41
$\begingroup$

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?

$\endgroup$
0
37
$\begingroup$

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, and these effects actually contribute to gold's color.

$^1$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.

$\endgroup$
14
  • 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? $\endgroup$ – Chiral Anomaly Jun 17 at 4:03
  • 16
    $\begingroup$ 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? $\endgroup$ – htmlcoderexe Jun 17 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 at 13:34
  • 13
    $\begingroup$ @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. $\endgroup$ – Michael Seifert Jun 17 at 14:56
  • 3
    $\begingroup$ @MaxW The inner shell is moving the fastest but that effects the shells above. For starters, the inner shell shrinks. $\endgroup$ – Loren Pechtel Jun 18 at 22:59
14
$\begingroup$

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,

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

as

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

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 at smaller $Z$.

$\endgroup$
1
  • 3
    $\begingroup$ Shouldn't the last sentence be: "Relativistic effects become important at larger Z." $\endgroup$ – MaxW Jun 21 at 19:39
10
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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