Chemistry ATOMIC STRUCTURE

Session Objectives 1.Dalton’s theory 2.Discovery of fundamental particles 3.Thomson’s model of an atom 4.Rutherford’s model 5.Concept of atomic number and mass number 6.Drawback of Rutherford’s model 7.Electromagnetic waves 8.Planck’s quantum theory 9.Bohr’s model

All matter is composed of atoms. All atoms of a given element are identical. Atom  Can not be cut. Indivisible and indestructible Dalton’s Theory – Atom Is Fundamental Particle Pre 1897

Cathode rays

Properties of cathode rays They are material particles as they produce mechanical motion in a small paddle wheel

Properties of Cathode Rays They are deflected from their path by electric and magnetic fields

Anode rays

Properties of anode Rays They travel in straight line They are deflected by electric and magnetic field The nature of anode rays depends upon the nature of gas e/m ratio for anode rays is not constant

Electron Positive sphere Thomson’s Model of an Atom

Rutherford experiment

Rutherford’s Experiment - Results A beam of  particles aimed at thin gold foil. Most of the particles passed through. Most of the space is empty A few came back Presence of concentrated mass at the centre Others deflected at various angles Repulsion between two +vely charged particles “ Like firing shells at paper handkerchief with few of them coming back.” - Ernst Rutherford

Rutherford’s Model Atom consist of two parts: (a)Nucleus:Almost the whole mass of the atom is concentrated in this small region (b)Extra nuclear part:this is the space around the nucleus in which electrons are revolving at high speeds in fixed path

Concept of atomic mass and atomic number Atomic number(Z)=number of protons Mass number(A)=number of protons+number of neutrons Entire mass of the atom is concentrated at the centre

Concept of atomic number and mass number For example: Mass number=number of protons+number of neutrons =23 Atomic number =number of protons =11

Concept of atomic number and mass number We express weight of an atom in terms of atomic mass unit (a.m.u). Mass of a proton=Mass of neutron =1 a.m.u(approx) Mass number=Atomic weight (expressed in a.m.u)

Drawback of Rutherford’s model

Atom thus collapses. Drawback of Rutherford’s model

Direction of propagation Electric field component Magnetic field component X Y Z Electric and magnetic fields are perpendicular to each other Electromagnetic waves

. (i)Wavelength: It is represented by Units: m, cm(10 -2 m), nm(10 -9 m), pm( m) or A 0 ( m). Characteristics of a wave Direction of Propogation of wave

(iv) Wave number: The number of waves present in 1 cm length. It is represented by. Its unit is cm -1. (iii) Velocity: The linear distance travelled by a crest or a trough in one second. Its unit is cm s -1. (ii) Frequency: The number of waves passes through a given point in 1 second. It is represented by. Its unit is Hertz or second -1. Characteristics of a wave

Electromagnetic spectrum

Radio city broadcasts on a frequency of 5,090 KHz.What is the wavelength of electromagnetic radiation emitted by the transmitter? Illustrative problem 1

Radiant energy is emitted or absorbed discontinuously in the form of quanta. Planck’s quantum theory

Questions

The ratio of the energy of a photon of 2000 wavelength radiation to that of 6000 radiation is (a) ¼ (b) 4 (b)½ (d) 3 Illustrative Problem 2 Hence, answer is (d). Solution:

Bohr’s model Positively charged nucleus Negatively charged electrons Stationary Orbit + h – h

Bohr’s Postulates Retained key features of Rutherford’s model. Concept of stationary circular orbits. Quantization of angular momentum. Energy emitted/absorbed when electrons jump from one orbit to another.

