Respuesta :
Answer :
(a) The volume of unit cell is, [tex]9.91\times 10^{-23}cm^3/\text{ unit cell}[/tex]
(b) The theoretical density of Ti is, [tex]4.81g/cm^3[/tex]
Explanation :
(a) First we have to calculate the volume of the unit cell.
Formula used :
[tex]V=6r^2c\sqrt {3}[/tex]
where,
V = volume of unit cell = ?
r = atomic radius = [tex]0.1445nm=1.445\times 10^{-8}cm[/tex]
conversion used : [tex](1nm=10^{-7}cm)[/tex]
Ratio of lattice parameter = c : a = 1.58 : 1
So, c = 1.58 a
And, a = 2r
c = 1.58 × 2r
Now put all the given values in this formula, we get:
[tex]V=6\times r^2\times (1.58\times 2r)\sqrt {3}[/tex]
[tex]V=6\times r^3\times (1.58\times 2)\sqrt {3}[/tex]
[tex]V=6\times (1.445\times 10^{-8}cm)^3\times (1.58\times 2)\sqrt {3}[/tex]
[tex]V=9.91\times 10^{-23}cm^3/\text{ unit cell}[/tex]
The volume of unit cell is, [tex]9.91\times 10^{-23}cm^3/\text{ unit cell}[/tex]
(b) Now we have to calculate the density of Ti.
Formula used for density :
[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]
[tex]\rho=\frac{Z\times M}{N_{A}\times V}[/tex] ..........(1)
where,
[tex]\rho[/tex] = density of Ti = ?
Z = number of atom in unit cell = 6 atoms/unit cell (for HCP)
M = atomic mass = 47.87 g/mol
[tex](N_{A})[/tex] = Avogadro's number = [tex]6.022\times 10^{23}atoms/mole[/tex]
[tex]a^3=V[/tex] = volume of unit cell = [tex]9.91\times 10^{-23}cm^3/\text{ unit cell}[/tex]
Now put all the values in above formula (1), we get:
[tex]\rho=\frac{6\times 47.87}{(6.022\times 10^{23})\times (9.91\times 10^{-23})}[/tex]
[tex]\rho=4.81g/cm^3[/tex]
The theoretical density of Ti is, [tex]4.81g/cm^3[/tex]
