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Chapter 21: Problem 5

BI0 Signal Propagation in Neurons. Neurons are components of the nervous system of the body that transmit signals as electrical impulses travel along their length. These impulses propagate when charge suddenly rushes into and then out of a part of the neuron called an axon. Measurements have shown that, during the inflow part of this cycle, approximately \(5.6 \times 10^{11} \mathrm{Na}^{+}\) (sodium ions) per meter, each with charge \(+e\) , enter the axon. How many coulombs of charge enter a \(1.5-\mathrm{cm}\) length of the axon during this process?

Short Answer

Expert verified
The total charge is approximately \(1.35 \times 10^{-9}\) C.

Step by step solution

01

Understand the Problem

We are given a density of sodium ions entering a length of a neuron (axon) per meter, and we need to determine the total charge that enters a specific length of the axon.
02

Identify Given Values

We know that the number of sodium ions entering per meter is approximately \(5.6 \times 10^{11} \) Na\(^+\). Each ion carries a charge of \(+e\), where \(e = 1.602 \times 10^{-19}\) C (the charge of a single proton). We also have a length of the axon given as \(1.5\) cm.
03

Convert Length to Meters

Convert the given length from centimeters to meters to match the unit of the ion density. \[ 1.5\text{ cm} = 0.015\text{ m} \]
04

Calculate the Total Number of Ions

Calculate the total number of sodium ions entering the \(1.5\text{ cm}\) axon:\[\text{Number of ions} = 5.6 \times 10^{11} \text{ ions/m} \times 0.015 \text{ m} = 8.4 \times 10^{9} \text{ ions}\]
05

Calculate the Total Charge

Multiply the total number of ions by the charge of a single ion to find the total charge:\[\text{Total charge} = 8.4 \times 10^{9} \text{ ions} \times 1.602 \times 10^{-19} \text{ C/ion} = 1.34568 \times 10^{-9} \text{ C}\]
06

Present Final Answer

The total charge that enters a 1.5 cm length of the axon is approximately \(1.35 \times 10^{-9}\) C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

The Role of Neuron Electrical Impulses
In the world of biology, neurons play a crucial role in transmitting information throughout the body. These cells communicate by using electrical impulses, crucial for activities ranging from muscle movement to sensory perception. Neurons have a unique ability to generate and propagate electrical signals along their length, using changes in voltage to send messages from one part of the body to another. These impulses are called action potentials, and they occur when a neuron becomes "excited" by certain stimuli.

An electrical impulse in a neuron starts at the cell's body and travels down through the axon to reach its terminal. This journey of the signal occurs as charged ions move in and out of the neuron's membrane. Such ion movement changes the electrical charge of the neuron momentarily, allowing the impulse to "jump" to the next cell.
  • Neurons are specialized to carry electrical signals swiftly.
  • Action potentials are the key events that signify electrical impulses.
  • The movement of charged particles is essential for impulse propagation.
Understanding Axon Charge Transfer
The axon is a long, slender projection of a neuron, responsible for transmitting electrical signals over distances within the nervous system. The charge transfer along the axon is a vital step in this process, facilitated by mechanisms that allow ions to move across the axonal membrane.

During signal propagation, specific channels open in the axon's membrane, allowing sodium ions (Na\(^+\)) to rush in, increasing the positive charge inside the axon. This results in a shift in voltage, termed depolarization, which rapidly spreads along the axon. Following this depolarization, potassium ions exit the cell, restoring the axon's resting state, a process known as repolarization. This cycle is what supports the transmission of an action potential along the axon.

The coordinated opening and closing of ion channels are crucial, ensuring that electrical impulses can travel quickly and efficiently. The balance of ions inside and outside the axon changes with each signal, coordinating the symphony of neuron activity that underlies our thoughts and actions.
  • Axons serve as the conduits for electrical signals in neurons.
  • The movement of sodium and potassium ions facilitates charge transfer.
  • Depolarization and repolarization are critical for signal propagation.
The Importance of Sodium Ions in Neurons
Sodium ions (Na\(^+\)) are integral to the function of neurons, especially in the generation and propagation of electrical impulses. These ions, carrying a positive charge, are responsible for the initial surge of positive voltage in neurons when a signal is initiated.

