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How to Proof c/a=1.633 in HCP?

  1. Dec 25, 2005 #1
    In the HCP Lattice in solid state physics,
    Can any one proof GEOMETRICALLY and in algebric way that the the ratio between the height and the constant of the lattice equals sqrt(8/3)=1.633?
    Maybe a is not the lattice constant, but really i need some one to explain for me everything about the HCP dimensions because it's gonna blow my brain while i'm having an exam after 8 days !!!!!!!!!!!
    What is the constant of this lattice? and which 2 points we are taking distances from to get this damn constant in HCP???
    Thanks
     
  2. jcsd
  3. Dec 25, 2005 #2
    This is proved using the hard sphere approximation ie. that the atoms at the lattice sites are spheres of radius r that just touch eachother. You don't need much besides some basic trigonometry to solve for the ratio. The easiest way to calculate it is to take a slice of the conventional unit cell with a triangular base and height of c/2.
     
  4. Dec 25, 2005 #3

    Astronuc

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    Inha's method is correct.

    In the hcp system, the basal plane has a hexagonal or triangular lattice with lattice parameter a. The lattice parameter in the normal to the basal plane, the basal pole direction, has the value 'c'.

    Mg, Ti, Be, Sc, Te, Co, Zn, Y, Zr, Tc, Ru, Gd, Tb, Py, Ho, Er, Tm, Lu, Hf, Re, Os, Tl

    Here is a reasonably nice representation - http://www.engr.ku.edu/~rhale/ae510/lecture2/sld013.htm

    Over the basal triangle, the atom in the next layer is centered over the center of the triangle, and the next basal plane layer aligns with the first.
     
  5. Dec 27, 2005 #4
    As an alternative, one can calculate the distance between close-packed planes in fcc, which is the same as in bcc.

    The plane distance is one third of the body diagonal of the fcc cubic unit cell.
     
  6. Dec 29, 2005 #5
    c/a + 1.6333

    This is the mathematical calculation ^ means to the power of

    c/2 = a/2 Then a^2 /2 = c^2/2

    a^2 + a^2
    ----- = (4R)^2
    2

    2a^2+a^2
    -------------- = 16 R^2
    2

    3a^2 = 2 *16 R^2

    a^2 = 2*16 R^2
    -----
    3

    a = 2* square root of 16 divided by square root of 3

    a = 8 / 3 = 1.6329 R
     
  7. Jan 26, 2006 #6
    The assumptions above are roughly correct, but we have to take into the considerations the Pythagorean Theorem.
    The Pythagorean Theorem states: b^2=a^2+c^2. as well as c/2 or half of hexagonal crystal structure as well as the cosine for 30 degree triangles generated by the hole in Hexagonal closed packed.
    So we write:
    c/a half of this c/2.

    Cos 30 degree
     
  8. Feb 9, 2009 #7
    Dear all,

    i have tried. the last post by Kouros Khamoushi almost worked...

    how ever i do not understand where some of the steps....

    1. c/2 = a/2??? how?

    2. 3a^2 = 2 *16 R^2???? where in the world did the factor of 3 come from on the LHS of this eqn.

    You are prob correct... however please clarify the steps as i am totally baffled.

    ps. inha: what shape of slice do u mean? also i hope it is from hexagonal lattice?


    Kind regards.
     
  9. Feb 9, 2009 #8

    Gokul43201

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    The post by KK is wrong.

    Start from scratch: consider a plane of 3 neighboring atoms forming a triangle. Place a 4th atom above this plane, in the middle of the triangle. Calculate away.

    If you have any trouble, show us what you've done and where exactly you are stuck.
     
  10. Feb 10, 2009 #9

    Thank you for the question. The Hexagonal closed packed structure has 6 corners, because each corner cosists of 1/6 of a sphere and the top and bottom faces each contanins 1/2 of the spheres. Therefore 12(1/6)+2(1/2)+3=6 spheres.

    Each diagonal has a distance of 2a. The radius of sphere is 2a=4r . r means radius.by simplyfication and solving for a so that r=a/2 or a=2r. in 3 atoms closed in diagonal of Hexagonal we have 3a. that means each a is equal to 2r.
    The assumptions above are roughly correct, but we have to take into the considerations the Pythagorean Theorem.
    The Pythagorean Theorem states: b^2=a^2+c^2. as well as c/2 or half of hexagonal crystal structure as well as the cosine for 30 degree triangles generated by the hole in Hexagonal closed packed.
    So we write:
    c/a half of this c/2. Cos 30 degree. We have to take into the consideration 6r in three neighboring atoms radius generated by closed packed which is equal to 3a. As you see this is correct. If one think is I am wrong please let me know the correct and complete mathematical calculation with proof.
     
  11. Feb 10, 2009 #10

    Gokul43201

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    Re: c/a + 1.6333

    We do not give away solutions to textbook problems, but we weill certainly show where someone has made a mistake.

    Clearly, this is wrong. If c=a and a=2R, then c=2R, which is not true.

    Then you do this:
    There are so many errors in this.

