1. CHEMISTRY
ATOMIC STRUCTURE
LECTURE # 1 TO 11
COURSE
LECTURE NOTES

2. Lecture : 1
Atomic structure
1. Introduction :
STRUCTURE OF ATOM

Rutherford'sModel Bohr's Model Wave mechanical model
Dalton’s concept of the indivisibility of the atom was completely discredited by a series of experimental
evidences obtained by scientists. Anumber of new phenomena were brought to light and man’s idea about
the natural world underwent a revolutionary change. the discovery of electricity and spectral phenomena
opened the door for radical changes in approaches to experimentation. It was concluded that atoms are
made of three particles: electrons, protonsand neutrons. These particlesare called the fundamental particles
of matter.
2. Earlier efforts to reveal structure of atom :
CATHODE RAYS - DISCOVERY OF ELECTRON
In 1859 Julius Plucker started the study of conduction of electricity through gases at low pressure
(10–4atm) in a discharge tube When a high voltage of the order of 10,000 volts or more wasimpressed across
the electrodes, some sort of invisible rays moved from the negative electrode to the positive electrode these
rays are called as cathode rays.
Cathode rays have the following properties.
(i) Path of travelling is straight from the cathode with a very high velocity
As it produces shadow of an object placed in its path
(ii) Cathode rays produce mechanical effects. If small paddle wheel is placed between the electrodes,
it rotates. This indicates that the cathode rays consist of material part.
(iii) When electric and magnetic fields are applied to the cathode rays in the discharge tube. The rays
are deflected thus establishing that they consist of charged particles. The direction of deflection
showed that cathode rays consist of negatively charged particles called electrons.
(iv) They produce a green glow when strike the glass wall beyond the anode. Light is emitted when they
strike the zinc sulphide screen.
(v) Cathode rays penetrate through thin sheets of aluminium and metals.
(vi) They affect the photographic plates
(vii) The ratio of charge(e) to mass(m) i.e. charge/mass is same for all cathode rays irrespective of the
gas used in the tube. e/m = 1.76 × 1011 Ckg–1
Thus, it can be concluded that electrons are basic constituent of all the atoms.

3. POSITIVE RAYS –DISCOVERY OF PROTON
The existance of positively charged particles in an atom was shown by E. Glodstein in 1886.He repeated the
same discharge tube experiments by using a perfoated cathode. It was observed that when a high potential
differece was applied between the electrodes, not only cathode rays were produced but also a new type of
rays were produced simultaneously from anode moving towards cathode and passed through the holes or
canal of the cathode. These rays are termed as canal ray or anode ray
Characteristics of anode rays are as follows.
(i) The e/m ratio of for these rays is smaller than that of electrons.
(ii) Unlike cathode rays, their e/m value is dependent upon the nature of the gas taken in the tube. For
different gases used in the discharge tube, the charge to mass ratio (e/m) of the positive particles
constituting the positive rays is different. When hydrogen gas is taken in the discharge tube, the e/
m value obtained for the positive rays is found to be maximum. Since the value of charge (e) on the
positive particle obtained from different gases is the same, the value of m must be minimum for the
positive particles obtained from hydrogen gas. Thus, the positive particle obtained from hydrogen
gas is the lightest among all the positive particles obtained from different gases. this particle is
called the proton.
(iii) They are capable to produce ionisation in gases.
(iv) They can produce physical and chemical changes.
Discovery of Neutron :
Later, a need was felt for the presence of electrically neutral particles as one of the constituent of
atom. These particles were discovered by Chadwick in 1932 by bombarding a thin sheet of Beryllium
with -particles, when electrically neutral particles having a mass slightly greater than that of the
protons were emitted. He named these particles as neutrons.
n
C
He
Be 1
0
12
6
4
2
9
4 



The NUCLEUS :
Electrons, protons & neutrons are the fundamental particles present in all atoms,(except hydrogen)
Particles Symbol Mass Charge Discoverer
Electron –1e0 or  9.1096 x 10-31 kg – 1.602 x l0–19 J.J. Thompson
Coulombs Stoney Lorentz 1887
0.000548 amu – 4.803 × 10–10 esu
Proton 1H1 1.6726 x 10–27 kg + 1.602 x 10–19 Goldstein
Coulombs Rutherford1907
1.00757 amu + 4.803 x 10–10 esu
Neutron 0n1 1.6749 x 10–27 kg neutral James Chadwick
1.00893 amu 0 1932
1 amu  1.66 × 10–27 kg

