Effects of the Nucleus on Atomic Structure

Overview

Until now we have discussed atoms as if the nuclei were point charges with no structure and infinite mass. Real nuclei have finite mass, possibly non-zero angular momentum, ${\bf {I}}$, and a charge which is spread out over a finite volume. As a result, they possess magnetic dipole moments and electric quadrupole moments coupled to the angular momentum, and possibly higher moments as well. All of these properties affect the atomic energy levels at a level about $10^{-5}$ rydberg. Here's a catalog of these effects:

Cause	Result	How Observed
Finite Mass	Mass Shift	Isotope shifts
Finite Volume of Charge	Volume Shift
Magnetic Dipole ($I\geq 1/2$)	Hyperfine structure	Energy splittings
Electric Quadrupole ($I\geq 1$)	Hyperfine structure

The first two effects produce only a small shift of the spectral line, and thus the only quantity accessible to measurement is the variation of the line position between different isotopes of the same element. (In atoms with only one or two electrons it may be possible to predict the position of a line with significant accuracy to deduce the isotope shift absolutely.) Laser spectroscopy makes it possible to measure isotope shifts to at least $10^{-9}$ Ry or $10^{-3}-10^{-4}$ of the shift. The moments of the nucleus are aligned with its spin and interact with the angular momentum of the rest of the atom. This couples the angular momenta J and I to the total angular momentum F: ${\bf {F}}={\bf {I}}+{\bf {J}}$, and leads to hyperfine splitting of atomic levels. The resulting hyperfine structure can be measured with almost limitless precision (certainly $<10^{-18}$ Rydberg) using the techniques of RF spectroscopy. Hyperfine transitions in Cs and H are currently the best available time and frequency standards. Generally speaking, magnetic dipole interactions predominate in atoms and electric quadrupole interactions in molecules. With the exception of the mass shift, the manifestations of nuclear structure in atomic spectra provide important information on the static properties of nuclei which are among the most precise information about nuclei. Unfortunately the great precision of the atomic measurements is generally lost in deducing information about nuclear structure because the core electrons affect the magnetic and electric interactions of the valence electrons with the nucleus.

Hyperfine interaction

Introduction

Hyperfine structure, as its name suggests, is extremely small on the scale of atomic interactions. To give an idea of its size, note that fine structure, which arises from relativistic effects and the spin- orbit interaction, is $O(\alpha ^{2})$ (i.e. of order $\alpha ^{2}$ on the scale of atomic interactions). The magnetic hyperfine interaction, which arises from the interaction of the nuclear magnetic moment with the surrounding electrons, is $O(\alpha ^{2}m_{e}/M_{p})$, which is approximately 2000 times smaller. Nevertheless, studies of hyperfine structure have played an important role in the determination of nuclear properties. Perhaps more relevant today is the role of hyperfine structure in many laser-atomic experiments, particularly those that involve manipulating atoms with light. This is because hyperfine structure affects the optical selection rules and the transfer of momentum and angular momentum to atoms by light. The magnetic hyperfine interaction is most important for atoms with an unpaired electron. Consequently, the alkali-metal atoms, which are the workhorses for laser-atomic physics, all display prominent hyperfine structure, sometimes to the despair of the experimenter.

The fact that the nucleus is a charge cloud with angular momentum suggests the possibility that it might possess magnetic and electric moments. Time reversal and parity invariance restrict the possible magnetic moments to dipole, octopole, ... and the possible electric moments to monopole ($Q=Ze$), quadrupole... The magnetic dipole and electric quadrupole interactions are dominant in the hyperfine interaction. The magnetic dipole moment can be measured only if the nucleus has $I\geq 1/2$, and it splits only those levels for which $J\geq$ 1/2. Similarly, the electric quadrupole interaction is observable only when $I$ and $J$ are both $\geq$ 1. Note that the units used in this section are Gaussian electro-static units (ESU).

