Question

Aluminum crystallizes in a cubic close-packed structure. Its metallic radius is $ 125\;pm $ .
(i) What is the length of the side of the unit cell?
(ii) How many unit cells are there in $ 1.00\;c{m^3} $ of aluminum?

Answer 1

Hint : In general, "closest packed structures" refers to the most densely packed or efficiently spaced crystal structure composition. If the atoms present in that crystal composition are spheres, they must be placed very closely in a way that maximizes the packing efficiency and minimizes the empty space volume, only then it becomes a structure that is closed packed.

Complete Step By Step Answer:
The cubic close packed structures commonly called CCP are one among the different types of closed packed structures.
Every cubic close packing arrangement is able to fill up efficiently $ 74 $ percent of volume.
Such a closed pack structure consists of three recurring surfaces of atoms in a hexagonal arrangement. Each atom makes contact with six atoms within its own level, then with three in the level above it, and with three in the level that is below it. As a result, the layer order is "a-b-c-a-b-c."
Since each atom within such an arrangement has $ 12 $ close neighbors, it has $ 12 $ as its coordination number.
(i) We are required to find the side length of a unit cell in CCP arrangement.
It is given that the radius ( $ r $ ) of the unit cell is $ 125\;pm $ .
Let the side be denoted as ‘ $ a $ ’.
Then we know that for any CCP structure the formula for the radius of it is;
 $ \Rightarrow r = \dfrac{a}{{2\sqrt 2 }} $ , where $ r = $ radius of unit cell
So the side length will be;
 $ \Rightarrow a = 2\sqrt 2 r $
Substituting the value of $ r $ in the formula gives;
 $ \Rightarrow a = 2\sqrt 2 \times (125)\;pm $
 $ \Rightarrow a = 353.5\;pm $
 $ \therefore $ Length of unit cell’s side ( $ a $ ) with the given radius $ = \;353.5\;pm $ .
(ii) Next we require the number of unit cells in the given volume which is unit volume.
For volume of $ 1 $ unit cell in aluminium;
 $ \Rightarrow $ Volume of $ 1 $ unit cell of aluminum $ = {a^3} $
 $ \Rightarrow $ Volume of $ 1 $ unit cell of aluminum $ = {(353 \times {10^{ - 10}})^3}c{m^3} $
 $ \Rightarrow $ Volume of $ 1 $ unit cell of aluminum $ = 442 \times {10^{ - 25}}c{m^3} $
Now number of unit cells in $ 1.00\;c{m^3} $ of aluminum will be;
 $ \Rightarrow $ Number of unit cells $ = \dfrac{{1\;c{m^3}}}{{442 \times {{10}^{ - 25}}\;c{m^3}}} $
 $ \Rightarrow $ Number of unit cells $ = 2.27 \times {10^{22}} $ unit cells
 $ \therefore $ In $ 1.00\;c{m^3} $ of aluminum, we have calculated that there will be $ 2.27 \times {10^{22}} $ unit cells.

Note :
Aluminum is found in a wide range of items. It is not toxic, it has a low density, resistant to corrosion and has a high thermal conductivity. It has no magnetism and it's a very malleable metal. Since aluminum is not particularly powerful on its own, it is frequently used with other metals as an alloy.