Cubic Lattice Atom Diameter Calculation

[Bob Ho]

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.

 

[Borek]

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.

 

[Blueshift5]

I would first commend the student for their attempt at solving the problem and for seeking clarification on the concept. I would then explain that the cell diagonal length in a body-centered cubic (BCC) lattice is equal to the face diagonal length of the cube, which is √3 times the length of the side of the cube. This is because in a BCC lattice, the atoms are located at the corners and the center of the cube, resulting in a diagonal arrangement. In other lattice structures, such as face-centered cubic (FCC) or simple cubic (SC), the atoms are arranged differently and therefore the cell diagonal length would be different. It is important to understand the differences between lattice structures and how they affect the arrangement and spacing of atoms. I would also recommend practicing more problems and consulting with a teacher or tutor for further clarification.