Question

Aluminum crystalizes in a cubic close-packed structure. Its metallic radius is 125 pm. How many unit cells are there in 1.00 $c{{m}^{3}}$ of aluminium?
A. 4.42 $\times {{10}^{22}}$
B. 2.36 $\times {{10}^{22}}$
C. 2.26 $\times {{10}^{22}}$
D. 3.92 $\times {{10}^{22}}$

Answer 1

There is a relation between the radius and the length of the side in the unit cell and it is as follows.
$r=\dfrac{a}{2\sqrt{2}}$
Here, r = radius and
a = length of the side in a unit cell.

Complete answer:
- In the question it is asked to calculate how many unit cells are there in an aluminium with a metallic radius of 125 pm.
- To find the number of unit cells first we should calculate the length of the unit cell by using the radius of the unit cell and it is as follows.
$r=\dfrac{a}{2\sqrt{2}}$
Here, r = radius = 125 pm
and
a = length of the side in a unit cell.

& r=\dfrac{a}{2\sqrt{2}} \\\ & a=r\times 2\sqrt{2} \\\ & a=125\times 2\sqrt{2} \\\ & a=353.5pm \\\ \end{aligned}$$ \- By using the length of the unit cell, we can calculate the volume of the one-unit cell and it is as follows. \- The formula to calculate the volume of the unit cell is ${{a}^{3}}$ . \- The volume of one unit cell $$\begin{aligned} & ={{(353.5\times {{10}^{-3}})}^{3}} \\\ & =442\times {{10}^{-25}}c{{m}^{3}} \\\ \end{aligned}$$ \- Therefore, the number of unit cells in aluminium in 1 $c{{m}^{3}}$ is as follows. $$\begin{aligned} & =\dfrac{1}{442\times {{10}^{-25}}} \\\ & =2.26\times {{10}^{22}}unit\text{ }cells \\\ \end{aligned}$$ \- So, the number of unit cells in cells are there in 1.00 $c{{m}^{3}}$ of aluminium is 2.26 $\times {{10}^{22}}$. **So, the correct option is C.** **Note:** We should know the length of the unit cell to calculate the number of unit cells present in a particular area. The length of the unit cell can be calculated by using the radius of the particular metal.