Mass effects

The mass difference (mass shift), which dominates the isotope shift of light elements.^[2] It is traditionally divided to a normal mass shift (NMS) resulting from the change in the reduced electronic mass, and a specific mass shift (SMS), which is present in multi-electron atoms and ions.

The NMS is a purely kinematical effect, studied theoretically by Hughes and Eckart.^[3] It can be formulated as follows:

In a theoretical model of atom, which has a infinitely massive nucleus, the energy (in wavenumbers) of a transition can be calculated from Rydberg formula:

$${\displaystyle {\tilde {\nu }}_{\infty }=R_{\infty }\left({\frac {1}{n^{2}}}-{\frac {1}{n'^{2}}}\right),}$$

where ${\displaystyle n}$ and ${\displaystyle n'}$ are principal quantum numbers, and ${\displaystyle R_{\infty }}$ is Rydberg constant.

However, for a nucleus with finite mass ${\displaystyle M_{N}}$, reduced mass is used in the expression of Rydberg constant instead of mass of electron:

$${\displaystyle {\tilde {\nu }}={\tilde {\nu }}_{\infty }{\frac {M_{N}}{m_{e}+M_{N}}}.}$$

For two isotopes with atomic mass approximately ${\displaystyle A'M_{p}}$ and ${\displaystyle A''M_{p}}$, the difference in the energies of the same transition is

$${\displaystyle \Delta {\tilde {\nu }}={\tilde {\nu }}_{\infty }\left({\frac {1}{1+{\frac {m_{e}}{A''M_{p}}}}}-{\frac {1}{1+{\frac {m_{e}}{A'M_{p}}}}}\right)\approx {\tilde {\nu }}_{\infty }\left[1-{\frac {m_{e}}{A''M_{p}}}\left(1-{\frac {m_{e}}{A'M_{p}}}\right)\right]\approx {\frac {m_{e}}{M_{p}}}{\frac {A''-A'}{A'A''}}{\tilde {\nu }}_{\infty }.}$$

The above equations imply that such mass shift is greatest for hydrogen and deuterium, since their mass ratio is the largest, ${\displaystyle A''=2A'}$.

The effect of the specific mass shift was first observed in the spectrum of neon isotopes by Nagaoka and Mishima.^[4]

Consider the kinetic energy operator in Schrödinger equation of multi-electron atoms:

$${\displaystyle T={\frac {p_{n}^{2}}{2M_{N}}}+\sum _{i}{\frac {p_{i}^{2}}{2m_{e}}},}$$

For a stationary atom, the conservation of momentum gives

$${\displaystyle p_{n}=-\sum _{i}p_{i}.}$$

Therefore, the kinetic energy operator becomes

$${\displaystyle T={\frac {\left(\sum _{i}p_{i}\right)^{2}}{2M_{N}}}+{\frac {\sum _{i}p_{i}^{2}}{2m_{e}}}={\frac {\sum _{i}p_{i}^{2}}{2M_{N}}}+{\frac {1}{M_{N}}}\sum _{i>j}p_{i}\cdot p_{j}+{\frac {\sum _{i}p_{i}^{2}}{2m_{e}}}.}$$

Ignoring the second term, the rest two terms in equation can be combined, and original mass term need to be replaced by the reduced mass ${\displaystyle \mu ={\frac {m_{e}M_{N}}{m_{e}+M_{N}}}}$, which gives the normal mass shift formulated above.

The second term in the kinetic term gives an additional isotope shift in spectral lines known as specific mass shift, giving

$${\displaystyle {\frac {1}{M_{N}}}\sum _{i>j}p_{i}\cdot p_{j}=-{\frac {\hbar ^{2}}{M_{N}}}\sum _{i>j}\nabla _{i}\cdot \nabla _{j}.}$$

Using perturbation theory, the first-order energy shift can be calculated as

$${\displaystyle \Delta E=-{\frac {\hbar ^{2}}{M}}\sum _{i>j}\int \psi ^{*}\nabla _{i}\cdot \nabla _{j}\psi \,d^{3}r,}$$

which requires the knowledge of accurate many-electron wave function. Due to the ${\displaystyle 1/M_{N}}$ term in the expression, the specific mass shift also decrease as ${\displaystyle 1/M_{N}^{2}}$ as mass of nucleus increase, same as normal mass shift.

