Dimensions of Unit Cells (Old NCERT) - Free MCQ Practice Test with solutions,

Simple cube has 1 atom

Therefore volume of cube i.e. a³ = 6.97 x 10^-23 cm³

(Molar mass = M) (1 molar of atom = M 6.023×10²³)( atom = M/Na)

The correct answer is Option B.
Mass of unit cell = number of atoms in unit cell × mass of each atom = z × m
Where, z = number of atoms in unit cell,
m = Mass of each atom
Mass of an atom can be given with the help of Avogadro number and molar mass as: 
Where, M = molar mass
N_A = Avogadro’s number
Volume of unit cell, V = a³
⇒ Density of unit cell = 
⇒ Density of unit cell = 

In a body centered cubic unit cell ,√3a = 4r a= 4r/√3 where a = side of the cube volume = a³ = (4r/√3)³.

Volume of unit cell, V = a3

=> Density of unit cell = mass of unit cell /volume of unit cell

The correct answer is option A
Unit cell is essentially considered a cube, so a cube has 3 axes. X,Y,Z so X represents length , Y represents breadth, Z represents height.

The face-centered cubic (fcc) has a coordination number of 12 and contains 4 atoms per unit cell. The body-centered cubic (bcc) has a coordination number of 8 and contains 2 atoms per unit cell. The simple cubic has a coordination number of 6 and contains 1 atom per unit cell.

In case of fcc lattice,number of atoms per unit cell,x=4 atoms

∴  Atomic mass of copper =63.1u

Length of the edge , a = 290 pm =290 x 10^-10 cm 
Volume of unit cell = ( 290 x 10^-10 cm )³ = 24.39 x 10^-24 cm³

Since it is bcc arrangement, 
Number of atoms in the unit cell, Z = 2 
Atomic mass of the element = 50 

Mass of the atom = atomic mass/ Avogadro number = M/N_o = 50/6.02 x 10²³

Mass of the unit cell = Z x M/N_o = 2 x 50/6.02 x 10²³ = 100/6.23 x 10²³
Therefore , density = mass of unit cell / volume of unit cell 
= 100/6.023 x 10²³ x 24.39 x 10^-24 = 6.81 g cm^-3

The total number of octahedral void(s) per atom present in a cubic close packed structure is 4. Besides the body centre, there is one octahedral void at the centre of each of the 12 edges. It is surrounded by six atoms, four belonging to the same unit cell (2 on the corners and 2 on face centres) and two belonging to two adjacent unit cells. Since each edge of the cube is shared between four adjacent unit cells, so is the octahedral void located on it. Only 1/4 ​th of each void belongs to a particular unit cell. Thus, in cubic close packed structure, octahedral void at the body-centre of the cube is 1.
12 octahedral voids located at each edge and shared between four unit cells=12× 1/4​=3
Total number of octahedral voids =4
We know that in ccp structure, each unit cell has 4 atoms. Thus, the number of octahedral voids is equal to 4/4=1.