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

Answer:

4.81 g / mol

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

$V_c = 6R^2 \times 3.16 R \times \sqrt3$

$V_c=6 \times 3.16 \times \sqrt3 \times (1.445 \times 10^{-8})^3$

$V_c= 9.91 \times 10^{-23}\ cm^3 / unit \ cell$

Density of Ti (theoretical), $\rho=\frac{n. A_{Ti}}{V_C.N_A}$

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

$\rho=\frac{6 \times 47.87}{9.91 \times 10^{-23} \times 6.023 \times 10^{23}}$

  = 4.81 g/mol

This is the theoretical density of titanium.

The value given in literature is 4.51 g/mol