In a $CsCl$ crystal,the interionic distance between $Cs^+$ and $Cl^-$ ions is:

  • A
    $a$
  • B
    $\frac{a}{2}$
  • C
    $\frac{\sqrt{3}a}{2}$
  • D
    $\frac{2a}{\sqrt{3}}$

Explore More

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 $7.2 \times 10^{-22} \ g$.

$A$ metal $M$ crystallizes into two lattices: face-centered cubic $(fcc)$ and body-centered cubic $(bcc)$ with unit cell edge lengths of $2.0 \ \mathring{A}$ and $2.5 \ \mathring{A}$ respectively. The ratio of densities of the $fcc$ lattice to the $bcc$ lattice for the metal $M$ is $...........$ (Nearest integer).

$A$ 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

$A$ metal crystallises in $bcc$ lattice with unit cell edge length of $300 \ pm$ and density $6.15 \ g \ cm^{-3}$. The molar mass of the metal is

Calculate the number of atoms in $1 \text{ g}$ of a metal that forms a $bcc$ crystal structure. Given that the product of density and unit cell volume is $\rho \times a^3 = 6.6 \times 10^{-22} \text{ g}$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo