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Homework Help: Introduction level Solid State - Mean free time/path & lattice spacing

  1. Apr 2, 2014 #1
    1. The problem statement, all variables and given/known data

    Silver has a density of 10.5E3 kg/m3 and a resistivity of 1.6E-8 Ω*m at room temperature. On the basis of the classical free electron gas model, and assuming that each silver atom contributes one electron to the electron gas, calculate the average time, Tau, between collisions of the electrons. Calculate the mean free path from tau and the electron's thermal velocity. How does the mean free path compare to the lattice spacing?

    2. Relevant equations

    3. The attempt at a solution

    I was able to do all of the calculations fairly easily, but I'm stuck on the lattice spacing. My professor told us that we could just look up the lattice spacing of a silver crystal for the sake of comparison, but I was unable to find it online. As my class just has a brief one or two chapter long introduction to solid state physics, we haven't touched on how to calculate this and I could not find it in my textbook. Since I cannot find the lattice spacing of a silver crystal online, I was just wondering if anybody knows how I can calculate the lattice spacing from the information that I am given in this problem. Is this sufficient information to calculate it myself without having to look it up? I haven't found anything in my textbook, but as this is just a few-chapter introduction to the basics of solid state physics, I was wondering if there is any way to calculate this that my book may not mention with such a brief introduction?

    Thanks in advanced.
  2. jcsd
  3. Apr 3, 2014 #2


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    Gold Member

    Well, to get an estimate for lattice spacing, you could probably use the density of silver, and the mass of a silver nucleus right? Most of the mass of silver is in the nucleus, so from the density ρ we can get the rough number density of nucleons n=ρ/m (where m is the mass of one silver nucleus) and from n, you can deduce a rough lattice spacing from d~(1/n)^(1/3).

    Using this method I get an answer of ~2.5 angstroms. Wolfram alpha gives the silver lattice spacings of ~2.4 angstroms, 2 anstroms, and 1.2 angstroms. So it seems like we were close (to one of them lol)...
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