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Chapter 21: Problem 38

In gaseous sodium chloride, chloride ions have one more electron than they have protons, and sodium ions have one more proton than they have electrons. These ions are separated by about \(0.24 \mathrm{nm}\). Suppose a free electron is located \(0.48 \mathrm{nm}\) above the midpoint of the sodium chloride molecule. What are the magnitude and the direction of the electrostatic force the molecule exerts on it?

Short Answer

Expert verified
Answer: The magnitude of the net electrostatic force is equal to the net vertical component: \(F_{net} = F_{vertical}\). The direction of the force is towards the sodium ion (Na+).

Step by step solution

01

Identify the known variables

We are given the following information: 1. Distance between the chloride ion (Cl-) and sodium ion (Na+): \(0.24 \mathrm{nm}\) 2. Distance between the free electron and the midpoint of the NaCl molecule: \(0.48 \mathrm{nm}\) 3. Charge of an electron or proton: \(e = 1.6 \times 10^{-19} \mathrm{C}\)
02

Convert distances to SI units

We need to work with distances in meters. We can convert the given distances as follows: 1. Distance between Cl- and Na+: \(0.24 \times 10^{-9} \mathrm{m}\) 2. Distance between the free electron and the midpoint of the NaCl molecule: \(0.48 \times 10^{-9} \mathrm{m}\)
03

Find the distance between each ion and the free electron

We need to find the distance (r) between the free electron and the Na+ ion (r1), and the distance between the free electron and the Cl- ion (r2) using Pythagorean theorem. Let's consider a right triangle, with the vertical side equal to the height of the free electron above the midpoint (\(0.48 \times 10^{-9} \mathrm{m}\)), and the horizontal side as half of the distance between Cl- and Na+ ions (\(0.12 \times 10^{-9} \mathrm{m}\)). \(r1 = \sqrt{(0.48 \times 10^{-9})^2 + (0.12 \times 10^{-9})^2}\) \(r2 = r1\)
04

Calculate the force exerted by each ion using Coulomb's law

Coulomb's Law is given by \(F = \dfrac{k q1 q2}{r^2}\), where \(k\) is the electrostatic constant (\(8.99 \times 10^{9} \mathrm{Nm^2C^{-2}}\)), \(q1\) and \(q2\) are the charges of the two particles, and \(r\) is the distance between them. Let's calculate the force exerted on the electron by the Na+ ion (F_Na) and the Cl- ion (F_Cl) separately: \(F_{Na} = \dfrac{k \cdot e \cdot e}{r1^2}\) \(F_{Cl} = \dfrac{k \cdot e \cdot (-e)}{r2^2}\) (since the chloride ion has a negative charge)
05

Calculate the net force on the free electron

Since both forces have equal magnitudes but opposite directions, we can find their horizontal and vertical components separately. The horizontal components will cancel out, because they are equal in magnitude but opposite in direction. So, let's find the net vertical component to determine the net force: \(F_{vertical} = F_{Na} \cdot \dfrac{(0.48 \times 10^{-9})}{r1} - F_{Cl} \cdot \dfrac{(0.48 \times 10^{-9})}{r2}\) \(F_{net} = F_{vertical}\), because the horizontal components cancel out.
06

Find the direction of the net force

The net force is equal to the vertical component of the force exerted by the Na+ ion, so its direction is the same as that of the force exerted by the Na+ ion, which is towards the Na+ ion. In conclusion, the magnitude of the net electrostatic force exerted by the sodium chloride molecule on the free electron is equal to the net vertical component, and the direction of the force is towards the sodium ion (Na+).

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Most popular questions from this chapter

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In solid sodium chloride (table salt), chloride ions have one more electron than they have protons, and sodium ions have one more proton than they have electrons. These ions are separated by about \(0.28 \mathrm{nm} .\) Calculate the electrostatic force between a sodium ion and a chloride ion.
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