I have nano powder material but i do not know the exact nm size, however it is understood nano powder consist of various different nm size, but if surface area of particle is known then can find the average nano size particles of particular nano powder material?
Abdul Rehman
'Nanopowder' is an oxymoron as there are not, and cannot be, free, discrete, independent particles < 100 nm in a powder.
If you have a SSA in m2/g then the conversion is to a mean size (Sauter Mean Diameter, SMD), for unit density, is D[3, 2] = 6/SSA (any good old fashioned particle text will show this derivation and the inherent assumptions of 3-fold symmetry) and will give the size directly in microns (10-6 m). So, at 25 m2/g, the (Sauter) mean particle size is 6/25 microns or 0.24 microns or 240 nm (i.e. sub-micron and not nano). See also the book chapter: Chapter Characterization of Nanomaterials
and the attachment and webinar (registration required):
Basic Principles of Particle Size Analysis
https://www.malvernpanalytical.com/en/learn/events-and-training/webinars/W170216BasicsParticleSize
Incidentally, Mike and Ron gave their names to the micron - it could have been the ronmike but was put in alphabetical order...
Abdul Rehman
'Nanopowder' is an oxymoron as there are not, and cannot be, free, discrete, independent particles < 100 nm in a powder.
If you have a SSA in m2/g then the conversion is to a mean size (Sauter Mean Diameter, SMD), for unit density, is D[3, 2] = 6/SSA (any good old fashioned particle text will show this derivation and the inherent assumptions of 3-fold symmetry) and will give the size directly in microns (10-6 m). So, at 25 m2/g, the (Sauter) mean particle size is 6/25 microns or 0.24 microns or 240 nm (i.e. sub-micron and not nano). See also the book chapter: Chapter Characterization of Nanomaterials
and the attachment and webinar (registration required):
Basic Principles of Particle Size Analysis
https://www.malvernpanalytical.com/en/learn/events-and-training/webinars/W170216BasicsParticleSize
Incidentally, Mike and Ron gave their names to the micron - it could have been the ronmike but was put in alphabetical order...
Please follow the link below:
Chapter Surface Area Analysis of Nanomaterials
Alan F Rawle Dear Sir Alan, thank you I got idea.. one more thin i would like to ask, a particular shape of nanoparticle also have specific size? I mean if particle shape changes than it also change in its size? or size will be same
Abdul Rehman
Depends what you mean by 'size' and 'shape'. There are many 'sizes' and 'shapes' for an irregular particle. In terms of volume or mass diameters then a shape (spherical) is assumed and if the volume or mass is constant, then the volume equivalent is the same. Look at the attachment and webinar. Note also we're concerned usually with distribution and this is the killer word. Mean or average values say nothing about a distribution.
Abdul Rehman
Imagine that nanoparticles have the shape of a ball, a cube, an ellipsoid, then their size will be respectively: radius or diameter, the size of the cube face, the values of the ellipse axes.
Dear Abdul Rehman ,
The value that you have is Q=SurfaceArea/Mass. Assuming identical spherical nanoparticles, as Mass=Volume x Density, it is easy to find that the radius of the nanoparticle is given by R=3/(Q x Density).
You need to know the density of your material to get a number. For instance, if those are silver nanoparticles, as the density of silver is 10.5 g/cm3, you obtain R=11nm. Of course, there is another assumption here: that the NP density is the same than that of the bulk. But at least you get a value for the NP radius.
Jose-Miguel Garcia-Martin Abdul Rehman
Yes, as I indicated in my answer the density of the particles should be taken into account [mean size (Sauter Mean Diameter, SMD), for unit density, is D[3, 2] = 6/SSA]. And, of course, diameter is twice the radius... I find it easier to describe the volume of a sphere as (pi/6)*d3 rather than the archaic (IMHO) (4/3)*pi*r3. As, for example, the volume of a 1 unit sphere is ~ 0.5 cubic units (pi/6 ~ 0.524). So the mass of a 1 unit diameter sphere of silver (your example) is [1/2]*10 or 5 cubic units (more exact 0.524*10.49 or 5.49 units)
Jose-Miguel Garcia-Martin You must be a biologist or protein chemist using radii! :-)) . Actually, sphericity isn't implied in the equation but rather 3-fold symmetry. The equation is correct for cubes too. Indeed, some catalyst engineers will use mean size = 5/SSA for supported metals implying a cube of the active catalyst (metal) with 5 faces exposed to the gas and 1 face inactive as it's 'glued' to the support.
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