I have now read on the Wikipedia pages for unbihexium, unbinilium, and copernicium that these elements will not behave similarly to their forebears because of “relativistic effects”. When I read about rutherfordium, it too brings up the relativistic effects, but only to say that it compared well with its predecessors, despite some calculations indicating it would behave differently, due to relativistic effects.

The dubnium page on Wikipedia says that dubnium breaks periodic trends, because of relativistic effects. The Wikipedia page on seaborgium doesn't even mention relativistic effects, only stating that it behaves as the heavier homologue to tungsten. Bohrium's Wikipedia page says it's a heavier homologue to rhenium.

So, what are these relativistic effects and why do they only take effect in superheavy nuclei? When I think of relativistic effects, I think speeds at or above $.9 c$ or near incredibly powerful gravitational forces. So, I fail to see how it comes into play here. Is it because the electrons have to travel at higher speeds due to larger orbits?

  • 1
    $\begingroup$ Yes. (Good guess) $\endgroup$
    – Al Brown
    Aug 13, 2021 at 2:00

4 Answers 4


When quantum mechanics was initially being developed, it was done so without taking into account Einstein's special theory of relativity. This meant that the chemical properties of elements were understood from a purely quantum mechanical description i.e., by solving the Schrödinger equation.

The more accurate models following that time, that do use special relativity, were found to be more consistent with experiment than compared with the ones that were used without special relativity.

So when they quote "relativistic effects" they are referring to chemical properties for elements calculated using special relativity.

Is it because the electrons have to travel at higher speeds due to larger orbits?

Changes to chemical properties of elements due to relativistic effects are more pronounced for the heavier elements in the periodic table because in these elements, electrons have speeds worthy of relativistic corrections. These corrections show properties that are more consistent with reality, than with those where a non-relativistic treatment is given.

A very good example of this would be the consideration of the color of the element gold, Au.

Physicist Arnold Sommerfeld calculated that, for an electron in a hydrogenic atom, its speed is given by $$v \approx (Zc)\alpha$$ where $Z$ is the atomic number, $c$ is the speed of light, and $$\alpha\approx\frac{1}{137}$$ is a (dimensionless) number called the fine structure constant or Sommerfeld's constant. For Au, since $Z= 79$, its outer shell electrons would be moving$^1$ at about $0.58c$. This means that relativistic effects will be pretty noticeable for gold$^2$, and these effects actually contribute to gold's color.

Interestingly, we also note from the above equation, that if $Z\gt 137$ then $v\gt c$ which would violate one of the postulates of special relativity, namely that no object can have a velocity greater than that for light. But it is also well known that no element can have atomic number $Z\gt 137$.

$^1$Electrons are not "moving around" a nucleus, but they are instead probability clouds surrounding the nucleus. So "most likely distances of electrons" would be a better term.

$^2$In the example of the element Gold, which has an electron configuration $$\bf \small 1s^2 \ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^6\ 5s^2\ 4d^{10}\ 5p^6\ 6s^1\ 4f^{14}\ 5d^{10}$$ relativistic affects will increase the $\bf \small 5d$ orbital distance from the nucleus, and also decrease the $\bf \small 6s$ orbital distance from the nucleus.

  • 1
    $\begingroup$ The estimate $v\approx (Zc)\alpha$ is valid for the ground state of a single-electron ion whose nucleus has $Z$ protons. Is it also valid for an outer-shell electron in a neutral atom with $Z$ electrons? If yes, then how would one deduce that? Is it because outer orbitals with low angular momentum (like S-orbitals) have a significant presence closer to the nucleus, where screening is less effective? $\endgroup$ Jun 17, 2021 at 4:03
  • 17
    $\begingroup$ I don't understand this fully - it basically says "electrons are moving at 0.58" - which causes relativistic effects - but then it says "not really moving", so how does this 0.58 come into play and how do things that don't "really move" have effects that are dependent on how fast things move? $\endgroup$ Jun 17, 2021 at 11:25
  • 3
    $\begingroup$ @htmlcoderexe I could be wrong, but my assumption is that the position with the highest probability changes over time, so that's what's moving with velocity $(Zc)\alpha$. $\endgroup$
    – chepner
    Jun 17, 2021 at 13:34
  • 14
    $\begingroup$ @htmlcoderexe: If you know the Schrödinger equation, you might have encountered the term $$-\frac{\hbar^2}{2m} \nabla^2 \psi.$$ This term is supposed to represent the kinetic energy of the particle, and it works because the momentum operator is $\vec{p} = i \hbar \vec{\nabla}$ and the kinetic energy is $K = p^2/2m$. But this expression for kinetic energy only applies when $v \ll c$; at relativistic speeds the kinetic energy would be expressed differently in terms of $p$, and the Schrödinger equation would change accordingly. $\endgroup$ Jun 17, 2021 at 14:56
  • 4
    $\begingroup$ @MaxW The inner shell is moving the fastest but that effects the shells above. For starters, the inner shell shrinks. $\endgroup$ Jun 18, 2021 at 22:59

