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[PoM] Cohesive energy in an argon crystal

  1. Dec 5, 2016 #1

    BRN

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

    Only hcp case is OK... what am I doing wrong?
     
  2. jcsd
  3. Dec 5, 2016 #2

    TSny

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    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".
     
  4. Dec 7, 2016 #3

    BRN

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    On my book these arguments are trated very bad. I'm confused a little...:confused:
    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...
     
  5. Dec 7, 2016 #4

    TSny

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    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
  6. Dec 7, 2016 #5

    TSny

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    You got lucky with the hcp case. An atom in hcp has how many nearest neighbors?
     
  7. Dec 8, 2016 #6

    BRN

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    True! I did not think about it!:oops:

    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?
     
  8. Dec 8, 2016 #7

    TSny

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    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
  9. Dec 8, 2016 #8

    BRN

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    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...
     
  10. Dec 8, 2016 #9

    TSny

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    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
  11. Dec 10, 2016 #10

    BRN

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    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
  12. Dec 10, 2016 #11

    TSny

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    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
  13. Dec 11, 2016 #12

    BRN

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    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...
     
  14. Dec 11, 2016 #13

    TSny

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    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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