Chapter 18, Problem 43P

Assume a length of axon membrane of about 0.10 m is excited by an action potential (length excited = nerve speed × pulse duration = 50.0 m/s × 2.0 × 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 Figure P18.43. Model the axon as a parallel-plate capacitor and take C = κϵ₀A/d and Q = C ΔV to investigate the charge as follows. Use typical values for a cylindrical axon of cell wall thickness d = 1.0 × 10⁻⁸ m, axon radius r = 1.0 × 10¹ μm, and cell-wall dielectric constant κ = 3.0. (a) Calculate the positive charge on the outside of a 0.10-m piece of axon when it is not conducting an electric pulse. How many K⁺ ions are on the outside of the axon assuming an initial potential difference of 7.0 × 10⁻² V? Is this a large charge per unit area? Hint: Calculate the charge per unit area in terms of electronic charge e per squared (Ų). An atom has a cross section of about 1 Ų (1 Å = 10⁻¹⁰ m). (b) How much positive charge must flow through the cell membrane to reach the excited state of + 3.0 × 10⁻² V from the resting state of −7.0 × 10⁻² V? How many sodium ions (Na⁺) is this? (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? (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?

Figure P18.43 Problem 43 and 44.

To determine

The positive charge on the outside of axon piece and number of $K^{+}$ ions.

Answer to Problem 43P

The positive charge on the outside of axon piece is $1.2\times {10}^{-9}\text{C}$. The number of $K^{+}$ ions is $7.5\times {10}^9$.

Explanation of Solution

Given info: The length of the axon membrane (l) is 0.10 m. Thickness of the cell wall is $d=1.0\times {10}^{-8}\text{m}$. Radius of the axon is $r=1.0\times {10}^1\text{μm}$. The cell wall dielectric constant is $\kappa =3$. The initial potential difference is $7.0\times {10}^{-2}\text{V}$.

Explanation:

Formula to calculate the charge is,

$Q=C\left(\Delta V\right)$

• C is the capacitance.
• $\Delta V$ is the potential difference.

Formula to calculate the capacitance is,

$C=\kappa \frac{{\epsilon }_{0}A}{d}$

• ${\epsilon }_0$ is the permittivity of free space.
• A is the cross sectional area.

The area A is given by,

$A=2\pi rl$

From the above equations,

$Q=\left(\kappa \frac{2\pi {\epsilon }_{0}rl}{d}\right)\left(\Delta V\right)$

Substitute $8.85\times {10}^{-12}{\text{m}}^{-3}{\text{kg}}^{-1}{\text{s}}^{\text{4}}{\text{A}}^{\text{2}}$ for ${\epsilon }_0$, $1.0\times {10}^1\text{μm}$ for r, 3 for $\kappa$, 0.10 m for l and $7.0\times {10}^{-2}\text{V}$ for $\Delta V$.

$\begin{array}{c}Q=\left(3\right)\left(8.85\times {10}^{-12}{\text{m}}^{-3}{\text{kg}}^{-1}{\text{s}}^{\text{4}}{\text{A}}^{\text{2}}\right)\left(\frac{2\pi }{1.0\times {10}^{-8}\text{m}}\right)\left(1.0\times {10}^1\text{μm}\right)\left(0.10\text{m}\right)\left(7.0\times {10}^{-2}\text{V}\right)\\ =\left(3\right)\left(8.85\times {10}^{-12}{\text{m}}^{-3}{\text{kg}}^{-1}{\text{s}}^{\text{4}}{\text{A}}^{\text{2}}\right)\left(\frac{2\pi }{1.0\times {10}^{-8}\text{m}}\right)\left(10\times {10}^{-6}\text{m}\right)\left(0.10\text{m}\right)\left(7.0\times {10}^{-2}\text{V}\right)\\ =1.2\times {10}^{-9}\text{C}\end{array}$

Formula to calculate the number of $K^{+}$ ions is,

$n=\frac{Q}{e}$

• e is the elementary charge.

