Question

Cell membranes contain channels that allow K+ ions to leak out. Consider a channel that has a diameter of 1.47 nm and a length of 12.0 nm.

If the channel has a resistance of 18.5 GΩ,

what is the resistivity ? of the solution in the channel?

Answer 1

Resistance of a wire like structure is given by:

R = rho*L/A

rho = resistivity of solution in the channel = ?

So

rho = R*A/L

R = resistance of channel = 18.5 G-ohm = 18.5*10^9 ohm

A = Cross-sectional Area of channel = pi*d^2/4

d = diameter of channel = 1.47 nm = 1.47*10^-9 m

L = length of channel = 12.0 nm = 12.0*10^-9 m

So,

rho = R*pi*d^2/(4*L)

rho = 18.5*10^9*pi*(1.47*10^-9)^2/(4*12.0*10^-9)

rho = resistivity of solution = 2.62 ohm.m

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