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Spreading a point nuclear electric charge over a small spherical nuclear charge distribution weakens the central attraction and raises electronic energy levels. In nonrelativistic first-order time-independent perturbation theory, a slowly varying $s$-state contact density gives the displayed shift. For a uniform sphere, $⟨r^2{⟩}_{\mathrm{ch}}=3R^2/5$, and integration of the difference between spherical and point Coulomb potentials gives $\Delta E_{1s}=(4/5)Z^4\mathrm{Ry}(R/a_0{)}^2$ for a hydrogen-like ion. A nonrelativistic $p$ wavefunction vanishes at the origin, suppressing its leading shift to order $R^4$. Heavy atomic nuclei require relativistic and realistic-charge-distribution corrections. An observed transition frequency shifts by the difference of the two level corrections divided by the Planck constant.

For the one-electron Coulomb potential $-KZ/r$ with fixed nucleus, the normalized spherically symmetric ground state is ${\psi }_Z=({\alpha }_Z^3/\pi {)}^{1/2}{e}^{-{\alpha }_{Z}r}$, where ${\alpha }_Z=Z/a_{\ast }$ and $a_{\ast }={\hslash }^2/(mK)$. Its radial probability density is $p_Z(r)=4{\alpha }_Z^3{r}^2{e}^{-2{\alpha }_{Z}r}$. The factorial integral givesMore generally $⟨r^q{⟩}_Z=\Gamma (q+3)/[2(2{\alpha }_Z{)}^q]$ for real $q>-3$, by the gamma function integral. The density transformation $p_Z(r)=Z{p}_1(Zr)$ expresses the full scaling, not just its mean. It applies to a hydrogen-like ion with one electron; the exact neutral helium atom radius additionally depends on the two-electron Coulomb interaction.

For a nonrelativistic point-nucleus hydrogen atom, $⟨r^2{⟩}_{nl}=a_0^2{n}^2[5n^2+1-3l(l+1)]/2$. Integrating the normalized radial wavefunction squared with measure $r^{4}dr$ gives $3a_0^2,42a_0^2,30a_0^2$ for $1s,2s,2p$. For a hydrogen-like ion, replace $a_0$ by $a_0/Z$ in the infinite-nuclear-mass model. These moments determine first-order shifts under quadratic potential energy perturbations.

The ground electron configuration of caesium has a single $6s$ electron, with $J=1/2$. Addition of angular momentum giving $F=3,4$ fixes the nuclear spin to $I=7/2$. With dimensionless angular momentum operators, write the Fermi contact interaction asThe two hyperfine levels therefore differ by $4A$. In the convention ${\mathbf{\mu }}_I=g_I{\mu }_B\mathbf{I}$ and $g_s\simeq 2$,This exposes a missing ingredient in a prediction from the supplied data: $Z$ and $g_I$ alone do not specify the valence electron contact density. Screening effect and penetration of the closed core strongly affect it. If $g_I$ instead denotes the total nuclear moment in units of ${\mu }_B$, it must first be divided by $I$ in the displayed convention.

An explicitly crude unscreened hydrogen-like ion model uses $|{\psi }_{6s}(0){|}^2=Z^3/(\pi 6^3{a}_0^3)$ and gives ${\nu }_0\simeq 5.76\times 10^{11}\mathrm{Hz}$, or ${\omega }_0\simeq 3.62\times 10^{12}{\mathrm{s}}^{-1}$. It describes neither the shielding nor the radial function of neutral caesium. Using a pure unit-charge $6s$ hydrogen orbital instead gives only $3.46\mathrm{MHz}$, illustrating why a unique useful estimate cannot follow from an unspecified effective charge.

For physical neutral ${}^{133}\mathrm{Cs}$, the independently known zero-magnetic field clock transition isThis measured frequency, used to define the second, is supplementary physical information, not a consequence of $Z=55$ and the rounded $g_I$ alone.^[ref] The direct electronic transition is in the microwave range and is a magnetic dipole transition, rather than a direct optical electric dipole transition. An optical implementation needs a two-field Raman transition.

Using the corrected weak-pulse Ramsey spectrum of part (b), the peak at $\delta =0$ has height $(\Omega {\tau }_p{)}^2$. The quantum interference fringes have spacing $2\pi /T$ in angular frequency, with zeros at $(2k+1)\pi /T$, and the sinc function envelope has its first zeros at $\delta =\pm 2\pi /{\tau }_p$. For $T\gg {\tau }_p$, the central peak has approximately full width at half maximum $\pi /T$ in angular frequency. The clock transition between the two $M_F=0$ states has no first-order Zeeman energy, although a quadratic one remains.

The finite nuclear size correction perturbation is unchanged for the same lead atomic nucleus. Its leading effect on one occupied $1s$ orbital is still proportional to that orbital's contact density,Other electrons screen the nuclear attraction and alter the self-consistent $1s$ wavefunction, usually reducing its contact density and hence its positive shift relative to the bare hydrogen-like ion. The reduction is modest for a deep $1s$ orbital: it penetrates well inside most other shells and still samples nearly the full $Z=82$ attraction. Replacing $Z$ by the neutral atom's asymptotic charge one would be inappropriate for this core orbital.

A neutral lead atom has two $1s$ electrons. Their contributions add in a frozen-orbital estimate of total atomic energy, roughly twice the per-electron shift, with relaxation and correlation corrections. An experimentally observed transition frequency again depends on the difference between the finite-size shifts of both participating many-electron quantum states. A quantitative neutral-lead result requires a screened relativistic many-electron calculation, not the one-electron formula with a single universally valid effective charge.

