An axon cable that connects the input and output ends of a human nerve cell is 5

Question

An axon cable that connects the input and output ends of a human nerve cell is 5.0 ×106 m in radius and 0.5 m long. The thickness of the membrane surrounding the fluid in the axon is 6 ×109 m, its resistivity is 1.6 ×107 m, and there is a 0.070-V potential difference across the membrane. Part A Determine the current through the membrane. Express your answer to two significant figures and include the appropriate units. I = SubmitMy AnswersGive Up Part B If the current is due to Na+ ions, how many of these ions leak across the membrane wall per second? Express your answer using two significant figures. N = SubmitMy AnswersGive Up Part C How does the current change if the membrane is twice as thick? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

A) thickness of membrane, L = 6 x 10^-9 m

Area, A = 2pi r l = 2 pi x 5 x 10^-6 x 0.5 = 1.57 x 10^-5 m^2

Resistance = resistivitty x Length / Area

= 1.6 x 10^7 x 6 x 10^-9 / 1.57 x 10^-5

= 6114.65 ohm

I = V/R = 0.07 / 6114.65 = 1.14 x 10^-5 A

B) I = q / t

so 1.14 x 10^-5 C charge per seocnd.

charge on single Na+ is 1.6 x 10^-19

no of ions = (1.14 x 10^-5 ) / (1.6 x 10^-19)

     = 7.15 x 10^13

C) if it is twice thick then resistance will doubled.

so current will be halved.

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