In molecules of gaseous sodium chloride, the chloride ion has one more electron than proton, and the sodium ion has one more proton than electron. These ions are separated by about \(0.24 \mathrm{nm}\). How much work would be required to increase the distance between these ions to \(1.0 \mathrm{~cm} ?\)

Short Answer

Expert verified
Answer: The work required to increase the distance is approximately \(4.61 \times 10^{-18}\) J.

Step by step solution

01

Identify important values and constants

The charges of Sodium (Na+) and Chloride (Cl-) can be calculated by multiplying the elementary charge with the number of excess protons or electrons. We also need to take note of the given initial distance and final distance between the ions, as well as the constant for electrostatic force. Elementary charge : \(e = 1.6 \times 10^{-19} \mathrm{C}\) Initial distance (r_initial) : \(0.24 \times 10^{-9} \mathrm{m}\) Final distance (r_final): \(1.0 \times 10^{-2} \mathrm{m}\) Electrostatic constant (k) : \(8.99 \times 10^{9} \mathrm{N \cdot m^2 / C^2}\)
02

Calculate the electrostatic force between the ions

Using Coulomb's law, we can calculate the electrostatic force between the sodium and chloride ions: \(F = k \frac{q_1 \cdot q_2}{r^2}\) Where \(q_1\) and \(q_2\) are the charges of the sodium and chloride ions, respectively, and r is the distance between them. Since the charges are just 1 elementary unit apart, we can substitute the value of the elementary charge: \(F = k \frac{e^2}{r^2}\)
03

Calculate the work required to increase the distance

To find the work required to increase the distance between the ions, we can integrate the force with respect to the distance r: \(W = \int_{r_{initial}}^{r_{final}} F dr = \int_{r_{initial}}^{r_{final}} k \frac{e^2}{r^2} dr\) After integrating, we get: \(W = -k e^2 \left(\frac{1}{r_{final}} - \frac{1}{r_{initial}}\right)\)
04

Plug in values and calculate the work

Now we can plug in the values we gathered in Step 1 and solve for the work: \(W = - (8.99 \times 10^{9} \mathrm{N \cdot m^2 / C^2}) (1.6 \times 10^{-19} \mathrm{C})^2 \left(\frac{1}{1.0 \times 10^{-2} \mathrm{m}} - \frac{1}{0.24 \times 10^{-9} \mathrm{m}}\right)\) After calculating: \(W \approx 4.61 \times 10^{-18} \mathrm{J}\) Therefore, the work required to increase the distance between the sodium and chloride ions from 0.24 nm to 1.0 cm is approximately \(4.61 \times 10^{-18}\) J.

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Key Concepts

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

Coulomb's Law
To fully grasp the exercise involving the separation of ions, it is essential to understand Coulomb's Law. This foundational principle in physics describes the electrostatic interaction between electrically charged particles. The law states that the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

The equation for Coulomb's Law is beautifully simple: \[ F = k \frac{q_1 \cdot q_2}{r^2} \]Where F is the electrostatic force between the charges, k is Coulomb's constant (approximately \(8.99 \times 10^9 \mathrm{N \cdot m^2 / C^2}\)), q_1 and q_2 are the amounts of the charges, and r is the distance separating them. Understanding Coulomb's Law is crucial because it lets you determine the strength of the force that holds ions together or pushes them apart, which is at the heart of the original exercise.
Electrostatic Force
The electrostatic force is a fundamental aspect of the interaction between two charged particles. Derived from Coulomb's Law, this force can either be attractive or repulsive depending on the charges of the interacting particles: opposite charges attract, while like charges repel.

Electrostatic force is what keeps electrons in atoms orbiting around the nucleus, and it's also what makes it possible for ions of opposite charge to form stable compounds. In the original exercise, the electrostatic force is what initially holds the sodium and chloride ions together, and overcoming this force requires work.The significance of electrostatic force in the context of the exercise is that you are effectively doing work against this force to separate the ions, which is why understanding how this force behaves with changing distance is essential to solving the problem.
Energy Calculations in Physics
Energy calculations are a vital part of physics, especially when considering work done on or by a system. Work is defined as the process of transferring energy to an object via the application of a force that causes the object to move.

In the context of our exercise, the work required to separate the ions is directly related to the electrostatic potential energy, which is energy stored due to the positions of charged objects relative to each other. The formula for work done in moving a charge in an electric field is:\[ W = \int F \, dr \]In this scenario, we are integrating the electrostatic force over the distance the ions are separated. The negative sign indicates that work is done against the electrostatic force. It’s the integration and understanding of how energy changes with the distance in an electrostatic field that allow us to calculate the work needed to increase the distance between the ions.
Elementary Charge
The elementary charge is a constant that represents the smallest unit of electric charge that is possible and is denoted by the symbol \(e\). This value is fundamental in physics because it is the charge of a single proton (or the negative of the charge of a single electron), and it plays a vital role in various physical phenomena and equations, including Coulomb's Law.

The value of the elementary charge is approximately \(1.6 \times 10^{-19}\) Coulombs (C). In our original problem, the charges of the sodium and chloride ions are multiples of the elementary charge, and this constant is critical for calculating the force and work associated with moving charged particles across a distance.

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Most popular questions from this chapter

Consider an electron in the ground state of the hydrogen atom, separated from the proton by a distance of \(0.0529 \mathrm{nm}\) a) Viewing the electron as a satellite orbiting the proton in the electrostatic potential, calculate the speed of the electron in its orbit. b) Calculate an effective escape speed for the electron. c) Calculate the energy of an electron having this speed, and from it determine the energy that must be given to the electron to ionize the hydrogen atom.

Suppose that an electron inside a cathode ray tube starts from rest and is accelerated by the tube's voltage of \(21.9 \mathrm{kV}\). What is the speed (in \(\mathrm{km} / \mathrm{s}\) ) with which the electron (mass \(=9.11 \cdot 10^{-31} \mathrm{~kg}\) ) hits the screen of the tube?

Two metal spheres of radii \(r_{1}=10.0 \mathrm{~cm}\) and \(r_{2}=\) \(20.0 \mathrm{~cm},\) respectively, have been positively charged so that both have a total charge of \(100, \mu C\) a) What is the ratio of their surface charge distributions? b) If the two spheres are connected by a copper wire, how much charge flows through the wire before the system reaches equilibrium?

The electric potential energy of a continuous charge distribution can be found in a way similar to that used for systems of point charges in Section \(23.6,\) by breaking the distribution up into suitable pieces. Find the electric potential energy of an arbitrary spherically symmetrical charge distribution, \(\rho(r) .\) Do not assume that \(\rho(r)\) represents a point charge, that it is constant, that it is piecewise-constant, or that it does or does not end at any finite radius, \(r\). Your expression must cover all possibilities. Your expression may include an integral or integrals that cannot be evaluated without knowing the specific form of \(\rho(r) .\) (Hint: A spherical pearl is built up of thin layers of nacre added one by one.)

Using Gauss's Law and the relation between electric potential and electric field, show that the potential outside a uniformly charged sphere is identical to the potential of a point charge placed at the center of the sphere and equal to the total charge of the sphere. What is the potential at the surface of the sphere? How does the potential change if the charge distribution is not uniform but has spherical (radial) symmetry?

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