Question

Copper crystallizes in face centred cubic lattice and has a density of $ 8.930g.c{m^{ - 3}} $ at $ 298K $ . Calculate the radius of the copper atom. (At mass of $ Cu = 63.55u,{N_A} = 6.02 \times 1023mo{l^{ - 1}} $ ).

Answer 1

Hint :In Face centred cubic lattice the total number of atoms per unit cell is 4, as there are 8 atoms at eight corners and 6 at face centres. Hence,
$ z = \dfrac{1}{8} \times 8 + \dfrac{1}{2} \times 6 \\
z = 1 + 3 \\
z = 4 $
Also, In Face centred cubic lattice the relation between edge length(a) and radius of atom(r) is $ a = 2\sqrt {2r} $ .

Complete Step By Step Answer:
From the data given in the question, we know that
 $z = 4 $
 $ a = 2\sqrt {2r} $
 $ {N_A} = 6.02 \times {10^{23}} $
 $ \rho = 8.930 $
 $ M = 63.55 $
Where ,
 $ z $ is the number of atoms in each unit cell
 $ a $ is the edge length
 $ {N_A} $ is the number of atoms in one mole
 $ \rho $ is the density of copper
 $ M $ is the mass of copper.
We also know that the formula to calculate the density are as follows:
 $ \rho = \dfrac{{z \times M}}{{{a^3} \times {N_A}}} $
Substituting the values into the formula we will get,
 $ 8.930 = \dfrac{{4 \times 63.55}}{{{a^3} \times 6.02 \times {{10}^{23}}}} \\
  {a^3} = \dfrac{{4 \times 63.55}}{{8.930 \times 6.02 \times {{10}^{23}}}} \\
  {a^3} = 4.728 \times {10^{ - 23}} $
We know that $ a = 2\sqrt {2r} $
 $ {(2\sqrt 2 r)^3} = 4.728 \times {10^{ - 23}} \\
  2\sqrt 2 r = \sqrt[3]{{4.728 \times {{10}^{ - 23}}}} \\
  2\sqrt 2 r = 3.616 \times {10^{ - 8}} \\
  r = \dfrac{{3.616 \times {{10}^{ - 8}}}}{{2\sqrt 2 }} \\
  r = 1.278 \times {10^{ - 8}}cm $
Hence, the radius of copper atoms is $ 1.278 \times {10^{ - 8}}cm $ .

Note :
A unit cell is an arrangement of similar atoms which forms a pattern. The density of a unit cell is the ratio of the mass of the cell to the volume of the cell and the mass is calculated as the product of total number of atoms and the individual mass of the atom.