I have silver nanoparticles of sizes 68 nm and 6 nm. I want to remove the particles of size 6 nm. In order to do that I want to use centrifugation. What I need to know is how to find the G force for the centrifuge.
As,the rate of separation in a suspension of particles by way of gravitational force mainly depends on the particle size and density. Generally, particles of higher density or larger size typically travel at a faster rate and at some point will be separated from particles less dense or smaller. Thus, the sedimentation of larger particles takes place and the G force, can be explained by the Stokes equation, which describes the movement of a sphere in a gravitational field.
In continuation of the recommendations Anupma Thakur. Play around with the formula for determining the radius of nanoparticles in a centrifugal field with different speeds and find yourself what you need.
As explained, Stokes' Law gives you the engine for calculating the increase in g needed to centrifuge different sizes. Note that the terminal velocity is proportional to d2 in Stokes equation, so smaller particles will require increasing times and g.
Be aware that Stokes' Law applies to solid spherical particles. Thus irregularity or minor porosity will mean that the particles will stay in suspension longer than predicted than straight line (or curved in a disc centrifuge) settling. Your 6 nm particles will remain in free suspension so the only hope is to drive down the 68 nm (how did you get this value?) particles. As a rough guide in a CF3 instrument, then 2300 g ~ 4500 rpm and 100 nm is precipitated. At 14000 g (11250 rpm) not even 10 nm is driven down.
Thank you very much for your reply, Rawle. I used DLS for size determination.
Just increase ‘g’ in the Stokes’ formula. Obviously the centrifugal force is related to ‘g’.
10 nm can never centrifuged down. I’ve provided a couple of practical answers above.
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