 Solution

ID: past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2019/paper-1/section-b/10/iv/b/solution

I = \frac{1}{2} \sum_{i} c_{i} z_{i}^{2}

, measured in

mol L^{- 1}

κ^{- 1} = (\frac{ε _{r} ε _{0} RT}{2 F ^{2} ( 1000 ) I})^{1/2} .

Taking water's relative permittivity as approximately

78.5

2 5^{\circ} C

I = 0.10 M

for NaCl. Therefore $κ^{- 1} ≃ 0.96 nm$ . This is an exponential screening length, not a hard outer edge of the diffuse layer. The factor of

1000

converts molarity into

mol m^{- 3}

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