Semi-classical analysis of the magnetic hyperfine interaction

The magnetic moment of the nucleus is generally expressed in terms of the nuclear magneton,

$\mu _{N}={\frac {e\hbar }{2M_{\rm {P}}c}}$

and the nuclear g-factor, $g_{I}$

$\mu =g_{I}\mu _{N}I$

In some cases it is convenient to express the nuclear moment in terms of the Bohr magneton. This is done by defning the g-factor $g_{i}^{\prime }$ by

$\mu =g_{I}^{\prime }\mu _{B}I$

To emphasize the fact that the nuclei are complex particles we note that the g-factors of the neutron and proton are

${\begin{aligned}g_{P}&=+5.586&g_{N}&=-3.826\end{aligned}}$

neither one of which is close to a simple integer. The magnetic moment of the nucleus couples to the magnetic field produced at the nucleus by the electrons in the atom. As a result ${\bf {J}}$ and ${\bf {I}}$ are coupled together to form ${\bf {F}}$, the total angular momentum of the entire atom

${\bf {F}}={\bf {J}}+{\bf {I}}$

The magnetic coupling between ${\bf {J}}$ and ${\bf {I}}$ adds a term to the Hamiltonian for the magnetic dipole hyperfine structure which is the interaction between a nucleus with magnetic moment ${\bf {\mu }}_{I}$, and the magnetic field ${\bf {B}}_{J}$ due to a single valence electron.

$H=-{\bf {\mu }}_{I}\cdot {\bf {B}}_{J}$

The electron's magnetic field is proportional to its angular momentum { ${\bf {J}}$ }, and so we can write

$H=ah{\bf {I}}\cdot {\bf {J}}$

Writing ${\bf {I}}\cdot {\bf {B}}_{J}$ as $({\bf {I}}\cdot {\bf {J}})({\bf {J}}\cdot {\bf {B}}_{J})/J^{2}$, we have

{EQ_cathree}

$ah=-{\frac {\mu _{I}}{I}}\langle {\frac {{\bf {J}}\cdot {\bf {B}}_{J}}{{\bf {J}}^{2}}}\rangle$

The constant $a$ is called the hyperfine coupling constant. By convention, it is written in units of frequency. There are two contributions to ${\bf {B}}_{J}$, orbital and spin: ${\bf {B}}_{J}={\bf {B}}_{L}+{\bf {B}}_{S}$. We shall first evaluate the fields classically. The magnetic moment of the electron is given by

${\bf {\mu }}_{e}=-g_{e}\mu _{B}{\bf {S}}$

where $\mu _{B}$ is the Bohr magneton. (The negative sign is taken by convention, so that $g_{e}\sim 2$ is a positive number.) Then,

${\bf {B}}_{L}(0)=\int {\frac {Id{\bf {s}}\times {\bf {\hat {r}}}}{{\bf {r}}^{2}}}\rightarrow -e{\bf {v}}\times {\left({\frac {-{\bf {r}}}{r^{3}}}\right)}=-2\mu _{B}{\frac {\bf {L}}{r^{3}}}$

${\bf {B}}_{S}(0)=-{\frac {1}{r^{3}}}{\left[{\bf {\mu }}_{e}-3({\bf {\mu }}_{e}\cdot {\bf {\hat {r}}}){\bf {\hat {r}}}\right]}=+g_{e}{\frac {\mu _{B}}{r^{3}}}{\left[{\bf {S}}-3({\bf {S}}\cdot {\bf {\hat {r}}}){\bf {\hat {r}}}\right]}$

{EQ_caseven}

${\bf {B}}_{J}(0)=-{\frac {2\mu _{B}}{r^{3}}}{\left[{\bf {L}}-{\frac {g_{e}}{2}}({\bf {S}}-3({\bf {S}}\cdot {\bf {\hat {r}}}){\bf {\hat {r}}})\right]}.$

We need to evaluate

$\langle {{\bf {J}}\cdot {\bf {B}}_{J}}\rangle =-{\frac {2\mu _{B}}{r^{3}}}[\ell (\ell +1)-S(S+1)+\langle {3({\bf {S}}\cdot {\bf {\hat {r}}}){\bf {\hat {r}}}\cdot ({\bf {L}}+{\bf {S}})}\rangle ]$

Using $\langle {({\bf {S}}\cdot {\bf {\hat {r}}})^{2}}\rangle =\langle {({\bf {\hat {\sigma }}}\cdot {\bf {\hat {n}}}/2)^{2}}\rangle =1/4$, ${\bf {\hat {r}}}\cdot {\bf {L}}=0$, and $S(S+1)=3/4$, we obtain

$\langle {\bf {J}}\cdot {\bf {B}}_{J}\rangle =-2\mu _{B}\langle {\frac {1}{r^{3}}}\rangle \ell (\ell +1)$

So after all this work, the field turns out to depend only on $\ell$. From Eq. \ref{EQ_cathree}, we obtain

