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Or could someone please walk me through the process of how you find the lattice constants. Thanks in advance

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- Thread starter JoshHolloway
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Or could someone please walk me through the process of how you find the lattice constants. Thanks in advance

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malawi_glenn

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Yep. It is semiconductor physics.

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Will someone please give me a link that explains how to find the "lattice constant" of certain atoms.

The "lattice constant" is not a property of atoms. It is a property of crystal lattices - i.e., periodic arrangements of atoms in three dimensions. Basically, the lattice constant is the length of periodicity of the lattice, i.e., the minimum distance at which the lattice repeats itself. For most crystals the lattice constant(s) is few angstroms.

Eugene.

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Gokul43201

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What are you given? If someone just names a crystalline element/compound and asks you to determine the lattice parameters, that's essentially a nonsense question (unless looking up a reference counts as "determining").Will someone please give me a link that explains how to find the "lattice constant" of certain atoms.

Or could someone please walk me through the process of how you find the lattice constants.

If you know the chemical composition, the pure state density and crystal structure, then you may be able to determine it's lattice parameters to some reasonable accuracy.

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I'll post an example of what I am talking about tommorow.

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malawi_glenn

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The lattice constant refers to the constant distance between unit cells in a crystal lattice. Lattices in three dimensions generally have three lattice constants, referred to as a, b, and c. However, in the special case of cubic crystal structures, all of the constants are equal and we only refer to a.

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It seems to me that you would need to know the radius of the atom to find the lattice constant.

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malawi_glenn

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Then the metod for obtaing in experimentally is by doing x-ray diffraction. This thing is covered in almost all basic Solid State physics books, such as Kittel for example.

Just google a bit if you want, there is plenty of sites that contains info. Other keywords is Bragg diffraction.

maybe this site has info too

http://freescience.info/index.php

Do you want to calculate it theoretically or experimentally? What are given in your problem etc..

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OK, here is an example:

Given:

The dimensions of a diamond structure unit cell are (length, width, height) = (a,a,a).

Find:

The lattice constant for the diamond structure.

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malawi_glenn

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why not "a" ?..

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a is not the distance of the two closest atoms.

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malawi_glenn

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your post #8

"The lattice constant refers to the constant distance between unit cells in a crystal lattice. Lattices in three dimensions generally have three lattice constants, referred to as a, b, and c. However, in the special case of cubic crystal structures, all of the constants are equal and we only refer to a."

Your post #11

"The dimensions of a diamond structure unit cell are (length, width, height) = (a,a,a)"

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I found that definition on the internet, so that is not the definition my book uses. In fact the lattice constant is not even defined in my book! (It is Neaman's Semiconductor device physics).

But I am trying to figure out the distance between the two closest atoms in a unit cell.

- #16

malawi_glenn

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well that depends on what definition you have of distance between to atoms, atoms are not "hard" objecs you know..

But if we assume they are hard spheres, then with equal radius, the closest distance should be 0 if as closed packed as possible. Atoms at 000 and (1/4)(1/4)(1/4) are in contact. Are you sure you are not looking for the second closest distance?

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Yeah, I am looking for the second closest distance, sorry.

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Gokul43201

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Typically, this refers to the center to center distance. For instance, in a simple cubic, the nearest neighbor spacing is the same as the lattice parameter, a; in a BCC, it is [itex]a \sqrt{3}/2 [/itex], etc.

well that depends on what definition you have of distance between to atoms, atoms are not "hard" objecs you know..

But if we assume they are hard spheres, then with equal radius, the closest distance should be 0 if as closed packed as possible. Atoms at 000 and (1/4)(1/4)(1/4) are in contact. Are you sure you are not looking for the second closest distance?

For any given crystal structure, you need to first draw the unit cell and the ratio of nearest-neighbor spacing to lattice parameter can be determined geometrically.

PS: Almost no one ever wants to know the second closest (next nearest neighbor) spacing.

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Yeah gokul, that is exactly what I am talking about.

- #20

Gokul43201

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So, for the case of the diamond structure (GaAs, GaN, Si, etc.), the answer is almost already in mg's post#16 : If you know that (0,0,0) and (1/4,1/4,1/4) are the locations of neighboring atoms, then the nearest neighbor spacing [itex]d_{nn}=a\sqrt{(1/4)^2 + (1/4)^2 +(1/4)^2 } = a\sqrt{3}/4 [/itex]

If you do not know the positions of atoms in the unit cell, then it's a few steps longer:

1. Refer to this picture: http://images.google.com/imgres?img...ges?q=diamond+cubic+&svnum=10&um=1&hl=en&sa=N

2. Notice that neighboring atoms are tetrahedrally arranged, with one atom at the vertex and the other at the centroid of the tetrahedron.

3. The centroid of any pyramidal/conical shape is located at 3h/4 from the vertex along the symmetry axis (h is the height).

4. The height of a tetrahedron in [itex]h=\sqrt{6}s/3 [/itex], where s is the side of the tetrahedron.

5. Finally, notice that the other vertices of the tetrahedron lies on face centers, so [itex]s=a/\sqrt{2} [/itex].

Put these together and see that you get the same result as above.

If you do not know the positions of atoms in the unit cell, then it's a few steps longer:

1. Refer to this picture: http://images.google.com/imgres?img...ges?q=diamond+cubic+&svnum=10&um=1&hl=en&sa=N

2. Notice that neighboring atoms are tetrahedrally arranged, with one atom at the vertex and the other at the centroid of the tetrahedron.

3. The centroid of any pyramidal/conical shape is located at 3h/4 from the vertex along the symmetry axis (h is the height).

4. The height of a tetrahedron in [itex]h=\sqrt{6}s/3 [/itex], where s is the side of the tetrahedron.

5. Finally, notice that the other vertices of the tetrahedron lies on face centers, so [itex]s=a/\sqrt{2} [/itex].

Put these together and see that you get the same result as above.

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For a simple cubic or body centred cubic lattice it isn't that hard, but I'm not sure how to calculate for a diamond lattice because of the different layout of atoms inside the lattice.

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