4. 3. ATOMIC MODELS :
(A) Thomson’s Model of the Atom :
An atom is electrically neutral. It contains positive charges (due to the presence of protons ) as well
as negative charges (due to the presence of electrons). Hence, J J Thomson assumed that an atom is a
uniform sphere of positive charges with electrons embedded in it.
(B) Rutherford’s Experiment :
1. Most of the -particles passed straight through the gold foil without suffering any deflection from
their original path.
2. A fewof them were deflected through small angles, while a very few were deflected to a large extent.
3. A very small percentage (1 in 100000) was deflected through angles ranging from 90° to 180°.
Rutherford’s nuclear concept of the atom.
(i) The atom of an element consists of a small positively charged ‘nucleus’ which is situated at the
centre of the atom and which carries almost the entire mass of the atom.
(ii) The electrons are distributed in the empty space of the atom around the nucleus in differnt concentric
circular paths, called orbits.
(iii) The number of electrons in orbits is equal to the number of positive charges (protons) in the nucleus.
Hence, the atom is electrically neutral.
(iv) The volume of the nucleus is negligibly small as compared to the volume of the atom.
(v) Most of the space in the atom is empty.
DRAWBACKS OF RUTHERFORD MODEL :
1. This was not according to the classical theory of electromagnetism proposed by maxwell. According
to this theory, every accelerated charged particle must emit radiations in the form of electromagnetic
waves and loses it total energy.
Since energy of electrons keep on decreasing, so radius of the circular orbits should also decrease
and ultimately the electron should fall in nucleus.
2. It could not explain the line spectrum of H-atom.
Home - Work
NCERT (Reading) 2.1, 2.1.1, 2.1.2, 2.1.3, 2.1.4, 2.2, 2.2.1, 2.2.2, 2.2.5
NCERT (Exercise) 2.35 - 2.40

5. Lecture : 2
5. PROPERTIES OF CHARGE :
1. Q = ne ( charge is quantized)
2. Charge are of two types :
(i) positive charge (ii) Negative Charge
e = –1.6 x 10–19
p = + 1.6 x 10–19C
This does not mean that a proton has a greater charge but it implies that the charge is equal and opposite.
Same charge repel each other and opposite charges attract each other.
3. Charge is a SCALAR Qty. and the force between the charges always acts along the line joining the
charges.
The magnitude of the force between the two charge placed at a distance ‘r’ is given by
FE =
0
4
1
 2
2
1
r
q
q
(electrical force)
4. If two charge q1 and q2 are sepreted by distance r then the potential energy of the two charge system is
given by.
The magnitude of the force between the two charge placed at a distance ‘r’ is given by
P.E. =
0
4
1
 r
q
q 2
1
5. If a charged particle q is placed on a surface of potential V then the potential energy of the charge is
q x V.
Estimation of closest distance of approach (derivation)
Ex. An -particle is projected from infinity with the velocity V0 towards the nucleus of an atom having atomic
number equal to Z then find out (i) closest distance of approach (R) (ii) what is the velocity of the -particle at
the distance R1 (R1 > R) from the nucleus.
Sol. +
m
V
1 2
3
R
R1
From energy conservation P.E1 + KE1 = P.E2 + KE2
 0 +
2
1
m V
2=
R
)
e
2
)(
Ze
(
K
+ 0
R = 2
2
V
m
KZe
4


(closest distance of approch)
Let velocity at R1 is V1.
From energy conservation P.E1 + KE1 = P.E3 + KE3
 0 +
2
1
m V
2=
1
R
)
e
2
)(
Ze
(
K
+
2
1
m V1
2

6. 4. Size of the nucleus :
The volume of the nucleus is very small and is only a minute fraction of the total volume of the atom. Nucleus
has a diameter of the order of 10–12 to 10–13 cm and the atom has a diameter of the order of 10–8 cm.
Thus, diameter (size) of the atom is 100,000 times the diameter of the nucleus.
The radius of a nucleus is proportional to the cube root of the number of nucleons within it.
R = R0 (A)1/3 cm
where R0 can be 1.1 × 10–13 to 1.44 × 10–13 cm ; A = mass number ; R = Radius of the nucleus.
Nucleus contains protons & neutrons except hydrogen atoms which does not contain neutron in the nucleus.
Atomic number and Mass number :
 Atomic number of an element
= Total number of protons present in the nucleus
= Total number of electrons present in the atom
 Atomic number is also known as proton number because the charge on the nucleus depends upon
the number of protons.
 Since the electrons have negligible mass, the entire mass of the atom is mainly due to protons and
neutrons only. Since these particles are present in the nucleus, therefore they are collectively called
nucleons.
 As each of these particles has one unit mass on the atomic mass scale, therefore the sum of the
number of protons and neutrons will be nearly equal to the mass of the atom.
 Mass number of an element = No. of protons + No. of neutrons.
 The mass number of an element is nearly equal to the atomic mass of that element. However, the
main difference between the two is that mass number is always a whole number whereas atomic
mass is usually not a whole number.
 The atomic number (Z) and mass number (A) of an element ‘X’ are usually represented alongwith the
symbol of the element as
e.g. Cl
,
Na 35
17
23
11 and so on.
1. sotopes : Such atomsof the same element having same atomic number but different massnumbers
are called isotopes.
H
,
H 2
1
1
1 and H
3
1 and named as protium, deuterium (D) and tritium (T) respectively. Ordinary hydrogen
is protium.
2. Isobars : Such atoms of different elements which have same massnumbers (and of course different
atomic numbers) are called isobars
e.g. .
Ca
,
K
,
Ar 40
20
40
19
40
18
3. Isotones : Such atoms of different elements which contain the same number of neutrons are called
isotones
e.g. .
O
,
N
,
C 16
8
15
7
14
6
4. Isoelectronic : The species (atoms or ions) containing the same number of electrons are called
isoelectronic.
For example, O2–, F–, Na+, Mg2+, Al3+, Ne all contain 10 electrons each and hence they are
isoelectronic.
Example. Complete the following table :
Particle Mass No. Atomic No. Protons Neutrons Electrons
Nitrogen atom – – – 7 7
Calcium ion – 20 – 20 –
Oxygen atom 16 8 – – –
Bromide ion – – – 45 36

