Explanation: Maximum value of incident angle can be 90 °for which sine is 1. Hence d = λ/2 ( n λ= 2.d.sin θ)

Orthorhombic lattice exists as all four types of unit cells. Besides primitive cell, monoclinic lattice exists only as end centered cell while tetragonal exists only as body centered lattices.

(b)
Calcium appears at eight corners, hence number of calcium atoms = 8/8 =1
Oxygen appears on six faces and two faces share one oxygen atom hence, number of oxygen atoms = 6/2 =3
Titanium appears at the centre, thus, number of Ti-atom = 1/1 =1
Thus, A is CaTi0₃ .
Let the oxidation number of Ti = x
CaTi0₃
(+2) + x + 3 ( - 2 ) = 0
2 + x - 6 = 0
x = + 4

(c)
Calcium appears at eight corners,hence number of calcium atoms = 8/8 =1
Oxygen appears on six faces and two faces share one oxygen atom hence, number of oxygen atoms = 6/2 =3
Titanium appears at the centre, thus, number of Ti-atom = 1/1 =1
Thus, A is CaTi0₃ .
Let the oxidation number of Ti = x CaTi0₃
(+2) + x + 3 ( - 2 ) = 0
2 + x - 6 = 0
x = + 4

(i) Triclinic, a b c and α βγ= 90 °Thus, (i)→(r)

(ii) Hexagonal, a = b c, α= β= 90 °,γ= 120 °Thus, (ii) →(s)

(iii) C u b ic ,a = b = c , α= β=γ= 90 °Thus, (iii) →(p)

(iv) Tetragonal, a = b c, α= β=γ= 90^o  

Thus, (iv) →(q)

(6) Edge length : a, b, c and angles between the edges :  α,β,γ

 Primitive unit cells are

(i) Cubic

(ii) Tetragonal

(iii) Orthorhombic

(iv) Hexagonal

(v) Rhombohedral

(vi) Monoclinic

(vii) Triclinic

A rhombohedral is a prism whose base is shaped as a non-square rhombus. Hence two of its six faces are rhombuses while others are rectangles. This results in a geometry for which two of the axial angles are right angles but the third is not.

(1) By Bragg 's equation,

n = 2  (second order diffraction)

(3)

Number of atoms in one unit cell of bcc structure =2
Thus, number of unit cells =   
Thus, y = 3