Question

Suppose that the resistance between the walls of a biological cell is 4.2 × 109 Ω. (a) What is the current when the potential difference between the walls is 75 mV? (b) If the current is composed of Na+ ions (q = +e), how many such ions flow in 0.74 s?

Answer 1

Answer:

(a) Current will be $17.857\times 10^{-12}A$

(b) Number of ions will be $8.258\times 10^6$

Explanation:

We have given that resistance of the biological cell $R=4.2\times 10^9ohm$

(a) We have given potential difference of 75 mV

So $V=75\times 10^{-3}volt$

From ohm's law we know that current is given by

$i=\frac{V}{R}=\frac{75\times 10^{-3}}{4.2\times 10^9}=17.857\times 10^{-12}A$

(b) We have given time t = 0.74 sec

We have to find the charge

We know that charge is given by Q = it, here i is current and t is time  

So charge will be $Q=17.857\times 10^{-12}\times 0.74=13.214\times 10^{-12}C$

So number of ions will be $n=\frac{13.214\times 10^{-12}}{1.6\times 10^{-19}}=8.258\times 10^6$