Answer:
A) i) LD_110 = √3/(4R√2)
ii) LD_111 = 1/(2R)
B)i) LD_110 = 2.4 × 10^(9) m^(-1)
ii) LD_111 = 4 × 10^(9) m^(-1)
Explanation:
A) i) To find linear density expression for BCC 110, first of all we will calculate the length of the vector using the length of the unit cell which is 4R/√3 and the cell edge length which is 4R. Thus, the vector length can now be calculated from this expression;
√((4R)² - (4R/√3)²)
This reduces to; 4R√(1 - 1/3) = 4R√(2/3)
Now, the expression for the linear density of this direction is;
LD_110 =
Number of atoms centered on (110) direction/vector length of 110 direction
In this case, there is only one atom centred on the 110 direction. Thus;
LD_110 = 1/(4R√(2/3))
LD_110 = √3/(4R√2)
ii) The length of the vector for the direction 111 is equal to 4R, since
all of the atoms whose centers the vector passes through touch one another. In addition, the vector passes through an equivalent of 2 complete atoms. Thus, the linear density is;
LD_111 = 2/(4R) = 1/(2R)
B)i) From tables, the atomic radius for iron is 0.124 nm or 0.124 x 10^(-9) m. Therefore, the linear
density for the [110] direction is;
LD_110 = √3/(4R√2) = √3/(4*0.124*10^(-9)(√2))
LD_110 = 2.4 × 10^(9) m^(-1)
ii) for the 111 direction, we have;
LD_111 = 1/(2R) = 1/(2*0.124*10^(-9))
LD_111 = 4 × 10^(9) m^(-1)
