Question

A metal has bcc structure and the edge length of its unit cell is 3.04 \[\mathop A\limits^o \]. The volume of the unit cell in \[c{m^3}\] will be:
A. \[1.6 \times {10^{21}}c{m^3}\]
B. \[2.81 \times {10^{ - 23}}c{m^3}\]
C. \[6.02 \times {10^{ - 23}}c{m^3}\]
D. \[6.6 \times {10^{ - 24}}c{m^3}\]

Answer 1

Hint:In the body-centered cubic unit cell the volume is given as a cube of edge length which is denoted by \[{(a)^3}\]. The edge length value is given in angstrom and we need to calculate the volume in the unit of \[c{m^3}\].

Complete step by step answer: Given,
Edge length of the unit cell is 3.04 \[\mathop A\limits^o \].
The smallest repeating unit present in the cubic crystal system is known as unit cell.
The body-centered cubic cell (BCC) is the unit cell. In a body- centered cubic cell, the atoms are present in each corner of the cube and at the center of the cube.
In Body centered cubic cell, 8 corners have 1/8 contribution per atom and one body center have 1 contribution per atom.
\[ \Rightarrow 8 \times \dfrac{1}{8} = 1\]atom
\[ \Rightarrow 1 \times 1 = 1\]atom
Thus, the total number of atoms present in a body centered cubic unit cell is 2.
The edge length is given by a.
As it is a cubic unit cell, the volume is given as \[{(a)^3}\].
As, the value of edge length is given in angstrom and we need to calculate the volume in \[c{m^3}\].
First convert the value in angstrom into meters.
3.04 \[\mathop A\limits^o \] = \[3.04 \times {10^{ - 10}}\]
Now, convert meters into centimeter
\[3.04 \times {10^{ - 10}}\]m = \[3.04 \times {10^{ - 8}}\]cm
Substitute the value in cm into the volume formula.
\[ \Rightarrow {(3.04 \times {10^{ - 8}}cm)^3}\]
\[ \Rightarrow 2.81 \times {10^{ - 23}}c{m^3}\]
Thus, the volume of the unit cell in \[c{m^3}\]will be \[2.81 \times {10^{ - 23}}c{m^3}\].
Therefore, the correct option is B.

Note:
Make sure to convert angstrom into meters and then convert meter to centimeter. 1 \[\mathop A\limits^o \] = \[1 \times {10^{ - 10}}\]m and 1 m = 100 cm.