$ah=+{\frac {g_{e}\mu _{I}\mu _{B}}{I}}\langle {\frac {1}{r^{3}}}\rangle {\frac {\ell (\ell +1)}{J(J+1)}}$

For a hydrogenic atom with $\ell >0$

$\langle {\frac {1}{r^{3}}}\rangle ={\frac {Z^{3}}{n^{3}}}{\frac {1}{\ell (\ell +1)(\ell +1/2)a_{0}^{3}}}$

Writing $\mu _{I}=g_{I}\mu _{N}I=g_{I}\mu _{B}(m_{e}/M_{p})I$, we finally obtain

$ah={\frac {g_{e}\mu _{B}Z^{3}g_{I}{\frac {m_{e}}{M_{p}}}}{n^{3}a_{0}^{3}}}{\frac {1}{(\ell +1/2)J(J+1)}}$

For hydrogen in the ground state, one obtains

$ah={\frac {4g_{e}}{3}}hcR\alpha ^{2}g_{I}{\frac {m_{e}}{M_{p}}}$

$S$-states require a separate treatment: $\left\langle 1/r^{3}\right\rangle$ diverges for $S$-states.

The orbital magnetic field is absent in $S$-states. However, the electron has finite probability of being at the origin and it must be regarded as a magnetic "cloud" with magnetization

${{\bf {M}}(r)}={\bf {\mu }}_{e}|{\Psi ({\bf {r}})}|^{2}$

The magnetization gives rise to a field at the origin

${\bf {B}}(0)={{\bf {H}}(0)}+4\pi {\bf {M}}(0)$

Decomposition of a spherically symmetric cloud of magnetization for finding the field at its center.

The magnetization can be viewed as the sum of a small uniform sphere at the origin, plus a hollow sphere containing the remainder of the magnetization. It is easily shown that the field due to the hollow sphere vanishes. However, the uniform sphere give rise to a finite value of ${\bf {H}}(0)$ due to an equivalent surface magnetic charge density

$\sigma _{m}={\bf {M}}(0)\cdot {\bf {\hat {n}}}={\bf {M}}(0)\cos \theta$

which acts as the source of { ${\bf {H}}$ }.

$H_{z}(0)=-\int {\frac {\sigma \cos \theta dS}{r^{2}}}=-{\frac {4\pi }{3}}M(0).$

Since ${\bf {M}}(0)={\bf {\mu }}_{e}\left|\psi (0)\right|^{2}=-g_{e}\mu _{B}{\bf {S}}\left|\psi (0)\right|^{2}$,

{EQ_caeighteen}

${\bf {B}}(0)=-{\frac {8\pi }{3}}g_{e}f\mu _{B}S\left|\Psi (0)\right|^{2}$

This $S$-state interaction is often called the "contact" (in the sense of touch) term. We can summarize these results by combining Eqs. \ref{EQ_caseven} and Eq. \ref{EQ_caeighteen} taking $g_{e}$ = 2:

{EQ_caninteen}

${\bf {B}}(0)=-2\mu _{B}{\left[{\frac {\bf {L}}{r^{3}}}-{\frac {\bf {S}}{r^{3}}}+{\frac {3({\bf {S}}\cdot {\bf {\hat {r}}}){\bf {\hat {r}}}}{r^{3}}}-{\frac {8}{3}}\pi \delta ({\bf {r}}){\bf {S}}\right]}$

The first three terms in the bracket average to zero in an $S$-state; the last term contributes only in an $s$-state.

Quantum mechanical treatment

The nucleus gives rise to the vector potential

${\bf {A}}_{I}={\frac {{\bf {\mu }}_{I}\times {\bf {r}}}{r^{3}}}$

The atomic Hamiltonian is

$H={\frac {1}{2m}}({\bf {p}}+e{\bf {A}}_{I})^{2}+V({\bf {r}})-{\bf {\mu }}_{e}\cdot (\nabla \times {{\bf {A}}_{I})}$

(We continue to use the convention that $e$ is the magnitude of the electron charge.) The last term describes the interaction of the electron and the magnetic field due to the nucleus. We can write $H=H_{0}+H_{\text{hf}}$, where, neglecting the term in $A_{I}^{2}$

