I recently found a publication that states extruded nanoparticles of roughly 120nm with +/- 1.21nm. This converts to a PDI of roughly 0.00011. Is this reliable? Or even possible?
I never saw any NPs owing so lowest PDI. I think it's impossible. If it does, what an amazing NP!
How do you define the polydispersity Index? If standart deviation devided by mean radius times 100, it would calculate in this case to roughly 1. This is very good, but not impossible.
Philip Born
The PDI reported in the Zetasizer software comes from the normalized second order term of the cumulant fitting. Assuming a Gaussian distribution the relation PDI = (sigma/peak mean)^2 holds.
The calculation is well-described in the international standards. For example ISO22412:2020 and ASTM E2490-09 (15).
http://www.materials-talks.com/blog/2015/03/31/pdi-from-an-individual-peak-in-dls/
For a standard polystyrene latex (e.g. ThermoFisher Scientific) then such a PDI value is possible. Also look at the NIST 100 nm standard (SRM 1963) then we have 101.8 nm ± 1.1 nm. See attached.
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