# Crystal questions

## Homework Statement

I'm calculating the distances between to nearest atoms from various planes an a cubic crystal. For example I know that the distance from the [111] plane to the nearest atom is for a fcc lattice

$$d=\frac{\sqrt{3}a}{3}$$

What I'm unsure about is where the 1/3 factor comes from. I know it has to do with the position of the plane to the atom but the actually calculation escapes me.

Can anyone explain it?

Also when I draw the [110] plane for fcc lattice and try to work out the nearest atom in that direction, I can't find the nearest atom. There doesn't seem to be anything in that direction. Where am I going wrong?

Thanks

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You have a 3-4-5 right triangle.

OlderDan
Homework Helper

## Homework Statement

I'm calculating the distances between to nearest atoms from various planes an a cubic crystal. For example I know that the distance from the [111] plane to the nearest atom is for a fcc lattice

$$d=\frac{\sqrt{3}a}{3}$$

What I'm unsure about is where the 1/3 factor comes from. I know it has to do with the position of the plane to the atom but the actually calculation escapes me.

Can anyone explain it?

Also when I draw the [110] plane for fcc lattice and try to work out the nearest atom in that direction, I can't find the nearest atom. There doesn't seem to be anything in that direction. Where am I going wrong?

Thanks
There is a decent diagram of an fcc (along with many other structures) here

http://www.keele.ac.uk/depts/ch/resources/xtal/xray.html [Broken]

For easier visualization, ignore the y label and make the origin the lower left corner, with x to the right, y into the page, and z vertical. The cluster of 6 atoms diagonally up and to the right from that lower left origin is in a [111] plane, and so is the next cluster of 6 atoms. The lower left and upper right corners are also in [111] planes. From the geometry you can see that the plane spacing is 1/3 the diagonal of the cube. The diagonal squared is 3a². The spacing is d = (1/3)*sqrt(3a²). If you are not convinced of the 1/3 diagonal, you can aways use the coordinates of the three corner points to find the equation of the plane and find its distance from the origin.

A [110] plane would be vertical in the diagram. One such plane would include the two front right corners and the two back left corners with the top and bottom face centers. There is a rectangle in this plane drawn in the diagram. One parallel plane would include the face centers of the front face and the left side face. Another would include the rear face center and the right side face center. The next planes farther out would include the z axis on one side and the two right rear points in the diagram. How far apart are these planes?

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I'm sorry I'm not visualising that last plane. I can recognise the vertical central plane and the right-rear face and left rear face planes.

So in general the idea is not to calculate the distance between atoms in the lattice but between parellel planes normal to the plane you cleaved?

OlderDan
Homework Helper
I'm sorry I'm not visualising that last plane. I can recognise the vertical central plane and the right-rear face and left rear face planes.

So in general the idea is not to calculate the distance between atoms in the lattice but between parellel planes normal to the plane you cleaved?
The planes parallel to the one vertical plane you are seeing only have two atoms showing in the single cube diagram. If you extended the diagram to several cubes, you would be able to see the planes. You don't really need to do that because you know those two atoms are in a plane, and that there is a plane parallel to the one you do see that contains those atoms.

I could not find a diagram of multiple cubes, but here is a reasonable alternative that might help you visualize the planes. If you are Java enabled, you can rotate the diagrams at this site

http://www.le.ac.uk/eg/spg3/atomic.html [Broken]

Sodium Chloride is an fcc structure as you can see if you ignore the blue atoms, but the blue atoms will help you see the plane you are missing. Even though the blue atoms are not in the basic fcc lattice, they are in the planes of that lattice. Move your cursor to the Socium Chloride figure and click once to select it. Now do a single rotation by pointing carefully at the middle of the very top red atom and draging slightly straight downward and then to the right until you see a rectangular outline with the largest blue atom in the middle of the figure. You are now seeing one edge of the cube between two faces looking in the direction of the diagonals of the top and bottom faces. Each vertical line in the figure is a [110] plane. If you click in the middle of the central blue atom, and drag it straight down, all the faces will be revealed and you will still see the vertical planes.

As for the distances, the wording of your post and the equation you posted suggested you ae calculating the distance from a plane to its nearest atoms. That is the same thing as the distance between planes. If you want to calculate the distance between nearest neighbor atoms, that is a different calculation. That would be half the diagonal of one face of the cube.

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The calculations I'm making involve step heights on surfaces. I'm given a cleaved surface in a specified driection i.e. [111] with a monatomic and asked to calculate the height of the step.

I'm assuming that since you cut through the plane drawn the distance is the plane-to-atom distance.

So on the NaCl diagram the distance I would need to calculate is from the blue-blue-blue line diagonally across the centre of the cube, to the red-red line connecting the front and left face-centred atoms?

OlderDan
Homework Helper
The calculations I'm making involve step heights on surfaces. I'm given a cleaved surface in a specified driection i.e. [111] with a monatomic and asked to calculate the height of the step.

I'm assuming that since you cut through the plane drawn the distance is the plane-to-atom distance.

So on the NaCl diagram the distance I would need to calculate is from the blue-blue-blue line diagonally across the centre of the cube, to the red-red line connecting the front and left face-centred atoms?
A monatomic step would be the distance between adjacent planes. I was playing with my photo editor and came up with some diagrams by editing images of the rotated cubes from that site I posted. The image on the left has the [111] planes coming at you and the one on the right has the [110] planes coming at you. I shaded planes that included face centered atoms, but there are planes at the right and left edges of the pictures as well, and one in the middle for the [110]. If you imagine a stack of cubes for the [110], there are face centered atoms above and below the cube shown that lie in the shaded planes. The total distance from left to right for the [111] figure is one cube diagonal, so the spacing is 1/3 of a cube diagonal, as we already discussed. The total distance from left to right for the [110] figure is one face diagonal. So what is the plane spacing?

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Well if I'm reading this right the planes seem to split the diagonal into 4.

Thus $$d=\frac{\sqrt{2}a}{4}$$

OlderDan
Homework Helper
Well if I'm reading this right the planes seem to split the diagonal into 4.

Thus $$d=\frac{\sqrt{2}a}{4}$$
Looks good to me.

I think my problem was I was calculating the distance between atoms instead of planes. When steps are actually successive planes or layers of atoms. It makes more sense to me now.

I swear ever since I started this course I've become more and more stupid than when I started the last.