7. Sol. For nitrogen atom.
No. of electron = 7 (given)
No. of neutrons = 7 (given)
 No. of protons = Z = 7
( atom is electrically neutral)
Atomic number = Z = 7
Mass No. (A) = No. of protons + No. of neutrons
= 7 + 7 = 14
For calcium ion.
No. of neutrons = 20 (Given)
Atomic No. (Z) = 20 (Given)
 No. of protons = Z = 20 ;
No. of electrons in calcium atom
= Z = 20
But in the formation of calcium ion, two electrons are lost from the extranuclear part according to the
equation Ca  Ca2+ + 2e– but the composition of the nucleus remains unchanged.
 No. of electrons in calcium ion
= 20 – 2 = 18
Mass number (A) = No. of protons + No. of neutrons
= 20 + 20 = 40.
For oxygen atom.
Mass number (A) = No. of protons + No. of neutrons
= 16 (Given
Atomic No. (Z) = 8 (Given)
No. of protons = Z = 8,
No. of electrons = Z = 8
No. of neutrons = A – Z = 16 – 8 = 8
For bromide ion.
No. of neutrons = 45 (given)
No. of electrons = 36 (given)
But in the formation of bromide ion, one electron is gained by extra nuclear part according to equation
Br + e–  Br – , But the composition of nucleus remains unchanged.
 No. of protons in bromide ion = No. of electrons in bromine atom = 36 – 1 = 35
Atomic number (Z) = No. of protons = 35
Mass number (A) = No. of neutrons + No. of protons = 45 + 35 = 80.
Ex. (NCERT) Calculate the number of protons, neutrons and electrons in Br
80
35 .
Sol. In this case, Br
80
35 , Z = 35, A = 80, species is neutral
Number of protons = number of electrons = Z = 35
Number of neutrons = 80 – 35 = 45.
Ex. (NCERT) The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively.
Assign the proper symbol to the species.
Sol. The atomic number is equal to number of protons = 16. The element is Sulphur (S).
Mass number = number of protons + number of neutrons = 16 + 16 = 32
Species is not neutral as the number of protons is not equal to electrons. It is anion (negatively
charged) with charge equal to excess electrons = 18 – 16 = 2. Symbol is –
2
32
16 S .
Home - Work
NCERT (Reading) 2.2.3, 2.2.4
NCERT (Exercise) 2.1 - 2.4, 2.22, 2.41 - 2.44
Sheet (Ex - 1) Part - I Section - A (1 - 4), Part - II Section - A (1 - 5)

8. LECTURE : 3
6. Electromagnetic wave radiation :
The oscillating electrical/magnetic field are electromagnetic radiations. Experimentally, the direction of oscillations
of electrical and magnetic field are prependicular to each other.

E = Electric field,

B = Magnetic field
Direction of wave propogation.
7. Some important characteristics of a wave :
 Wavelength of a wave is defined as the distance between any two consecutive crests or troughs. It is
represented by  (lambda) and is expressed in Å or m or cm or nm (nanometer) or pm (picometer).
1 Å = 10– 8 cm = 10–10 m
1 nm = 10– 9 m, 1 pm = 10–12 m
 Frequency of a wave is defined as the number of waves passing through a point in one second. It is
represented by  (nu) and is expressed in Hertz (Hz) or cycles/sec or simply sec–1 or s–1.
1 Hz = 1 cycle/sec
 Velocity of a wave is defined as the linear distance travelled by the wave in one second. It is represented by
v and is expressed in cm/sec or m/sec (ms–1).
 Amplitude of a wave is the height of the crest or the depth of the trough. It is represented by ‘a’ and is
expressed in the units of length.
 Wave number is defined as the number of waves present in 1 cm length. Evidently, it will be equal to the
reciprocal of the wavelength. It is represented by  (read as nu bar).