{EQ_qmt3}

$H_{\text{hf}}={\frac {e}{2m}}({\bf {p}}\cdot {\bf {A}}_{I}+{\bf {A}}_{I}\cdot {\bf {p}})+g_{e}\mu _{B}{\bf {S}}\cdot (\nabla \times {\bf {A}}_{I})$

We can take ${\bf {A}}_{I}$ to obey the Coulomb gauge $\nabla \cdot {\bf {A}}_{I}=0$. Then the first term in Eq. \ref{EQ_qmt3} gives

${\bf {p}}\cdot {\bf {A}}_{I}={\frac {{\bf {p}}\cdot ({\bf {\mu }}_{I}\times {\bf {r}})}{r^{3}}}={\frac {{\bf {\mu }}_{I}\cdot ({\bf {r}}\times {\bf {p}})}{r^{3}}}={\frac {\hbar {\bf {\mu }}_{I}\cdot {\bf {L}}}{r^{3}}}$

and

$H_{\text{hf}}^{\text{orb}}={\frac {2\mu _{B}{\bf {\mu }}_{I}\cdot {\bf {L}}}{r^{3}}}$

The second term can be transformed using

$\nabla \times {\left[{\frac {{\bf {\mu }}_{I}\times {\bf {r}}}{r^{3}}}\right]}=-\nabla \times {\left[{\bf {\mu }}_{I}\times \nabla {\frac {1}{r}}\right]}$

Using this result, the second term in Eq. \ref{EQ_qmt3} can be written

$H_{\text{hf}}^{\text{spin}}=g_{e}\mu _{B}{\left[({\bf {S}}\cdot \nabla )({\bf {\mu }}_{I}\cdot \nabla {\frac {1}{r}})-{\frac {1}{3}}{\bf {S}}\cdot {\bf {\mu }}_{I}\nabla ^{2}{\frac {1}{r}}\right]}-{\frac {2}{3}}g_{e}\mu _{B}{\bf {S}}\cdot {\bf {\mu }}_{I}\nabla ^{2}{\frac {1}{r}}$

In a spherical state the average of the first factor in the bracketed term becomes

$\langle {({\bf {S}}\cdot \nabla )({\bf {\mu }}_{I}\cdot \nabla {\frac {1}{r}})}\rangle =\langle {\left(S_{x}\mu _{x}{\frac {\partial ^{2}}{\partial x^{2}}}+S_{y}\mu _{y}{\frac {\partial ^{2}}{\partial y^{2}}}+S_{z}\mu _{z}{\frac {\partial ^{2}}{\partial z^{2}}}\right)}\rangle$

plus cross terms which average to zero. Consequently,

$\langle {({\bf {S}}\cdot \nabla )({\bf {\mu }}_{I}\cdot \nabla {\frac {1}{r}})}\rangle ={\frac {1}{3}}{\bf {S}}\cdot {\bf {\mu }}_{I}\nabla ^{2}{\frac {1}{r}}$

which is cancelled by the second factor. For states that vanish at the origin,

$({\bf {S}}\cdot \nabla )({\bf {\mu }}_{I}\cdot \nabla {\frac {1}{r}})=-{\frac {{\bf {S}}\cdot {\bf {\mu }}_{I}}{r^{3}}}+{\frac {3({\bf {S}}\cdot {\hat {r}})({\bf {\mu }}_{I}\cdot {\hat {r}})}{r^{3}}}$

Taking $g_{e}=2$, we have

$H_{\text{hf}}^{\text{spin}}=-2\mu _{B}{\left[{\frac {{\bf {S}}\cdot {\bf {\mu }}_{I}}{r^{3}}}-{\frac {3({\bf {S}}\cdot {\hat {r}})({\bf {\mu }}_{I}\cdot {\hat {r}})}{r^{3}}}\right]}-16{\frac {\pi }{3}}\mu _{B}{\bf {S}}\cdot {\bf {\mu }}_{I}\delta ({\bf {r}}).$

where it is understood that the first term vanishes in $S$ states. We finally obtain

$H_{\text{hf}}=2\mu _{B}{\frac {\mu _{I}}{I}}{\bf {I}}\cdot {\left[{\frac {1}{r^{3}}}({\bf {L}}-{\bf {S}}+3({\bf {S}}\cdot {\bf {\hat {r}}})){\bf {\hat {r}}}-{\frac {8\pi }{3}}{\bf {S}}\delta ({\bf {r}})\right]}$

Writing this as $H_{hf}=ah{\bf {I}}\cdot {\bf {J}}$, we obtain the result found by the classical argument, Eq. \ref{EQ_caninteen}.