    1. 3a^2 = 32R^2 => a=3.266R (not 1.633R)
    2. 8/3 = 2.667 (where did this come from and why did the R disappear?)
     
  12. Feb 10, 2009 #11
    "There are so many errors in this.

    1. 3a^2 = 32R^2 => a=3.266R (not 1.633R)
    2. 8/3 = 2.667 (where did this come from and why did the R disappear?)"

    >>First of all you forget about a/2 yes a=3.265, but half of it is 1.6329R.
    R is there and is not disappear. It is the radius of atom.
    >>2. first you should divid 8/3 and tak the square roots of this that will give you
    2.667. What is square roots of 2.667? The answer is 1.633. This on also is right.

    "Clearly, this is wrong. If c=a and a=2R, then c=2R, which is not true.
    Then you do this:"
    >>I have emphasis on c in Hexagonal plus cos of 30 degree which is clarifying the position of c. This is also correct. My solution is right.
     
  13. Feb 10, 2009 #12
    C=1.633 a (eq 1). We know a=2R if we substatute this to equation we get
    C=1.633(2R) which is equal to C= 3.266 . According to your own consumption

    ""There are so many errors in this.
    1. 3a^2 = 32R^2 => a=3.266R (not 1.633R) "" >>Half of a is 1.633R

    a=3.266R if C=3.266R according to given equation above c/a=1.633R We know a=2 it is a definition for radius of Hexagonal which is 2 times R. so that c/2=1.633R the result is
    C=3.266R . It shows that c=a, therefore we can write c/2=a/2.
     
  14. Feb 10, 2009 #13

    Gokul43201

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    You are simply producing more and more contradictions:

    1. According to you, c/a=1 (posts #5, #12). But also, in the first line of post #12, you write c/a=1.633. Both can not be true.

    2. According to posts #12 (top line), and post #9, you say a=2R, but in post #5, you say a=1.6329R as well as 3a^2=32R^2. Any two of these three statements are in contradiction with each other.

    And as I said before, we do not allow members to post solutions to standard textbook problems, so your attempting to do so would be in violation of the guidelines. If you wish, I can write a proof and send it to you by private message, since I know this is not your homework.
     
  15. Feb 11, 2009 #14
    Thank you for your answer.
    "1. According to you, c/a=1 (posts #5, #12). But also, in the first line of post #12, you write c/a=1.633. Both can not be true."

    >>We should not forget the homework or question is: "How to Proof c/a=1.633 in HCP?"
    if this is not true or so that all the given problem is wrong. What should we proof?

    According to your post #13 you are comparing my step of solutions without taking into the consideration the correlation between them. a=2R is from "Ibach and Luth Problem 2.7 page 13 The Hexagonal close packed structure" and c/a=1.633R is from your given problem on top of your page.

    However, mathematical problems can have different solutions.Would you please send me your solution.

    Yours sincerely

    Kouros Khamoushi
     
  16. Feb 11, 2009 #15
    I have solved the completely this problem, but your editor does not have the possiblity to draw a picture or triangle and write the equations. If you have any suggestion please let me know.

    Best regards

    Kouros Khamoushi
     
  17. Feb 13, 2009 #16

    Gokul43201

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    Naturally, c/a = 1.633... is correct, which means c/a=1 is not correct.

    Any set of equations you write must be mathematically consistent with each other, irrespective of context. The manner in which you have written things is extremely misleading because you write equations that contradict each other, and the bases for the equations are not clearly explained.

    I do not doubt that there are different ways to solve the problem. I suspect that a big problem here may be miscommunication due to your writing style. If you write clearly, using unambiguous mathematical and physical relationships, then we may be able to determine clearly if there is a problem. Part of the issue here may be one of language, but a bigger part of it is the poorly written equations.

    I shall send you a derivation by Private Message, later today. If I forget, feel free to PM me a reminder.

    Edit: PM sent.
     
    Last edited: Feb 13, 2009
  18. Feb 13, 2009 #17
    Hi , I'll answer later on.
     
    Last edited: Feb 13, 2009
  19. Feb 14, 2009 #18
    As I said I have complete solution of this very simple problem. Your system does not let me to wirite: 1. Equation
    2. Draw a triangle

    Mathematic problem does not need Enlish language, because mathematic is a logical language. Therefore by equations one can proof that.

    Best regards,

    Kouros Khamoushi
     
  20. Feb 14, 2009 #19

    Gokul43201

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    Yes, we do have a capability for writing equations (see my signature). But as I said before, if you do attempt to write out a complete solution, you will be in violation of the forum guidelines. Third, even in some mathematics problems (especially applied math), you definitely need to use non-mathematical language to connect your mathematical symbols and expressions with the physical picture. When you write a bunch of equations without defining any of the variables, it is completely useless. And finally, a logical solution must not contain logical contradictions. Your attempts so far reveal several contradictions.

    That is all I have to say.
     
  21. Feb 14, 2009 #20
    I think this discussion will never get anywhere.
    Best Wishes
     
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