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Chapter 1: Problem 42

Potential energy in a one-dimensional crystal * * Calculate the potential energy, per ion, for an infinite 1 D ionic crystal with separation \(a\); that is, a row of equally spaced charges of magnitude \(e\) and alternating sign. Hint: The power-series expansion of \(\ln (1+x)\) may be of use.

Short Answer

Expert verified
The potential energy, per ion, for an infinite 1 D ionic crystal is 0.

Step by step solution

01

Simplification and assume constants

Let's denote the number of ions as \(n\). The charge of each ion is alternating and can be written as \((-1)^ne\). Let's assume all the constants including the Coulomb constant \(k\) and the charge \(e\) are 1 for simplification since they won't affect the final solution. This simplifies our calculation.
02

Determine potential energy between ions

The potential energy \(V\) between two ions can be calculated using Coulomb's Law, multiplying by \(-1\) if the charges are the same and \(1\) if the charges are different. Hence the potential energy \(V\) between ions \(n\) and \(m\) is \((-1)^{n+m}/|n-m|\).
03

Sum the total potential energy

We then sum the potential energy for every pair of ions to get the total potential energy. This can be done by summing over all integer pairs \((n, m)\) where \(n < m\). We can use the formula for the sum of an infinite geometric series \(1 + x + x^2 + ... = 1/(1-x)\), for |x| < 1. So the sum is \(2\ln (2)\).
04

Average the potential energy per ion

The exercise asks for the potential energy per ion. Hence we need to compute the average potential energy per ion, which gives the potential energy per ion in an infinite crystal. Since the sum of potential energy is \(2\ln (2)\), the potential energy per ion would be \(2\ln (2)/n\). But here \(n\) refers to an infinite number of ions, so we get \(0\) potential energy per ion in the infinite limit.

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Key Concepts

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

Coulomb's Law
In the context of crystals, understanding Coulomb's Law is pivotal. This fundamental principle governs the interactions between particles with electric charge. Put simply, it states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Mathematically, Coulomb's Law can be expressed as \( F = k \frac{q_1 q_2}{r^2} \), where \(F\) is the force, \(k\) is Coulomb's constant, \(q_1\) and \(q_2\) are the charges, and \(r\) is the separation distance between the charges. While the law focuses on force, it directly relates to potential energy, \(V\), which is the energy due to the positions of the charges relative to each other: \(V = -\frac{F}{q} \cdot r\) if \(F\) and \(q\) are considered of the same sign. Understanding this law is vital for analyzing the potential energy within ionic crystals, where charged ions arrange in a regular pattern and interact electromagnetically.

Infinite Geometric Series
When calculating potential energy in an infinite crystal structure, one often encounters a summation that forms an Infinite Geometric Series. This series is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio \(r\). The series looks like \(a + ar + ar^2 + ar^3 + ...\), with \(a\) being the first term.

The sum of an infinite geometric series can be infinitely large, but under certain conditions鈥攕pecifically when the absolute value of the common ratio \(|r| < 1\)鈥攖he series converges to a finite value. This sum is given by the formula \(S = \frac{a}{1-r}\). The power of this formula allows us to solve for the potential energy in the crystal lattice, even with an infinite number of ions, because it sidesteps the need to sum an infinite number of terms individually. This simplifies the analysis considerably and provides a neat, finite value for the infinite lattice as seen in the exercise provided.
Potential Energy Calculation
The calculation of potential energy in an ionic crystal brings together Coulomb's Law and the properties of an infinite geometric series. It requires understanding the interactions of an infinite array of ions with alternating charges and applying mathematical techniques to account for an infinite sum.

In the step-by-step solution provided, potential energy is first considered between two individual ions using a simplified version of Coulomb's Law. By pairing up ions and then summing up the potential energies of all possible pairs within the crystal, we can establish the total potential energy of the system. Importantly, by employing the infinite geometric series, we extract a concise expression for this total potential energy.

Finally, to find the average potential energy per ion, which the original exercise asks for, we would typically divide the total potential by the number of ions. However, as the crystal is infinitely large, when you try to average the potential energy by dividing by an infinite number, the result interestingly approaches zero. This conveys a crucial concept in physics鈥攚hile individual interactions in a large system can be complex and significant, the overall effect per particle can become negligible in an infinite system.

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Most popular questions from this chapter

Intersecting sheets ** (a) Figure \(1.49\) shows the cross section of three infinite sheets intersecting at equal angles. The sheets all have surface charge density \(\sigma .\) By adding up the fields from the sheets, find the electric field at all points in space. (b) Find the field instead by using Gauss's law. You should explain clearly why Gauss's law is in fact useful in this setup. (c) What is the field in the analogous setup where there are \(N\) sheets instead of three? What is your answer in the \(N \rightarrow \infty\) limit? This limit is related to the cylinder in Exercise 1.68.

Building a sheet from rods ** An infinite uniform sheet of charge can be thought of as consisting of an infinite number of adjacent uniformly charged rods. Using the fact that the electric field from an infinite rod is \(\lambda / 2 \pi \epsilon_{0} r\), integrate over these rods to show that the field from an infinite sheet with charge density \(\sigma\) is \(\sigma / 2 \epsilon_{0}\)

Field from a hemisphere ** (a) What is the electric field at the center of a hollow hemispherical shell with radius \(R\) and uniform surface charge density \(\sigma\) ? (This is a special case of Problem \(1.12\), but you can solve the present exercise much more easily from scratch, without going through all the messy integrals of Problem 1.12.) (b) Use your result to show that the electric field at the center of a solid hemisphere with radius \(R\) and uniform volume charge density \(\rho\) equals \(\rho R / 4 \epsilon_{0}\)

Gravity vs. electricity (a) In the domain of elementary particles, a natural unit of mass is the mass of a nucleon, that is, a proton or a neutron, the basic massive building blocks of ordinary matter. Given the nucleon mass as \(1.67 \cdot 10^{-27} \mathrm{~kg}\) and the gravitational constant G as \(6.67 \cdot 10^{-11} \mathrm{~m}^{3} /\left(\mathrm{kg} \mathrm{s}^{2}\right)\), compare the gravitational attraction of two protons with their electrostatic repulsion. This shows why we call gravitation a very weak force. (b) The distance between the two protons in the helium nucleus could be at one instant as much as \(10^{-15} \mathrm{~m}\). How large is the force of electrical repulsion between two protons at that distance? Express it in newtons, and in pounds. Even stronger is the nuclear force that acts between any pair of hadrons (including neutrons and protons) when they are that close together.

Escaping field lines * Charges \(2 q\) and \(-q\) are located on the \(x\) axis at \(x=0\) and \(x=a\), respectively. (a) Find the point on the \(x\) axis where the electric field is zero, and make a rough sketch of some field lines. (b) You should find that some of the field lines that start on the \(2 q\) charge end up on the \(-q\) charge, while others head off to infinity. Consider the field lines that form the cutoff between, these two cases. At what angle (with respect to the \(x\) axis) do these lines leave the \(2 q\) charge? Hint: Draw a wisely chosen Gaussian surface that mainly follows these lines.
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