Answer to Question #153648 in Electric Circuits for Jerick

Question

Sodium ions (Na⁺) are flowing through a cylindrical ion channel, which has a diameter of 0.85 µm and is 5 µm long. There is a potential difference of 225 mV between the ends of the channel. The sodium ions have a drift velocity through the channel of 0.015 m s⁻¹ and in a period of 1 ms a total of 15×10⁶ ions exit the channel.

(a) What total charge exits the channel in a period of 1 ms?

(b) What is the current in the ion channel?

(c) What is the ‘resistance’ of the channel to the flow of sodium ions?

(d) What is the number density, n, of ions in the channel?

(e) How many sodium ions are in the channel at any one time?

Answer 1

Solution:

a)

$Q=qnV$

$I=qnAv=\dfrac{Q}{change \;in\;time}=\dfrac{length}{drift \;velocity}$

$n=\dfrac{N}{V}$

$q=1.6*10^{-19}$

$N=15*10^{6}$

$V=I*Plr^2=(5µm )(PI)(4.25*10^{-7})=2.8*10^{-18}$

by plugging in this value to the equation of Q, I will get:

$Q=2.4*10^{-12}$

b)

$I=\dfrac{Q}{t}=\dfrac{2.4*10^{-12}}{0.001}$

$I=2.4*10^{-9}$

t= 1ms = 0.001 sec

c)

$V=IR$

$V=(5µm )(PI)(4.25*10^{-7})^2=2.8*10^{-18}$

$R=\dfrac{V}{I}=\dfrac{2.8*10^{-18}}{2.4*10^{-9}}$

$R=1.167*10^{-9}$

d)

$density=\dfrac{m}{V}=\dfrac{22.98\;\frac{g}{mole}\cdot15*10^6}{2.8*10^{-18}}$

$d=123*10^{24}$

e)

$n=\dfrac{N}{V}=\dfrac{15*10^{6}}{2.8*10^{-18}}$

$n=5.35*10^{24}$