Say you have a monatomic substance simulated with 100 atoms on the computer simu

Question

Say you have a monatomic substance simulated with 100 atoms on the computer simulation program Atoms-In-Motion. You see that it is in its solid phase and when all KE is removed and the temperature reads 0 K, the PE of these 100 atoms is - 400 x 10-21 J. You therefore know that the bond energy of these 100 atoms is -400 x 10-21 J.

(a) The well depth, , for these atoms is 0.8 x 10-21 J. Determine the average number of nearest-neighbors for each atom.

(b) Why do you think that the average number of nearest-neighbors per atom turns out to be less than 12 for this sample of 100 atoms?

Explanation / Answer

Determine the average number of nearest-neighbors for each atom.

n = E/(A.e)
n, number of nearest-neighbors for each atom
e, well depth (0.8x10-21 J)
E, bond energy of 100 atoms (400x10-21 J)
A, number of atoms

Solving the equation gives that the number of nearest-neighbors for each atom is 5 (n = 5)

A number of nearest-neighbors per atom less than 12 suggests the 100 atoms are of the same type and are arranged in a simple cubic unit.

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