The number of ways of placing M atoms on the interstices of a lattice

  • Thread starter Thread starter ppy
  • Start date Start date
  • Tags Tags
    Atoms Lattice
ppy
Messages
64
Reaction score
0
Hi,

N atoms are arranged to lie on a simple cubic crystal lattice. Then
M of these atoms are moved from their lattice sites to lie at the
interstices of the lattice, that is points which lie centrally between the
lattice sites. Assume that the atoms are placed in the interstices in a
way which is completely independent of the positions of the vacancies.
Show that the number of ways of taking M atoms from lattice sites
and placing them on interstices is W = (N!/M!(N − M)!)2 if there
are N interstitial sites where displaced atoms can sit.

I literally do not know where to start with this question I know W is the number of ways of choosing M atoms from N atoms but I don't really know where to go with this.

Help would be great.

Thanks :)
 
Physics news on Phys.org
What does the 2 mean? W = (N!/M!(N − M)!)2?

This can be expressed as (N choose M)2. And it does exactly what the name suggests: (N choose M) calculates the number of ways to choose M atoms out of N (to decide which atoms you move), and you get the same factor again for the choice of the interstitials.
 
mfb said:
What does the 2 mean? W = (N!/M!(N − M)!)2?

This can be expressed as (N choose M)2. And it does exactly what the name suggests: (N choose M) calculates the number of ways to choose M atoms out of N (to decide which atoms you move), and you get the same factor again for the choice of the interstitials.

Hi yes the 2 should be a squared. This question is worth 6 marks. What would you have to do to get so many marks other than writing what you suggested?
 
I have no idea about the scale of the marks, but writing that down in a more ordered way should be sufficient.
 
ppy said:
N atoms are arranged to lie on a simple cubic crystal lattice. Then
M of these atoms are moved from their lattice sites to lie at the
interstices of the lattice, that is points which lie centrally between the
lattice sites. Assume that the atoms are placed in the interstices in a
way which is completely independent of the positions of the vacancies.
Show that the number of ways of taking M atoms from lattice sites
and placing them on interstices is W = (N!/M!(N − M)!)2 if there
are N interstitial sites where displaced atoms can sit.Thanks :)
From which book you have taken this question?
 
This thread is from 2013, it is unlikely that OP is still around.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top