Answer:
[tex]4\ R=\sqrt 3\ a[/tex]
Explanation:
Given that
Lattice constant = a
Radius of unit cell cell =R
Atom is in BCC structure.
In BCC unit cell (Body centered cube)
1.Eight atoms at eight corner of cube which have 1/8 part in each cube.
2.One complete atom at the body center of the cube
So the total number of atoms in the BCC
Z= 1/8 x 8 + 1 x 1
Z=2
In triangle ABD
[tex]AB^2=AD^2+BD^2[/tex]
[tex]AB^2=a^2+a^2[/tex]
[tex]AB=\sqrt 2\ a[/tex]
In triangle ABC
[tex]AC^2=AB^2+BC^2[/tex]
AC=4R
BC=a
[tex]AB=\sqrt 2\ a[/tex]
So
[tex]16R^2=2a^2+a^2[/tex]
[tex]4\ R=\sqrt 3\ a[/tex]
So the relationship between lattice constant and radius of unit cell
[tex]4\ R=\sqrt 3\ a[/tex]
