3
$\begingroup$

I read somewhere that the energy is around $U=-\frac{3\alpha I}{4r^6}$ where $\alpha$ is the polarizability and $I$ is the ionization energy but I'm obtaining way too large energies for a lithium crystal.

The shortest distance between two lithium atoms in the crystal is $304 pm$

The first ionization energy of lithium is $520.2 kJ/mol$

I've found various polarizability constants online in papers but I can't really figure them out or the units they are using.

Also, I suspect that I want the dispersion force between $\ce{Li^+}$ ions and not neutral $\ce{Li}$ atoms.

$\endgroup$
2
$\begingroup$

Polarisability as used here has dimensions of volume. The London formula is $$U \approx -\frac{3}2\frac{I_aI_b}{I_a+I_b}\frac{\alpha _a \alpha _b}{r^6}$$ for two species a and b so you just need to square $\alpha$.

$\endgroup$
  • $\begingroup$ So, for a polarizability constant of $24\times 10^{-30}$ which I'm not sure is correct I got a total energy of 290kJ/mol which still seems too high. Is this correct? $\endgroup$ – Steven Stewart-Gallus Jul 25 '16 at 22:22
  • $\begingroup$ still seems very high $\endgroup$ – porphyrin Jul 26 '16 at 5:58
  • 1
    $\begingroup$ having looked at some values the polarisability of 24 seems very large, atoms seem to have values not bigger than 10 although I have not done a thorough check on this. $\endgroup$ – porphyrin Jul 26 '16 at 8:45
  • $\begingroup$ The source at arxiv.org/abs/quant-ph/0506106 says about 24 and the source says other sources say 22 and 24. Maybe they mean a different kind of polarizability? $\endgroup$ – Steven Stewart-Gallus Jul 26 '16 at 19:01
  • $\begingroup$ Volume polarisability is correct. I had a look again and found values for benzene of 25 .10$^{-24}$ ml so that is consistent with your values in m$^3$. Carbon tetrachloride has a value 10.5 for example. ( I had only looked at nitrogen (1.73) and water (1.44), before, apologies). An interaction energy of approx 24000 cm$^{-1}$ is still huge, its as large as an electronic excitation, but I can't see what wrong :( The ionisation energy and bond length seem ok . $\endgroup$ – porphyrin Jul 27 '16 at 10:59

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.