Final answer:
The atomic radius of barium in a body-centered cubic structure with an edge length of 5.025 Å is found to be approximately 1.454 Å by using the Pythagorean theorem to relate the body diagonal to the edge length and solving for the radius.
Step-by-step explanation:
Finding the Atomic Radius in a Body-Centered Cubic Unit Cell
The atomic radius of an element within a crystal structure can be calculated if the edge length of the unit cell is known. Barium crystallizes in a body-centered cubic (BCC) lattice, which means that each corner of the unit cell has an atom, and there is one atom at the very center. To find the atomic radius (r) of barium from the given edge length (5.025 Å), we need to use the geometry of the BCC lattice.
In a BCC lattice, the body diagonal (the diagonal that passes through the center of the cube) connects opposite corners through the center atom. This diagonal is constituted from the radii of three atoms: two halves of the corner atoms' radii and the full diameter of the central atom, which equals four radii (4r). This forms a right-angled triangle within the cube where the body diagonal is the hypotenuse, and the sides are edges of the cube with length a.
We can use the Pythagorean theorem to relate the body diagonal to the edge length: a² + a² + a² = (4r)². This simplifies to 3a² = (4r)². To find r, the atomic radius, we solve for r as follows:
r = a / sqrt(3) / 2
Substituting the edge length of 5.025 Å into the formula, we get:
r = 5.025 Å / sqrt(3) / 2 ≈ 1.454 Å
The atomic radius of barium in this BCC structure is, therefore, approximately 1.454 Å.