Question

How many lithium atoms are present in a unit cell with an edge length of $3.5\mathrm{\AA }$ and density of $0.53\mathrm{g}/{\mathrm{cm}}^{-3}$? (Atomic Mass Of $\mathrm{Li}=6.94$):

• A. 1
• B. 2
• C. 4
• D. 6

Answer 1

The correct option is B

2

The explanation for correct option:

Option (B):

Step 1: Given data:

For the lithium atoms present in the unit cell, the data provided are;

Edge Length (a)= $3.5\mathrm{\AA }$,

Density =$0.53\mathrm{g}/{\mathrm{cm}}^{-3}$

Atomic Mass Of Li =$6.94$

Step 2: Formula for calculatingthe number of atoms present in a unit cell:

We know the formula for the density of the crystal cell is given by the expression;

$\mathrm{density}=\frac{\mathrm{Z}\times \mathrm{m}}{{\mathrm{N}}_{\mathrm{A}}\times {\mathrm{a}}^3}$

Where Z = the number of atoms in a unit cell

m = the molecular mass of the atom,

N= the Avogadro's number,

a =is the edge length of the unit cell.

Hence, from this, the expression for calculating the number of atoms can be given as:

$\mathrm{Number}\mathrm{of}\mathrm{atoms}\left(\mathrm{z}\right)=\frac{\mathrm{density}\times {\mathrm{N}}_{\mathrm{A}}\times {\mathrm{a}}^3}{\mathrm{m}}$

Step 3: Calculatingthe number of atmos of Lithium present in the unit cell:

By putting the given values and applying the given formula we can get ;

$$\begin{gathered} \mathrm{number}\mathrm{of}\mathrm{atoms}\left(\mathrm{z}\right)=\frac{\mathrm{density}\times {\mathrm{N}}_{\mathrm{A}}\times {\mathrm{a}}^3}{\mathrm{m}} \\ =\frac{0.53\times 6.023\times {10}^{23}\times {\left(3.8\times {10}^{-8}\right)}^3}{6.94} \\ =2 \end{gathered}$$

Hence, the total number of Li atoms present in the unit cell is given as; 2.