Question

Determine size of Cs as a compared Li atom, if their densities $1.86g/cc$ and $0.53g/cc$ respectively.

Answer 1

Hint: We must assume first about three crystalline in the same manner. Then we would find the atomic mass of both cesium and lithium. Then we would find the edge length of the both Lithium and Cesium. Later will compare both of them to find the ratio of edge length Cesium to Lithium. It gives us the relation between the edge lengths of both.

Complete step by step answer:
Step1: First of all we cannot find the relation until we do not assume that both Lithium and Cesium are crystal lined in the same manner which would say that $(Z = same)$ .
Step2. The atomic mass of the Cesium is $133$ . The atomic mass of Lithium mass is 7.
Step3. We have given two atoms in question, Cesium and Lithium.
The density of Cesium is $1.86g/cc$ .
The density of Lithium is $0.53g/cc$ .
Step4. The formula of density is
$Density = \dfrac{{Z \times Mass}}{{{N_A} \times {a^3}}}$ , the Z is the number of atoms per unit cell, ${N_A}$ is Avogadro number, ‘a’ is the length of the edge.
Now let us talk about the edge of Cesium.
$ \Rightarrow 1.86 = \dfrac{{Z \times 133}}{{{N_A} \times a_1^3}}$ , ${a_1}$ is the edge length of Cesium. $eq1$
$ \Rightarrow 0.53 = \dfrac{{Z \times 7}}{{{N_A} \times a_2^3}}$ , ${a_2}$ is the edge length of Lithium $eq2$
Step5.
Now compare both the equations: $\dfrac{{eq1}}{{eq2}}$
$ \Rightarrow $$\dfrac{{1.86a_1^3}}{{133}} = \dfrac{{0.53a_2^3}}{7}$
$
   \Rightarrow \dfrac{{a_1^3}}{{a_2^3}} = 5.41 \\
    \\
$
$ \Rightarrow \dfrac{{{a^1}}}{{{a^2}}} = 1.756$
Step6. So the ratio of the edge lengths is $1.756$ . So the edge length of Cs is $1.756$ times of Li.

Note:
Lattice is a three-dimensional array of points on which a crystal is built. A crystal lattice is made up of a very large number of unit cells. The unit cell can be viewed as a three dimensional structure with more than one atoms. We can even determine the volume of a unit cell by knowing the dimensions of the unit cell.