Assume a length of axon membrane of about 0.10 m is excited by an action potential (length excited = nerve speed x pulse duration = 50.0 m/s x 0.0020 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= x A/d and Q = CAV to investigate the charge as follows. Use typical values for a cylindrical axon of cell wall thickness d = 2.0 x 10-8 m, axon radius r = 1.6 x 10¹ μm, and cell-wall dielectric constant x = 2.9. Positive charge layer Negative { charge layer + + External fluid Axon wall membrane + Internal fluid - Axon radius= r Ⓡ (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 10-² V.) X Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. C How many K+ ions are on the outside of the axon assuming an initial potential difference of 7.0 x 10-² V? K+ ions How many sodium ions (Na+) is this? Na+ 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 (A2). An atom has a cross section of about 1 A² (1 A = 10-10 m). (Compare to normal atomic spacing of one atom every few A.) O Yes O No (b) How much positive charge must flow through the cell membrane to reach the excited state of +3.0 x 10-2 V from the resting state of -7.0 x 10-² V? с (c) If it takes 2.0 ms for the Nations to enter the axon, what is the average current in the axon wall in this process? μA (d) How much energy does it take to raise the potential of the inner axon wall to +3.0 x 10-2 V, starting from the resting potential of -7.0 x 10-2 V? (Assume that no energy is required to first raise the potential to 0 V from the resting potential of -7.0 x 10-² V.)
Assume a length of axon membrane of about 0.10 m is excited by an action potential (length excited = nerve speed x pulse duration = 50.0 m/s x 0.0020 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= x A/d and Q = CAV to investigate the charge as follows. Use typical values for a cylindrical axon of cell wall thickness d = 2.0 x 10-8 m, axon radius r = 1.6 x 10¹ μm, and cell-wall dielectric constant x = 2.9. Positive charge layer Negative { charge layer + + External fluid Axon wall membrane + Internal fluid - Axon radius= r Ⓡ (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 10-² V.) X Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. C How many K+ ions are on the outside of the axon assuming an initial potential difference of 7.0 x 10-² V? K+ ions How many sodium ions (Na+) is this? Na+ 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 (A2). An atom has a cross section of about 1 A² (1 A = 10-10 m). (Compare to normal atomic spacing of one atom every few A.) O Yes O No (b) How much positive charge must flow through the cell membrane to reach the excited state of +3.0 x 10-2 V from the resting state of -7.0 x 10-² V? с (c) If it takes 2.0 ms for the Nations to enter the axon, what is the average current in the axon wall in this process? μA (d) How much energy does it take to raise the potential of the inner axon wall to +3.0 x 10-2 V, starting from the resting potential of -7.0 x 10-2 V? (Assume that no energy is required to first raise the potential to 0 V from the resting potential of -7.0 x 10-² V.)
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Chapter1: Units, Trigonometry. And Vectors
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
Transcribed Image Text:Assume a length of axon membrane of about 0.10 m is excited by an action potential (length excited = nerve speed x pulse duration = 50.0 m/s x 0.0020 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 = ke A/d and Q = CAV to investigate the charge as follows. Use typical values for a cylindrical axon of cell wall thickness d =
2.0 × 10-8 m, axon
radius r = 1.6 × 10¹ µm, and cell-wall dielectric constant x = 2.9.
Positive
charge
layer
Negative
charge
layer
External fluid
Axon wall membrane
Internal fluid
Axon radius = r
No
+
(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 10-² v.)
d
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. C
How many K+ ions are on the outside of the axon assuming an initial potential difference of 7.0 × 10-² V?
K+ ions
How many sodium ions (Na+) is this?
Na+ 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 (Ų). An atom has a cross section of about 1 Ų (1 Å = 10−¹0 m). (Compare to normal atomic spacing of one atom every few Å.)
Yes
(b) How much positive charge must flow through the cell membrane to reach the excited state of +3.0 x 107 V from the resting state of -7.0 × 10-² V?
C
(c) If it takes 2.0 ms for the Nations to enter the axon, what the average current in the axon wall in this process?
μA
(d) How much energy does it take to raise the potential of the inner axon wall to +3.0 × 10-2 V, starting from the resting potential of -7.0 × 10-² V? (Assume that no energy is required to first raise the potential to 0 V from the resting potential of -7.0 × 10-² V.)
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