Suppose that the resistance between the walls of a biological cell is 6.8 × 109 ω. (a) what is the current when the potential difference between the walls is 69 mv? (b) if the current is composed of na+ ions (q = +e), how many such ions flow in 0.86 s?
(a) We can find the current flowing between the walls by using Ohm's law: [tex]I= \frac{\Delta V}{R} [/tex] where [tex]\Delta V=69 mV=0.069 V[/tex] is the potential difference and [tex]R=6.8\cdot 10^9 \Omega[/tex] is the resistance. Substituting these values, we get [tex]I=1.01 \cdot 10^{-11} A[/tex]
(b) The total charge flowing between the walls is the product between the current and the time interval: [tex]Q=I \Delta t[/tex] The problem says [tex]\Delta t=0.86 s[/tex], so the total charge is [tex]Q=(1.01\cdot 10^{-11} A)(0.86 s)=8.73 \cdot 10^{-12} C[/tex]
The current consists of Na+ ions, each of them having a charge of [tex]e=1.6 \cdot 10^{-19} C[/tex]. To find the number of ions flowing, we can simply divide the total charge by the charge of a single ion: [tex]N= \frac{Q}{e} = \frac{8.73 \cdot 10^{-12}C}{1.6 \cdot 10^{-19}C} = 5.45 \cdot 10^7 ions [/tex]
