Question

1) A hypothetical metal has the simple cubic crystal structure shown in Figure 3.3. If its atomic weight is 79.4 g/mol and the atomic radius is 0.187 nm, compute its density.

2)Iron (Fe) undergoes an allotropic transformation at 912°C: upon heating from a BCC (α phase) to an FCC (γ phase). Accompanying this transformation is a change in the atomic radius of Fe—from R_BCC = 0.12584 nm to R_FCC = 0.12894 nm—and, in addition, a change in density (and volume). Compute the percentage volume change associated with this reaction. Indicate a decreasing volume by a negative number.

Answer 1

1) For a simple cubic system, 8 atoms occupy 8 corners of a cube and side of this cube is equal to twice the radius. SInce each atom is shared between 8 cells, the effective number of atoms per unit cell is 8 x 1/8 = 1

2 x r = a

2 x 0.187 nm = 0.374 nm = a

density = mass / volume

= ( 79.4 amu x 1.66 x 10^-24 kg ) / ( 0.372 x 10^-9 )³ m³

denisty = 2519.49 kg/m³ = 2.51 gm/cm³

2) For BCC the relation between a and r is   and given r_bcc =0.12584 nm , a_bcc = 0.2906 nm

For FCC the relation between a and r is     and given r_fcc = 0.12894 nm, a_fcc = 0.3647 nm

V_bcc - V_fcc= a³_bcc - a³_fcc = -0.0239 nm³ ( change in volume )

% change in volume = ( change in volume / vol of bcc ) x 100

= (-0.0239 / 0.2906³ ) x 100 = - 97.38 %