1

9. If  is expressed in cm,  will have the units cm–1.
Relationship between velocity, wavelength and frequency of a wave. As frequency is the number of
waves passing through a point per second and  is the length of each wave, hence their product will give the
velocity of the wave. Thus
v =  × 
Order of wavelength in Electromagnetic spectrum
Cosmic rays <  – rays < X-rays < Ultraviolet rays < Visible < Infrared < Micro waves < Radio waves.
Ex. (NCERT)
The Vividh Bharati station of All India Ratio, Delhi, broadcasts on a frequency of 1,368 kHz (kilo hertz).
Calculate the wavelength of the electromagnetic radiation emitted by transmitter. Which part of the
electromagnetic spectrum does it belong to ?
Sol. The wevelength, , is equal to c/, where c is the speed of electromagnetic radiation in vacuum and  is the
frequency. Substituting the given values, we have



c
= 1
3
1
8
s
10
1368
ms
10
00
.
3




= 219.3 m
Ex. (NCERT)
The wavelength range of the visible spectrum extends from violet (400 nm) to red (750 nm). Express these
wavelengths in frequencies (Hz). (1 nm = 10–9 m)
Sol. Frequency of violet light
m
10
400
ms
10
00
.
3
c
9
1
8







 = 7.50 × 1014 Hz
Frequency of red light
m
10
750
ms
10
00
.
3
c
9
1
8







 = 4.00 × 1014 Hz
The range of visible spectrum is from 4.0 × 1014 to 7.5 × 1014 Hz in terms of frequency units.
Ex. (NCERT)
Calculate (a) wavenumber and (b) frequency of yellow radiation having wavelength 5800 Å.
Sol. (a) Calculation of wavenumber )
(  = 5800 Å = 5800 × 10–8 cm = 5800 × 10–10 m
m
10
5800
1
1
10





 = 1.724 × 106 m–1 = 1.724 × 104 cm–1
(b) Calculation of the frequency ()
m
10
5800
s
m
10
3
c
10
1
8







 = 5.172 × 1014 s–1
Home - Work
NCERT (Reading) 2.3, 2.3.1
NCERT (Exercise) 2.5, 2.7, 2.45

10. Lecture : 4
Particle Nature of Electromagnetic Radiation :Planck's Quantum Theory
Some of the experimental phenomenon such as diffraction and interference can be explained by the wave
nature of the electromagnetic radiation. However, following are some of the observations which could not be
explained
(i) the nature of emission of radiation from hot bodies (black - body radiation)
(ii) ejection of electrons from metal surface when radiation strikes it (photoelectric effect)
Black Body Radiation :
When solids are heated they emit radiation over a wide range of wavelengths.
The ideal body, which emits and absorbs all frequencies, is called a black body and the radiation emitted by
such a body is called black body radiation. The exact frequency distribution of the emitted radiation (i.e.,
intensity versus frequency curve of the radiation) from a black body depends only on its temperature.
The above experimental results cannot be explained satisfactorily on the basis of the wave theory of light.
Planck suggested that atoms and molecules could emit (or absorb) energy only in discrete quantities and
not in a continuous manner.
Photoelectric Effect :
When certain metals (for example Potassium, Rubidium, Caesium etc.) were exposed to a beam of light
electrons were ejected as shown in Fig.
The phenomenon is called Photoelectric effect. The results observed in this experiment were :

11. (i) The electrons are ejected from the metal surface as soon as the beam of light strikes the surface, i.e., there
is no time lag between the striking of light beam and the ejection of electrons from the metal surface.
(ii) The number of electrons ejected is proportional to the intensity or brightness of light.
(iii) For each metal, there is a characteristic minimum frequency, 0
(also known as threshold frequency)
below which photoelectric effect is not observed. At a frequency  > 0
, the ejected electrons come out with
certain kinetic energy. The kinetic energies of these electrons increase with the increase of frequency of the
light used.
When a photon of sufficient energy strikes an electron in the atom of the metal, it transfers its energy
instantaneously to the electron during the collision and the electron is ejected without any time lag or delay.
Greater the energy possessed by the photon, greater will be transfer of energy to the electron and greater the
kinetic energy of the ejected electron. In other words, kinetic energy of the ejected electron is proportional to
the frequency of the electromagnetic radiation. Since the striking photon has energy equal to h and the
minimum energy required to eject the electron is h0
(is also called work function, W0
) then the difference in
energy (h – h0
) is transferred as the kinetic energy of the photoelectron. Following the conservation of
energy principle, the kinetic energy of the ejected electron is given by the equation
h = h0
+
2
1
me
2
where me
is the mass of the electron and v is the velocity associated with the ejected electron.
Example 1.
The therehold frequency 0
for a metal is 6 × 1014
s–1
. Calculate the kinetic energy of an electron emitted when
radiation of frequency  = 1.1 × 1015
s–1
hits the metal.
Sol. K.E. =
2
1
me
V2
= h ( – 0
)
 K.E. = (6.626 × 10–34
) (1.1 × 1015
– 6 × 1014
)
 K.E. = (6.626 × 10–34
) (5 × 1014
)
= 3.313 × 10–19
J
Example 2.
When electromagnetic radiation of wavelength 310 nm fall on the surface of Sodium, electrons are emitted
with K.E. = 1.5 eV. Determine the work function (W0
) of Sodium.
Sol. h =
3100
12400
= 4 eV
2
1
me
V2
= 1.5 eV
 h0
= W0
= h –
2
1
me
V2
= 4 – 1.5 = 2.5 eV
Ex. (NCERT)
When electromagnetic radiation of wavelength 300 nm falls on the suface of sodium, electrons are emitted
with a kinetic energy of 1.68 × 105
J mol–1
. What is the minimum energy needed to remove an electron from
sodium ? What is the maximum wavelength that will cause a photoelectron to be emitted ?
Sol. The energy (E) of a 300 nm photon is given by h = hc /  =
m
10
300
s
m
10
0
.
3
s
J
10
626
.
6
9
1
8
34







