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# Sodium ions (Na+) are ﬂowing through a cylindrical ion channel, which has a diameter of 0.85 µm and is 5 µm long. ?

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?

### 2 Answers

- Dr WLv 72 months agoFavorite Answer
hints

.. (a).. 1 Na+ ion has a charge of 1.6x10^-19 C. you have 15x10^6 ions / ms

.. .. . . simple conversion to get C/ms

.. (b).. current has units of amps. 1 C = 1 amp/1s

.. (c).. V = IR

.. (d).. you know linear flowrate, you can easily convert to volumetric flowrate

.. . .. . .and from ions exiting, you can calculate density.

.. (e).. assume stead state. you know density, multiply by volume.

- 2 months ago
Solution:

a)

Q=qnV

I=qnAv=Q/change in time

n= V/N

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= Q/t=(2.4*10^{-12})/(0.001)

Note:t= 1ms = 0.001 sec

I=2.4*10^{-9}

c)

V=IR

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

R= V/I=(2.8*10^{-18})/(2.4*10^{-9})

R=1.167*10^{-9}

d)

density=m/V

(23*15x10^6)/(2.8*10^{-18})

d=123*10^{24}

e)

n=N/V=(15*10^{6})/(2.8*10^{-18})

n=5.35*10^{24}