Substitute $1.2\times {10}^{-9}\text{C}$ for Q and $1.6\times {10}^{-19}\text{C}$ for e.

$\begin{array}{c}n=\frac{1.2\times {10}^{-9}\text{C}}{1.6\times {10}^{-19}\text{C}}\\ =7.5\times {10}^9\end{array}$

The corresponding charge per unit area is low. This is because the magnitude of charge is very less.

Conclusion:

The positive charge on the outside of axon piece is $1.2\times {10}^{-9}\text{C}$. The number of $K^{+}$ ions is $7.5\times {10}^9$

To determine

The positive charge on the outside of axon piece and number of $N{a}^{+}$ ions at the excited state.

Answer to Problem 43P

The positive charge on the outside of axon piece is $1.7\times {10}^{-9}\text{C}$. The number of $N{a}^{+}$ ions is $1.0\times {10}^{10}$.

Explanation of Solution

Given info: The length of the axon membrane (l) is 0.10 m. Thickness of the cell wall is $d=1.0\times {10}^{-8}\text{m}$. Radius of the axon is $r=1.0\times {10}^1\text{μm}$. The cell wall dielectric constant is $\kappa =3$. The potential difference of the excited state is $3.0\times {10}^{-2}\text{V}$.

Explanation:

From (a),

$Q^{\text{'}}=\left(\kappa \frac{2\pi {\epsilon }_{0}rl}{d}\right)\left(\Delta V\right)$

The total positive charge that passes through membrane is,

$\Delta Q=Q^{\text{'}}-Q$

Therefore,

$\Delta Q=\left(\kappa \frac{2\pi {\epsilon }_{0}rl}{d}\right)\left(\Delta V\right)-Q$

Substitute $1.2\times {10}^{-9}\text{C}$ for Q, $8.85\times {10}^{-12}{\text{m}}^{-3}{\text{kg}}^{-1}{\text{s}}^{\text{4}}{\text{A}}^{\text{2}}$ for ${\epsilon }_0$, $1.0\times {10}^1\text{μm}$ for r, 3 for $\kappa$, 0.10 m for l and $3.0\times {10}^{-2}\text{V}$ for $\Delta V$.

$\begin{array}{c}\Delta Q=\left(3\right)\left(8.85\times {10}^{-12}{\text{m}}^{-3}{\text{kg}}^{-1}{\text{s}}^{\text{4}}{\text{A}}^{\text{2}}\right)\left(\frac{2\pi }{1.0\times {10}^{-8}\text{m}}\right)\left(1.0\times {10}^1\text{μm}\right)\left(0.10\text{m}\right)\left(3.0\times {10}^{-2}\text{V}\right)-\left(1.2\times {10}^{-9}\text{C}\right)\\ =\left(3\right)\left(8.85\times {10}^{-12}{\text{m}}^{-3}{\text{kg}}^{-1}{\text{s}}^{\text{4}}{\text{A}}^{\text{2}}\right)\left(\frac{2\pi }{1.0\times {10}^{-8}\text{m}}\right)\left(10\times {10}^{-6}\text{m}\right)\left(0.10\text{m}\right)\left(3.0\times {10}^{-2}\text{V}\right)-\left(1.2\times {10}^{-9}\text{C}\right)\\ =1.7\times {10}^{-9}\text{C}\end{array}$

Formula to calculate the number of $N{a}^{+}$ ions is,

$n=\frac{\Delta Q}{e}$

Substitute $1.7\times {10}^{-9}\text{C}$ for $\Delta Q$ and $1.6\times {10}^{-19}\text{C}$ for e.

$\begin{array}{c}n=\frac{1.7\times {10}^{-9}\text{C}}{1.6\times {10}^{-19}\text{C}}\\ =1.0\times {10}^{10}\end{array}$

The corresponding charge per unit area is low. This is because the magnitude of charge is very less.

Conclusion:

The positive charge on the outside of axon piece is $1.7\times {10}^{-9}\text{C}$. The number of $N{a}^{+}$ ions is $1.0\times {10}^{10}$.

