Homework Help: [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.