# 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?

• Less phonon scattering from the 'impurity' C13 atoms. I'm not equipped right now to dig up the original paper, but the predicted results matched quite well with reality. Mar 9 '17 at 12:46

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).