= 6.626 × 10–19
J

12. The energy of one mole of photons
= 6.626 × 10–19
× 6.022 × 1023
= 3.99 × 105
J mol–1
The minimum energy needed to remove a mole of electrons from sodium = (3.99 – 1.68) 105
J mol–1
= 2.31 × 105
J mol–1
The minimum energy for one electron = 1
23
1
5
mol
electrons
10
022
.
6
mol
J
10
31
.
2




= 3.84 × 10–19
J
This corresponds to the wavelength 
E
hc

 =
m
10
84
.
3
s
m
10
0
.
3
s
J
10
626
.
6
19
1
8
34







= 517 nm.
(This corresponds to green light)
Ex. (NCERT)
The thereshold frequency 0
for a metal for a metal is 7.0 × 1014
s–1
. Calculate the kinetic energy of an
electron emitted when radiation of frequency  = 1.0 × 1015
s–1
hits the metal.
Sol. According to Einstein’s equation Kinetic energy - 1/2 me
v2
= h( – 0
) = (6.626 × 10–34
J s) (1.0 × 1015
s–1
– 7.0
× 10–14
s–1
) = (6.626 × 10–34
J s) × (3.0 × 1014
s–1
) = 1.988 × 10–19
J
8 . QUANTUM THEORY OF LIGHT :
The smallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiation is called
as quantum of light.
According to Planck, the light energy coming out from any source is always an integral multiple of a smallest energy
value called quantum of light.
Let quantum of light be = E0(J), then total energy coming out is = nE0 (n = Integer)
Quantum of light = Photon ( Packet or bundle of energy)
Energy of one photon is given by
E0 = h (- Frequency of light)
h = 6.625 x 10–34 J-Sec (h - Planck const.)
E0 =

hc
(c - speed of light)
( - wavelength)
Order of magnitude of Eo = 10
8
34
10
10
10



= 10–16 J
One electron volt (e.v.) : Energy gained by an electron when it is accelerated from rest through a potential difference
of 1 volt.
Note : Positive charge always moves from high potential to low potential and –ve charge always. moves from low
potential to high potential if set free.
From Energy conservation principle,
P.E.i + K.E.i = P.E.f + K.E.f
(– e) 0 + 0 = (– e)(1V) +
2
1 2
f
mV
K.E. =
2
1 2
f
mV = e(1 volt)

13. If a charge ‘q’ is accelerated through a potential dirrerence of ‘V’ volt then its kinetic energy will be increased by q.V.
1eV = 1.6 x 10–19 C x 1 volt
 1eV = 1.6 x 10–19J
Ex. (NCERT)
Calculate energy of one mole of photons of radiation whose frequency is 5 × 1014 Hz.
Sol. Energy (E) of one photon is given by the expression E = h
h = 6.626 × 10– 34 J s
= 5 × 1014 s–1 (given)
E = (6.626 × 10–34 J s) × (5 × 1014 s–1) = 3.313 × 10–19 J
energy of one mole of photons = (3.313 × 10–19 J) × (6.022 × 1023 mol–1) = 199.51 KJ mol–1
Ex. (NCERT)
A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per
second by the bulb
Sol. Power of the bulb = 100 watt = 100 J s–1
Energy of one photon E = h = hc/=
m
10
400
s
m
10
3
s
J
10
626
.
6
9
–
1
–
8
34
–




