Question

Titanium has an HCP unit cell for which the ratio of the lattice parameters c/a is 1.58. If the radius of the Ti atom is 0.1445 nm, calculate the density of Ti and compare it with the literature value of 4.51 g/cm3 . The atomic mass of titanium is 47.87 g/mol.

Answer 1

4.81 g / mol

Step-by-step explanation:

Given :

Titanium has an HCP unit cell

Radius of titanium, R = 0.1445 nm

Unit cell volume,
`$V_c= 6R^2 C\sqrt3$`

But for Ti, c/a = 1.58

So, c = 1.58 a

And a = 2R or c = 3.16 R

`$V_c = 6R^2 * 3.16 R * \sqrt3$`

`$V_c=6 * 3.16 * \sqrt3 * (1.445 * 10^(-8))^3 $`

`$V_c= 9.91 * 10^(-23)\ cm^3 / unit \ cell $`

Density of Ti (theoretical),
`$\rho=(n. A_(Ti))/(V_C.N_A) $`

For HCP, n = 6 atoms per unit cell and atomic mass = 47.87 g/mol

`$\rho=(6 * 47.87)/(9.91 * 10^(-23) * 6.023 * 10^(23)) $`

= 4.81 g/mol

This is the theoretical density of titanium.

The value given in literature is 4.51 g/mol