Spherical particles of a protein of density 2.0 g/cm3 are shaken up in a solution of 20°C water. The solution is allowed to stand for 1.0 h. If the depth of water in the tube is 3.7 cm, find the radius of the largest particles that remain in solution at the end of the hour.

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  1. ρ =2 g/cm³= 2•10³kg/m³
    The terminal speed of the particle is
    v<h/t =0.037/3600=1.0228•10^-5 m/s.
    The terminal speed of the particle of density ρ and radius R which is falling down in the water od density ρ1 =1000 kg/m³ and viscosity η is
    v=2•R²•g• (ρ - ρ1)/9•η.
    R=sqrt[9• η•v/2•g•(ρ - ρ1)] =
    = sqrt[9•0.001•1.0228•10^-5/2•9.8•1•10³]=
    =2.17•10^-6 m =2.17 μm

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