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Artemon [7]
2 years ago
5

Consider an infinite lattice with coordination number z in which every site is occupied by a molecule. (As a reminder, the coord

ination number is the number of neighbors each site has). Each molecule can be in one of two states: an 'up' state, and a 'down' state. A system interacts only with adjancet particles that share its state (i.e. ups interact with ups and downs interact with downs). Each interaction of this kind contributes an interaction energy of -epsilon to the system's energy. Using a mean-field approach, compute the following. (a) The fraction of molecules in the up state as a function of reduced temperature kT/(epsilon*z) (b) The mean energy per site as a function of reduced temperature. Here are some hints. In doing this, you will want to use the perturbation (mean field) approach retaining only F0 and F1. To obtain F0, just compute the entropy of the lattice in the absence of interactions. To obtain F1, compute the energy that is then introduced to the system by the attractions, using the configurational probabilities from the reference (noninteracting state). For this second part, you will have to use the idea I introduced in class that in a non-interacting lattice the site probability is just given by a mole fraction. When you determine the fraction of molecules in the up state, the easiest way will be to use the criterion that F is at a minimum at equilibrium, i.e. dF/dX = 0. In this case, you will want to minimize with respect to the fraction of up or down sites. This equation will require some numerical solution near the end. Note also that below some temperature, there should be two solutions to this minimization, in terms of two possible equilibrium values of the mole fraction of up sites (or down sites). This leads to the final part of this problem: (c) Determine the reduced temperature below which two different fractions of 'up' molecules are possible at the same temperature.
Engineering
1 answer:
ratelena [41]2 years ago
4 0
I am just questing to point a ok ok ok this might not be right
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