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, (a) determine the unit cell volume, and (b) calculate the density of Ti and compare it with the literature value

Answer 1

The unit cell volume of the crystal is $V_c = 9.9084*10^-^2^3 cm^3 / unit cell$ and the density of titanium is calculated as 4.71g/cm^3 while the literature value is 4.5 g/cm^3

Data;

• radius = 0.1445 nm
• c/a = 1.58
• A = 46.88 g/mol

Unit Cell Volume

The unit cell volume can be calculated as

$V_c = 6R^2c\sqrt{3}$

let's substitute the values into the formula

$V_c = 6 * (1.445*10^-^8)^2 * 1.58 8 * a * \sqrt{3} \\a = 2R\\V_c = 6 * (1.445*10^-^8)^2 * 1.58* 2 * 1.445*10^-^8 * \sqrt{3}\\ V_c = 9.984*10^-^2^3 cm^3 / unit cell$

The unit cell volume of the crystal is $V_c = 9.9084*10^-^2^3 cm^3 / unit cell$

Density of Ti

The density of titanium can be calculated as

$\rho = \frac{nA}{V_c N_a}$

• n = 6 for hcp
• A = 46.88 g/mol
• Na = Avogadro's number

let's substitute the values into the formula

$\rho = \frac{nA}{V_c Na}\\ \rho = \frac{6*46.88}{9.9084*10^-^2^3* 6.023*10^2^3} \\\rho = 4.71 g/cm^3$

The density of titanium is calculated as 4.71g/cm^3 while the literature value is 4.5 g/cm^3

Learn more on crystal lattice here;

brainly.com/question/6610542