Suppose that the resistance between the walls of a biologica

Suppose that the resistance between the walls of a biological cell is 3.0 × 109 . (a) What is the current when the potential difference between the walls is 68 mV? (b) If the current is composed of Na+ ions (q = +e), how many such ions flow in 0.63 s?

Solution

(a). Resistance R = 3 x 10⁹ and Potential difference V = 68 mV = 0.068 V

Current is, i = V / R = 0.068/3 x 10⁹ = 2.267 x 10 ^-11 A

(b). Time t = 0.63 s, so the charge is,

                 Q = It = 1.428e-11 C

Charge of electron e = 1.6 x 10 ^-19 C

Number of ions flow in 0.63 s is N = Q/e = 8.925e+7 ions