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 :(
3 answers
jishka won't let me post the website, but when you google the question its the second result.
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
The length of a diagonal of a face is .2 nm * sqrt 2
so half of that is .1 sqrt 2 = .141 nm
0.15
1 answer