Assume a length of axon membrane of about 0.10 m is excited by an action potenti

Question

Assume a length of axon membrane of about 0.10 m is excited by an action potential (length excited = nerve speed ½ pulse duration = 50.0 m/s ½ 2.0 x 10-2 V? x 10-2 V, starting from the resting potential of ?7.0 x 10-2 V? C How many sodium ions (Na+) is this? Na+ ions (c) If it takes 2.0 ms for the Na+ ions to enter the axon, what is the average current in the axon wall in this process? ?A (d) How much energy does it take to raise the potential of the inner axon wall to +3.0 x 10-2 V from the resting state of ?7.0 X 10-2 V? K+ ions Is this a large charge per unit area? Hint: Calculate the charge per unit area in terms of electronic charge e per angstrom squared (?2). An atom has a cross section of about 1 ?2 (1 ? = 10?10 m). (Compare to normal atomic spacing of one atom every few ?.) (b) How much positive charge must flow through the cell membrane to reach the excited state of +3.0 x 10-2 V.) How many K+ ions are on the outside of the axon assuming an initial potential difference of 7.0 (a) Calculate the positive charge on the outside of a 0.10-m piece of axon when it is not conducting an electric pulse. (Assume an initial potential difference of 7.0 X 101 ?m, and cell-wall dielectric constant ? = 2.3. X 10-8 m, axon radius r = 2.0 x 10-3 s = 0.10 m). In the resting state, the outer surface of the axon wall is charged positively with K+ ions and the inner wall has an equal and opposite charge of negative organic ions, as shown in the figure below. Model the axon as a parallel-plate capacitor and take C = ??0A/d and Q = C?V to investigate the charge as follows. Use typical values for a cylindrical axon of cell wall thickness d = 2.0

Explanation / Answer

a) The area of each surface of this axom membrane is

A = L (2 pi r) = 0.1 * (2pi * 2 * 101 * 10-6) = 1.257 * 10-5 m2

C = K eoA / d = 2.3 * (8.85 * 10-12) * (1.257 * 10-5) / (2 * 10-8)

C = 1.279 * 10-8 F

1 ) In resting state ,the charge on the outer surface of the membrane is Qi = C Vi

= 1.279 * 10-8  * 7 * 10-2

Qi = 8.96 * 10-10 C

N(K) = Qi / e = 8.96 * 10-10 / (1.6 * 10-19)

= 5.6 * 109 K+ ions

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b) In the resting state ,the charge on the inner surface of the membrane is Qi =-8.96 * 10-10 C

Qf = C Vf =1.279 * 10-8 * 3 * 10-2 = 3.837 * 10-10 C

The total positive charge which must pass through the membrane to produce the excited state

delta Q = Qf - Qi = 3.837 * 10-10 - (-8.96 * 10-10)

delta Q = 1.28 * 10-9 C

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c)i) Number of Na+ ions =delta(Q) / e =1.28 * 10-9 / 1.6 * 10-19 = 7.998 * 109 Na+ ions

ii) I=delta(Q) / delta(t) =1.28 * 10-9 / 2 * 10-3 = 0.64 muA

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d) W = 0.5 C V2 = 0.5 * 1.279 * 10-8 * (3 * 10-2)2

W = 5.76 * 10-12 J

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