Hyperfine structure at zero magnetic field

The Hamiltonian at zero magnetic field is

$H=ah\;{\bf {I}}\cdot {\bf {J}}$

The total angular momentum is ${\bf {F}}={\bf {I}}+{\bf {J}}$. In zero or low field, $F$ and $m_{F}$ are good quantum numbers. A "good" quantum number is the eigenvalue of an operator that (approximately) commutes with the Hamiltonian, so that it is not perturbed by simple time evolution. At zero field, for instance, eigenfunctions of $H$ are eigenfunctions of ${\bf {F}}={\bf {I}}+{\bf {J}}$. Physically, ${\bf {I}}$ and ${\bf {J}}$ are tightly coupled by the $ah{\bf {I}}\cdot {\bf {J}}$ interaction: they precess about each other, and about ${\bf {F}}$. Using ${\bf {F}}^{2}=({\bf {I}}+{\bf {J}})^{2}$, we obtain

$\langle {{\bf {I}}\cdot {\bf {J}}}\rangle =[F(F+1)-J(J+1)-I(I+1)]/2$

and for the energy levels

$W(F,m)={\frac {ah}{2}}[F(F+1)-J(J+1)-I(I+1)].$

Note that $m=\left\langle F_{z}\right\rangle$ remains a good quantum number at all fields. (The symbols $F$ and $I$ are used both for operators and eigenvalues: the meaning is clear from the context). $F$ has values $|I-J|$, $|I-J+1|\cdots |I+J|$. The interval between adjacent terms is

$W(F)-W(F-1)=haF$

This result, known as the Land\'{e} Interval Rule, played an important role in the early development of the theory of hyperfine structure. Furthermore, it is easy to show that

$\sum _{F}(2F+1)W(F)=0$

so that the "center of gravity" of a hyperfine multiplet is zero.

Electric quadrupole interaction

If the nucleus does not have a spherically symmetric charge distribution, it probably has a non-zero electric quadrupole moment

$Q={\frac {1}{e}}\int d^{3}r\rho ({\bf {r}})[3z^{2}-r^{2}]$

which is $<0$ for an oblate charge distribution. In contrast to the nuclear magnetic dipole, which is predominantly determined by the unpaired nucleons, $Q$ is sensitive to collective deformations of the nucleus. Some nuclei are observed with 30 % differences between polar and equatorial axes, so $Q$ can be comparable to $\left\langle r^{2}\right\rangle$ , i.e.~Failed to parse (unknown function "\unit"): {\displaystyle \unit{\approx 10^{-24}}{\centi\metre\squared}} .

Classically, the quadrupole interaction energy is

Failed to parse (unknown function "\fract"): {\displaystyle E_2= - e Q \fract{del E_z}{del z} \fract{1}{8}(3 {\rm cos}^2\beta-1) }

The interaction energy of the quadrupole moment $Q$ with the electron can be found by expanding the term $\left|{\bf {r}}_{e}-{\bf {r}}_{N}\right|^{-1}$ in spherical harmonics and evaluating the resulting expressions in terms of Clebsch-Gordon coefficients. The resulting energy shifts are then

$E_{\text{hf}}^{\text{Q}}=BC(C+1)$

where

$B={\frac {3(Q/a_{0}^{2})}{8I(2I-1)J(J+1)}}\langle {\frac {1}{r^{3}}}\rangle R_{\infty }$

and

$C=[F(F+1)-J(J+1)-I(I+1)]$

[Note that ${\bf {I}}\cdot {\bf {J}}/\hbar ^{2}$, which was involved in $H_{\text{mag}}^{\text{hf}}$, is equal to $C/2$].

Note: The preceding expressions, like the corresponding ones for the magnetic interactions, have several significant omissions. The most important are relativistic corrections and core shielding corrections. Calculations of core shielding have been made by Sternheimer \cite{Sternheimer1967}, and the quadrupole shielding by the core is sometimes prefixed by his name.

