I chanced upon this 1D chain Mass Impurity model:

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At the end of all the derivations, it concludes that

Case 1: $ 0 < M_0 < M$

The impurity is lighter than the host atoms. The frequencies are shifted slightly upwards. An additional frequency is higher than the band. This is the localized mode. We can infer that when there are more impurity atoms, a band of localized modes will form.

Case 2: $ M_0 > M$

The impurity is heavier than the host. The frequencies are shifted downwards (unevenly). There are no localized modes. Due to the uneven shift, a resonance is created where the density of states (DOS) within the band is enhanced somewhere within the band.

I am curious about the effect of this on thermal conductivity. But I am not too sure how do I interpret the conclusions of the derivation.

Does it suggest that an impure atom with a larger mass is better at scattering phonons? Or would it be the other way?


1 Answer 1


There's not enough information to say.

It looks like your model has spring-like (harmonic) potentials between neighboring atoms, but phonon scattering requires anharmonic potentials. Harmonic potentials mean that the superposition principle still holds, so the phonons just pass right through each other without scattering.

Since you have harmonic potentials, phonon scattering simply doesn't exist even with the impurities. If you add in anharmonic potentials, the details of the scattering rates depend on which potentials you use, so I can't make any predictions.

That said, localized modes play an important role in trapping energy; energy that does get scattered into a localized mode gets stuck there for a while, and that reduces the thermal conductivity in a way that scattering into a non-localized mode doesn't. See for example: http://dx.doi.org/10.1063/1.4913826


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