At rest, a neuron maintains a higher concentration of sodium ions outside its membrane. When a stimulus reaches a neuron, sodium channels open, allowing sodium ions to flood into the cell. This rapid inflow creates an electrical spike known as depolarization, marking the start of the action potential.

The precise control of sodium ion movement is essential for proper neuronal communication. It ensures signals are transmitted accurately and efficiently, maintaining the body's well-being. Moreover, the movement of these ions is part of a bigger picture involving other ions like potassium, which play a complementary role in resetting the neuron's electrical state after a signal has passed.
  • Sodium ions are key players in neuron signal initiation.
  • The controlled entry of sodium ions leads to depolarization.
  • Neuronal health relies on the careful regulation of sodium ion flow.

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Most popular questions from this chapter

Two very large parallel sheets are 5.00 \(\mathrm{cm}\) apart. Sheet \(A\) carries a uniform surface charge density of \(-9.50 \mu \mathrm{C} / \mathrm{m}^{2},\) and sheet \(B,\) which is to the right of \(A,\) carries a uniform charge density of \(-11.6 \mu \mathrm{C} / \mathrm{m}^{2}\) . Assume the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) 4.00 \(\mathrm{cm}\) to the right of sheet \(A\) (b) 4.00 \(\mathrm{cm}\) to the left of sheet \(A ;(c) 4.00 \mathrm{cm}\) to the right of sheet \(B\) .

A point charge is at the origin. With this point charge as the source point, what is the unit vector \(r\) in the direction of (a) the field point at \(x=0, \quad y=-1.35 \mathrm{m}\) ; (b) the field point at \(x=12.0 \mathrm{cm}, y=12.0 \mathrm{cm} ;(\mathrm{c})\) the field point at \(x=-1.10 \mathrm{m},\) \(y=2.60 \mathrm{m} ?\) Express your results in terms of the unit vectors \(\hat{\boldsymbol{\imath}}\) and \(\hat{\boldsymbol{J.}}\)

CP Two positive point charges \(Q\) are held fixed on the \(x\) -axis at \(x=a\) and \(x=-a .\) A third positive point charge \(q,\) with mass \(m,\) is placed on the \(x\) -axis away from the origin at a coordinate \(x\) such that \(|x| << a\) . The charge \(q,\) which is free to move along the \(x\) -axis, is then released. (a) Find the frequency of oscillation of the charge \(q .\) (Hint: Review the definition of simple harmonic motion in Section \(14.2 .\) Use the binomial expansion \((1+z)^{n}=1+n z+n(n-1) z^{2} / 2+\cdots,\) valid for the case \(|z|<1 .\) (b) Suppose instead that the charge \(q\) were placed on the \(y\) -axis at a coordinate \(y\) such that \(|y| < < a\) , and then released. If this charge is free to move anywhere in the \(x y\) -plane, what will happen to it? Explain your answer.

(a) An electron is moving east in a uniform electric field of 1.50 \(\mathrm{N} / \mathrm{C}\) directed to the west. At point \(A,\) the velocity of the electron is \(4.50 \times 10^{5} \mathrm{m} / \mathrm{s}\) toward the east. What is the speed of the electron when it reaches point \(B, 0.375 \mathrm{m}\) east of point \(A\) ? (b) A proton is moving in the uniform electric field of part (a). At point \(A,\) the velocity of the proton is \(1.90 \times 10^{4} \mathrm{m} / \mathrm{s},\) east. What is the speed of the proton at point \(B\) ?

\(\mathrm{A}-5.00\) -nC point charge is on the \(x\) -axis at \(x=1.20 \mathrm{m.}\) A second point charge \(Q\) is on the \(x\) -axis at \(-0.600 \mathrm{m} .\) What must be the sign and magnitude of \(Q\) for the resultant electric field at the origin to be (a) 45.0 \(\mathrm{N} / \mathrm{C}\) in the \(+x\) -x-direction, (b) 45.0 \(\mathrm{N} / \mathrm{C}\) in the - \(x\) -direction?
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