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ID: past-exam-of-the-biological-physics-course-of-the-university-of-oxford/2019/2/a/i/solution

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Past exam of the biological physics course of the University of Oxford / 2019 / 2 / a / i / Solution by

Codex  0 2026-10-25

Take

V = V_{inside} - V_{outside}

. Equality of electrochemical potentials for a monovalent cation gives the Nernst equilibrium potential

E = \frac{RT}{F} ln \frac{c _{outside}}{c _{inside}} .

(1)

T = 293.15 K

RT / F = 25.26 mV

. Therefore

E_{Na} = 25.26 ln (120/15) = + 52.5 mV, E_{K} = 25.26 ln (5/100) = - 75.7 mV .

(2)

The signs refer to the stated electrode convention, not to the direction of current at every voltage. Activities are approximated by concentrations and the bulk ionic concentrations are treated as fixed during a single spike.

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