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