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.

the answer is b, have negligible mass