\inf |x{|}^{4}F(x)>0$. Dilate around points approaching each extremum. Local compactness gives contact with $b|x{|}^{-4}$ from below and $h|x{|}^{-4}$ from above. The maximum principle and the equation imply

$$4\pi k{b}^{3/2}\le 12b,4\pi k{h}^{3/2}\ge 12h.$$

Thus $b\le A_{\ast }\le h$, forcing $F=A_{\ast }|x{|}^{-4}$. No radiality of the quantum state was used.

Convergence on retained observations is still not convergence of expected electron counts. A localized $L^2$ count bound supplies uniform integrability: if $Y_s$ is a rescaled annular mass and $G_s$ the retained event, then

$$\mathbb{E}[Y_s{1}_{G_s^c}]\le \Vert Y_s{\Vert }_2\mathbb{P}(G_s^c{)}^{1/2}⟶0.$$

Shrinking packet supports compare smoothed mass with raw counts on slightly contracted and enlarged annuli. A separate dyadic estimate bounds the rescaled mass beyond radius $M$ by $C{M}^{-3}$. It prevents mass from escaping to infinity while passing from annuli to exterior regions.

Consequently, along admissible sequences, $s^3{T}_Z(as)\to b_{\mathrm{TF}}^3/a^3$ for fixed $a>0$. Here admissibility includes $s\to 0$, $Z{s}^3\to \infty$, and a source-dependent barrier threshold $Z\ge Z_{\mathrm{bar}}(s)$. With $s=m^{-1/3}$, choose $a_{-}<b_{\mathrm{TF}}<a_{+}$. Eventually the tail at $a_{-}s$ exceeds $m$, while that at $a_{+}s$ is below $m$. The radius is squeezed between these two radii.

For each fixed sufficiently large $m$, the barrier condition removes only a finite prefix of nuclear charges. It therefore leaves the inner limsup and liminf unchanged. Taking those limits first and then $m\to \infty$ proves precisely the stated radius theorem, uniformly over ground-state choices. (OpenAI, 2026b)

🏷️ Key steps and proof outlines

Unequal shifts let the field exchange its reaction for real density

The molecular induction maintains a distributional lower bound for a comparison field: the inner region contains the actual conditional density, while the outer region still contains a convex nonlinear reaction $f_y(u)$. To enlarge the inner region from scale $r$ to $2r$, the source takes a local maximum of

$$u_1=u-{\lambda }_1(2r{)}^{-4},u_2=H-{\lambda }_2(2r{)}^{-4}-\Pi ,$$

together with an exterior barrier where needed. Here $H$ is the observed screened field, $\Pi$ penalizes cap failures, and ${\lambda }_1>{\lambda }_2>0$ are fixed with explicit margins.

Proof outline. If $u_1$ is near the maximum in the newly gained layer, the unequal shifts force $H$ below $u$ by a definite amount. The inverse density–field comparison then rules out density larger than $f_y(u)/(4\pi )$: the old reaction can pay for the new actual density. Conversely, if $u_2$ is active in the remaining exterior, the other inverse comparison rules out density smaller than $f_y(u_2)/(4\pi )$. Its Poisson equation pays for the reaction there. Failed comparisons are added to a nonnegative error density.

Only after this replacement does the proof forget the current observations. Conditional expectation preserves the density by the tower property, and convexity gives $\mathbb{E}[f_y(u^{\prime })\mid \mathcal{F}]\ge f_y(\mathbb{E}[u^{\prime }\mid \mathcal{F}])$. The error is summed in expected integral, not by requiring a simultaneous good event at every spatial point.

If each scale costs at most $CM{r}^{\beta }$ with $\beta >0$, dyadic summation costs $O(M{r}_{\ast }^{\beta })$ independently of the number of scales. At the terminal scale a global Coulomb comparison and a positive far-sphere mean yield

$$N-Z<CM{r}_{\ast }^{-3}+o(M).$$

Along a hypothetical sequence with excess $K_{\mathrm{exc}}/M\to \infty$, choose $r_{\ast }^3=B_{0}M/K_{\mathrm{exc}}$. The inequality becomes $K_{\mathrm{exc}}<(C/B_0)K_{\mathrm{exc}}+o(M)$, contradicting large fixed $B_0$. This explains how the local comparison becomes a position-uniform molecular bound. (OpenAI, 2026c)

A positive price supplies a quantitative exterior cutoff

For a minimizing sector at price $\lambda =s^{-4}$, comparison with the preceding sector gives an ionization gap at least $s^{-4}$. The source first proves that its unpriced energy offset $D_0$ is $O(s^{-7})$ and its electron deficit is $O(s^{-3})$. Indeed, with $z=s^3(Z-N)\ge 0$, screening and the neutral competitor give an inequality of the form $z\le C(1+\sqrt{z+C{s}^7})$, which bounds $z$.

