A metal crystallises in two cubic phases, fcc | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The density of a unit cell is given by the formula $ho = \frac{Z 	imes M}{N_A 	imes a^3}$, where $Z$ is the number of atoms per unit cell, $M$

Vedclass · Accepted Answer

$1.26$

Vedclass · Answer

(B) The density of a unit cell is given by the formula $ho = \frac{Z 	imes M}{N_A 	imes a^3}$, where $Z$ is the number of atoms per unit cell, $M$ is the molar mass, $N_A$ is Avogadro's number, and $a$ is the edge length.For $fcc$, $Z_{fcc} = 4$ and $a_{fcc} = 3.5 \ \mathring{A}$.For $bcc$, $Z_{bcc} = 2$ and $a_{bcc} = 3 \ \mathring{A}$.The ratio of densities is $\frac{ho_{fcc}}{ho_{bcc}} = \frac{Z_{fcc}}{Z_{bcc}} 	imes \left(\frac{a_{bcc}}{a_{fcc}}ight)^3$.Substituting the values: $\frac{ho_{fcc}}{ho_{bcc}} = \frac{4}{2} 	imes \left(\frac{3}{3.5}ight)^3 = 2 	imes \left(\frac{6}{7}ight)^3 = 2 	imes \frac{216}{343} = \frac{432}{343} \approx 1.26$.

$A$ metal crystallises in two cubic phases, $fcc$ and $bcc$ with edge lengths $3.5 \ \mathring{A}$ and $3 \ \mathring{A}$ respectively. The ratio of densities of $fcc$ and $bcc$ is approximately

Similar Questions

If the edge length of a unit cell of $Li$ is $351 \, pm$, what is the radius of $Li$ (in $pm$)?

Find the radius of a metal atom in a simple cubic unit cell having an edge length of $334.7 \ pm$. (in $pm$)

What is the volume of $1$ mole of a crystalline solid having unit cell edge length $16 \times 10^{-8} \ cm$,if its unit cell contains $24$ molecules?

Calculate the volume of the unit cell when a metal having a density of $1 \ g \ cm^{-3}$ and a molar mass of $23 \ g \ mol^{-1}$ crystallizes to form a $bcc$ structure.

Calculate the density of an element with molar mass $27 \ g \ mol^{-1}$ having $4$ atoms in a unit cell with edge length $405 \ pm$. (in $g \ cm^{-3}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module