FNT11 - FNT #11: Modes and equipartition Damien Martin...

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FNT #11: Modes and equipartition Damien Martin November 27, 2006 Question 1 You have a monatomic substance simulated with 100 atoms on the computer programme atoms in motion. The bond energy of these 100 atoms is - 400 × 10 - 21 J because when KE=0, PE = - 400 × 10 - 21 J. a) The well-depth for atom-atom potential is ε = 0 . 8 × 10 - 21 J. Determine the average number of nearest neighbours. The bond energy is (taking into account only nearest neighbour interactions) E bond = - N bond 2 ε The number of bonds is therefore N bonds = - E bond ε = +400 × 10 - 21 J 0 . 8 × 10 - 21 J = 500 bonds The number of nearest neighbours is given by N bonds = (# nearest neighbours) 2 N atoms Rearranging # nearest neighbours = 2 N bonds N atoms = (2)(100 bonds) 500 atoms = 10 bonds/atom b) Why is this less than 12? It is true that the packing structure has 12 nearest neighbours. However, this just means that the atoms in the centre have 12 nearest neighbours, while those on the edge have significantly fewer. The statement that we should be making is that the average number of nearest neighbours is 10. 1
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c) How do the equations change from mono-solid/liquid to gas? The equations in question (which are valid in solids and liquids) are PE = E bond + 1 2 E thermal KE = 1 2 E thermal It is important to remember where this equation comes from. We determined that the thermal component just came from equipartition. We have 3 KE modes per atom and 3 PE modes per atom, and since each mode has the same energy we see half the E thermal goes into KE and half goes into PE. The PE also has an extra contribution from the bond energy.
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This note was uploaded on 07/31/2008 for the course PHY 7A taught by Professor Pardini during the Winter '08 term at UC Davis.

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FNT11 - FNT #11: Modes and equipartition Damien Martin...

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