Baphomet
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Measuring Mean Particle Size
If one possesses a suspension or colloid with unknown average particle size, what kind of apparatus could be rigged-up to measure it?
I'm thinking along the lines of shooting a laser though a glass tube filled with the colloid and measuring the refraction
[Edited on 28-9-2007 by Baphomet]
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Ozone
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You would be measuring the scattering, or most conveniently, the light output orthogonal to the source. The amount of scattering is dependant both on
the size and quantity (and if equivalent, mass) of the particles and the wavelength of your laser. Blue scatters more than red, but has a tendency to
be absorbed (ergo, doing other things).
Check out dynamic light scattering (DLS, multiple angle light scattering, MALS), Rayleigh scattering, etc.
You would shoot the laser into your sample and, in the most trivial case, measure the light output at 90°. You would need (to be accurate) to split
the beam and measure the incident light simultaneously.
So far as I know,
O3
-Anyone who never made a mistake never tried anything new.
--Albert Einstein
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Baphomet
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Thanks for your help! After looking up light scattering I was able to find this patent that sounds very similar to what you are describing
http://www.freepatentsonline.com/4274740.html
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DrP
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Light scattering is the most modern way (I think), but you could also get a rough idea by putting your sample through a SEM or TEM with a graticule
near your sample. This will allow you to 'check' your light scattering results as I not sure how well light scattering takes into account the particle
size distribution. At least you will be able to actually see how consistantly your particles sized.
Regards,
P.
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Baphomet
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Thanks.. might see if the uni down the road could run that for me.
I've had an idea: what if it's possible to obtain some prepared colloids of known average particle size (e.g. one each of 1 micron, 10 micron, 100
micron) and use these to calibrate the homemade device?
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microcosmicus
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You could also measure you particles using elutriation.
The idea behind this method is that since your particle
is in a viscous medium, it will not fall freely, but
will attain a terminal velocity (think parachute), which
will depend on the size of the particle. The only apparatus
required is a ruler and a clock --- you simply time how long
it takes for the particles to fall a certain distance inside
the fluid.
To elucidate the elutration and obtain particle size from
measured velocity, we make use of some hydrodynamics. The
buoyant force on the particle may be obtained by a formula
of Archimedes, the drag of the fluid by a formula of Stokes.
When the particle is coasting at terminal velocity, its
acceleration is zero so the gravitational, buoyant, and drag
forces acting upon it must balance each other, so we obtain
the following formula:
h V = (2/9) (d1 - d2) g r^2
Here, the symbols have the following meaning:
V -- terminal velocity
d1 -- density of particle
d2 -- density of medium
g -- acceleration due to gravity
r -- radius of particle
h -- coefficient of viscosity
Let me illustrate this with a real-life example. Upon mixing
manganese sulfate and sodium hypochlorite, I precipitated
some manganese dioxide according to the reaction
Mn2+ + ClO– + 2OH– → MnO2 + H2O + Cl–
(By the way, the reason for doing this was to synthesize some
sodium permanganate(upon further oxidation, the dioxide
turns into permangante) as illustrated on W. Oelen's
"Science made Alive" website and discussed here in the
permanganate thread --- DerAlte's idea of extracting with
cold dry acetone sounds quite clever.) Anyway, I noted that
the precipitate dropped out of the liquid at a velocity of
about a millimeter per minute. Looking up some data and
converting to common units, we then have the following
values for the quantities appearing in the above equation:
V -- 0.0015 cm/sec
d1 -- 5.0 g/cc
d2 -- 1.0 g/cc
g -- 980 cm/sec^2
h -- 0.01 g/cm.sec
Solving for r, we conclude that the particle has a radius of
1.4E-4 cm, a.k.a. 1.4 microns.
[Edited on 31-12-2007 by microcosmicus]
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