Order of magnitude of hyperfine structure

If one examines the magnetic hyperfine structure in Eq.~\ref{EQ_caten}, it is obviously quite similar to the expression for the fine structure expression in Chapter 3. The ratio is

${\frac {E_{\text{mag}}^{\text{hfs}}}{E^{\text{fs}}}}\approx g_{I}\left({\frac {m}{M_{p}}}\right){\frac {1}{Z}}$

which is typically $10^{-3}$ to $10^{-4}$. For a neutral atom, one can estimate

Failed to parse (unknown function "\unit"): {\displaystyle \frac{E_\text{mag}^\text{hfs}}{\hbar} = \unit{(\ell + 3/4 )^{-3}}{\giga\hertz} }

with a factor of 10 spread in either direction. The quadrupole interaction is generally considerably smaller. An estimate is

Failed to parse (unknown function "\unit"): {\displaystyle \frac{B}{\hbar} \approx \unit{0.01 Z (\ell + 1/2 )^{-3}}{~GHz} }

Thus one generally expects that magnetic hyperfine structure dominates electric hyperfine structure in atoms. The opposite is generally true in molecules for two reasons: unpaired electrons are relatively rare, and the molecular binding mechanism can create large electric field gradients at the sites of the nuclei. In concluding this discussion of hyperfine structure in atoms I would like to point out that the preceding formulae are, except in hydrogenic atoms, only approximations and never permit one to extract the nuclear dipole or quadrupole moment with the full accuracy of laser spectroscopy experiment---let alone R.F.\ spectroscopy experiments. Thus $A$ and $B$ in the combined hyperfine energy formula

$E^{\text{hfs}}=E_{\text{mag}}^{\text{hfs}}+E_{\text{el}}^{\text{hfs}}={\frac {1}{2}}AC+BC(C+1)$

(with $C$ from Eq.~\ref{EQ_equadrfour}) should be regarded primarily as empirical constants from the standpoint of atomic physics. Even if the problems of connecting $A$ and $B$ with the nuclear moments could be solved, the principal result would be better measurements of nuclear properties.

Isotope Effects

When comparing the spectral lines originating from atoms whose nuclei differ only in the number of neutrons (eg. different isotopes of the same element), effects due to the finite mass and volume of the nucleus become apparent. Even neglecting hyperfine structure (by taking the center of gravity of the observed splitting), the spectral lines of the different isotopes vary slightly in position---generally at the many parts per million level. The difference between the lines of the various isotopes is referred to as the isotope shift: it is observed to have both positive (heavier isotope has higher energy spacing) and negative values. General speaking, light ($A<40$) elements have positive frequency shift whereas heavy elements ($A>60$) have negative shifts. This reflects the contribution of two distinct physical processes to the shift; the finite mass shift (almost always positive), and the nuclear volume shift (almost always negative). These will be discussed separately.

Mass effect

The origin of the mass effect is obvious from the Bohr energy level formula

$E_{n}=E_{n}^{0}\left({\frac {M}{m+M}}\right)\approx E_{n}^{0}\left(1-{\frac {m}{M}}\right)$

where the term involving $m/M$ comes from solving the two body system of electron (mass $m$) and nucleus (mass $M$) using the relative coordinate and associated reduced mass. Obviously increasing $M$ increases $E_{n}$. In two (or more) electron atoms the situation becomes more complicated due to the relative motion of the electrons. It would, for example, be possible to arrange the electrons symmetrically on opposite sides of the nucleus in which case there would be zero isotope effect. The virial theorem assures us that the mean value of the kinetic energy equals the negative of the total energy, so if we treat the nuclear motion as a perturbation on a fixed nucleus solution, the mass effect will be:

$\Delta E_{n,M}={\frac {-p^{2}}{2M}}=-{\frac {1}{2M}}{\left[\sum _{i}{\bf {p}}_{i}\right]^{2}}=-{\frac {m}{M}}\left[\underbrace {{\frac {1}{2m}}\sum p_{i}^{2}} _{\rm {NormalShift}}+\underbrace {{\frac {1}{2m}}\sum _{i\not =j}{\bf {p}}_{i}\cdot {\bf {p}}_{j}} _{\rm {SpecificShift}}\right]$

The first term is called the normal shift since (using the virial theorem again) it is

$\Delta E_{n,M}^{\text{Normal}}=-{\frac {m}{M}}E_{n}^{0}$

The second term is called specific because it depends on the atom's quantum state. A discussion can be found in Sobel'man (pp. 224-6). $\Delta E^{\text{Specific}}=0$ unless there are two or more valence electrons. For electronic configuration specified by quantum numbers $n,s,\ell$, Sobel'man finds:

