Answer:
n/A = 1*10^10 /cm^2
Explanation:
a) Each open ion can be considered as a resistor in the circuit. Then, you have that the equivalent resistance is:
[tex]\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+...=\frac{n}{R}\\\\R_{eq}=\frac{R}{n}[/tex] (1)
(you assume the resistance are in parallel)
R: resistance of an axon's membrane = 1×10^ 11Ω.
Req: equivalent resistance = ?
n: number density of open ion channels
Next, you take into the Ohm's law:
[tex]I=\frac{V}{R_{eq}}[/tex] (2)
I: current
V: voltage acroos the membrane = 50mV
you replace the equation (1) into the equation (2):
[tex]I=\frac{Vn}{R}[/tex] (3)
But yo know that the current is also
I = JA (4)
J: current density = 5mA/cm^2
A: cross sectional area of the axon's membrane
You replace equation (4) into the equation (3), an d you solve for n/A
[tex]JA=\frac{Vn}{R}\\\\\frac{n}{A}=\frac{JR}{V}[/tex]
Then, you replace the values of the parameters:
[tex]\frac{n}{A}=\frac{JR}{V}=\frac{(5mA/cm^2)(1*10^{11}\Omega)}{50mV}\\\\\frac{n}{A}=1*10^{10}\frac{ions}{cm^2}[/tex]
hence, the density of ion channels is 1*10^10 /cm^2