Calculate the number of unit cells in $1 \ cm^3$ volume of an element if unit cell edge length is $4.0 \times 10^{-8} \ cm$.

  • A
    $1.56 \times 10^{22}$
  • B
    $2.63 \times 10^{22}$
  • C
    $3.34 \times 10^{22}$
  • D
    $4.50 \times 10^{22}$

Explore More

Similar Questions

Polonium (atomic mass = $209$) crystallises in a simple cubic structure with a density of $9.32 \ g \ cm^{-3}$. Its lattice parameter (in $pm$) is closest to:

In a body-centered cubic $(BCC)$ unit cell of a metal,the edge length is $4 \times 10^{-10} \ m$. The diameter of the atom will be:

Calculate the molar mass of an element with density $2.7 \ g \ cm^{-3}$ that forms $fcc$ structure. $\left[a^3 \cdot N_{A}=40 \ cm^3 \ mol^{-1}\right]$

The number of atoms in $4.5 \ g$ of a face-centred cubic crystal with edge length $300 \ pm$ is (Given: Density $= 10 \ g \ cm^{-3}$ and $N_A = 6.022 \times 10^{23}$)

$Fe$ crystallizes in a $bcc$ lattice. If the edge length is $286.65 \,pm$ and the density is $7.874 \,g/cm^3$, calculate the Avogadro number. $(Fe = 55.845 \,u)$

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