To control remote electrons, use a shell weight $S_u={\sum }_i\chi (|x_i|/u{)}^2$, mean $m_u$, and enlarged-shell count $N_u$. Testing the eigen-equation yields

$$s^{-4}{m}_u\le C{u}^{-2}{N}_u+\mathbb{E}\sum _i\chi (|x_i|/u{)}^2{B}_{u/8}(x_i).$$

Bias the state by $\sqrt{S_u/m_u}$. Its additional energy cost is at most $C{u}^{-2}{N}_u/m_u$, because the gradient of a Euclidean norm is bounded by the norm of the component gradients. A spatial observation determining $S_u$ leaves the conditional field unchanged under this bias. Offset-sensitive screening therefore controls the last term without assuming independence between the selected particle and the core.

For $u\ge R_{0}s$, choose $R_0$ large enough to absorb the terms proportional to $m_u$. The remaining estimate is $m_u\le C{s}^4(u^{-2}+u^{-3/2})N_u$. The rough annular count bound makes its dyadic sum vanish after multiplying by $s^3$. Thus the priced limiting density has a genuine exterior cutoff, a needed boundary condition for identifying the ionization coefficient. (OpenAI, 2026a)

Transferring the neutral barrier requires canceling the nucleus first

For the radius companion, the limiting screened field can initially have either sign. To transfer a positive comparison field $U_{\ell }$ to $F_{\ell }$, subtract a Coulomb potential $L_{\ell }$ of the accumulated error and the small density-comparison error. The nucleus and inner density cancel in the distributional difference, leaving

$$\Delta (U_{\ell }-F_{\ell }-L_{\ell })\ge 1_{\{|x|\ge a_{\ell }\}}(f(U_{\ell })-f(F_{\ell })).$$

On a fixed outer sphere of radius $S$, subtract an additional constant $c_S=O(S^{-4})$. The positive part of $U_{\ell }-F_{\ell }-L_{\ell }-c_S$ belongs to $H_0^1$ on the whole ball, including the origin after cancellation. Testing the inequality with that positive part gives a nonpositive squared gradient on the left and a nonnegative reaction term on the right. Hence the positive part vanishes.

The error potential tends to zero at every fixed nonzero point: its total mass is small, and its density is uniformly small near that point, which separately controls the Coulomb singularity. First take the sequence limit, then $S\to \infty$. This supplies the positive lower bound needed for the nonradial dilation rigidity argument above. (OpenAI, 2026b)

🏷️ Connections

• Sharp Lieb–Thirring inequalities studies spectral-moment bounds. Here a coarse kinetic density bound is only a preliminary estimate; the exact Thomas–Fermi coefficient comes from matched local semiclassical comparisons.
• Anderson localization and delocalization also separates finite-scale estimates from a spectral conclusion. The present Coulomb problem instead concerns interacting fermions and sector energy differences.

🏷️ Manuscripts and verification scope

• Uniform excess charge for Coulomb molecules and the outer radius of neutral atoms — pinned manuscript and companion source.
• Generalized ionization energies for full Coulomb atoms — pinned manuscript and companion source.
• Generalized outer-electron radii of neutral Coulomb atoms — pinned manuscript and companion source.

The family scope page describes selected formalization coverage. Only the generalized-energy and generalized-radius companions are listed. The energy selection covers the joint large-ionization limit on the full antisymmetric Sobolev form domain, with the Thomas–Fermi variational coefficient and both iterated limits; fixed-number convergence and ground-state attainment are outside that statement. The radius selection covers the two iterated limits for every choice of normalized neutral ground states, whose existence is an input. The molecular excess-charge and uniform first-ionization conclusions are not listed. No Lean build, axiom audit, or manuscript-to-formal-statement equivalence check was performed for these notes.

OpenAI collection provides prerequisites and the reading sequence; coverage register records completed and pending families.

🐻  OpenAI 2026a. Generalized ionization energies for full Coulomb atoms.

🐻  OpenAI 2026b. Generalized outer-electron radii of neutral Coulomb atoms.

🐻  OpenAI 2026c. Uniform excess charge for Coulomb molecules and the outer radius of neutral atoms.