Question

Which type cubic lattice have the dhkl ratio is 1 : 1.414: 0.577 ?
(a) Body centred cubic
(b) Simple cubic
(C) Face centred cubic
(d) Hexagonal cubic​

Answer 1

Answer:hexangula cubic

Explanation:because containing one per side

Answer 2

Option (a) Body centered cubic

Explanation:

In crystals, the distance between planes which are in parallel condition are termed as $\text{d}_\text{hkl}$.

Where d - distance

$\text{h} = \frac{\text{a}}{\text{intercept of the plane along x-axis}}$

$\text{k} = \frac{\text{b}}{\text{intercept of the plane along y-axis}}$

$\text{l} = \frac{\text{c}}{\text{intercept of the plane along z-axis}}$

$\text{d}_\text{hkl} = \frac{\text{a}}{\sqrt{\text{h}^2 + \text{k}^2 + \text{l}^2}}$

In the above relation, 'a' is the length of the cube sides, h, k and l are the Miller indices of a plane.

The spacing of three planes 100, 110, 111 of Body centered cubic lattice is

$\text{d}_1_0_0= \frac{1}{2}$

$\text{d}_1_1_0= \frac{1}{\sqrt{2}}$

$\text{d}_1_1_1= \frac{1}{2\sqrt{3}}$

The $\text{d}_\text{hkl}$ ratio for body centered cubic lattice is

$\text{d}_1_0_0 : \text{d}_1_1_0 : \text{d}_1_1_1 = \frac{1}{2} : \frac{1}{\sqrt{2}} : \frac{1}{2\sqrt{3}}$

$\text{d}_1_0_0 : \text{d}_1_1_0 : \text{d}_1_1_1 = 1 : \sqrt{2} : \frac{1}{\sqrt{3}}$

$\text{d}_1_0_0 : \text{d}_1_1_0 : \text{d}_1_1_1 = 1 : 1.414 : 0.577$

Therefore, the $\text{d}_\text{hkl}$ ratio of the body centered cubic lattice is 1 : 1.44 : 0.577.  

To Learn More ...

1.  Which type of cubic lattice has maximum no of atoms per unit cell?

https://brainly.in/question/3953657

2. Explain with the help of diagram about Three types of cubic cells.

https://brainly.in/question/6328437