A crystalline solid consists of atoms stacked up in a repeating lattice structure. The atoms of a crystal reside at the corners of cubes of side

L = 0.200 nm. One piece of evidence for the regular arrangement of atoms comes from the flat surfaces along which a crystal separates, or cleaves, when it is broken. Suppose this crystal cleaves along a face diagonal. Calculate the spacing, d, between two adjacent atomic planes that separate when the crystal cleaves.

i have absolutely no clue how to do this question, i googled it to see if any help would come up and i found the following website, but i just don't understand their explanation. also the axes of their diagrams are flipped as well, which confuses things even further. please please help :(

Well, I have four Kleenex boxes piled on top of each other on my desk 2 below and 2 above looking at the ends and have drawn parallel diagonals on the faces. It looks to me like the distance between splitting planes is half the length of a diagonal of the face. (From the center of a face to the corner)

The length of a diagonal of a face is .2 nm * sqrt 2
so half of that is .1 sqrt 2 = .141 nm

jishka won't let me post the website, but when you google the question its the second result.

0.15

I can definitely help you with this problem! Let's break it down step by step.

First, we need to understand what a face diagonal is. In a crystal, the face diagonals are the lines that connect opposite corners of a face. For example, if you take a cube and draw a line from one corner to the opposite corner of the adjacent face, you would be drawing a face diagonal.

The next concept we need to understand is atomic planes. In a crystalline solid, the atoms are arranged in a repeating lattice. These lattices can be thought of as layers of atoms stacked on top of each other. Each layer of atoms forms an atomic plane.

Now, when a crystal cleaves along a face diagonal, it means that the cleavage separates two adjacent atomic planes. We want to calculate the spacing, denoted as d, between these two atomic planes.

To find the spacing between atomic planes, we need to use the concept of lattice parameters. In this problem, the given lattice parameter is L = 0.200 nm. The lattice parameter represents the distance between the atoms in the same plane.

Here's how you can calculate the spacing, d:

1. Draw a cube to represent the crystalline solid.

2. Identify the face diagonal along which the crystal cleaves. Let's call this diagonal as "d1".

3. Identify the atomic plane adjacent to the face diagonal. Let's call this plane as "p1".

4. To find the spacing between p1 and the next atomic plane (let's call it "p2"), we can use the relationship between the face diagonal (d1) and the lattice parameter (L) of the cube. The diagonal of a face of the cube can be calculated using the Pythagorean theorem, which gives:

d1 = √(2L^2)

Substitute the given value of L to find the value of d1.

5. Now, since d1 connects two corners of the cube, it passes through one atomic plane (p1) and intersects the next atomic plane (p2). The spacing between p1 and p2 is equal to d1 divided by the number of atomic planes it passes through. In this case, d1 intersects two atomic planes (p1 and p2).

So, the spacing between p1 and p2 (represented as d) can be calculated as:

d = d1 / 2

Substitute the value of d1 that you calculated earlier to find the value of d.

By following these steps, you should be able to calculate the spacing, d, between two adjacent atomic planes when a crystal cleaves along a face diagonal.

I hope this explanation helps you understand the process better! Let me know if you have any further questions.