= 4.969 × 10–19 J
Number of photons emitted
J
10
969
.
4
s
J
100
19
–
1
–

= 2.012 × 1020 s–1
Ex. If a charged particle having charge of 2e on being accelerated by 1 volt, its K.E. will be increased by ?
Sol. K.E. = (2e) . (1V)
= 2 x 1eV
= 2 x 1.6 x 10–19
J
= 3.2 x 10–19
J
Ex. A charged particle having a charge +3e is projected towards +ve plate, from –ve plate with K.E.i
= 12eV What
is the minimum potential that should be applied between the plates so that the charged particle cannot strike
the +ve plate ?
Sol. K.E.i
+ P.E.i
= P.E.f
+ K.E.f
(+3e) 0 + 12eV = 3e(V) + 0
V =
e
3
eV
12
= 4Volts
s
Ex. Number of photons emitted by a bulb of 40 watt in 1 minute with 50% efficiency will be approximately
( = 6200 Å, hc = 12400 eV Å)
(A) 7.5 × 1020 (B) 3.75 × 1020 (C) 3.75 × 1019 (D*) 3.75 × 1021
Sol. E =
6200
12400
= 2eV.
.
2 × 1.6 × 10–19 × n = 40 × 60 × 0.5
n = 3.75 × 1021
Ex. Bond energy of Br2
is 194 kJ mole–1
. The minimum wave number of photons required to break this bond is (h
= 6.62 x 10–34
Js, c = 3 x 108
m/s)
(A) 1.458 × 1023
m–1
(B*) 1.620 × 106
m–1
(C) 4.86 × 1014
m–1
(D) 1.45 × 107
m–1
Sol.

hc
x NA
= 194 x 103
 =

1
= 23
8
34
3
10
x
6
x
10
x
10
x
63
.
6
10
x
194

~ 1.62 x 106
m–1
Home - Work
NCERT (Reading) 2.3.2, 2.4
NCERT (Exercise) 2.6, 2.8 - 2.12, 2.46 - 2.54
Sheet (Ex - 1) Part - I Section - B ( 5 - 13)
Part - II Section - B ( 6 - 13)

14. Lecture : 5
9 . BOHR’S ATOMIC MODEL
Bohr’s model is applied only on one electron species like H, He+, Li++, Be+++ etc.
The important postulates of Bohr model of an atom are
(a) Electron revolves around the nucleus in a fixed circular orbit of definite energy. As long as the electron
occupy a definite energy level, it does not radiate out energy i.e. it does not lose or gain energy. These orbits
are called stationary orbits.
(b) Electron revolves only in those orbits whose angular momentum (mvr) is an integral multiple of the factor
h/2(where ‘h’ is Planck’s constant)
mvr = n

2
h
where :
m = mass of the electron v = velocity of the electron
n = number of orbit in which electron revolves i.e. n = 1, 2, 3 ........
r = radius of the orbit.
(c) The energy is emitted or absorbed only when the electron jumps from one energy level to another. It may
jump from higher energy level to a lower level by the emission of energy and jump from lower to higher energy
level by absorption of energy.
This amount of energy emitted or absorbed is given by the difference of the energies of the two energy levels
concerned.
Mathematical forms of Bohr’s Postulates :
Calculation of the radius of the Bohr’s orbit : Suppose that an electron having mass ‘m’ and charge ‘e’
revolving around the nucleusof charge ‘Ze’ (Z is atomic number & e = charge) with a tangential/linear velocity
of ‘v’. Further consider that ‘r’ is the radius of the orbit in which electron is revolving.
According to Coulomb’s law, the electrostatic force of attraction (F) between the moving electron and nucleus
is –
F = 2
2
r
KZe
where : K = constant =
0
4
1

= 9 x 109 Nm2/C2
and the centrifugal force F =
r
mv2
For the stable orbit of an electron both the forces are balanced.
i.e
r
mv2
= 2
2
r
KZe
then v2 =
mr
KZe2
......... (i)
From the postulate of Bohr,
mvr=

2
nh
 v =
mr
2
nh

+
V

15. On squaring v2 = 2
2
2
2
2
r
m
4
h
n

........ (ii)
From equation (i) and (ii)
mr
KZe2
= 2
2
2
2
2
r
m
4
h
n

On solving, we will get
r = 2
2
2
2
mKZe
4
h
n

On putting the value of e , h , m, the radius of nth Bohr orbit is given by :
rn = 0.529 x
Z
n2
Å
Ex. Calculate radius ratio for 2nd
orbit of He+
ion & 3rd
orbit of Be+++
ion.
Sol. r1
(radius of 2nd
orbit of He+
ion) = 0.529 







2
22
Å
r2
(radius of 3rd
orbit of Be+++
ion) = 0.529 







4
32
Å
Therefore
2
1
r
r
=
4
/
3
529
.
0
2
/
2
529
.
0
2
2


=
9
8
Ex. If the radius of second orbit of Li2+
ion is x then find the radius of Ist
orbit of He+
in terms of x.
Sol. r1 (radius of 2nd orbit of Li2+) = 0.529 ×
3
22
Å = x (given) ... (i)
 r2 (radius of 1st orbit of He+) = .0529 ×
2
12
=
2
1
4
x
3
 = 





8
x
3
Calculation of velocity of an electron in Bohr’s orbit :
Angular momentum of the revolving electron in nth orbit is given by
mvr =

