Final answer:
In a body-centered cubic crystal structure, the edge length a and the atomic radius r are related by the formula r = a√3/4. This is derived from the geometry of the unit cell and the application of the Pythagorean theorem to the body diagonal.
Step-by-step explanation:
For a body-centered cubic (BCC) crystal structure, the relationship between the unit cell edge length a and the atomic radius r can be understood by examining the geometry of the unit cell. In a BCC structure, the atoms are positioned at each corner of a cube, and there is also an atom at the very center of the cube. The body diagonal of the cube connects two opposite corners of the cube and passes through the central atom, and this diagonal is equivalent to four atomic radii (the radius of one corner atom, twice the radius of the center atom, and the radius of the opposite corner atom).
By applying the Pythagorean theorem to the cube, we see that the body diagonal d can be found from d = √(a2 + a2 + a2) = a√3. Since the diagonal d is also equal to 4r (the sum of the radii along the body diagonal), we can set these two equal to each other to find the relationship between a and r: a√3 = 4r, which gives r = a√3/4. This formula allows us to calculate the atomic radius if we know the length of the edge of the unit cell, and vice versa.