I've been skimming over the literature and have come into some conflicting information about the optical properties of partially deuterated polymers. From what I've seen, in general deuterated polymers have less attenuation of visible and IR than their non-deuterated equivalents. But I've seen some papers remarking that partially deuterated polymers also have improved optical properties while others reporting that partial deuteration increases opacity versus non-deuterated equivalents.

Is there some sort of general rule, or does it depend on the polymer in question? The underlying mechanisms governing transparency in polymers, beyond the basics such as crystallinity, are something I haven't studied in depth.


1 Answer 1


Deuteration lowers the vibrational frequencies C-H to C-D, but does not affect C-C etc vibrations. Thus some lowering of vibrational frequencies will make the ir region slightly more transparent by bunching more frequencies together. In the UV deuteration hardy affects the spectrum and so here the polymer should be pretty much the same. Unless there is something in the processing of polymers that is improved by deuteration I can't think of anyhthing else.


Having read your comments & looked at the data and thought more about it I now think that its the Franck-Condon factors that are responsible. These measure the strength of a transition and for electronic transitions in visible and uv are all proportional to $exp(-\alpha (\Delta R^2/2))$ where $\alpha = \sqrt(\mu k)/\hbar$ where $k$ is the force constant, (same for CH and CD bonds) and $\mu$ is reduced mass 1.7u for CD and 0.9u for CH so the deuterated bonds will be less intense as the exponential will be smaller for CD than CH. ($\Delta R$ is the displacement between levels in ground state and excited state). In the ir the intensity is proportional to the transition frequency multiplied by the square of $ \int \psi_i x \psi _{i+1} dx$ where x is the displacement from equilibrium bond length and $\psi$ a vibrational wavefunction with i quanta. Evaluating the integral shows that a transition is proportional to $\nu ^3$ where $\nu $ is the transition frequency between levels with i to i+1 quanta. The intensity is thus reduced by substituting D for H $via$ $\nu = \frac {1}{2\pi}\sqrt{\frac{k}{\mu}}$.

As CH/CD vibrations are of higher frequency than CC and the effect should be most pronounced at higher ir frequencies and proportional to the amount of deuterated species. Typically deuterated transition frequencies are 0.715 of the protonated values compared to the expected $1/\sqrt(2)$ =0.707, considered to be due to effects off cubic and quartic terms in the potential. This means that the protonated transitions are 2.7 times more intense than deuterated at approx 2900 cm$^{-1}$, the CH frequency in methane.

  • $\begingroup$ Thanks for giving an answer. :) However, it's not just a slightly higher transparency (vis. and IR), it's a quite massive difference. For example, page 10, figure 2 here for attenuation rates in PMMA fiber: tinyurl.com/j9sgmun - d8 is fully deuterated, while d5 is partially deuterated. I'm having trouble finding the reference that I read earlier that talked about reduced transparency in some cases, unfortunately... I think it had to do with changing the crystal size. Or maybe I misread it... $\endgroup$
    – KarenRei
    Jul 25, 2016 at 22:58
  • $\begingroup$ I'm just trying to get a handle on what one can "expect" with X% deuteration of a transparent, high degree of crystallinity plastic or X% deuteration of a transparent, amorphous plastic, where X is, say, 0%, 2%, 20%, or 100%. $\endgroup$
    – KarenRei
    Jul 25, 2016 at 23:03
  • $\begingroup$ Excellent answer. But I think you're still missing one aspect (I'll be choosing your answer regardless, I'm just trying to get a grip on the complete picture :) ). A major factor concerning transparency in polymers is the degree and nature of its crystallinity. Low crystallinity polymers tend to be transparent while high crystallinity polymers tend to be opaque, as light scatters at the crystal/amorphous boundaries. But crystal size matters, too, as the scattering (Bragg scattering, right?) is limited by wavelength. How might deuteration alter crystallinity (size and degree)? $\endgroup$
    – KarenRei
    Jul 26, 2016 at 23:01
  • $\begingroup$ Scattering in general is due to the different refractive index of one medium compared to another, e.g. polymer beads are visible in a beaker, but fill with water, or a solvent with a closer refractive index to the beads and they 'disappear'. So if the crystalline defects have a different refractive index than the bulk polymer then you expect scattering. I'm speculating now as I've no experience in polymer fabrication, but the deuterated polymers may be better for a simple reason; they are expensive to make so the starting material may be purer and more care may be taken over its manufacture. $\endgroup$
    – porphyrin
    Jul 27, 2016 at 16:51
  • $\begingroup$ Diffraction really does require some order in the material, such as a $periodic$ change in refractive index. Light is not diffracted from crystals as are x-rays. Only if a sufficient number of the crystals were ordered in a periodic way in your polymer would diffraction occur. A random order and orientation would just give light scattering. $\endgroup$
    – porphyrin
    Jul 27, 2016 at 17:00

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