For a simple cubic system,find the ratio of t | vedclass

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

Question

(A) The interplanar distance $d$ for a cubic system is given by the formula: $d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}$For the planes $(100)$,$(110)

Vedclass · Accepted Answer

$1 : 1/\sqrt{2} : 1/\sqrt{3}$

Vedclass · Answer

(A) The interplanar distance $d$ for a cubic system is given by the formula: $d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}$For the planes $(100)$,$(110)$,and $(111)$,the distances are:$d_{100} = \frac{a}{\sqrt{1^2 + 0^2 + 0^2}} = a$$d_{110} = \frac{a}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{a}{\sqrt{2}}$$d_{111} = \frac{a}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{a}{\sqrt{3}}$Therefore,the ratio $d_{100} : d_{110} : d_{111}$ is:$a : \frac{a}{\sqrt{2}} : \frac{a}{\sqrt{3}} = 1 : \frac{1}{\sqrt{2}} : \frac{1}{\sqrt{3}}$

For a simple cubic system,find the ratio of the interplanar distances between the $(100)$,$(110)$,and $(111)$ planes.

Similar Questions

Copper crystallises in $fcc$ structure with a unit cell length of $361 \ pm$. What is the radius of copper atom? ............ $pm$

What is the number of atoms in $1000 \ g$ of an $FCC$ crystal? Given density $= 10 \ g \ cm^{-3}$ and edge length $= 200 \ pm$.

$A$ metal crystallises in two phases,one as $fcc$ and other as $bcc$ with unit cell edge lengths of $3.5 \ \mathring{A}$ and $3.0 \ \mathring{A}$ respectively. The ratio of density of $fcc$ and $bcc$ phases approximately is

Calculate the density of a metal which forms a simple cubic structure with an edge length of unit cell $336 \ pm$. ($90 \ g$ of metal contains $2.64 \times 10^{23}$ atoms) (in $g \ cm^{-3}$)

The distance between two nearest $Ag$ atoms is $2.5 \times 10^{-8} \ cm$. The density of its crystal is .......... $g/cm^3$. $(Ag = 108 \ g/mol)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module