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)
Area Fraction 
At
Area of planeinunit cell ( Ap )
2
Ao   r
2
1
BCC crystal  4r  3a
2
2
 3a   3a 
r1  
  

4
6

 

5 a 2
Ao   r 
48
2
1
OC is the centre to corner distance = Body diagonal  2 = 3a/2
 3a   3a 
3a
h  
  
 
6
 2   3 
r1  r 2  h 2
2


3 2

Ap  12 
Sin
60


a
2a ) 
 ( face diagonal
2


2
BC (111) plane to corner distance = Body diagonal  3 = 3a/3
Looking parallel to the (111) plane
2
 3a   3a 
5 2
r  
  
  a
48
 4   6 
2
1
1 2  r 2 3 a 2
Ac  3(  r ) 

6
2
32
5 a 2 3 a 2
3  19
 5
Ao 

  a2      a2
48
32
 48 32  96
19 2
a
At 96
19
Area Fraction 


 0.717
Ap
3 2 48 3
a
2
3a
2
3a
3