Final answer:
The atomic radius of niobium in a body-centered cubic unit cell with an edge length of 457.3 pm is calculated as approximately 147.9 pm using the formula r = (√3 * a) / 4.
Step-by-step explanation:
The student has asked how to calculate the atomic radius of niobium, which crystallizes in a body-centered cubic (bcc) unit cell with an edge length of 457.3 pm. In a bcc unit cell, the diagonal across the body of the cube is equal to four radii (4r) and also equal to the square root of three times the edge length (√3a), where 'a' is the edge length. Therefore, the atomic radius 'r' can be calculated using the formula:
r = (√3 * a) / 4
Substituting the given edge length into the formula:
r = (√3 * 457.3 pm) / 4
After calculating, the atomic radius of niobium is found to be approximately 147.9 pm. Consequently, for the question regarding barium's atomic radius in a bcc structure with an edge length of 5.025 Å, one would use the same formula, plugging in the appropriate edge length for barium.