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The buoyancy frequency, also called the Brunt–Väisälä frequency, is the angular frequency of a small vertical displacement restored by buoyancy in a stably stratified fluid. Take $z$ upward, gravitational acceleration $-g\hat{\mathbf{z}}$, and a background in hydrostatic equilibrium,Displace a small gas element upward by $\xi$, sufficiently slowly for its pressure to match its surroundings but without heat exchange. For this adiabatic process, $p/{\rho }^{\gamma }$ is constant along the element's motion, so its mass density changes by $\rho {\gamma }^{-1}(d\ln p/dz)\xi$. The surrounding mass density changes by $(d\rho /dz)\xi$. The element's excess mass density is consequentlyIts acceleration is $\ddot{\xi }=-g\Delta \rho /\rho =-N^2\xi$. ThusFor a gas of fixed composition obeying the ideal gas law, $\rho \propto p/T$. The adiabatic buoyancy frequency of an ideal gas is thereforewhere $c_p=\gamma \mathcal{R}/(\gamma -1)$ is the specific heat capacity at constant pressure, and $\mathcal{R}=p/(\rho T)$ is the specific gas constant. Here $N^2$ has units ${\mathrm{s}}^{-2}$; $N$ is an angular frequency, not a cyclic frequency. Positive $N^2$ gives simple harmonic motion, negative $N^2$ gives convective instability, and $N^2=0$ is marginal. The temperature form assumes fixed composition; the preceding density form is more general.

Use plane wave perturbations proportional to $e^{i(k_{x}x+k_{z}z-\omega t)}$, with amplitudes $\hat{u},\hat{w},\hat{\pi },\hat{\varrho }$. The linear equations becomeTo eliminate pressure without dividing by $k_x$, multiply the vertical momentum equation by $k_x$ and subtract $k_z$ times the horizontal momentum equation:Use incompressible flow and the equation for mass density to obtain ${\omega }^2{\rho }_0(k_x^2+k_z^2)\hat{w}=-g{\rho }_0^{\prime }{k}_x^2\hat{w}$. Hence, for a nontrivial internal gravity wave,For stable density stratification, ${\rho }_0^{\prime }<0$ and $N=\sqrt{-g{\rho }_0^{\prime }/{\rho }_0}$ is the buoyancy frequency. Then $0\le {\omega }^2\le N^2$ and the normal modes oscillate rather than grow. For ${\rho }_0^{\prime }>0$, a mode with $k_x\ne 0$ has ${\omega }^2<0$ and grows exponentially: denser fluid lies above lighter fluid. Uniform density, ${\rho }_0^{\prime }=0$, is marginal for this restoring mechanism.

The case $k_x=0$, $k_z\ne 0$, has zero frequency in the displayed relation. Incompressible flow then requires $\hat{w}=0$, so it is a degenerate non-propagating case rather than a finite-frequency vertical oscillation. These equations concern the incompressible model of part (b); its $N^2$ does not contain the adiabatic compression correction needed in part (a).

For $N^2>0$, write $K=\sqrt{k_x^2+k_z^2}$ and choose either signed branch $\omega =sN{k}_x/K$, with $s=\pm 1$. These two branches cover the roots of the squared dispersion relation. The group velocity of an internal gravity wave isIf instead the positive-frequency root is chosen for both signs of $k_x$, set $s=+1$ and multiply the displayed components by $\mathrm{sgn}(k_x)$, away from the cusp $k_x=0$.

Taking the dot product with the wave vector givesThus the group velocity, which describes the travel of a narrow wave packet and its energy, is perpendicular to the wave vector, and hence perpendicular to the phase velocity ${\mathbf{c}}_p=\omega \mathbf{k}/K^2$. This is the internal-wave phase and group velocity geometry. It also follows from phase-group orthogonality for degree-zero dispersion: the frequency depends only on the direction of $\mathbf{k}$, not its magnitude. For $k_z=0$, the group velocity is zero, consistent with the stationary extremum $|\omega |=N$.

