Suppose that the resistance between the walls of a biological cell is 4.9 × 10^9 ohms (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 0.45 s?
R=4.9•10⁹ Ω
V=68•10⁻³V
I=V/R=68•10⁻³/4.9•10⁹=1.39•10⁻¹¹ A
I=q/t =Ne/t
N=It/e=1.39•10⁻¹¹•0.45/1.6•10⁻¹⁹=3.9•10⁷
To find the current (I) when the potential difference (V) between the walls of a biological cell is given, you can use Ohm's Law, which states that I = V/R, where R is the resistance.
(a) To calculate the current when the potential difference is given as 68 mV (0.068 V) and the resistance is 4.9 × 10^9 ohms, substitute these values into Ohm's Law:
I = V/R
I = 0.068 V / 4.9 × 10^9 ohms
Now, calculate the value:
I ≈ 1.39 × 10^-11 Amperes
So, the current is approximately 1.39 × 10^-11 Amperes.
(b) To determine the number of Na+ ions (q = +e) that flow in 0.45 seconds, you need to know the charge of a single Na+ ion (q) and the current (I). The charge of a single Na+ ion is +e, where e is the elementary charge (1.6 × 10^-19 Coulombs).
To find the number of Na+ ions, you can use the formula:
Number of ions = (Current × Time) / Charge
Substitute the values into the formula:
Number of ions = (1.39 × 10^-11 A × 0.45 s) / (1.6 × 10^-19 C)
Now, calculate the value:
Number of ions ≈ 3.9 × 10^7 ions
So, approximately 3.9 × 10^7 Na+ ions flow in 0.45 seconds.