Suppose that the resistance between the walls of a biological cell is 3.9 * 109 Ω. (a) What is the current when the potential difference between the walls is 84 mV? (b) If the current is composed of Na + ions (q = + e), how many such ions flow in 0.73 s?

(a) 2.154*10⁻¹¹ A.

(b) 98300000.

Explanation:

(a)

Using Ohm's law,

V = IR ... Equation 1

Where V = Potential difference between the walls, I = current, R = Resistance between the walls.

Make I the subject of the equation

I = V/R ... Equation 2

Given: V = 84 mV, = 0.084 V, R = 3.9*10⁹ Ω.

Substitute into equation 2

I = 0.084 / (3.9*10⁹)

I = 2.154*10⁻¹¹ A.

(b)

I = q/t

q = It ... Equation 1

Where q = quantity of electric charge, t = time.

Given: I = 2.154*10⁻¹¹ A, t = 0.73 s.

q = 2.154*10⁻¹¹*0.73

q = 1.572*10⁻¹¹ C.

The charge on an electron e = 1.6*10⁻¹⁹ C

n = q/e

where n = number of ions.

n = 1.572*10⁻¹¹/1.6*10⁻¹⁹

n = 9.83*10⁷

n = 98300000.