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

An element with $BCC$ structure has an edge length of $500 \ pm$. If its density is $4 \ g \ cm^{-3}$, find the atomic mass of the element.

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:

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

The edge length of a unit cell of a $BCC$ structure is $352 \ pm$. What is the radius of the atoms (in $pm$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module