Question

Titanium (ti) has an hcp crystal structure, a density of 4.51 g/cm3, and the atomic weight for ti, ati= 47.9 g/mol.(a)what is the volume of its unit cell in cubic meters?(b)if the c/a ratio is 1.58, compute the values of cand
a.

Answer 1

(a)

The volume of unit cell is given by:

$\rho = \frac{nA}{V_cN_A}$

where $\rho$ is density, n is the number of atoms in the given HCP crystal, A is the atomic weight of titanium, $N_A$ is the Avogadro's number.

Putting the values in the formula:

$4.51 = \frac{6\times 47.9}{V_C\times 6.022\times 10^{23}}$

$V_C = 1.058 \times 10^{-22} cm^{3}/unit cell$

Thus, the volume of unit cell is $V_C = 1.058 \times 10^{-28} m^{3}/unit cell$

(b)

The edge length of HCP is given by:

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

where R is the atomic radius of the atom and c is the height of hexagon.

Substituting $R = \frac{a}{2}$ in the above equation of $V_C$:

$V_C = 6(\frac{a}{2})^{2}c\sqrt{3}$ =  $\frac{3\sqrt{3}a^{2}c}{2}$

Substituting the values:

$1.058\times 10^{22} = \frac{3\sqrt{3}a^{2}\times 1.58a}{2}$

$a = 2.96 \times 10^{8} cm = 0.296 nm$

$c = 1.58 a$

$c = 1.58\times 0.296 = 0.468 nm$

Answer 2

Answer:

(a) $V_C=1.058x10^{-28}\frac{m^3}{UnitCell}$

(b) $a=2.96x10^{-10}m\\c=4.67x10^{-10}m$

Explanation:

Hello,

(a) In this case, we apply the following formula in order to solve for the volume per unit cell, $V_C$:

$\rho =\frac{a*M}{V_C*N_A}$

Now, a accounts for the atoms into the HCP which are 6, M accounts for the atomic mass of titanium and $N_A$ for the Avogadro number, thus:

$V_C=\frac{a*M}{\rho *N_A}=\frac{6*47.9g/mol}{4.51g/cm^3*6.022x10^{23}UnitCell/mol}=1.058x10^{-22}\frac{cm^3}{UnitCell}*\frac{1m^3}{100^3cm^3}=1.058x10^{-28}\frac{m^3}{UnitCell}$

(b) Here, for the HCP:

$V_C=6cR^2\sqrt{3}$

We know that $R=a/2$ and $c/a=1.58$, thus:

$V_C=6c(a/2)^2\sqrt{3}\\V_C=6a*1.58(a/2)^2\sqrt{3}\\V_C=16.4a^3/4\\a=\sqrt[3]{\frac{4*1.058x10^{-28}\frac{m^3}{UnitCell}}{16.4} } \\a=2.96x10^{-10}m\\c=1.58a\\c=1.58*2.96x10^{-10}m\\c=4.67x10^{-10}m$

Best regards.