# [PoM] Cohesive energy in an argon crystal

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1. Dec 5, 2016

### BRN

Hey guys! I come back to solve a new problem with your help!

1. The problem statement, all variables and given/known data

Use the Lennard-Jones function:
$$V(R)=4\epsilon[(\frac{\sigma}{R})^{12}- (\frac{\sigma}{R})^6]$$
like as a adiabatic potential energy model in function of separetion R between cople of argon nuclei. With $\epsilon=1.68*10^{-21}$, in the first neighbors approximation, calculates the coesive energy for 1 mole of crystal argon with these structures: fcc, bcc, hcp and sc. For fcc case, estimates the effect of the interaction with second neighbors.

(Solution:fcc→6070 J/mol; bcc→4047 J/mol; hcp→6070 J/mol; sc →3035 J/mol; fcc(including 2nd neighbor interaction at fixed lattice spacing)→7398 J/mol; fcc(including 2nd neighbor interaction at optimal lattice spacing)→7627 J/mol)

3. The attempt at a solution
The atom disposition in the lattice is due to the minimum energy condiction, where atoms are spaced by a minimum radius $R_{0}$. This occurs when $\frac{\partial V}{\partial R}=0$:
$$\frac{\partial }{\partial R}[4\epsilon[(\frac{\sigma}{R})^{12}- (\frac{\sigma}{R})^6]]=4\epsilon \sigma^6[-12\sigma^6R^{-13}+6R^{-7}]=0\Rightarrow R_0^6=2\sigma^6$$
then
$$V(0)=4\epsilon[(\frac{\sigma}{R_0})^{12}- (\frac{\sigma}{R_0})^6]=4\epsilon[\frac{\sigma^{12}}{(2\sigma^6)^2}- \frac{\sigma^6}{2\sigma^6}]=-\epsilon$$
The coesive energy is $E_c=-NN_AV(0)$, with N number of atoms in the lattice.

So
• sc case: $N= (\frac{1}{8}*8)=1 \Rightarrow E_c=1011.71 [J/mol]$
• fcc case: $N= (\frac{1}{2}*6+\frac{1}{8}*8)=4 \Rightarrow E_c=4046.85 [J/mol]$
• bcc case: $N= (1+\frac{1}{8}*8)=2 \Rightarrow E_c=2023.42 [J/mol]$
• hcp case: $N= (3+\frac{1}{2}*2+\frac{1}{6}*12)=6 \Rightarrow E_c= 6070.13 [J/mol]$
For each atom in fcc lattice, there are 12 2nd neighbors, then:
$$N= (\frac{1}{2}*6+\frac{1}{8}*8+\frac{1}{4}*12)=7 \Rightarrow E_c= 7081.18 [J/mol]$$

Only hcp case is OK... what am I doing wrong?

2. Dec 5, 2016

### TSny

N is the "number of atoms in the lattice"?
• Here it looks like you are treating N as representing the number of atoms in a unit cell. But is that what you need for this problem? Remember, the question asks for the cohesive energy per mole assuming "first neighbors approximation".

3. Dec 7, 2016

### BRN

On my book these arguments are trated very bad. I'm confused a little...
I have used "lattice" term because sc structure is a unit cell too, but the others not. Maybe I should have written "number of atoms in the structure". Maybe I'm wrong?

Ok, but i don't understand how to calculate the number of first neighbors atoms, because if i consider:
• sc case: 6 first neighbors atoms;
• fcc case: 12 first neighbors atoms;
• bcc case:8 first neighbors atoms.
then, i have the correct solutions only if i divide by 2 these numbers of atoms. Why?

Then, the fact that for hcp case i have the correct value using the number of atoms in the cell, it still confuses me more...

4. Dec 7, 2016

### TSny

If atom A is a nearest neighbor of atom B, then atom B is a nearest neighbor of atom A.

Last edited: Dec 7, 2016
5. Dec 7, 2016

### TSny

You got lucky with the hcp case. An atom in hcp has how many nearest neighbors?

6. Dec 8, 2016

### BRN

True! I did not think about it!

Therefore, $E_c=-\frac{1}{2}NN_AV(0)$ with N number of first neighdors, then:
• sc case: $N=6 \Rightarrow E_c=3035.13 [J/mol]$
• fcc case: $N=12 \Rightarrow E_c=6070.27 [J/mol]$
• bcc case: $N=8 \Rightarrow E_c=4046.85 [J/mol]$
• hcp case: $N=12 \Rightarrow E_c=6070.27 [J/mol]$
For fcc case, each atom has 14 2nd neighbors an then $E_c=7081.98 [J/mol]$

in hcp cell there are $(\frac {1}{2}*5+\frac {1}{3}*12)=6.5$ atoms, each atom has 12 first neighbors and 14 2nd neighbors. It's right?

