chemistry atomic structure session objectives 1.dalton’s theory 2.discovery of fundamental...

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

High voltage

Gas atlow pressure

Cathoderays To vacuum

pum pCathode

– +Anode

Discharge tube experim ent-Production of 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

+

–

ZnSCoating

Perforatedcathode

H gas insideat low pressure

2

Properties of anode Rays

They travel in straight lineThey are deflected by electric and magnetic fieldThe nature of anode rays depends upon the nature of gas

e/m ratio for anode rays is not constant

Subatomic

particlesSymbol Unit charge Unit mass

Charge in

CoulombMass in amu

Proton

Neutron

Electron Negligible

11H

10n

01e

-19+1.60 × 10

-19+1.60 × 10 -45.489×10

1.008665

1.0078251

1

-1

-1

0 0

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

aN2311

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

XY

Z

Electric and magnetic fields areperpendicular to each other

Electromagnetic waves

.

(i)Wavelength: It is represented by

Units:

m, cm(10-2m), nm(10-9m), pm(10-12m) or A0(10-10m).

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-1.1

λ

(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

c

8

3

3 105090 10

20.589 10

58.9 m

s/m103isc 8

Radiant energy is emitted or

absorbed discontinuously in the

form of quanta.

Planck’s quantum theory

34

cE h h hc

Wavelength

Frequency

Wave number

h Plank 's cons tant

6.626 10 J s

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

0

A

0

A

Illustrative Problem 2

hcE h

1 21 2

hc hcE E

1 2

2 1

E

E

0

0

6000 A3

2000 A

Hence, answer is (d).

Solution:

Bohr’s model

Positivelychargednucleus

Negativelychargedelectrons

StationaryOrbit

+ h

– h

Bohr’s Postulates

Retained key features of Rutherford’s model.

Concept of stationary circular orbits.

Quantization of angular momentum.

nhmvr

2

Energy emitted/absorbed when electrons jump from one orbit to another.

fiE E E

h

Bohr’s model

Bohr’s postulates

energy of electron

radius of various orbits

velocity of electron

+

r

Nucleus

electron

2

a 2

kZeF

r (i)

Calculation of radius of Bohr orbit

According to coulomb’s law

centrifugal force

2

c

muF

r (ii)

F Fa c2 2

2

kZe mur r

2

2 kZeu

mr

Calculation of radius of Bohr orbit

nhmur

2

Bohr’s postulate

nhu

2 mr

2 22

2 2 2n h

u4 m r

For hydrogen Z=1

2

0nr 0.529 A

Z

22 kZe

umr

2 2

22 2 2n h

u4 m r

222

222

rm4

hn

mr

kZe

22

22

mkZe4

hnr

For n=1,Z=1 , k = 9109 Nm2/C2

Calculation of radius of Bohr orbit

u is the velocity with which the electron revolves in

an orbit

nhmur

2

(1)

2 2

2

kZe murr

(2)

Dividing (1) by (2),we get:

Calculation of velocity of electron

nh

kZe2u

2 u is in m/s

Number of revolutions per second

velocity of electroncircumference of an orbit

Calculation of number of revolutions

Total energy(T.E) P.E K.E

21K.E mu

2

2kZeP.E

r

22

2

1 kZeT.E mu

2 r

Calculation of energy of an electron

We know that

2 2

2

mu kZer r

22 kZe

mur

2 2kZe kZeE

2r r

2kZe2r

Substituting the value of r we get

22

2422

n hn

mkeZ2E

P.E. = 2K.E.K.E. = -Total energy

Bohr’s model

Bohr’s postulates

8n

Zv 2.18 10 cm / sec

n

2 0

nn

r 0.529 AZ

2

n 2

219

2

ZE 13.6 eVper atom

n

Z21.8 10 J per atom

n

Questions

Illustrative Problem 3

The energy of the electron in the second and third Bohr orbits of the hydrogen atom is -5.42 X 10-12 and –2.41 X 10-12 respectively. Calculate the wavelength of the emitted radiation, when the electron drops from third to second orbit.

Solution

2 1E E E

According to Planck’s quantum theory

12 125.42 10 ( 2.41 10 )ergs 123.01 10 ergs.

E h

cE h

27h 6.6 10 ergs

Solution

56.6 10 cm 3 06.6 10 A

0 81A 10 cm

27 8

126.62 10 3 10

3.01 10

ch

E

Class Test

Class Exercise - 1

Which of the following fundamentalparticles are present in the nucleusof 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 almostthe 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) 1.672 × 10–24 g(b) 1.672 × 10–25 g(c) 1.672 × 1025 g(d) 1.672 × 1026 g

Solution

Hence, answer is (a).

The mass of the proton is 1.672 × 10–24 g

Class Exercise - 3

Which of the following is not truein 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 wouldhave 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 thenucleus.

Class Exercise - 5

Calculate and compare the energiesof two radiations which havewavelengths 6000Å and 4000Å (h = 6.6 x 10-34 J s, c = 3 x 108 m s-1)

Solution

34 8 1

1 7hc 6.6 10 Js 3 10 ms

E6 10 m

= 3.3 x 10-19 J

34 8 1

2 76.6 10 Js 3 10 ms

E4 10 m

= 4.9 x 10-19 J

191

192

E 3.3 10 JE 4.95 10 J

= 0.666 : 1

Class Exercise - 6

Why only very few a-particles aredeflected back on hitting a thingold foil?

Solution

Due to the presence of a very small centre inwhich the entire mass is concentrated.

Class Exercise - 7

Explain why cathode rays areproduced only when the pressurein the discharge tube is very low.

Solution

This is happened because at higher pressureno electric current flows through the tubeas gases are poor conductor of electricity.

Class Exercise - 8

If a neutron is introduced into thenucleus of an atom, it would resultin 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 weightof the nucleus, thus addition of neutron wouldresult in increase in the atomic weight.

Class Exercise - 9

The concept of stationary orbits liesin 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 emittedor absorbed only when the electron jumps fromone stationary orbit to another.

Class Exercise - 10

What is the energy possessed by1 mole of photons of radiationsof frequency 10 × 1014 Hz?

Solution

E = hE = 6.6 × 10–34 × 10 × 1014

E = 66 × 10–20 = 6.6 × 10–19 joules

energy of 1 mole of photons = 6.6 × 10–19 × 6.023 × 1023

= 39.7518 × 104

= 397.518 kJ/mol

Class test1.The radius of hydrogen atom in groundstate is 5.3x10–11m. It will have a radius of 4.77A after colliding with an electron. The principal quantum numberof the atom in the excited state is

(a) 2 (b) 4 (c)3 (d)5

2

n o

210 11

2

nSince r r x

Z

n4.77 10 5.3 10 (for hydrogen atom)

1

n 9

n 3

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