Cubic Lattice Atom Diameter Calculation

AI Thread Summary
Barium metal has a body-centered cubic lattice structure with a density of 3.50 g/cm³. The mass of one unit cell is calculated to be 4.56 x 10^-22 g, leading to a volume of 1.303 x 10^8 pm³. The cell side length is determined to be 506.9 pm, and the atomic radius is derived from the relationship between the cell diagonal and the side length, yielding a radius of 219.5 pm. The discussion highlights the geometric relationships in different lattice structures and emphasizes the importance of visualizing atomic arrangements to understand these calculations. Understanding these geometric principles is crucial for solving similar problems in crystallography.
Bob Ho
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Homework Statement


Barium metal crystallizes in a body-centred cubic lattice. The density of the metal is 3.50gcm^-3. Calculate the radius(in pm) of a barium atom.

M(Ba)=137.3g/mol , NA = 6.022x10^23/mol

The Attempt at a Solution


For 1 unit cell: m=2x137.3/6.022x10^23
=4.56x10^-22g

V=m/p
=4.56x10^-22/3.5
=1.303x10^8pm^3

Cell side length a=v^(1/3)=506.9pm

From here I am confused, the answer is that the cell diagonal length =4r=√3a
Atomic radius r=√3/4a=219.5pm

Can anyone please explain to me how the cell diagonal length varies with different structures, or anything that can help me, thanks.
 
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It is a (relatively) simple geometry. General approach is to find a plane that goes through atoms centers, draw the atoms and see what is known. You will usually end with some mix of triangles and rectagles or parallelograms where you know some of the distances and you have to find others.
 
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