# How is the lattice contant derived in a bcc unit cell?

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.

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?

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 $a=4r/\sqrt{3}$.