The isotopic shift (also called isotope shift) is the shift in various forms of spectroscopy that occurs when one nuclear isotope is replaced by another.

NMR spectroscopy

In NMR spectroscopy, isotopic effects on chemical shifts are typically small, far less than 1 ppm, the typical unit for measuring shifts. The ¹
H NMR signals for ¹
H
₂ and ¹
H²
H ("HD") are readily distinguished in terms of their chemical shifts. The asymmetry of the signal for the "protio" impurity in CD
₂Cl
₂ arises from the differing chemical shifts of CDHCl
₂ and CH
₂Cl
₂.

Vibrational spectra

Isotopic shifts are best known and most widely used in vibration spectroscopy, where the shifts are large, being proportional to the ratio of the square root of the isotopic masses. In the case of hydrogen, the "H-D shift" is (1/2)^1/2 ≈ 1/1.41. Thus, the (totally symmetric) C−H and C−D vibrations for CH
₄ and CD
₄ occur at 2917 cm⁻¹ and 2109 cm⁻¹ respectively.^[1] This shift reflects the differing reduced mass for the affected bonds.

Atomic spectra

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.^[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:

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

${\displaystyle n}$

${\displaystyle n'}$

${\displaystyle R_{\infty ))$

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

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

${\displaystyle A''=2A'}$

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:

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

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

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

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

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

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

${\displaystyle 1/M_{N))$

${\displaystyle 1/M_{N}^{2))$

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

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

${\displaystyle -{\frac {Ze^{2)){(4\pi \epsilon _{0})r))}$

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

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

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

${\displaystyle \psi _{nlm}=R_{nl}(r)Y_{lm}(\theta ,\phi )}$

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

${\displaystyle R}$

${\displaystyle r\leq R}$

${\displaystyle R_{nl}(r)\approx R_{nl}(0)}$

${\displaystyle r\approx 0}$

${\displaystyle l=0}$

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

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 ${\displaystyle \delta R}$:

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