Question

A metal crystallises in two cubic phases , fcc and bcc whose unit cell lengths are $3.5\;$A and $3.0$ A respectively. The ratio of density of fcc and bcc is:
(A)$2$
(B) $1.26$
(C) $3.34$
(D) $1.8$

Answer 1

Hint: BCC is known as body centred unit cell . It is a type of arrangement in which atoms are present at the corners as well as the body centre. The number of atoms in body centred unit cell. A face centred unit cell contains atoms at the corners as well as the face centres.

Complete step-by-step solution:We know that for a face centred unit cell that each atom located at the face centre is shared between two adjacent unit cells and only half of each atom belongs to the unit cell. In the question it is given that cell length of face centred cubic cell is equal to $3.5\;$A. And
Cell length of body centred cubic is equal to $3.0$ A. We have to calculate the densities of the face centred cubic cell and body centred cubic cell.
We know that for a unit cell of mass M , number of atoms per unit cell Z ,edge length a and Avogadro number ${N_A}$, we get
 $d = \dfrac{{ZM}}{{{N_A}{a^3}}}$
For face centred cubic cell effective number of atoms is equal to four and similarly for body centred cubic unit cell it is two.
So ${d_{fcc}} = \dfrac{{4 \times {M_W}}}{{{{(3.5 \times {{10}^{ - 8}})}^3} \times {N_A}}}$
 $\Rightarrow {d_{bcc}} = \dfrac{{2 \times {M_W}}}{{{{(3 \times {{10}^{ - 8}})}^3} \times {N_A}}}$
 $\Rightarrow \dfrac{{{d_{fcc}}}}{{{d_{bcc}}}} = \dfrac{{4 \times {3^3}}}{{{{(3.5)}^3} \times 2}} = 1.26$
Hence the ratio of density of fcc and bcc is $1.26$

Therefore the correct answer is option (B).

Note:Packing efficiency is the percentage of total space filled by the constituent particles in the unit cell. Packing efficiency = Packing Factor x 100. A vacant space not occupied by the constituent particles in the unit cell is called void space. The fraction of void space = 1 – Packing Fraction.The packing efficiency for fcc unit cell is $78\% $ and for bcc it is $68\% $.