Question

Tungsten and molybdenum both have the BCC crystal structure, and Mo forms a substitutional solid solution for all concentrations at room temperature. Compute the unit cell edge length for a 84 wt% W - 16 wt% Mo alloy. The room-temperature density and atomic weight of W are 19.3 g/cm3 and 183.85 g/mol, the room-temperature density and atomic weight of Mo are 10.22 g/cm3 and 95.94 g/mol, respectively.

Answer 1

Answer:

the unit cell edge length is 3.1585 × 10⁻⁸ cm

Explanation:

Given the data in the question;    

density = 100 / [ (84/19.3) + (16/10.22) = 100 / 5.917889 = 16.8979 g/cm³

so Aug density p_aug = 16.8979 g/cm³

Aug Atomic weight = 100 / [ (84/183.85 ) + (16/95.94) = 100 / 0.623665

= 160.3424 g/mol

now

volume = ( no. of atoms in BCC × Aug Atomic weight) / ( Aug density × Avogadro's number)

we substitute

volume = ( 2 × 160.3424 ) / ( 16.8979 × 6.023 × 10²³ )

volume =  320.6848 / 1.017760517 × 10²⁵

volume = 3.15088 × 10⁻²³ cm³

Now, unit cell edge length will be;

= $Volume^{1/3}$

= $(3.15088 * 10^{-23})^{1/3}$

= 3.1585 ×  10⁻⁸ cm  

Therefore, the unit cell edge length is 3.1585 × 10⁻⁸ cm