Define the dimensionless right-hand side of part (d) by $F(\kappa )$. Then ${\omega }^2=-4\pi G\rho F(\kappa )$. Because $I_0^{\prime }(\kappa )=I_1(\kappa )>0$ and $I_0(\kappa )>0$ for $\kappa >0$, its sign is the sign of $I_0(\kappa )K_0(\kappa )-1/2$. The original plot is positive for sufficiently long wavelengths and negative at short wavelengths; the nonzero crossing is approximately ${\kappa }_c\simeq 1.07$. Numerically solving the displayed equation gives ${\kappa }_c\simeq 1.0667$.

For $0<\kappa <{\kappa }_c$, ${\omega }^2<0$. Write $\omega =i\sigma$ with $\sigma =\sqrt{4\pi G\rho F}>0$. The time factor $e^{-i\omega t}$ then contains an exponentially growing solution. For $\kappa >{\kappa }_c$, ${\omega }^2>0$ and the ideal inviscid modes oscillate stably, without damping. At the crossing the frequency is zero and the mode is marginal. In terms of the axial wavelength ${\lambda }_z=2\pi /k$,The growth rate tends to zero as $kR\to 0^{+}$, so the very longest wavelengths are not the fastest growing. The source plot has its maximum near $kR\simeq 0.58$, corresponding to ${\lambda }_z\simeq 11R$ and $\sigma \simeq 0.25\sqrt{4\pi G\rho }$. Thus a characteristic fragmentation spacing is several cylinder diameters; a quantitative nonlinear fragment mass is beyond this linear stability calculation.

For unstable modes the physical displacement must be interpreted as $\mathrm{Re}[ϵe^{i(kz-\omega t)}]$ with complex amplitudes handled consistently throughout. The paper's real-part notation is harmless for the squared dispersion relation, but it should not be read as imposing real $\omega$ in a gravitational instability calculation. The exactly uniform mode $k=0$ is excluded from the finite-$k$ derivation; the limit of the curve alone does not establish a separate volume-preserving uniform-radius mode.

The changing mass quadrupole moment produces an inspiral gravitational wave signal whose frequency and amplitude increase toward merger. The chirp mass controls the leading frequency evolution, while higher-order phasing and the merger carry information about binary mass ratio and spins. Neutron stars also introduce tidal effects. These signals motivate a search with an interferometer such as LIGO; the supplied ranges quantify the paper's proposed detection prospects, without assuming any particular later observing-run performance.

Use the supplied comoving distance $D_C=5421$ Mpc. In a flat background, $D_A=D_C/(1+z)=1807$ Mpc and $D_L=(1+z)D_C=16263$ Mpc. The observed wavelength is stretched by $1+z$, soCare is needed with a luminosity per frequency. For corresponding emitted and observed frequencies, $d{\nu }_{\mathrm{em}}=(1+z)d{\nu }_{\mathrm{obs}}$. Combining that bandwidth relation with the bolometric definition of $D_L$ givesThis is the spectral luminosity at the rest-frame frequency corresponding to 200 nm. A value at some different rest wavelength would require the quasar's spectral shape. The factor is different for a density per unit wavelength; interchanging these two conventions would give an incorrect redshift factor.

The projected quasar-to-cloud separation isFor the usual transverse-separation estimate, take the cloud to lie at the same line-of-sight proper distance as the quasar. Since the cloud and galaxy are gravitationally bound, the local source-to-cloud path does not acquire the redshift factor $1+z=3$ that separates that system from us. Neglect internal relative Doppler shifts and attenuation. The observer in the cloud seesThis uses the quasar's isotropic intrinsic emission at the same rest frequency; anisotropic radiation or obscuration would modify it.

An angular separation only fixes $R_{\perp }$. If $s$ is a source-to-cloud displacement toward us, the true local distance is $r=\sqrt{R_{\perp }^2+s^2}$ and the general flux is $L_{\nu }/(4\pi r^2)$. The quoted transverse value is therefore an upper estimate at fixed projected separation. Equality of cosmological redshift and gravitational binding do not by themselves measure the small internal line-of-sight displacement; a peculiar velocity can also mask it.

A hydrogen recombination line is a discrete spectral line produced when radiative recombination captures an electron into an excited bound level and a subsequent radiative cascade emits photons. The free-to-bound capture itself produces a continuum; the later bound-to-bound transitions produce the recombination lines. For example, H-alpha is the $n=3\to 2$ transition in the Balmer series.

