Question

I got parts a-c, but not d

A crystal is a periodic lattice of positively and negatively charged one. The goal of the problem is to show that the potential energy person is negative. This is one condition for a crystal to be a stable configuration; otherwise, would be energetically favorable for the fons to be infinitely for aport thich would definitegrate the crystal. Although in general a crystal can have a completed 3D crangement of fone, here we will consider a ID crystal 68 a simpler erample. Problem 3 (12 points): Consider an infinite one-dimensional crystal of alternating ions with charges te and separated by distance : te -e te -E te te -e te -e - In the figure above, we have labeled some of the charges by 1, 1+1, etc. (a) The total potential energy of the infinite lattice is where U, is the potential energy between charges 1 and 1. Determine Vo, defined as the term in parentheses in Eq. (1), 1- ŜU. (2) and show that it is the same for every charge 1. You may leave your answer expressed as an infinite sum. (3 points) (b) In your answer for part (a), you should encounter an infinite sum -1) 1-5+ 7+3 6 It turns out this sum is equal to In(2). Prove this result by Taylor expanding the function (1) = In(1+2) and setting = 1. Use this result to evaluate the sum in part (a). (3 points) (c) Argue that Vo is the potential energy per fon and evaluate Uo using your answers from parts (a) and (b). Hint: the final answer for the potential energy per ion is Vo = -e In (2)/(41€d). (3 points) (d) Repeat parts (a)-(e) for the following 1D crystal, consisting of semi-alternating ions What is the potential energy per ion for this crystal? Which configuration is more stable, the alternating crystal or the semi-alternating crystal? (3 points) Hist: It is easier to set up the problem if you start with the alternative form for potential energy of point charges, so that UU

Answer 1

Answer: Part d> K d * ....te te -e e te te ere te te... from Part (C) wet get to = -e2 ine? 4R&d u = ez – ezz o 40 cod 4ñod U₂ = -e² - e2 41€ d) 47820) 4AEd Uz = 0 4 = e2 4RER b =0 -e2 416o3d Now botal energy Uz vitu2 + Ug+ -- u= -22 si-Ź + 3...] u= -e2 in2 4ÃO" for & voj = 0o = -e2 - ez + 02 + ez +.... 40&d 2047€0 3d47€ 4d4RE0 no = - [1 + Ž - 13 - 14] void late] Usemi alternating = rebaYuternating - 2e2lna and valternating < Usemi-alternating so alternating coystal is more stable