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Chapter 3: Problem 147

Use Coulomb's law, $$V=\frac{Q_{1} Q_{2}}{4 \pi \epsilon_{0} r}=2.31 \times 10^{-19} \mathrm{J} \cdot \mathrm{nm}\left(\frac{Q_{1} Q_{2}}{r}\right)$$ to calculate the energy of interaction, \(V\), for the following two arrangements of charges, each having a magnitude equal to the electron charge.

Short Answer

Expert verified
Case 1 (Two charges with the same sign): \[V = 2.31 \times 10^{-19} J \cdot nm \left(\frac{Q_1 Q_2}{r}\right) = 2.31 \times 10^{-19} J \cdot nm \left(\frac{e^2}{r}\right)\] Case 2 (Two charges with opposite signs): \[V = 2.31 \times 10^{-19} J \cdot nm \left(\frac{Q_1 Q_2}{r}\right) = 2.31 \times 10^{-19} J \cdot nm \left(\frac{-e^2}{r}\right)\]

Step by step solution

01

Case 1: Two charges with the same sign

Let's consider two charges with the same sign, both positive or both negative. We can denote the charges as \(Q_1 = e\) and \(Q_2 = e\). To find the energy of interaction, we can use Coulomb's law formula: \[V = \frac{Q_1 Q_2}{4 \pi \epsilon_{0} r}\] Plugging in the values, we get: \[V = \frac{e^2}{4 \pi \epsilon_{0} r}\] Since both charges have the same sign, the energy of interaction is positive.
02

Case 2: Two charges with opposite signs

Now, let's consider two charges with opposite signs. We can denote the charges as \(Q_1 = e\) and \(Q_2 = -e\). To find the energy of interaction, we can use Coulomb's law formula: \[V = \frac{Q_1 Q_2}{4 \pi \epsilon_{0} r}\] Plugging in the values, we get: \[V = \frac{-e^2}{4 \pi \epsilon_{0} r}\] Since one charge is positive and the other is negative, the energy of interaction is negative. For both arrangements, we can represent the energy of interaction in terms of the given unit factor: \[V = 2.31 \times 10^{-19} J \cdot nm \left(\frac{Q_1 Q_2}{r}\right)\] By substituting the expression for each case, we can find the energy of interaction in terms of \(J \cdot nm\).

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Most popular questions from this chapter

A polyatomic ion is composed of \(\mathrm{C}, \mathrm{N},\) and an unknown element \(X\). The skeletal Lewis structure of this polyatomic ion is \([\mathrm{X}-\mathrm{C}-\mathrm{N}]^{-} .\) The ion \(\mathrm{X}^{2-}\) has an electron configuration of [Ar]4s^{2} 3 d ^ { 1 0 } 4 p ^ { 6 } . \text { What is element X? Knowing the identity of } X, complete the Lewis structure of the polyatomic ion, including all important resonance structures.

Write the formula for each of the following compounds: a. diboron trioxide b. arsenic pentafluoride c. dinitrogen monoxide d. sulfur hexachloride

Rank the following bonds in order of increasing ionic character: \(\mathrm{N}-\mathrm{O}, \mathrm{Ca}-\mathrm{O}, \mathrm{C}-\mathrm{F}, \mathrm{Br}-\mathrm{Br}, \mathrm{K}-\mathrm{F}\)

The most common exceptions to the octet rule are compounds or ions with central atoms having more than eight electrons around them. \(\mathrm{PF}_{5}, \mathrm{SF}_{4}, \mathrm{ClF}_{3},\) and \(\mathrm{Br}_{3}^{-}\) are examples of this type of exception. Draw the Lewis structure for these compounds or ions. Which elements, when they have to, can have more than eight electrons around them? How is this rationalized?

Elements in the same family often form oxyanions of the same general formula. The anions are named in a similar fashion. Give the names of the oxyanions of selenium and tellurium: \(\mathrm{ScO}_{4}^{2-}, \mathrm{SeO}_{3}^{2-}, \mathrm{TeO}_{4}^{2-}, \mathrm{TeO}_{3}^{2-}\)
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