For the neutral hydrogen absorbing an ionizing stellar continuum, the principal bound-free transition is photoionization from the $1s$ ground state into the continuum. Most neutral hydrogen is in its ground state, and the photoionization cross section is largest near its $13.6\mathrm{eV}$ threshold, falling approximately as ${\nu }^{-3}$ above it. Dominance of the population-weighted opacity is intended here; this does not mean that every excited-state microscopic cross section is smaller at every frequency.

In an ionization-bounded, optically thick nebula, direct radiative recombination into $1s$ emits an ionizing photon that promptly photoionizes another hydrogen atom. One ion disappears and another is created: there is no net removal of an ion from the global budget. Case B recombination therefore uses ${\alpha }_B={\sum }_{n\ge 2}{\alpha }_n$, with local recycling of ground-state recombination photons. This is the on-the-spot approximation, rather than an unconditional property of radiative recombination: if those photons escape, direct capture does produce net neutralization and a Case A treatment is appropriate.

Retain the consistent ${\Theta }_{\mathrm{phys}}$ of d. Across a sufficiently thin layer, the neutral hydrogen column density is $N_{\mathrm{HI}}=n_H(1-{\zeta }_{\mathrm{in}})\Delta r+O(\Delta r^2)$. The direct stellar continuum acquires optical depth ${\tau }_{\nu }=N_{\mathrm{HI}}{\sigma }_{\nu }$. Define the photon-weighted second momentExpanding the attenuating exponential in the photoionization rate givesThe second cross section moment appears because the radiation both undergoes absorption and is weighted by the photoionization cross section in computing the rate.

In a local plane-parallel approximation that suppresses geometrical dilution, setThen local photoionization equilibrium becomes ${\Theta }_{\mathrm{phys}}{\zeta }^2=\Gamma (1-\zeta )$, soThis gives the requested algebraic dependence, with the corrected definition of $\Theta$. Besides $\Delta r\ll r_{\mathrm{in}}$, linear attenuation requires $N_{\mathrm{HI}}{\sigma }_{\nu }\ll 1$ over the frequencies contributing appreciably to the rate. In particular $\Gamma$ must remain positive.

The source's $\Gamma$ neglects $-2\Delta r/r_{\mathrm{in}}$, which is also first order in layer thickness. A consistent spherical first-order answer uses ${\Gamma }_{\mathrm{sph}}=\Gamma -2\Delta r/r_{\mathrm{in}}$ in the same root. Neglecting dilution relative to absorption additionally requires $n_H(1-{\zeta }_{\mathrm{in}})⟨{\sigma }^2⟩/⟨\sigma ⟩\gg 2/r_{\mathrm{in}}$. As a concrete limiting test, with negligible absorption the source's $\Gamma =1$ predicts no change, while inverse-square dilution still decreases $q$ and thus $\zeta$.

The Doppler factors for the two discrete blobs are ${\mathcal{D}}_{\pm }=[\gamma (1\mp \beta \cos \theta ){]}^{-1}$. Their ratio is $(1+x)/(1-x)=77/33=7/3$. Lorentz-invariant specific intensity means $I_{\nu }/{\nu }^3$ is unchanged; with $\nu =\mathcal{D}{\nu }^{\prime }$ and the same rest-frame power-law spectrum,The brightness ratio at the common observed frequency therefore givesThe stated identical intrinsic specific intensities must refer to comparable intrinsic states. If the blobs fade with their proper ages, equal observer times sample different ages and need that correction. A steady continuous jet has a different boosting exponent, $2+\alpha$; the discrete ejecta and the specified specific intensity transformation justify $3+\alpha$ here.

For optically thin synchrotron radiation from an isotropic electron distribution $dN/dE\propto E^{-p}$, the characteristic emitted frequency scales as $B{E}^2$ and the total single-particle power as $B^2{E}^2$. Thus $j_{\nu }\propto E^{-p}{E}^2/(d\nu /dE)\propto {\nu }^{-(p-1)/2}$; integration over the full single-particle spectrum preserves this exponent. HenceThe test-particle strong nonrelativistic diffusive shock acceleration prediction is $p=(r+2)/(r-1)=2$ for compression ratio $r=4$, giving $\alpha =0.5$. The inferred spectrum is steeper. If interpreted literally within that uncooled formula, $p=2.50$ would correspond to $r\simeq 3$, but cooling, escape, and other acceleration conditions can also steepen it.

