Energy of a crystal in thermal equilibrium

  • #1

Homework Statement

In a monatomic crystalline solid each atom can occupy either a regular lattice site or an interstitial site. The energy of an atom at an interstitial site exceeds the energy of an atom at a lattice site by an amount ε. Assume that the number of interstitial sites equals the number of lattice sites, and also equals the number of atoms N.

b)What is the temperature of the crystal in this state, if the crystal is in thermal equilibrium?

Homework Equations


The Attempt at a Solution

This is question 2.1 taken from Franz Mandl. Statistical Thermodynamics. Part a) asked me to find the total entropy which was fine. However in part b if you consider the total entropy and total energy of the entire system and use the 2nd formula i posted you will get n/N = exp(-ε/2kbT) (assuming ε >> kbT ) where n is the number of atoms in the interstitial site. Thats the answer given at the back of the book. In my first attempt however I divided my system into two subsystems (subsystem i of atoms in interstitial sites and subsystem r of atoms in regular sites). The entropy of each system is the same because each statistical weight is the same so the entropy of the subsystem i is half the total entropy found part a. The fact that I divided the entropy by two meant that my final answer was nearly the same as the answer at the back of the book but without the 2 in the exponent. I'm just trying to understand why my method of subdividing the system was not correct.

Answers and Replies

  • #2
Why would expect that to be right? The lattice positions and the interstitial positions are energy states available to all of the atoms. You are splitting the population up and counting them as two separate isolated populations based on nothing other than the energy state they achieved. In a gas you wouldn’t split the population into those above a certain temperature and those below a certain temperature and then calculate the entropy of the two populations like they were unrelated.