Many metals pack in cubic unit cells. The density of a metal and length of the u

Question

Many metals pack in cubic unit cells. The density of a metal and length of the unit cell can be used to determine the type for packing. For example, iron has a density of 7.87 g/cm3 and a unit cell side length a of 2.87 A. (1 A = 1, x 10-8 cm.) (a) How many iron atoms are in exactly 1 cm? atoms (b) How many unit cells are in exactly 1 cm3? unit cells (c) How many iron atoms are there per unit cell? atom(s) (d) The atoms/unit cell suggests that iron packs as a Select.unit cell. Knowledge of the type of packing and side length a of the unit cell will allow calculation of the atomic radius, using one of the equations below. You may need to draw the unit cell to determine which formula to use to calculate the atomic radius. (bd) - a + (fd) (bd)' = a' + 2a2 (bd) =13 a = 4r =a+ 3 a 4

Explanation / Answer

Answer – We are density of Fe = 7.87 g/cm3 , a = 2.87 Ao = 2.87*10-8cm

a) Atoms are in 1 cm3

We need to calculate the average mass of one atom of Fe

= 55.845 g.mol-1 / 6.023*1023 atoms.mol-1

= 9.27*10-23 g/atom

Atoms are in 1 cm3 = 7.87 g. cm-3/ 9.27*10-23 g/atom

                                 = 8.49*1022 atoms /cm3

b) Unit cell in 1 cm3

Volume of the unit cell –

V = (a)3

        = (2.87*10-8cm)3

       = 2.36*10-23 cm3

So, unit cell in 1 cm3

= 1 cm3 /2.36*10-23 cm3

= 4.23*1022 unit cell .

c) Atoms per unit cell

we calculated atoms and unit cell , so

= 8.49*1022 atoms / 4.23*1022 unit cell

= 2 atoms per unit cell

d) We know when there are 2 atoms per unit cell then there is body centered cubic.

So iron packs as body-centered unit cell.

e) We know the radius foe the body-centered cubic cell is

d2 + (d 2)2 = (4r)2

d2 + 2d2 = 16r2

3d2 = 16r2

r2 = 3d2 / 16

r = 3 (d / 4)

   = 3 (2.87 / 4)

   = 1.24 Ao

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