The observed Galactic cosmic ray differential energy spectrum below the knee is approximately $E^{-2.7}$, steeper still; its formal synchrotron radiation equivalent would be $\alpha \simeq 0.85$. Thus $p\simeq 2.50$ lies between the simplest source prediction and the propagated cosmic ray slope. The comparison is qualitative: most measured cosmic rays are nuclei, whereas the radio slope here probes electrons, and propagation and radiative losses alter the spectra differently.

The specific intensity $I_{\nu }$ is energy transported per unit time, projected area, solid angle and frequency. The path coordinate $s$ increases along the ray. The emission coefficient $j_{\nu }$ is emitted energy per unit time, volume, solid angle and frequency, whereas the absorption coefficient ${\alpha }_{\nu }$ has units of inverse length. Thus $j_{\nu }ds$ adds specific intensity and ${\alpha }_{\nu }{I}_{\nu }ds$ removes it. Stimulated emission can be included as negative absorption.

Define the optical depth between two points by ${\tau }_{\nu }(s_1,s_2)={\int }_{s_1}^{s_2}{\alpha }_{\nu }(u)du$. Multiplying the radiative transfer equation by the integrating factor $\exp [{\tau }_{\nu }(s_0,s)]$ and integrating givesEach local contribution is attenuated only by the material between its emission point and the observer. Where ${\alpha }_{\nu }\ne 0$, the source function is $S_{\nu }=j_{\nu }/{\alpha }_{\nu }$ and the same formal solution of the radiative transfer equation is $I_{\nu }(\tau )=I_{\nu }(0)e^{-\tau }+{\int }_0^{\tau }{S}_{\nu }(t)e^{-(\tau -t)}dt$. A constant source function gives $I_{\nu }=I_{\nu }(0)e^{-\tau }+S_{\nu }(1-e^{-\tau })$.

With the spontaneous emission term suppressed, $j_{\nu }=0$. A uniform absorption coefficient over length $\ell$ gives directlyThe normalized specific intensity deficit is unchanged by conversion between wavelength and frequency, since both numerator and denominator acquire the same Jacobian determinant. The Boltzmann distribution and the Einstein coefficients give $n_2{B}_{21}/(n_1{B}_{12})=e^{-h{\nu }_{12}/(k_B{T}_e)}\ll 1$. Consequently, in the optically thin limit,Here $N_1=n_1\ell$ is the column density, and $\nu \simeq {\nu }_{12}$ has been used in slowly varying factors. This linear relation between equivalent width and column density fails once the central optical depth becomes appreciable.

The magnetic Lorentz force is perpendicular to velocity, so it does no work and $\gamma$ remains constant over an orbit when radiative losses are neglected. For the transverse momentum, $p_{\perp }=\gamma m_e{v}_{\perp }$ and $|{\dot{\mathbf{p}}}_{\perp }|={\omega }_g{p}_{\perp }=eB{v}_{\perp }$. HenceSince $10\mu \mathrm{G}=10^{-9}\mathrm{T}$,This is the gyromotion frequency, not the much higher characteristic synchrotron radiation frequency.

One explanation is an upper electron energy imposed by the accelerator: finite age, finite confinement size, escape, or a balance of acceleration and losses limits $\gamma$ to ${\gamma }_{\max }$. Since the synchrotron critical frequency scales as ${\nu }_c\propto B{\gamma }^2\sin \alpha$, the absence of higher-energy electrons produces a corresponding high-frequency cutoff, broadened by the single-particle spectrum.

The other is synchrotron cooling, with inverse Compton scattering often contributing: $\dot{\gamma }=-b{\gamma }^2$ gives $t_{\mathrm{cool}}=1/(b\gamma )$. High-energy electrons disappear fastest, so a population of age $t$ develops a cooling scale ${\gamma }_b\sim 1/(bt)$ and ${\nu }_b\propto B/(b^2{t}^2)$. A burst population can develop a strong cutoff. With continuous power-law injection, the simplest steady cooling model instead steepens the electron energy index by one and the synchrotron radiation spectral index by $1/2$, giving a break rather than necessarily an exponential cutoff. The detailed spectral shape therefore distinguishes finite acceleration from aging and injection history.

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.

