$A$ metal with an atomic radius of $141.4 \, pm$ crystallises in the face-centred cubic structure. The volume of the unit cell in $pm^3$ is $.... . \times 10^7$.

  • A
    $2.74$
  • B
    $2.19$
  • C
    $6.40$
  • D
    $9.20$

Explore More

Similar Questions

$A$ metal crystallises in a face-centred cubic $(FCC)$ structure with a metallic radius of $\sqrt{2} \ \mathring{A}$. The volume of the unit cell (in $m^{3}$) is:

Calculate the number of unit cells in $3 \ g$ of a metal that crystallises in a simple cubic unit cell with an edge length of $336 \ pm$. (Density of metal $= 9.4 \ g \ cm^{-3}$)

Calculate the molar mass of a metal having a density of $22.24 \ g \ cm^{-3}$,which crystallizes to form a unit cell containing $4$ particles. Given $a^3 = 5.6 \times 10^{-23} \ cm^3$.

$A$ solid has a $bcc$ structure. If the distance of nearest approach between two atoms is $1.73 \, \mathring{A}$, the edge length of the cell is ........... $pm$.

For a $Cr$ crystal crystallizing in a $bcc$ structure, the edge length of the unit cell is $287 \, pm$. What is the density of the crystal in $\text{g/cm}^3$? $(Cr = 51.99 \, \text{g/mol})$

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