(a)
The volume of unit cell is given by:
[tex] \rho = \frac{nA}{V_cN_A} [/tex]
where [tex] \rho [/tex] is density, n is the number of atoms in the given HCP crystal, A is the atomic weight of titanium, [tex] N_A [/tex] is the Avogadro's number.
Putting the values in the formula:
[tex] 4.51 = \frac{6\times 47.9}{V_C\times 6.022\times 10^{23}} [/tex]
[tex] V_C = 1.058 \times 10^{-22} cm^{3}/unit cell [/tex]
Thus, the volume of unit cell is [tex] V_C = 1.058 \times 10^{-28} m^{3}/unit cell [/tex]
(b)
The edge length of HCP is given by:
[tex] V_C = 6R^{2}c\sqrt{3} [/tex]
where R is the atomic radius of the atom and c is the height of hexagon.
Substituting [tex] R = \frac{a}{2} [/tex] in the above equation of [tex] V_C [/tex]:
[tex] V_C = 6(\frac{a}{2})^{2}c\sqrt{3} [/tex] = [tex] \frac{3\sqrt{3}a^{2}c}{2} [/tex]
Substituting the values:
[tex] 1.058\times 10^{22} = \frac{3\sqrt{3}a^{2}\times 1.58a}{2} [/tex]
[tex] a = 2.96 \times 10^{8} cm = 0.296 nm [/tex]
[tex] c = 1.58 a [/tex]
[tex] c = 1.58\times 0.296 = 0.468 nm [/tex]
Answer:
(a) [tex]V_C=1.058x10^{-28}\frac{m^3}{UnitCell}[/tex]
(b) [tex]a=2.96x10^{-10}m\\c=4.67x10^{-10}m[/tex]
Explanation:
Hello,
(a) In this case, we apply the following formula in order to solve for the volume per unit cell, [tex]V_C[/tex]:
[tex]\rho =\frac{a*M}{V_C*N_A}[/tex]
Now, a accounts for the atoms into the HCP which are 6, M accounts for the atomic mass of titanium and [tex]N_A[/tex] for the Avogadro number, thus:
[tex]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}[/tex]
(b) Here, for the HCP:
[tex]V_C=6cR^2\sqrt{3}[/tex]
We know that [tex]R=a/2[/tex] and [tex]c/a=1.58[/tex], thus:
[tex]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[/tex]
Best regards.