Specify the radiation normalization first. Let $u_{\nu }$ be radiation energy density per ordinary frequency, with SI units $\mathrm{J}{\mathrm{m}}^{-3}\mathrm{H}{\mathrm{z}}^{-1}$. Then $A_{21}$ is the spontaneous emission rate per upper-state chemical atom, while $B_{12}{u}_{\nu }$ and $B_{21}{u}_{\nu }$ are the induced absorption and stimulated emission rates, evaluated at a sharp transition frequency. For a finite spectral line profile, replace $u_{\nu }$ by its profile-weighted average.

The Einstein coefficients obeyHere $g_i$ are the statistical weights of the two energy levels, and the field is unpolarized and isotropic. The units and physical dimensions areThe apparently time-independent dimension of $B$ follows because a per-hertz density includes an extra unit of time. If instead $B$ multiplies the mean intensity $J_{\nu }$, use $B^{(J)}=(4\pi /c)B$ and $A/B^{(J)}=2h{\nu }_0^3/c^2$. These distinct normalizations must not be mixed in the radiative transfer equation.

Use the energy-density convention of part (a), a normalized common spectral line profile $\int \varphi (\nu )d\nu =1$, and isotropic spontaneous emission. For a narrow spectral line centered at ${\nu }_0$,The factor $1/c$ follows from $u_{\nu }=4\pi J_{\nu }/c$: angle-integrated absorption removes $c{\kappa }_{\nu }{u}_{\nu }$ of energy per volume and time. The emission coefficient is per solid angle, whereas the absorption coefficient has units of inverse length.

The Boltzmann distribution in thermal equilibrium gives $n_2/n_1=(g_2/g_1)e^{-h{\nu }_0/(k_{B}T)}$. Dividing the two expressions and using the relations between Einstein coefficients,Thus the source function equals the Planck function at the transition frequency, and in the usual narrow-line approximationWith an exactly common finite-width $\varphi$ and constant line-center Einstein coefficients, their ratio is constant across the profile; exact equality to a varying $B_{\nu }$ at every frequency is only a narrow-line approximation. Exact thermal radiative transfer requires frequency-resolved detailed balance, which supplies the corresponding small profile corrections. This distinction does not alter the intended equilibrium result.

To derive fermionic Einstein-coefficient relations, first consider an idealized emitter that can exchange one ultrarelativistic fermion with the bath, with zero chemical potential. Let ${\mathcal{D}}_{\nu }=d4\pi {\nu }^2/c^3$ be its density of states per volume and ordinary frequency, with $d$ populated helicity states; for one neutrino helicity $d=1$, whereas the corresponding photon count is two. The Fermi-Dirac distribution givesPauli blocking changes the final-state factor in downward emission from the bosonic $1+f$ to $1-f$. Define positive coefficients $B_{12},C_{21}$ and $A_{21}=C_{21}h\nu {\mathcal{D}}_{\nu }$. The upward rate per lower emitter is $B_{12}{u}_{\nu }$, and the downward rate per upper emitter isThermal detailed balance, $n_1{B}_{12}{u}_{\nu }=n_2(A_{21}-C_{21}{u}_{\nu })$ with $n_2/n_1=(g_2/g_1)e^{-h\nu /(k_{B}T)}$, yieldsFor example, substituting $e^{h\nu /(k_{B}T)}=x$ reduces the balance equation to $B_{12}=(g_2/g_1)C_{21}$ because $x{f}_{\nu }=1-f_{\nu }$.

If one insists on the original additive notation $A_{21}+B_{21}{u}_{\nu }$, its formal coefficient is negative:A negative correction means inhibited emission into occupied modes, not a negative total decay rate; $0\le f_{\nu }\le 1$ ensures positivity. The dimensions are those of part (a), and the factor of two relative to the photon density is a helicity count, independent of the statistical sign.

There is a physical qualification: ordinary electronic energy levels of the same chemical atom have the same fermion parity, and cannot interconvert by emission or absorption of a single neutrino. A realizable single-fermion emitter must change the relevant particle-number and parity sector, with matching conservation laws. For ordinary atomic levels, neutrino-pair processes would require a different calculation. A bath with nonzero chemical potential would also require the corresponding chemical-equilibrium population ratio. The calculation above is the intended single-mode statistical analogy under explicit assumptions.

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.

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.