Question

titanium has an hcp unit cell which has a ratio of the lattice parameter c/a = 1.58. if the radius of Ti is 0.1442nm,(a) determine the unit cell volume and (b) calculate the density​

Answer 1

Answer:

Volume:

If a solid body is kept under the three-dimension space then the space taken by a body gives the volume of that particular object. Volume is generally denoted by letter (V) and it is measured in cubic units under the SI system. The volume determines how much a body will take space to fit into that particular three-dimension space.

The titanium has a

H

C

P

unit cell for which lattice parameter

c

/

a

is

1.58

The radius of the...

See full answer below.

Answer 2

Given: The ratio of lattice parameter $(c/a)=1.58$

Radius of Titanium $=0.1442nm$

To find: (a) The unit cell volume (b) The density

Solution:

The ratio of lattice parameter $(c/a)=1.58$

Radius of Titanium $=0.1442nm$

(a) The formula for volume is;

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

It is given that for Ti, $c=1.58$, $a=2R$, ∴$c=3.16R$

∴ $V_C=(6)*(3.16)*R^3\sqrt{3}$

⇒ $V_C=(6)*(3.16)*(\sqrt{3})[1.445*10^{-8}cm]^3$

⇒ $V_C=9.91*10^{-23}cm^3/unit-cell$

∴ The volume of the unit cell is $V_C=9.91*10^{-23}cm^3/unit-cell$

(b) The formula for the density is given by,

$d=\frac{nA_T_i}{V_CN_A}$

where $n$ is the number of atoms per unit cell, $N_A$ is the Avogadro number.

$d=\frac{6*47.87}{9.91*10^{-23}*6.023*10^23}$

$d=4.81g/cm^3$

∴ The density of the unit cell of Titanium is $4.81g/cm^3$