# Solid State

## NCERT Exercise Solution (part 3)

Question – 1.11 – Silver crystallizes in fcc lattice. If edge length of the cell is 4.07 x 10^{– 8} cm and density is 10.5 g cm^{– 3}. Calculate the atomic mass of silver.

Solution:

Question – 1.12 - A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound? What are the coordination numbers of P and Q?

Solution:

Given, Atoms of Q are at the corners of the cube and P at the body-centre.

Number of atoms of P in one unit cell = 1

Question – 1.13 - Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm^{–3}, calculate atomic radius of niobium using its atomic mass 93 u.

Solution:

Thus atomic radius of niobium = 14.31 nm

Question – 1.14 - If the radius of the octahedral void is r and radius of the atoms in close packing is R, derive relation between r and R.

Solution:

In the given figure, let an sphere having centre ‘O’ is fitted in the octahedral void.

As given, radius of the sphere fitted in the octahedral void = r

And radius of the atoms in close packing = R

This is the required relation between r and R.

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Solid State - Introduction

Solid State - Crystal Lattice

Solid State - Close Packing Structure

Solid State - Efficiency of Close Packing Structure - 1

Solid State - Efficiency of Close Packing Structure - 2

Solid State - Imperfections in Solids - Crystal Defects

Solid State - Electrical and Magnetic Properties

Solid State - NCERT In Text Solution 1

Solid State - NCERT In Text Solution 2

Solid State - NCERT In Text Solution 3

Solid State - NCERT Exercise Solution (Part 1)

Solid State - NCERT Exercise Solution (Part 2)

Solid State - NCERT Exercise Solution (Part 4)

Solid State - NCERT Exercise Solution (Part 5)