Question

Lead metal has a density of $11.34{\text{ g c}}{{\text{m}}^{ - 3}}$ and crystallizes in a face-centred lattice.
Choose the correct alternatives.
This question has multiple correct alternatives:
(A) The volume of one-unit cell is ${\text{1}}{\text{.214}} \times {\text{1}}{{\text{0}}^{ - 22}}{\text{ c}}{{\text{m}}^3}$
(B) The volume of one-unit cell is ${\text{1}}{\text{.214}} \times {\text{1}}{{\text{0}}^{ - 19}}{\text{ c}}{{\text{m}}^3}$
(C) The atomic radius of lead is ${\text{175 pm}}$
(D) The atomic radius of lead is ${\text{155}}{\text{.1 pm}}$

Answer 1

For a unit cell, you can use the following expression for the density, to obtain the volume of the unit cell.
$V{\text{ = }}\dfrac{{Z \times M}}{{{N_A} \times \rho }}$
You can calculate the edge length of the cubic unit cell by taking the cube root of the volume of the unit cell. You can then obtain the radius of the lead atom by using the following expression.
${\text{r = }}\dfrac{{\sqrt 2 }}{4}{\text{a}}$

Complete step by step answer:
For a unit cell, write the expression for the density, to obtain the volume of the unit cell.
$\rho {\text{ = }}\dfrac{{Z \times M}}{{{N_A} \times V}}$
Here, $\rho$ is the density of the unit cell, $Z$ is the number of atoms present in one unit cell, $M$ is the molecular weight of lead, ${N_A}$ is the Avogadro’s number and $V$ is the volume of the unit cell.
Rearrange the above expression to obtain an expression for the volume of the unit cell.
$V{\text{ = }}\dfrac{{Z \times M}}{{{N_A} \times \rho }}$
Substitute values in the above expression and obtain the value for the volume of the unit cell.

{\Rightarrow V \text{ = }}\dfrac{{4 \times 207}}{{6.023 \times {{10}^{23}} \times 11.34}} \\\ \Rightarrow V= 1.214 \times {10^{ - 22}}{\text{ c}}{{\text{m}}^3} \\\ $$ Hence, the volume of one-unit cell is $${\text{1}}{\text{.214}} \times {\text{1}}{{\text{0}}^{ - 22}}{\text{ c}}{{\text{m}}^3}$$ The option (A) is the correct option and the option (B) is the incorrect option. Calculate the edge length of the cubic unit cell by taking the cube root of the volume of the unit cell. $$a{\text{ = }}{\left( V \right)^{1/3}} \\\ \Rightarrow a= {\left( {1.214 \times {{10}^{ - 22}}{\text{ c}}{{\text{m}}^3}} \right)^{1/3}} \\\ \Rightarrow a= 4.95 \times {10^{ - 8}}{\text{ cm}} \\\\$$ Write the expression for the relationship between the radius of the lead atom and the edge length of the unit cell. $${\text{r = }}\dfrac{{\sqrt 2 }}{4}{\text{a}}$$ Here, $${\text{r}}$$ is the radius of the lead atom and $${\text{a}}$$ is the edge length. $${\text{r = }}\dfrac{{\sqrt 2 }}{4}{\text{a}} \\\ {\text{ = }}\dfrac{{\sqrt 2 }}{4} \times \left( {4.95 \times {{10}^{ - 8}}{\text{ cm}}} \right) \\\ \Rightarrow r = 1.75 \times {10^{ - 8}}{\text{ cm}} \\\ \Rightarrow r= 175 \times {10^{ - 10}}{\text{ cm}} \\\\$$ Convert the unit from cm to pm. $${\text{r }} = 175 \times {10^{ - 10}}{\text{ cm }} \times {\text{ }}\dfrac{{{\text{1 pm}}}}{{1 \times {{10}^{ - 10}}{\text{ cm }}}} \\\\$$ $\Rightarrow r= 175 pm $ Hence the atomic radius of lead is 175 pm. Thus, the option (C) is the correct option and the option (D) is the incorrect option. _**Therefore the correct options are (A) and (C).**_ **Note:** The density of the unit cell is the ratio of the total mass of the constituent particles of the unit cell to the total volume of the unit cell. To obtain volume occupied by the constituent particles of the unit cell, you should multiply the mass of one constituent particle (atom or ion) with the total number of constituent particles present in a one-unit cell. You can obtain the mass of one constituent particle by dividing the molecular weight with Avogadro’s number.