Question

Aluminium crystallizes in a cubic close packed structure. Its metallic radius is $125pm$

A What is the length of the side of the unit cell?

B How many unit cells are there in $1.00c{m}^3$ of aluminum?

Answer 1

Given: Metallic radius of aluminium

$=125pm=125\times {10}^{-12}m$

$\text{Length of the side of the unit cell}$

We know, for cubic close-packed structure:

$a=2\sqrt{2}r$ ............ $\left(1\right)$

Where, ${}^{\prime }{r}^{\prime }$ is the radius of sphere and ${}^{\prime }{a}^{\prime }$ be the edge length of the cube.

Substituting the value of ${}^{\prime }{r}^{\prime }$ in equation $\left(1\right)$, we get

$a=2\times 1.414\times 125=353.55pm\approx 354pm$


Hence, length of the side of the unit cell is $354pm$.

$\text{Final answer:}$ Length of the side of the unit cell is $354pm$.

Given: Metallic radius of aluminium

$=125pm=125\times {10}^{-12}m$

$\text{Number of unit cells}$

Volume of one unit cell $=a^3=(354pm{)}^3$

$=4.4\times {10}^{7}p{m}^3$

$[1pm={10}^{-12}m={10}^{-10}cm]$

$=4.4\times {10}^7\times {10}^{-30}c{m}^3$

$=4.4\times {10}^{-23}c{m}^3$

Number of unit cells $=\frac{\text{Total volume of unit cell}}{\text{volume of one unit cell}}$

Therefore, number of unit cells in $1.00c{m}^3$

$=\frac{1}{4.4\times {10}^{-23}}=2.27\times {10}^{22}$

Hence, number of the unit cells in $1.00c{m}^3$ of aluminium are $2.27\times {10}^{22}$.

$\text{Final answer}:$ Number of the unit cells in $1.00c{m}^3$ of aluminum are $2.27\times {10}^{22}$.