Volume effects

The volume difference (field shift) dominates the isotope shift of heavy elements. This difference induces a change in the electric charge distribution of the nucleus. The phenomenon was described theoretically by Pauli and Peierls.^[5][6][7] Adopting a simplified picture, the change in an energy level resulting from the volume difference is proportional to the change in total electron probability density at the origin times the mean-square charge radius difference.

For a simple nuclear model of an atom, the nuclear charge is distributed uniformly in a sphere with radius ${\displaystyle R=r_{0}A^{1/3}}$, where A is the atomic mass number, and ${\displaystyle r_{0}\approx 1.2\times 10^{-15}\ {\text{m}}}$ is a constant.

Similarly, calculating the electrostatic potential of an ideal charge density uniformly distributed in a sphere, the nuclear electrostatic potential is

$${\displaystyle V(r)={\begin{cases}{\dfrac {Ze^{2}}{(4\pi \epsilon _{0})2R}}\left({\dfrac {r^{2}}{R^{2}}}-3\right),&r\leq R,\\-{\dfrac {Ze^{2}}{(4\pi \epsilon _{0})r}},&r\geq R.\end{cases}}}$$

When the unperturbed Hamiltonian is subtracted, the perturbation is the difference of the potential in the above equation and Coulomb potential ${\displaystyle -{\frac {Ze^{2}}{(4\pi \epsilon _{0})r}}}$:

$${\displaystyle H'={\begin{cases}{\dfrac {Ze^{2}}{(4\pi \epsilon _{0})2R}}\left({\dfrac {r^{2}}{R^{2}}}+{\frac {2R}{r}}-3\right),&r\leq R,\\0,&r\geq R.\end{cases}}}$$

Such a perturbation of the atomic system neglects all other potential effect, like relativistic corrections. Using the perturbation theory (quantum mechanics), the first-order energy shift due to such perturbation is

$${\displaystyle \Delta E=\langle \psi _{nlm}|H'|\psi _{nlm}\rangle .}$$

The wave function ${\displaystyle \psi _{nlm}=R_{nl}(r)Y_{lm}(\theta ,\phi )}$ has radial and angular parts, but the perturbation has no angular dependence, so the spherical harmonic normalize integral over the unit sphere:

$${\displaystyle \Delta E={\frac {Ze^{2}}{(4\pi \epsilon _{0})2R}}\int _{0}^{R}|R_{nl}(r)|^{2}\left({\frac {r^{2}}{R^{2}}}+{\frac {2R}{r}}-3\right)r^{2}\,dr.}$$

Since the radius of nuclues ${\displaystyle R}$ is small, and within such a small region ${\displaystyle r\leq R}$, the approximation ${\displaystyle R_{nl}(r)\approx R_{nl}(0)}$ is valid. And at ${\displaystyle r\approx 0}$, only the s sublevel remains, so ${\displaystyle l=0}$. Integration gives

$${\displaystyle \Delta E\approx {\frac {Ze^{2}}{(4\pi \epsilon _{0})}}{\frac {R^{2}}{10}}|R_{n0}(0)|^{2}={\frac {Ze^{2}}{(4\pi \epsilon _{0})}}{\frac {2\pi }{5}}R^{2}|\psi _{n00}(0)|^{2}.}$$

The explicit form for hydrogenic wave function, ${\displaystyle |\psi _{n00}(0)|^{2}={\frac {Z^{3}}{\pi a_{\mu }^{3}n^{3}}}}$, gives

$${\displaystyle \Delta E\approx {\frac {e^{2}}{(4\pi \epsilon _{0})}}{\frac {2}{5}}R^{2}{\frac {Z^{4}}{a_{\mu }^{3}n^{3}}}.}$$

In an real experiment, the difference of this energy shift of different isotopes ${\displaystyle \delta E}$ is measured. These isotopes have nuclear radius difference ${\displaystyle \delta R}$. Differentiation of the above equation gives the first order in ${\displaystyle \delta R}$:

$${\displaystyle \delta E\approx {\frac {e^{2}}{(4\pi \epsilon _{0})}}{\frac {4}{5}}R^{2}{\frac {Z^{4}}{a_{\mu }^{3}n^{3}}}{\frac {\delta R}{R}}.}$$

This equation confirms that the volume effect is more significant for hydrogenic atoms with larger Z, which explains why volume effects dominates the isotope shift of heavy elements.