$\Delta E_{ns,n^{\prime }\ell }^{\text{specific}}=(1-2S){\frac {m}{M}}{\frac {3f_{ns,n^{\prime }\ell }}{2}}\hbar \omega _{n^{\prime }s^{\prime },n\ell }$

where $3f_{ns,n^{\prime }\ell }$ is the oscillator strength (see Chapter~\ref{ch:E-field}). Thus the specific shift has opposite signs for $S=1$ and $S=0$ states---a reflection of the fact that the specific isotope shift is closely related to the exchange interaction. Eq.\ \ref{EQ_mefour} also reflects the general result that $\Delta E^{\text{Specific}}=0$ unless the two electrons are connected by an allowed dipole transition (otherwise $f$ will vanish). Furthermore the specific isotope shift is of the same order of magnitude as the normal isotope shift: for $f>2/3$, in fact, it can be larger (reversing the sign of the mass dependence of the isotope effect.) The preceding discussion shows that the fractional energy shift of a level due to the mass of the nucleus decreases rapidly with increasing mass of the nucleus. The normal part of this shift has a variation in the fractional magnitude due to a change $\Delta M$ in the mass of the isotope of

${\frac {\Delta E_{n,M+\Delta M}^{\text{Normal}}-\Delta E_{n,M}^{\text{Normal}}}{E_{n}}}={\left({\frac {m}{M+\Delta M}}-{\frac {m}{M}}\right)}={\frac {m}{M}}{\left({\frac {\Delta M}{M}}\right)}$

which decreases as $M^{-2}$, reaching 10 parts per million for a nucleus with $A=54$ (assuming $\Delta M$ = 1).

Volume effect

Inside the nucleus, the electrostatic potential no longer behaves like $Ze/r$, but is reduced from this value. If the valence electron(s) penetrate significantly into this region (eg. for $S$ electrons) then its energy will rise, relative to the value for a point nucleus, because of this reduced potential. Adding neutrons to the nucleus generally spreads out the charge distribution, causing a further rise in its energy. This reduction in the binding energy results in a decrease of the transition energy and therefore to a negative mass shift (assuming that the $S$ state is the lower energy state involved in the transition). \begin{figure} \centering

\caption{Simplified example of volume effect.}

\end{figure} For an $S$ state, the density of the electron probability distribution at the nucleus is given by the semi-empirical Fermi-Segr\'{e} formula \cite{Fermi1933}:

$\left|\Psi _{s}(0)\right|^{2}={\frac {Z_{a}^{2}Z}{\pi a_{0}^{2}n^{*3}}}\left(1+{\left|{\frac {\partial \delta _{s}}{\partial n}}\right|}\right)$

where $\delta _{s}$ is the quantum defect and $Z_{a}e$ the charge of the atomic core. Combining this with a model of the nuclear charge cloud results (Sobel'man p. 229) in the nuclear volume correction to the energy (of an $S$ electron):

$\Delta E_{n}^{V}=Z_{a}^{2}{\frac {R_{\infty }}{n^{*3}}}\left(1+\left|{\frac {\partial \delta _{s}}{\partial n}}\right|\right)C$

with

$C={\frac {4(\gamma +1)}{[\Gamma (2\gamma +1)]^{2}}}B(\gamma )\left({\frac {2Zr_{0}}{a_{0}}}\right)^{2\gamma }{\frac {\delta r_{0}}{r_{0}}}$

where

$\gamma =[1-\alpha ^{2}Z^{2}]^{1/2}$

$\Gamma$ is the gamma function $\Gamma (N+1)=N!$, $B(\gamma )$ is a factor which depends on the nuclear charge distribution. For a charged shell

$B(\gamma )=(2\gamma +1)^{-1}$

and for a uniform charge

$B(\gamma )=(2\gamma +1)^{-1}{\left({\frac {3}{2\gamma +3}}\right)}$

The nuclear radius is taken as (for atomic number $A$)

Failed to parse (unknown function "\unit"): {\displaystyle r_0 = \unit{1.15 \times 10^{-15} A^{1/3}}{\meter} }

so that

${\frac {\delta r_{0}}{r_{0}}}={\frac {\delta A}{3A}}$

There are obviously a number of assumptions in these equations, and they should not be expected to work as well as expressions for the nuclear mass shift. Sobel'man states that the observed shift is generally 1/2 to 3/4 of the one given above except for non-spherical nuclei (eg. rare-earth nuclei) which have anomalously large shifts.