2
nh
v =
mr
2
nh

......... (iii)
put the value of ‘r’ in the equation (iii)
then, v = 2
2
2
2
h
mn
2
K
mZe
4
nh



v =
nh
K
Ze2


on putting the values of , e-, h and K
velocity of electron in nth orbit vn = 2.18 x 106 x
n
Z
m/sec v  Z ; v 
n
1
T, Time period of revolution of an electron in its orbit =
v
r
2
f, Frequency of revolution of an electron in its orbit =
r
2
v


16. Calculation of energy of an electron :
The total energy of an electron revolving in a particular orbit is
T.E. = K.E. + P.E.
where :
P.E. = Potential energy , K.E. = Kinetic energy , T.E. = Total energy
The K.E. of an electron =
2
1
mv2
and the P.E. of an electron = –
r
KZe2
Hence, T.E. =
2
1
mv2 –
r
KZe2
we know that,
r
mv2
= 2
2
r
KZe
or mv2 =
r
KZe2
substituting the value of mv2 in the above equation :
T.E. =
r
2
KZe2
–
r
KZe2
= –
r
2
KZe2
So, T.E. = –
r
2
KZe2
substituting the value of ‘r’ in the equation of T.E.
Then T.E. = –
2
KZe2
x 2
2
2
2
h
n
m
Ze
4
= – 2
2
2
4
2
2
h
n
K
m
e
Z
2
Thus, the total energy of an electron in nth orbit is given by
T.E. = En = – 2
2
4
2
h
k
me
2








2
2
n
z
... (iv)
Putting the value of m,e,h and  we get the expression of total energy
En = – 13.6 2
2
n
Z
eV n T.E.  ; Z  T.E. 
T.E. =
2
1
P.E.
T.E. = – K.E.
Note : - The P.E. at the infinite = 0
The K.E. at the infinite = 0
Conclusion from equation of energy :
(a) The negative sign of energy indicates that there is attraction between the negatively charged electron and
positively charged nucleus.
(b) All the quantities on R.H.S. in the energy equation [Eq. iv] are constant for an element having atomic
number Z except ‘n’ which is an integer such as 1,2,3, etc . i.e. the energy of an electron is constant as long
as the value of ‘n’ is kept constant.
(c) The energy of an electron is inversely proportional to the square of ‘n’ with negative sign.
Ex. (NCERT)
What are the frequency and wavelength of a photon emitted during a transition from n = 5 state to the n = 2
state in the hydrogen atom ?
Sol. Since ni = 5 and nf = 2, this transition gives rise to a spectral line in the visible region of the Balmer series.
E = 2.18 × 10–18 J 





2
2
2
1
–
5
1
= – 4.58 × 10–19 J

17. It is an emission energy
The frequency of the photon (taking energy in terms of magnitude) is given by
 =
h
E

=
Js
10
626
.
6
J
10
58
.
4
34
–
19
–


= 6.91 × 1014 Hz
 =

c
=
Hz
10
91
.
6
ms
10
0
.
3
14
1
–
8


= 434 nm
Ex. (NCERT)
Calculate the energy associated with the first orbit of He+. What is the radius of this orbit?
Sol. En = 2
2
18
–
n
Z
)
J
10
18
.
2
( 
atom–1
For He+, n = 1, Z = 2
E1 = – 2
18
–
1
)
J
10
18
.
2
( 
= – 8.72 × 10–18 J
The radius of the orbit is given by equation
rn =
Z
)
n
(
9
.
52 2
pm =
Z
n
)
nm
0529
.
0
( 2
Since n = 1, and Z = 2
rn =
2
1
)
nm
0529
.
0
( 2
= 0.02645 nm
Ex. Calculate energy ratio for 3rd
orbit of Li++
ion & 2nd
orbit of Be+++
ion.
Sol. E1
for 3rd
orbit of Li++
ion
= – 13.6 × 2
2
3
3
ev/atm
E2
for 2nd
orbit of Be+++
ion
= – 13.6 × 2
2
2
4
ev/atm
4
1
E
E
2
1
 Ans.
Ex. Radius of two different orbits in a H like sample is 4R and 16R respactively then find out the ratio of the
frequency of revolution of electron in these two orbits.
Sol.
2
1
r
r
= 2
2
2
1
n
529
.
0
n
529
.
0
=
R
16
R
4
 2
2
2
1
n
n
=
4
1
or
2
1
n
n
=
2
1
Now,
2
1
f
f
=
2
2
1
1
r
2
/
v
r
2
/
v


= 3
1
3
2
n
n
=
2
1
2
n
n








=
3
1
2






= 8 : 1
Home - Work
Sheet (Ex - 1) Part - I : Section - C (14 - 18)
Part - II : Section - C (14 - 18)

