Question

If calcium crystallises in bcc arrangement and the radius of the Ca atom is 96 pm, then the volume (in \[{{\rm{m}}^{\rm{3}}}\]) of the unit cell of Ca is:
A. \[{\rm{10}}{\rm{.9 \times 1}}{{\rm{0}}^{{\rm{ - 36}}}}\]
B. \[{\rm{10}}{\rm{.9 \times 1}}{{\rm{0}}^{{\rm{ - 30}}}}\]
C. \[{\rm{21}}{\rm{.8 \times 1}}{{\rm{0}}^{{\rm{ - 30}}}}\]
D. \[{\rm{21}}{\rm{.8 \times 1}}{{\rm{0}}^{{\rm{ - 36}}}}\]

Answer 1

Hint: A lattice or a crystal lattice is a three-dimensional structure of ions, atoms, and molecules in space. A body-centred cubic lattice is a lattice that has eight atoms on the corners and one atom within the body. For a body-centred unit cell, the relationship between edge length(a) and atomic radius(r) is \[{\rm{r = }}\frac{{\sqrt 3 }}{{\rm{4}}}{\rm{a}}\].

Formula Used: \[{\rm{r = }}\frac{{\sqrt 3 }}{{\rm{4}}}{\rm{a}}\]; where a = edge length of the unit cell and r = radius of the atom

Complete Step by Step Solution:
We are given that calcium has a bcc or body-centred cubic lattice.

In this question, the following are the given values:-
- The radius of the Ca atom = 96 pm
We have to find the volume of the unit cell.
- To find out the volume we have to find out the edge length first.
We know that the relationship between edge length (a) and atomic radius (r) is \[{\rm{r = }}\frac{{\sqrt 3 }}{{\rm{4}}}{\rm{a}}\].
So, \[96pm = \frac{{\surd 3}}{4}a\]
\[ \Rightarrow a = \frac{{96 \times 4}}{{1.732}}\]
So, a = 221.7pm

The lattice is a cube, hence, the volume is \[{{\rm{a}}^{\rm{3}}}\].
Hence, volume
\[ = {\left( {221.7 \times {{10}^{ - 12}}m} \right)^3}\]
\[ = 10.89 \times {10^{ - 30}}{m^3}\]
\[ = 10.9 \times {10^{ - 30}}{m^3}\]
So, the volume of the unit cell is \[ = 10.9 \times {10^{ - 30}}{m^3}\].
So, option B is correct.

Note: While attending to the question, one must note the correct unit in each step of the calculation. It must be pointed out that the radius of the atom is given in pm and the volume must be in \[{m^3}\]. So, one must convert the given radius into m. So, one must know how to convert pm into m. \[1pm = {10^{ - 12}}m\].