In a body-centred cubic $(bcc)$ lattice of potassium, the correct relation between the atomic radius $(r)$ of potassium and the edge-length $(a)$ of the cube is:

  • A
    $r = \frac{a}{\sqrt{2}}$
  • B
    $r = \frac{a}{\sqrt{3}}$
  • C
    $r = \frac{\sqrt{3}}{2} a$
  • D
    $r = \frac{\sqrt{3}}{4} a$

Explore More

Similar Questions

$A$ metal has $BCC$ structure with edge length of unit cell $400 \ pm$. The density of the metal is $4 \ g \ cm^{-3}$. What is the molar mass of the metal?

An element has a density of $6.8 \ g \ cm^{-3}$ and crystallizes in a $bcc$ structure with a unit cell edge length of $290 \ pm$. The number of atoms in $200 \ g$ of the element is:

Difficult
View Solution

Calculate the number of unit cells in $0.4 \ g$ of metal if the product of density and volume of the unit cell is $1.2 \times 10^{-22} \ g$.

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 $= 9.4 \ g \ cm^{-3}$)

$A$ metal crystallises with a face-centred cubic $(fcc)$ lattice. The edge length of the unit cell is $408 \, pm$. The diameter of the metal atom is ............. $pm$.

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