Why are isotopically pure diamonds 50% more thermally conductive than other diamonds?

A laboratory grown type-IIa diamond (no nitrogen defects) has a lambda of $\pu{1800-2200 W/mK}$, but an isotopically pure diamond of $\ce{^{12}C}$ can have up to $\pu{3320 W/mK}$.


Why are $\ce{^{12}C}$ diamonds so much more thermally conductive than diamonds with $\approx \pu{1.1\%}$ $\ce{^{13}C}$ atoms?


Sourcing for my question:  

<https://www.sciencedirect.com/science/article/pii/092596359290197V?via%3Dihub>   

<https://adsabs.harvard.edu/abs/1990PhRvB..42.1104A>




From [*THERMAL CONDUCTIVITY OF NATURAL AND ISOTOPICALLY ENRICHED DIAMOND - EFFECT OF NEUTRON IRRADIATION, D.P. White, Department of Physics, Merrimack College, N. Andover, MA 01845*](http://web.ornl.gov/sci/physical_sciences_directorate/mst/fusionreactor/pdf/dec2001/07.3%20-%20134-138%20WHITE.pdf): 



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> The thermal conductivity was calculated using the Callaway method. This method takes into account the effects of intrinsic three-phonon normal (N) processes. These processes do not create thermal resistance directly but affect the thermal resistance by transferring phonons from frequencies where they are not scattered to frequencies where scattering is more efficient. 
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> In order to explain the large increase (originally much larger than expected) in terms of isotope scattering it was necessary to include the effects of N processes. Subsequent analyses of the thermal conductivity of enriched diamond have included N processes. Wei et.al. used the Callaway expression to fit experimental thermal conductivity data on diamond with the natural isotope concentration and isotopically enriched (0.1% $\ce{^{13}C}$) diamond over a wide temperature range (100-1000K).
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