Isotope shifts in atomic spectra are minute differences between the electronic energy levels of isotopes of the same element. They are the focus of a multitude of theoretical and experimental efforts due to their importance for atomic and nuclear physics. If atomic spectra also have
hyperfine structure, the shift refers to the
center of gravity of the spectra. From a nuclear physics perspective, isotope shifts combine different precise
atomic physics probes for studying
nuclear structure, and their main use is nuclear-model-independent determination of charge-radii differences. Two effects contribute to this shift:
Mass effects The mass difference (mass shift), which dominates the isotope shift of light elements. 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. 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: \tilde{\nu}_\infty = R_\infty \left(\frac{1}{n^2} - \frac{1}{n'^2} \right), where n and n' are principal quantum numbers, and R_\infty is
Rydberg constant. However, for a nucleus with finite mass M_N, reduced mass is used in the expression of Rydberg constant instead of mass of electron: \tilde{\nu} = \tilde{\nu}_\infty \frac{M_N}{m_e + M_N}. For two isotopes with
atomic mass approximately A' M_p and A'' M_p, the difference in the energies of the same transition is \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, A'' = 2A'. The effect of the specific mass shift was first observed in the spectrum of neon isotopes by
Nagaoka and Mishima. Consider the
kinetic energy operator in
Schrödinger equation of multi-electron atoms: T = \frac{p_n^2}{2M_N} + \sum_i \frac{p_i^2}{2m_e}, For a stationary atom, the conservation of momentum gives p_n = -\sum_i p_i. Therefore, the kinetic energy operator becomes 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 \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 \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 \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 1/M_N term in the expression, the specific mass shift also decrease as 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. 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 R = r_0 A^{1/3}, where
A is the atomic mass number, and 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 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 -\frac{Ze^2}{(4\pi\epsilon_0)r}: H' = \begin{cases} \dfrac{Ze^2}{(4\pi\epsilon_0)2R} \left( \dfrac{r^2}{R^2} + \dfrac{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 \Delta E = \langle \psi_{nlm} | H' | \psi_{nlm} \rangle. The wave function \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: \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 R is small, and within such a small region r \leq R, the approximation R_{nl}(r) \approx R_{nl}(0) is valid. And at r \approx 0, only the
s sublevel remains, so l = 0. Integration gives \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, |\psi_{n00}(0)|^2 = \frac{Z^3}{\pi a_\mu^3 n^3}, gives \Delta E \approx \frac{e^2}{(4\pi\epsilon_0)} \frac{2}{5} R^2 \frac{Z^4}{a_\mu^3 n^3}. In a real experiment, the difference of this energy shift of different isotopes \delta E is measured. These isotopes have nuclear radius difference \delta R. Differentiation of the above equation gives the first order in \delta R: \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. ==See also==