So this is not a coincidence, but is in fact, rather fundamental. With the Heisenberg Uncertainty Principle:

$$ \sigma_p\sigma_x\ge \frac{\hbar}2$$

if a particle is confined to a space less than:

$$ \Delta x = \frac 1 2 \left(\frac{\hbar}{mc}\right)$$

then the uncertainty in energy becomes enough to create a particle antiparticle pair. That is "fully relativistic". At half that energy, we can say "relativistic effects are important". That is when the confinement is:

$$ \Delta x = \frac{\hbar}{mc} = \bar{\lambda}_c $$

which is the reduced Compton wavelength of the particle. It is a function of the (inverse) mass scaled by fundamental constants.

Because the proton mass is so much larger than the electron mass, we can discuss the hydrogen-like atom as if the reduced mass were basically $m_e$. With that, the Schrödinger equation is an eigenvalue equation relating the kinetic energy and potential energy with binding:

$$V(r) = \frac {Ze^2} {4\pi\epsilon_0}\times \frac 1 r$$

which can be rewritten in terms of the dimensionless fine structure constant,

$$\alpha =\frac 1{4\pi\epsilon_0}\frac{e^2}{\hbar c}\approx \frac 1{137}$$


$$ V(r) = Z\frac{\hbar c}r = \bar{\lambda}_cmc^2\frac Z r$$

The radial coordinate scales as $1/Z$, and indeed, the ground state solution has (see: Bohr radius, $a_0$) size:

$$ \Delta x= \frac{a_0}Z = \frac{\bar{\lambda}_c}{Z\alpha}$$

So the condition that things become relativistic is that $Z\alpha$ is approaching $1$, or in super-heavy nuclei.

At $Z=137$, relativity suggest "sparking of the vacuum". That is, the electric field near the nucleus is so strong that there is enough energy to create an electron-positron pair, which will quench the field.

That is fully relativistic. Relativistic effects become important even at a significantly smaller $Z$ than that. As a very general rule of thumb, they should be always considered starting at about half the absolute theoretical limit, i.e. for $Z\gtrsim 70$. See the example of Au ($Z=79$) in another answer to this very question by @josephh, for why the relativistic correction is required to explain the gold's color.

  • 4
    $\begingroup$ Shouldn't the last sentence be: "Relativistic effects become important at larger Z." $\endgroup$
    – MaxW
    Jun 21, 2021 at 19:39
  • $\begingroup$ @MaxW I read the whole paragraph, and took the “fully relativistic," probably practically unachievable regime of $Z=137$ as the comparison reference, i.e. $Z<137$, thus “smaller” is correct in this reading. But you are right, the last sentence may be read both ways, and could benefit from a little clarification. I'll submit an edit. $\endgroup$
    – kkm
    Feb 25 at 4:21

The Wikipedia page for copernicium has a hyperlink describing what "relativistic effects" means in this context: relativistic quantum chemistry.

These effects actually come into play well before the superheavy synthetic elements described in the question.

Two distinctive phenomena that can be explained by relativistic quantum chemistry include why gold has its characteristic yellowish color (instead of being grayish like other metals) and why lead, but not tin, can be used to build car batteries.


So, what are these relativistic effects and why do they only take effect in superheavy nuclei?

This is wrong. Relativistic effects do take effect on light atoms as well, and on every particle on the universe actually, it all depends on what the resolution of your measurements are. It's just that the effects are more pronounced in heavier atoms because higher bonding energies imply in higher "speeds".

For atoms, two major relativistic effects usually are taken into account: kinetic energy relativistic correction and spin–orbit coupling, which you can obtain a priori by calculating with Dirac's equation instead of the Schrödinger equation.

For example, the hydrogen atom has a non-corrected ground state energy of $-13.60569$ eV, and the relativistic correction due to kinetic energy is of order $-9 \times 10^{-4}$ eV.


Not the answer you're looking for? Browse other questions tagged or ask your own question.