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QUESTION: 1

If an element crystallizes as a simple cube, what is the volume of an element provided its density is 1.5 g/cm^{3} and atomic mass of the element is 63?

Solution:

Simple cube has 1 atom

Therefore volume of cube i.e. a^{3} = 6.97 x 10^{-23} cm^{3}

QUESTION: 2

Mass of an atom in an unit cell is calculated by which of the following formula if M is the molar mass:

Solution:

(Molar mass = M) (1 molar of atom = M 6.023×10^{23})( atom = M/Na)

QUESTION: 3

In the formula to calculate the density of a unit cell what is z ?

Solution:

QUESTION: 4

Volume of a body centred cubic unit cell is

Solution:

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

QUESTION: 5

Density of a unit cell is equal to mass of unit cell divided by

Solution:

**Volume of unit cell, V = a3**

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

QUESTION: 6

Length a in unit cell is along

Solution:

QUESTION: 7

The number of atoms (z) in simple cubic unit cell is

Solution:

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.

QUESTION: 8

X-ray diffraction studies show that copper crystallizes in an fcc unit cell with cell edge of 3.608 x 10^{-8} cm. In a separate experiment, copper is determined to have a density of 8.92 g/cm^{3}, calculate the atomic mass of copper.

Solution:

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

∴ Atomic mass of copper =63.1u

QUESTION: 9

An element with bcc geometry has atomic mass 50 and edge length 290 pm. The density of unit cell will be

Solution:

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} cm^{3}

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 10^{23}

Mass of the unit cell = Z x M/N_{o} = 2 x 50/6.02 x 10^{23} = 100/6.23 x 10^{23}

Therefore , density = mass of unit cell / volume of unit cell

= 100/6.023 x 10^{23} x 24.39 x 10^{-24} = 6.81 g cm^{-3}

QUESTION: 10

The number of octahedral voids (s) per atom present in a cubic close-packed structure is

Solution:

No of atoms per unit cell is 4,,it means it contain 4 octahedral voids and 8 tetrahedral voids

Explanation:

**Total no of octahedral voids per atom is 1**

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