Question

1. Consider the case where a liquid is confined between two walls a distance D apart with one wall located at r--D/2 and the other at x- D/2. The walls carry a surface charge density of ? and the liquid has only counter-ions to the surface charge. The Poisson-Boltzman equation (PBE) for this case is where is the electric potential that gives the electric field E- d '/dr, no is the nunber density of counter-ions of valency at x = 0, e = 1.602 × 10-19 C is the fundamental charge, kB 1.38x 10-23 J/K is Boltzmann's constant. Here, e-???, where ?? 8.85x 10-12 F/m and ? is the dielectric constant of the liquid (e.g. ?water-80) (a) Consider the problem where there are two walls a distance D apart with one wall located at x =-D/2 and the other at x = D/2. Show by direct substitution that the exact solution to the PBE is given by kBT ze where 2K2 = K2 = (ze)210 where K is the Debye screening length. (b) Starting from Eq1), the Poiss-Boltzman equation, expand the exponential out to linear order and neglect higher order terms, which which is valid when zev/kBT<1 You should obtain an equation of the form where A and B are constants. Show that by defining a new potential ?-?-kBT/ze that this equation can be rewritten in the form dar2 Find expressions for the constants A, B, and C. Then use this equation to find a solution for ?| and ?. What is the spatial distribution of ions as a function of x? (c) What is the Debye length for a 0.1 M NaCl solution?

Answer 1

Given differential equation: - d^2 psi/dx^2 = -1/elementof zen_0 e^- ze psi/k_B T psi to be substituted: psi(x) = kB^T/ze ln(cos^2kx) where 2k^2 = (ze)^2 n_0/ elementof k_b T therefore d psi/dx = k_B T/ze d/dx (ln cos^2k x) = kB^T/ze(1/cos^2 kx) (- 2 cos kx sin kx) (k) = -2k k_B T/ze tan kx therefore d^2 psi/dx^2 = -2k k_B t/ze (sec^2 kx)(k) = -2k^2 k_B t/ze sec^2 kx Now 2k^2 = (ze)^2 n_0/elementof k_B T therefore d^2 psi/dx^2 = -(ze)^2 n_0/elementof k_B t k_B T/ze sec^2 kx = -ze n_0/elementof sec^2 kx \r\n Now RHS = -1/elementof ze n_0 exp[- ze k_B T/ze ln(cos^2 kx)/k_B T] = -1/elementof ze n_0 exp [- ln cos^2 kx] = -ze n_0/elementof (1/cos^2 kx) = -ze n_0/elementof sec^2 kx hence LHS = RHS given differential equation d^2 psi/dx^2 = -1/elementof ze n_0 e^- ze psi/k_B T