A nonrelativistic $2p$ wavefunction vanishes linearly at the origin. Its probability density is therefore proportional to $r^2$, so its first-order finite nuclear size correction shift scales as $R^4$, rather than the $R^2$ result for $1s$. It is positive but much smaller.

More explicitly, the normalized hydrogen-like ion radial function obeys $R_{21}^2(r)\simeq Z^5{r}^2/(24a_0^5)$ near the origin. Since the spherical harmonic has unit angular norm,ThusThis is about $2.4\mu \mathrm{eV}$ in the nonrelativistic model. At large $Z$, the relativistic $2p_{1/2}$ Dirac spinor has an $s$-type small component and a much stronger nuclear sensitivity than this nonrelativistic argument predicts; $2p_{3/2}$ remains much more suppressed. Neglecting spin-orbit interaction corrections would miss that distinction.

Put $k_Z=Z{e}^2/(4\pi {ϵ}_0)$. Gauss's law for a uniform spherical nuclear charge distribution, with the electrostatic potential matched at $R$, gives the electronic potential energySubtracting the point-nucleus Coulomb potential leaves the nonnegative perturbationThe normalized nonrelativistic hydrogen-like ion $1s$ wavefunction has $|\psi (r){|}^2=Z^3{e}^{-2Zr/a_0}/(\pi a_0^3)$. Since $ZR/a_0\ll 1$, approximate it inside the atomic nucleus by $|\psi (0){|}^2$. First-order time-independent perturbation theory then givesDefine $E_H=-Z^2\mathrm{Ry}$, as the paper specifies for the hydrogen-like ion, rather than the neutral-hydrogen value. Using $\mathrm{Ry}=e^2/(8\pi {ϵ}_0{a}_0)$ givesto leading order in $ZR/a_0$. For $A=208$, $R\simeq 7.11\mathrm{fm}$, $ZR/a_0\simeq 0.0110$, andThe finite nuclear charge radius makes the energy level less bound. The quoted frequency is a level shift divided by $h$: a line frequency shifts by $(\Delta E_u-\Delta E_l)/h$, so a decay into this shifted $1s$ level alone has the opposite sign.

The formula is a nonrelativistic, small-nuclear charge radius estimate. Keeping the exponential in the first-order integral gives $\Delta E_{1s}=\Delta E_{\mathrm{leading}}[1-5ZR/(6a_0)+O((ZR/a_0{)}^2)]$, about a one-percent correction here. More seriously, $Z{\alpha }_{\mathrm{fs}}\simeq 0.598$ for lead, so relativistic contraction is substantial. A precise physical prediction needs a finite-nucleus Dirac equation calculation; the boxed estimate is the model requested, not precision heavy-ion spectroscopy.

The one-electron species are ${\mathrm{C}}^{5+}$ and ${\mathrm{N}}^{6+}$, so there is no electron–electron screening and the nuclear charges remain $Z=6$ and $7$.Within the stated hydrogen-like ion model, $2s$ and all three $2p$ atomic orbitals of a given species have the same energy. The energy zero is a separated electron and the residual nucleus.

The $2s$ atomic orbital is spherically symmetric and has one radial node. A real $2p$ atomic orbital has two lobes of opposite quantum phase, one atomic-orbital nodal plane through the nucleus and no radial node. The three spatial orientations $2p_x,2p_y,2p_z$ remain degenerate in a spherically symmetric atom when external fields and fine structure are neglected.

In a many-electron atom, $2s$ penetrates the screened core more effectively than $2p$, experiences a stronger attraction and is lower in energy: ${\epsilon }_{2s}<{\epsilon }_{2p}$. The hydrogen-like ion degeneracy therefore does not survive screening. Both valence orbital types are generally more tightly bound and spatially contracted in nitrogen than in carbon, because of the larger effective nuclear charge. These are central-field orbital statements; the detailed atomic multiplet structure also involves electron repulsion and Hund's rules.

For a hydrogen-like ion, the Coulomb potential is stronger by the nuclear atomic number $Z$, and the energy levels scale as $Z^2$:Here $Z$ is the number of protons in the atomic nucleus; the ion has exactly one electron, so its net electric charge is $(Z-1)e$. The zero of energy is a separated stationary electron and atomic nucleus at infinite distance. This is the nonrelativistic hydrogenic approximation using the supplied $R_H$; a more precise isotope-specific result also corrects the reduced mass.

The radial probability density includes the spherical-shell volume factor:Consequently the radial expectation value isKeeping $m$ and $K$ fixed, the hydrogenic ground-state radius scaling is therefore $⟨r⟩\propto 1/Z$: the $Z=2$ state has half the mean radius of the $Z=1$ state. Equivalently, with $a_{\ast }={\hslash }^2/(mK)$, the mean radii are $3a_{\ast }/2$ and $3a_{\ast }/4$. For the physical electron-nucleus Coulomb potential, $a_{\ast }$ is the Bohr radius of the fixed-nucleus hydrogen atom model.

The exact $Z=2$ comparison in this one-electron model describes the hydrogen-like ion ${\mathrm{He}}^{+}$. A neutral helium atom has two electrons; its exact size also depends on their mutual Coulomb interaction and is not determined by this one-electron calculation. The stated half-radius conclusion is the prediction of the model supplied in the paper.