Gold crystallises in fcc lattice. The edge le | vedclass

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

Question

(B) For an $fcc$ lattice,the edge length $a = 4 \ \mathring{A} = 4 	imes 10^{-8} \ cm$.$1$. The closest distance between atoms $(x)$ in an $fcc$ latt

Vedclass · Accepted Answer

$2\sqrt{2}, 20.52$

Vedclass · Answer

(B) For an $fcc$ lattice,the edge length $a = 4 \ \mathring{A} = 4 	imes 10^{-8} \ cm$.$1$. The closest distance between atoms $(x)$ in an $fcc$ lattice is given by $x = \frac{a}{\sqrt{2}}$.$x = \frac{4}{\sqrt{2}} = 2\sqrt{2} \ \mathring{A} \approx 2.828 \ \mathring{A}$.$2$. The density $(y)$ is given by the formula $d = \frac{Z 	imes M}{N_A 	imes a^3}$.For $fcc$,$Z = 4$. Given $M = 197 \ g \ mol^{-1}$,$N_A = 6 	imes 10^{23} \ mol^{-1}$,and $a = 4 	imes 10^{-8} \ cm$.$y = \frac{4 	imes 197}{6 	imes 10^{23} 	imes (4 	imes 10^{-8})^3} = \frac{788}{6 	imes 10^{23} 	imes 64 	imes 10^{-24}} = \frac{788}{38.4} \approx 20.52 \ g \ cm^{-3}$.Thus,$x = 2\sqrt{2}$ and $y = 20.52$.

Similar Questions

Calculate the number of atoms in $5.4 \ g$ of a metal forming an $fcc$ structure,given that the unit cell volume $\left(a^3\right)$ multiplied by density $\left(\varrho\right)$ is $7.2 \times 10^{-22} \ g$.

$CsBr$ has $bcc$ structure with edge length $4.3 \ \mathring{A}$. The shortest interionic distance in between $Cs^{+}$ and $Br^{-}$ (in $\mathring{A}$) is :

The formula for the determination of the density of a unit cell is:

Metal $M$ crystallizes in $fcc$ lattice. If the edge length of the unit cell is $4.077 \times 10^{-8} \ cm$ and the density is $10.5 \ g \ cm^{-3}$,the atomic mass of the metal is:

Bragg's law is given by the equation

Vedclass Test Series

Exam Paper Generator

Online Exam Module