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.

1.15×105 A

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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.

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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.

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1.15×105 A

Explanation / Answer

the current through membrane is

I 1= V/R = V/ rho t/ pi r^2

= V pi r^2/ rho t

= 0.070-V * pi (5.0 ×106 m)^2 /1.6 ×10^7 m, 6 ×109 m

=5.74 * 10 ^-11 A

(b)

N = I/q = 5.74 * 10 ^-11 A/1.6 * 10 ^-19 C= 358750000

(c)

I2 = V/ 2rho t/ pi r^2

I2 = I1/2 =  5.74 * 10 ^-11 A/2 =  2.87* 10 ^-11 A

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