7. Dec 8, 2016

### TSny

For number of atoms per cell and number of nearest neighbors (coordination number), see
http://chem.libretexts.org/Core/Phy...olids/Crystal_Lattice/Closest_Pack_Structures

As far as 2nd nearest neighbors, I think there are only 2. That makes 14 as the sum of nearest and 2nd nearest neighbors. But I didn't look this up to make sure. [EDIT: No, I overlooked the actual 2nd nearest neighbors. I believe there are six 2nd nearest neighbors. So, a total of 18 nearest and 2nd nearest neighbors.]

Last edited: Dec 8, 2016
8. Dec 8, 2016

### BRN

Damn! I still wrong!

Ok, in hcp cell there are $(3+\frac {1}{2}*2+\frac {1}{6}*12)=6$ atoms, each atom has 12 first neighbors and 2 2nd neighbors.

But fcc structure has 6 2nd neigbors, and then the interaction with 2nd neighbors is $E_c=3035.13 [J/mol]$, wrong...

9. Dec 8, 2016

### TSny

For the last part of the question where it asks for the energy for the fcc including 2nd nearest neighbors, I don't get the right answer.

I might not be interpreting the question correctly. I'm not sure how to interpret "including 2nd nearest neighbors with fixed lattice". Suppose we let $R_1$ be the distance to nearest neighbors and $R_2$ the distance to 2nd nearest neighbors. If I'm not mistaken, $R_2 = \sqrt{2} R_1$ for fcc. Verify this.

My interpretation of "fixed lattice" is to let $R_1 = R_0 = 2^{1/6}\sigma$ for the nearest neighbors and $R_2 = \sqrt{2} R_1$ for the 2nd nearest neighbors. But this doesn't give me the given answer of 7398 J/mol. [EDIT: After correcting a mistake, I now get 7380 J/mol, which is close.]

My interpretation of the "optimal lattice spacing" is to let $R_1 = r$ and $R_2 = \sqrt{2} r$, where r is chosen to minimize the total energy for nearest plus 2nd nearest interactions. But it did not lead to the given answer. (I'm getting 7400 J/mol.)

So, I'm not sure I can be of much help here.

Last edited: Dec 8, 2016
10. Dec 10, 2016

### BRN

Yes, $R_2 = \sqrt{2} R_1$, as shown here[/PLAIN] [Broken]

But I can not get your same result...
with $R_2=\sqrt{2} R_1 =\sqrt [6] {16}\sigma$, I have

$$V(R_2)=4\epsilon[\frac{\sigma^{12}}{R_2^{12}}- \frac {\sigma^{6}}{R_2^{6}}]=4\epsilon[\frac {\sigma^{12}} {(\sqrt [6] {16}\sigma)^{12}} - \frac {\sigma^{6}}{(\sqrt [6] {16}\sigma)^{6}}]=-0.2499\epsilon \approx -\frac {1}{4} \epsilon$$

and then

$E_c=-\frac{1}{2}NN_AV(R_2)=758.78 [J/mol]$

Last edited by a moderator: May 8, 2017
11. Dec 10, 2016

### TSny

When I checked my work again, I realized another calculational error. I'm now getting about 6780 J/mol for the "fixed lattice" case. So it's no longer close to the given answer.
OK, but instead of 0.2499 I get 0.2343

If I use 0.2343 instead of 0 .2499, I get Ec = -711 J/mole. But this is just the contribution from the 2nd neighbors. You also need to include the -6070 J/mole form the first neighbors. This gives a total of Ec = - 6780 J/mole.

I can't see where we're going wrong.

Last edited by a moderator: May 8, 2017
12. Dec 11, 2016

### BRN

Here is used Morse potential, but cohesion energy is calculate in same way our.

Error will be in the L-J potential for 2nd neighbors...

13. Dec 11, 2016

### TSny

Thanks for the link. Equations 11 through 13 seem to confirm our approach, including 12 nearest neighbors, 6 next nearest neighbors, and $r_2 = \sqrt{2} r_1$.

I still can't see an error in how we handle the next nearest neighbors. The next nearest neighbors do not seem to change $E_c$ very much.

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