A crystal structure has a face-centered cubic (FCC) arrangement of atoms A and atoms B occupy all the octahedral voids. If the atoms A are removed, what is the resulting lattice type for atoms B?
Simple Cubic
Body-centered Cubic
FCC
Hexagonal Close Packed
Related Questions
A pure crystalline substance, on being heated gradually, first forms a turbid looking liquid and then the turbidity completely disappears. This behavior is the characteristic of substances forming
isomeric crystals
liquid crystals
isomorphous crystals
allotropic crystals
The edge of unit cell of fcc Xe crystal is 620 pm. The radius of Xe atom is
Which has no rotation of symmetry?
Hexagonal
Orthorhombic
Cubic
Triclinic
Lithium borohydride crystallizes in an orthorhombic system with 4 molecule per unit cell. The unit cell dimensions are . If the molar mass is 21.76, then the density of crystals is :
None of these
A crystal system has $a
eq b
eq c\alpha = \beta = \gamma
eq 90^\circ$. What is the Bravais lattice associated with this crystal system?
Primitive Triclinic
Rhombohedral
Base-Centered Orthorhombic
Body-Centered Tetragonal
The unit cell with dimensions $\alpha = \beta = \gamma = 90^\circ ,a = b
e c$ is
Cubic
Triclinic
Hexagonal
Tetragonal
A crystal with and $\alpha = \beta = \gamma
eq 90^\circ$ belongs to which crystal system?
Cubic
Tetragonal
Rhombohedral
Orthorhombic
A hypothetical crystalline substance exhibits a unique crystal structure with the following unit cell dimensions: $a = b
eq c\alpha = \beta = 90^\circ\gamma = 120^\circ$. This substance most likely belongs to which crystal system?
Trigonal
Orthorhombic
Monoclinic
Hexagonal
For orthorhombic system which one is correct
$a
e b
e c,\alpha = \beta = \gamma = {90^0}$
$a = b = c,\alpha
e \beta
e \gamma
e {90^0}$
$a
e b = c,\alpha = \beta = \gamma
e {90^0}$
$a
e b
e c,\alpha
e \beta
e \gamma
e {90^0}$
The crystal system of a compound with unit cell dimensions isβΆ
Cubic
Hexagonal
Orthorhombic
Rhombohedral