Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 114

The potential difference across a cell membrane from outside to inside is initially at \(-90 \mathrm{mV}\) (when in its resting phase). When a stimulus is applied, Na" ions are allowed to move into the cell such that the potential changes to \(+20 \mathrm{mV}\) for a short amount of time. (a) If the membrane capacitance per unit area is $1 \mu \mathrm{F} / \mathrm{cm}^{2},\( how much charge moves through a membrane of area \)0.05 \mathrm{cm}^{2} ?\( (b) The charge on \)\mathrm{Na}^{+}\( is \)+e$ How many ions move through the membrane?

Short Answer

Expert verified
How many sodium ions (Na+) move through the membrane, given that the charge on Na+ is +e? Answer: The amount of charge that moves through the membrane is 5.5 x 10^-9 C. Approximately 3.4 x 10^10 sodium ions move through the membrane.

Step by step solution

01

Calculate the change in potential difference

First, calculate the change in potential difference (voltage) by finding the difference between the final voltage (+20 mV) and the initial voltage (-90 mV): ΔV = V_final - V_initial ΔV = (20 - (-90)) mV ΔV = 110 mV
02

Convert potential difference to volts

Since we need to work in standard units (SI units), we need to convert the potential difference from millivolts (mV) to volts (V): ΔV = 110 mV * (1 V / 1000 mV) = 0.110 V
03

Calculate the charge moved through the membrane

Now, we can use the capacitance formula to find the charge (Q) that moves through the membrane: Q = C * ΔV Given the membrane capacitance per unit area (C/A) is 1 μF/cm² and the area (A) is 0.05 cm², we can find the total capacitance (C) as follows: C = (C/A) * A C = (1 μF/cm²) * (0.05 cm²) C = 0.05 μF Now, convert the capacitance to standard units (farads) by multiplying by 1x10⁻⁶ C = 0.05 μF * (1 F / 10⁶ μF) = 5 x 10^-8 F Finally, calculate the charge (Q): Q = C * ΔV Q = (5 x 10^-8 F) * (0.110 V) Q = 5.5 x 10^-9 C Thus, the charge that moves through the membrane is 5.5 x 10^-9 C.
04

Calculate the number of sodium ions

Now that we have found the charge that moves through the membrane, we can find the number of sodium ions (N) by dividing the total charge (Q) by the charge of a single Na+ ion (+e = 1.6 x 10^-19 C): N = Q / e N = (5.5 x 10^-9 C) / (1.6 x 10^-19 C/ion) N ≈ 3.4 x 10^10 ions Thus, approximately 3.4 x 10^10 sodium ions move through the membrane.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

By rewriting each unit in terms of kilograms, meters, seconds, and coulombs, show that \(1 \mathrm{N} / \mathrm{C}=1 \mathrm{V} / \mathrm{m}\)
In capacitive electrostimulation, electrodes are placed on opposite sides of a limb. A potential difference is applied to the electrodes, which is believed to be beneficial in treating bone defects and breaks. If the capacitance is measured to be \(0.59 \mathrm{pF},\) the electrodes are \(4.0 \mathrm{cm}^{2}\) in area, and the limb is \(3.0 \mathrm{cm}\) in diameter, what is the (average) dielectric constant of the tissue in the limb?
A tiny hole is made in the center of the negatively and positively charged plates of a capacitor, allowing a beam of electrons to pass through and emerge from the far side. If \(40.0 \mathrm{V}\) are applied across the capacitor plates and the electrons enter through the hole in the negatively charged plate with a speed of \(2.50 \times 10^{6} \mathrm{m} / \mathrm{s}\) what is the speed of the electrons as they emerge from the hole in the positive plate?
Point \(P\) is at a potential of \(500.0 \mathrm{kV}\) and point \(S\) is at a potential of \(200.0 \mathrm{kV} .\) The space between these points is evacuated. When a charge of \(+2 e\) moves from \(P\) to \(S\), by how much does its kinetic energy change?
Suppose you were to wrap the Moon in aluminum foil and place a charge \(Q\) on it. What is the capacitance of the Moon in this case? [Hint: It is not necessary to have two oppositely charged conductors to have a capacitor. Use the definition of potential for a spherical conductor and the definition of capacitance to get your answer. \(]\)
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Electricity

Read Explanation

Solid State Physics

Read Explanation

Radiation

Read Explanation

Fluids

Read Explanation

Famous Physicists

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free