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Show for the body-centered cubic crystal structure that the unit cell edge length a and the atomic radius r are related through a=4r/sqrt3

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Final answer:

In a body-centered cubic crystal structure, the unit cell edge length a and atomic radius r are related by the equation a=4r/√3, derived from the body diagonal of the cube representing a right-angled triangle according to the Pythagorean theorem.

Step-by-step explanation:

A body-centered cubic (BCC) crystal structure has atoms at each corner of a cube and one atom at the center of the cube. The relationship between the unit cell edge length a and the atomic radius r is important for understanding the geometry of the crystal structure. In a body-centered cubic structure, a diagonal can be drawn through the cube, passing through the center atom and connecting two opposite corners. This body diagonal can be seen as the hypotenuse of a right-angled triangle whose sides are defined by the edges of the unit cube.

According to the Pythagorean theorem, the diagonal d of the cube can be calculated using the formula d² = a² + a² + a², which simplifies to d² = 3a². Because the diagonal d passes through the center of the lattice and touches two corners, it equals four atomic radii (d = 4r). Put mathematically, 4r² = 3a². By further simplifying, we get that the edge length a is related to the atomic radius r by the equation a = 4r/√3.

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