# Distance between two atoms that lie along a body diagonal

It is given that two atoms lie along the body diagonal of a unit cell (crystalline solid). The length of such cubic cell is $a=\mathrm{0.336\,nm}$. My book says that we should take the coordinates of the two atoms to be $(0, 0, 0)$ and $(1, 1, 1)$. By my intuition is that if we place the cube at the origin then one atom should be at $(0, 0, 0)$ but the other atom should be at $(0.336, 0.336, 0.336)$. After this we can simply find out the required distance using $$l = a[(x_2-x_1)^2 + (y_2-y_1)^2 +(z_2-z_1)^2]$$ My doubt is that what should be the second coordinate? Should it be $(0.336, 0.336, 0.336)$ or $(1, 1, 1)$? If $(1, 1, 1)$ then why?

• Both answers are right. It's just that they use different units. – Ivan Neretin Nov 4 '16 at 19:35
• @IvanNeretin Can you please elaborate? I mean how both of them are correct? $(0.336, 0.336, 0.336)$ is using nanometre. What unit is $(1, 1, 1)$ using? – Shuvam Shah Nov 4 '16 at 19:41
• Side of the unit cell, of course. It is pretty natural when working with crystals. – Ivan Neretin Nov 4 '16 at 19:56
• It might be worth noting for the future that in may texts and papers on x-ray crystallography the coordinates are often given as fractions of unit cell length i.e. $x/b, y/b, z/c$ and not in regular cartesian coordinates because a unit cell need not be cubic or rectangular. The values of $a, b, c$ are then given separately together with the angles between them. – porphyrin Nov 5 '16 at 8:50