To determine

The average current in the axon wall.

Answer to Problem 43P

The average current in the axon wall is $\text{0}\text{.83}\text{μA}$.

Explanation of Solution

Given info: The length of the axon membrane (l) is 0.10 m. Thickness of the cell wall is $d=1.0\times {10}^{-8}\text{m}$. Radius of the axon is $r=1.0\times {10}^1\text{μm}$. The cell wall dielectric constant is $\kappa =3$. The potential difference of the excited state is $3.0\times {10}^{-2}\text{V}$. Time taken for the $N{a}^{+}$ ions to enter the axon is ( $\Delta t$) 2.0 ms.

Explanation:

Formula to calculate the average current is,

$I=\frac{\Delta Q}{\Delta t}$

Substitute $1.7\times {10}^{-9}\text{C}$ for $\Delta Q$ and 2.0 ms for $\Delta t$.

$\begin{array}{c}I=\frac{1.7\times {10}^{-9}\text{C}}{2.0\text{ms}}\\ =\frac{1.7\times {10}^{-9}\text{C}}{2.0\times {10}^{-3}\text{s}}\\ =8.3\times {10}^{-7}\text{A}\\ \text{=}\text{0}\text{.83}\text{μA}\end{array}$

Conclusion:

The average current in the axon wall is $\text{0}\text{.83}\text{μA}$.

To determine

The energy required.

Answer to Problem 43P

The energy required is $7.5\times {10}^{-12}\text{J}$.

Explanation of Solution

Given info: The length of the axon membrane (l) is 0.10 m. Thickness of the cell wall is $d=1.0\times {10}^{-8}\text{m}$. Radius of the axon is $r=1.0\times {10}^1\text{μm}$. The cell wall dielectric constant is $\kappa =3$. The potential difference of the excited state is $3.0\times {10}^{-2}\text{V}$. Time taken for the $N{a}^{+}$ ions to enter the axon is ( $\Delta t$) 2.0 ms.

Explanation:

Formula to calculate the energy required is,

$E=\frac{1}{2}C{\left(\Delta V\right)}^2$

From the (a),

$C=\kappa \frac{2\pi {\epsilon }_{0}rl}{d}$

Therefore,

$\begin{array}{c}E=\frac{1}{2}\left(\kappa \frac{2\pi {\epsilon }_{0}rl}{d}\right){\left(\Delta V\right)}^2\\ =\left(\kappa \frac{\pi {\epsilon }_{0}rl}{d}\right){\left(\Delta V\right)}^2\end{array}$

Substitute $8.85\times {10}^{-12}{\text{m}}^{-3}{\text{kg}}^{-1}{\text{s}}^{\text{4}}{\text{A}}^{\text{2}}$ for ${\epsilon }_0$, $1.0\times {10}^1\text{μm}$ for r, 3 for $\kappa$, 0.10 m for l and $3.0\times {10}^{-2}\text{V}$ for $\Delta V$.

$\begin{array}{c}E=\left(3\right)\left(8.85\times {10}^{-12}{\text{m}}^{-3}{\text{kg}}^{-1}{\text{s}}^{\text{4}}{\text{A}}^{\text{2}}\right)\left(\frac{\pi }{1.0\times {10}^{-8}\text{m}}\right)\left(1.0\times {10}^1\text{μm}\right)\left(0.10\text{m}\right){\left(3.0\times {10}^{-2}\text{V}\right)}^2\\ =\left(3\right)\left(8.85\times {10}^{-12}{\text{m}}^{-3}{\text{kg}}^{-1}{\text{s}}^{\text{4}}{\text{A}}^{\text{2}}\right)\left(\frac{\pi }{1.0\times {10}^{-8}\text{m}}\right)\left(10\times {10}^{-6}\text{m}\right)\left(0.10\text{m}\right){\left(3.0\times {10}^{-2}\text{V}\right)}^2\\ =7.5\times {10}^{-12}\text{J}\end{array}$

Conclusion:

The energy required is $7.5\times {10}^{-12}\text{J}$.