 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2/section-d/25/b/ii/solution

Here

r ≫ λ_{D}

, so the thin-double-layer result gives

μ_{e} = \frac{( 8.85 \times 1 0 ^{- 12} ) ( 78.5 ) ( 0.045 )}{1 0 ^{- 3}} = 3.13 \times 1 0^{- 8} m^{2} V^{- 1} s^{- 1},

and therefore $v = 3.13 \times 1 0^{- 5}$ m s $^{- 1}$ , along the field for the positive zeta potential.

In the opposite limit

r ≪ λ_{D}

μ_{e} = 2 ϵ ζ / (3 η)

, giving $2.08 \times 1 0^{- 8}$ m $^{2}$ V $^{- 1}$ s $^{- 1}$ , two-thirds of the thin-layer value at the same

ζ

. Both asymptotic expressions are independent of radius at fixed zeta potential; the crossover depends on

r / λ_{D}

. These classical approximations neglect double-layer relaxation and strong-potential corrections.

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