Bohr’s model Bohr’s postulates energy of electron radius of various orbits velocity of electron

+ r Nucleus electron (i) Calculation of radius of Bohr orbit According to coulomb’s law

centrifugal force (ii) Calculation of radius of Bohr orbit Bohr’s postulate

For hydrogen Z=1 For n=1,Z=1, k =Nm 2 /C 2 Calculation of radius of Bohr orbit

u is the velocity with which the electron revolves in an orbit (1)(2) Dividing (1) by (2),we get: Calculation of velocity of electron u is in m/s

Number of revolutions per second Calculation of number of revolutions

Calculation of energy of an electron

We know that Substituting the value of r we get P.E. = 2K.E. K.E. = -Total energy

Bohr’s model Bohr’s postulates

Questions

Illustrative Problem 3 The energy of the electron in the second and third Bohr orbits of the hydrogen atom is X and –2.41 X respectively. Calculate the wavelength of the emitted radiation, when the electron drops from third to second orbit.

Solution According to Planck’s quantum theory

Solution

Class Test

Class Exercise - 1 Which of the following fundamental particles are present in the nucleus of an atom? (a) Alpha particles and protons (b) Protons and neutrons (c) Protons and electrons (d) Electrons, protons and neutrons Solution The nucleus of an atom is positively charged and almost the entire mass of the atom is concentrated in it. Hence, it contains protons and neutrons. Hence, answer is (b).

Class Exercise - 2 The mass of the proton is (a) × 10 –24 g (b) × 10 –25 g (c) × g (d) × g Solution Hence, answer is (a). The mass of the proton is × 10 –24 g

Class Exercise - 3 Which of the following is not true in case of an electron? (a) It is a fundamental particle (b) It has wave nature (c) Its motion is affected by magnetic field (d) It emits energy while moving in orbits Solution Hence, answer is (d). An electron does not emit energy while moving in orbit. This is so because if it would have done that it would have eventually fallen into the nucleus and the atom would have collapsed.

Class Exercise - 4 Positive charge of an atom is (a) concentrated in the nucleus (b) revolves around the nucleus (c) scattered all over the atom (d) None of these Solution Hence, answer is (a). Positive charge of an atom is present entirely in the nucleus.

Class Exercise - 5 Calculate and compare the energies of two radiations which have wavelengths 6000Å and 4000Å (h = 6.6 x J s, c = 3 x 10 8 m s -1 ) Solution = 3.3 x J = 4.9 x J = : 1

Class Exercise - 6 Why only very few a-particles are deflected back on hitting a thin gold foil? Solution Due to the presence of a very small centre in which the entire mass is concentrated.

Class Exercise - 7 Explain why cathode rays are produced only when the pressure in the discharge tube is very low. Solution This is happened because at higher pressure no electric current flows through the tube as gases are poor conductor of electricity.

Class Exercise - 8 If a neutron is introduced into the nucleus of an atom, it would result in the change of (a) number of electrons (b) atomic number (c) atomic weight (d) chemical nature of the atom Solution Hence, answer is (c). Neutrons contribute in a major way to the weight of the nucleus, thus addition of neutron would result in increase in the atomic weight.

Class Exercise - 9 The concept of stationary orbits lies in the fact that (a) Electrons are stationary (b) No change in energy takes place in stationary orbit (c) Electrons gain kinetic energy (d) Energy goes on increasing Solution Hence, answer is (c). When an electron revolves in a stationary orbit, no energy change takes place. Energy is emitted or absorbed only when the electron jumps from one stationary orbit to another.

Class Exercise - 10 What is the energy possessed by 1 mole of photons of radiations of frequency 10 × Hz? Solution E = h E = 6.6 × 10 –34 × 10 × E = 66 × 10 –20 = 6.6 × 10 –19 joules  energy of 1 mole of photons = 6.6 × 10 –19 × × = × 10 4 = kJ/mol

Class test 1.The radius of hydrogen atom in ground state is 5.3x10 –11 m. It will have a radius of 4.77A after colliding with an electron. The principal quantum number of the atom in the excited state is (a) 2(b) 4(c)3(d)5 Solution Hence, answer is (c).

Thank you

Electromagnetic waves  Light is an oscillating electro-magnetic field.  Oscillating electric field generates the magnetic field and vice-versa. Electric and magnetic fields are perpendicular to each other