Titanium aluminide (TiAl) is an important intermetallic compound that can be use

Question

Titanium aluminide (TiAl) is an important intermetallic compound that can be used at high temperatures. Ti atoms are located at the corners of a tetragonal unit cell and at two faces of the unit cell; Al atoms are located at four of the faces of the unit cell. The lattice parameters are a_o = 0.399 nm and c_o = 0.407 nm. The atomic mass of titanium is 47.90 g/mol and that of aluminum is 26.98 g/mol. The radius of titanium atom is 0.145 nm and that of aluminum atom is 0.143 nm. Calculate the (a) volumetric density of the titanium aluminide, and (b) atomic packing factor.

Explanation / Answer

a) No. of Ti atoms in TiAl unit cell = (8 X 1/8) + (2 X 1/2) = 2

No. of Al atoms in TiAl unit cell = 4 X 1/2 = 2

Total No. of atoms in TiAl unit cell = Ti atoms + Al atoms = 2+2 = 4

Mass of Ti atoms in TiAl unit cell = (No. of Ti atoms in TiAl unit cell X Atomic mass of Ti) / Avagadro's Number

= (2 X 47.90) / 6.022 X 1023 = 15.91 X 10-23 g

Mass of Al atoms in TiAl unit cell = (No. of Al atoms in TiAl unit cell X Atomic mass of Al) / Avagadro's Number

= (2 X 26.98) / 6.022 X 1023 = 8.96 X 10-23 g

Mass of TiAl unit cell = Mass of Ti atoms + Mass of Al atoms = (15.91 X 10-23) + (8.96 X 10-23)

=  24.87 X 10-23 g

Lattice parameter or side of tetragonal, a0 = 0.399 nm = 0.399 X 10-7 cm

Lattice parameter or side of tetragonal, c0 = 0.407 nm = 0.407 X 10-7 cm

Volume of tetragonal or TiAl = a0 X a0 X c0 = 0.399 X 10-7 X 0.399 X 10-7 X 0.407 X 10-7 (In tertragonal,a0=a0 c0)

Volume of TiAl = 6.479 X 10-23 cm3

Volumetric density of TiAl = Mass of TiAl / Volume of TiAl = 24.87 X 10-23 g / 6.479 X 10-23 cm3 = 3.839 g/cm3

b) Atomic Packing Factor = [No. of atoms X Volume of atoms] / Volume of the unit cell

Atomic Packing Factor = [(No. of Ti atoms X Volume of Ti atom) + (No. of Ti atoms X Volume of Ti atom)] / Volume of TiAl

Radius of Ti atom = 0.145 nm = 0.145 X 10-7 cm, Radius of Al atom = 0.143 nm = 0.143 X 10-7 cm

Volume of Ti atom = 4/3 r3 = 4/3 X 3.14 X (0.145 X 10-7)3 = 1.276 X 10-19 cm3

Volume of Al atom = 4/3 r3 = 4/3 X 3.14 X (0.143 X 10-7)3 = 1.224 X 10-19 cm3

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