Question

The density of krypton (face centered cubic cell) is \[3.19{\text{ g / cc}}\]. What is the radius of an atom?
(Atomic weight of \[{\text{Kr}} = 84\] )
A.\[0.198{\text{ nm}}\]
B.\[0.466{\text{ nm}}\]
C.\[0.206{\text{ nm}}\]
D.\[0.225{\text{ nm}}\]

Answer 1

Hint: For this we must use the formula for calculation of radius in a unit cell. We will calculate the number of atoms present in the face centered unit cell and will put in the formula along with other known variables. We need to use the relation between edge length and radius to find the final value.
Formula used:
\[{\text{d}} = \dfrac{{{\text{Z}} \times {\text{M}}}}{{{{\text{N}}_{\text{a}}} \times {{\text{a}}^3}}}\]
Here d is density, Z is number of atoms in a unit cell, M is molecular weight, \[{{\text{N}}_{\text{a}}}\] is Avogadro number and a is the edge length of atom in a unit cell.

Complete step by step answer:
A face centered unit cell is that unit cell in which atoms are present on the corners as well as the center of each face in a cube. There are 8 corners and 6 faces in a cube. The contribution of atoms on the corner is one eighth and the contribution of the atoms present on the face is half. Hence the total number of atoms in a unit cell will be:
\[{\text{N}} = 8 \times \dfrac{1}{8} + 6 \times \dfrac{1}{2} = 4\]
Hence the number of atoms comes out to be 4 in a face centered unit cell.
Molecular weight of krypton is given to us as 84.
\[{{\text{N}}_{\text{a}}}\] is a constant whose value is \[{\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{23}}\]
Rearranging the formula we will get,
\[{{\text{a}}^3} = \dfrac{{{\text{Z}} \times {\text{M}}}}{{{{\text{N}}_{\text{a}}} \times {\text{d}}}}\]
Now we will substitute the values,
\[{{\text{a}}^3} = \dfrac{{4 \times 84}}{{{\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{23}} \times 3.19}}\]
\[ \Rightarrow {{\text{a}}^3} = 175 \times {10^{ - 24}}{\text{c}}{{\text{m}}^3}\]
Taking cube root both sides we will get,
\[{\text{a}} = 5.59 \times {10^{ - 8}}{\text{cm}}\]
For a face centered unit cell the relation between the edge length and radius, r of atom is
\[\sqrt 2 \times a = 4{\text{r}}\]
Substituting the value we will get,
\[{\text{r}} = \dfrac{{\sqrt 2 \times 5.59 \times {{10}^{ - 8}}}}{4}{\text{cm}}\]
Solving this, we get:
\[{\text{r}} = 0.198 \times {10^{ - 9}}{\text{ cm}} = 0.198{\text{ nm}}\]

Hence the correct option is A.

Note:
There are various types of unit cell such as simple cubic unit or primitive unit cell in which atoms are only present at corners, body centered unit cell, face centered unit cell and edge centered unit cell.