Say you have a monatomic substance simulated with 100 atoms

Say you have a monatomic substance simulated with 100 atoms on the computer simulation program Atoms-ln-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 Times 10^-21 J. You therefore know that the bond energy of these 100 atoms is -400 Times 10^-21 J. The well depth, e, for these atoms is 0.8 Times 10^-21 J. Determine the average number of nearest-neighbors for each atom (refer to Activity 3.2.3 \"Connecting the Particle Model of Bond Energy to Macroscopic Empirical Values\" 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?

Solution

the average number of nearest-neighbors for each atom:

n = E/(A.e)
Here, e=0.8x10-21 J
E =400x10-21 J
A= 100

Thus, n =5

This is less than12 suggests the 100 atoms

Because they are same type

and are arranged in a simple cubic unit.