Areal fraction of atoms on (111) plane of BCC crystal What is the true areal fraction of atoms lying in the (111) plane of a BCC crystal? Areal fraction (area occupied by intersection of atoms with planes  area of the plane) can be calculated in two ways.  Usually, we include only the atoms whose centre lies on the plane (in the areal fraction calculation).  However, in reality some atoms may intersect the plane ‘partially’ (e.g. the body centre atom with the (111) plane). In the example here we include the ‘partially intersecting atom’ in the areal fraction calculation.

Areal fraction of atoms on (111) plane of BCC crystal What is the true areal fraction of atoms lying in the (111) plane of a BCC crystal? The areal fraction (area occupied by atoms: area of the plane) of the (111) plane in BCC crystal is 3/16 = 0.34 (→ taking into account the atoms whose centre of mass lie on the (111) plane). However the (111) plane partially intersects the atom in the body centre position (as shown in the figure below). We now make a calculation of the areal density of atoms taking into account this partial intersection. Video: (111) plane in BCC crystal

Let us consider the view parallel to the (111) plane (perpendicular to the [111] direction) → the (111) plane is the blue line C is the corner of the unit cell O is the centre of the unit cell and the atom at body centering position r is the radius of the atom OC is the centre to corner distance CB is the distance from the corner of the unit cell to the (111) plane Looking parallel to the (111) plane

Area occupied by atoms (At) = Area occupied by corner atoms (Ac) + Area occupied by atom at the centre (Ao) OC is the centre to corner distance = Body diagonal  2 = 3a/2 BC (111) plane to corner distance = Body diagonal  3 = 3a/3 Looking parallel to the (111) plane