2
$\begingroup$

I have a text saying:

The atoms touch along a $\langle111\rangle$ direction and this is referred to as the close-packed direction. The lattice parameter $a=4r/\sqrt{3}$ and the spacing of atoms along $\langle110\rangle$ directions is $a\sqrt{2}$.

I am trying to verify this lattice constant $a$. On the picture below $a$ is shown. The left image is the bcc unit cell and the right a $(110)$ plane (indicated in green to the left).

On the right is an arrow showing a close-packed direction where the atoms meet, as the text says.

enter image description here

In deriving $a$, I take a right-angled triangle like the one marked blue above. $a$ is the hypetenuse, and since the atoms meet (according to the text) along the catheti (each of the other two legs), these would each be 2 times the radius, $2r$.

From Pythagoras:

$$a^2=(2r)^2+(2r)^2\quad\Leftrightarrow\quad a=\sqrt{4r^2+4r^2}=\sqrt{8r^2}=\sqrt{8}r\approx2.83r$$

So not the same answer as $a=4r/\sqrt{3}\approx 2.30 r$.

Am I mixing up the directions? Or where is the mistake?

$\endgroup$
3
$\begingroup$

The point of the calculation is to derive a relationship between the lattice constant $a$ and the atomic radius $r$; you can't "derive" either but you can relate them to each other.

The key to the relationship is the diagonal that joins opposite vertices of the unit cell, going through the midpoint, whose length is given by (the three-dimensional) Pythagoras's theorem as $$L=\sqrt{a^2+a^2+a^2}=\sqrt{3}a.$$ This length crosses through half of the atom in one vertex, the full length of the midpoint atom, and half of the atom in the other vertex, and since you're guaranteed that the atoms touch in the $⟨111⟩$ direction then this completely covers the length of the diagonal, giving you $$L=r+2r+r=4r.$$ Putting the two together you get $a=4r/\sqrt{3}$.

$\endgroup$
2
  • $\begingroup$ I see, that is a clear derivation, thank you. Do you by any chance see the mistake in my try? $\endgroup$
    – Steeven
    Aug 4 '16 at 18:11
  • $\begingroup$ @Steeven To be honest with you I can't make either heads or tails of the workings in the question. $\endgroup$ Aug 4 '16 at 18:52
2
$\begingroup$

Your mistake is to assume that the blue triangle has a right angle. The triangle is not in the horizontal plane. The atom on the right is above the other two. The two blue lines that you consider as sides of a right angle are actually segments of the body diagonals. The angle between them is about 109 degrees.

$\endgroup$
1
  • $\begingroup$ Welcome to Physics.SE! I suggest the following: 1) Take the tour! 2) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. 3) If you have a good question, ask it! Just search for duplicates and follow the help center rules. $\endgroup$ Aug 4 '16 at 20:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.