Sodium ions (Na+) are flowing thorugh a cylindrical ion channel which has a diam

Question

Sodium ions (Na+) are flowing thorugh a cylindrical ion channel which has a diameter of 0.85 um and is 5 um 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 and in a period of 1 ms a total of 15 x 10^6 ions exit the channel

a)What total charge exits the channel in a time 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?

Thank you ahead of time! :)

Explanation / Answer

(a)Q=qnV I

=qnAv=Q/change in time change in time = length/drift velocity

n=N/V

so for a. using the formula for

Q: q=1.6*10-19

N=15*106

V=l*PIr2

V= (5 um)(PI)(4.25*10-7)2=2.8*10-18

plugging in this value to the equation of Q,

: Q=2.4*10-12

(b) I=Q/t we know t = length/velocity so

i plug in giving me

=(5 um)(0.015 m/s)= 7.2*10^-9.= 2.4*10^-9..

so now that i have t, i just use the

(c) V=IR

V= (5 um)(PI)(4.25*10^-7)2=2.8*10-18

I=2.4*10-9..

R=V/I

=2.8*10^-18/2.4*10^-9..

=1.166 X1027

(d) D=M/V

=22.989769 g/mol x 15*106/2.8*10-18

=123.15x1024

(e) n=N/V

=15*10^6 / 2.8*10^-18

=5.35x1024

there may be calculation mistake but concept is right

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