18. Lecture : 6
Failures / limitations of Bohr’s theory:
(a) He could not explain the line spectra of atoms containing more than one electron.
(b) He also could not explain the presence of multiple spectral lines.
(c) He was unable to explain the splitting of spectral lines in magnetic field (Zeeman effect) and in electric
field (Stark effect)
(d) No conclusion was given for the principle of quantisation of angular momentum.
(e) He was unable to explain the de-Broglie’s concept of dual nature of matter.
(f) He could not explain Heisenberg’s uncertainty principle.
10. Energy Level Diagram :
(i) Orbit of lowest energy is placed at the bottom, and all other orbits are placed above this.
(ii) The gap between two orbits is proportional to the energy difference of the orbits.
-0.85 eV
-13.6eV
-3.4eV
-1.51eV
n=
n=4
n=3
n=2
n=1
10.2eV
12.1eV
Energy level diagram of H-atom
0 eV 
11. DEFINITION VALID FOR SINGLE ELECTRON SYSTEM :
(i) Ground state :
Lowest energy state of any atom or ion is called ground state of the atom It is n = 1.
Ground state energy of H–atom = – 13.6 ev
Ground state energy of He+ on = – 54.4 ev
(ii) Excited State :
States of atom other than the ground state are called excited states :
n = 2 first excited state
n = 3 second excited state
n = 4 third excited state
n = n + 1 nth excited state
(iii) Ionisation energy (E) :
Minimum energy required to move an electron from ground state to
n =  is called ionisation energy of the atom or ion.
onisation energy of H–atom = 13.6 ev
onisation energy of He+ ion = 54.4 ev
onisation energy of Li+2 ion = 122.4 ev
(iv) Ionisation Potential (.P.) :
Potential difference through which a free electron must be accelerated from rest, such that its kinetic energy
becomes equal to ionisation energy of the atom is called ionisation potential of the atom.
.P. of H atom = 13.6 V, .P. of He+ on= 54.4 V
(v) Excitation Energy :
Energy required to move an electron from ground state of the atom to any other state of the atom is called
excitation energy of that state.
Excitation energy of 2nd state = excitation energy of 1st excited state = 1st excitation energy = 10.2 ev.
(vi) Excitation Potential :
Potential difference through which an electron must be accelerated from rest to so that its kinetic energy
become equal to excitation energy of any state is called excitation potential of that state.
Excitation potential of third state = excitation potential of second exicited state = secondexcitation potential =
12.09 V.

19. (vii) Binding Energy ‘or’ Seperation Energy :
Energy required to move an electron from any state to n =  is called binding energy of that state.
Binding energy of ground state = .E. of atom or on.
Ex. Calculate ionisation energy for Li++
ion.
Sol. I.E.H
= 
E – E1
= 0 – (– 13.6) = + 13.6 ev/atom
E  Z2
I.E for Li++
= (ZLi++
)2
× IEH
= 9 × 13.6 = 122.4 ev Ans.
Ex. Calculate 2nd
excitation energy for He+
ion.
Sol. n1
= 1 n2
= 3 Z = 2
E1
= – 13.6 × 22
= – 54.4 ev E3
= – 13.6 × 2
2
3
2
ev = – 6.04 ev
E = E3
– E1
= – 6.04 – (–54.4 ev) = 47.36 ev Ans.
Ex. Binding Energy of H like system corresponding to second excited state is given by 13.5 eV. then
(a) Identify the sample
(b) Ionisation energy of sample
(c) Energy of photon required to cause the transition from 2nd
to 3rd
excited state.
(d) Wavelength of the emitted photon when electron fall from 1st
excited state to ground state.
Ans. (A)
1.5 Z2
= 13.5
Z2
=
5
.
1
5
.
13
= 9
Z = 3
Ans. Li++
(B) I.E. of sample = 13.6 × 32
= 122.4 ev.
(C) n = 2 to n = 4
1
2
3
4
Energy of photon needed = (E)2  4
(3.4 – 0.85) 32
2.55 × 9
––––––––––––
22.95 eV
(d) (E)2  1
= 10.2 Z2
10.2 × 32
= 10.2 × 9 = 91.8 eV
 =
8
.
91
12400
= 135.1 Å
Ex. If st
excitation potential of H like sample is 15V, find : The .E. and nd
excitation potential of sample.
Sol. st
excitation energy  15 eV (E)
E = – 2
2
n
Z
6
.
13
E1
= –
1
Z
6
.
13 2
(E0
= 13.6 Z2
= IE)

20. E2
= – 2
2
2
Z
6
.
13
 E1
= – E0
 E2
= –
4
E0
E = E2
– E1
= E0
– 





4
E0
= 15eV
4
E
3 o
= 15
Eo
=
3
4
15 
= 20eV
IE = 20eV
E1
= –
1
Z
6
.
13 2
E3
= – 2
2
3
Z
6
.
13
(E)1 – 3
= Eo
–
9
Eo
=
9
8
Eo
=
9
8
× 20 = 17.75 eV.
.
2nd
excitation potential = 17.75 V.
Home - Work
NCERT (Reading) 2.4.2
NCERT (Exercise) 2.13, 2.14, 2.15, 2.16, 2.18, 2.19, 2.34
Sheet (Ex - 1) : Part - I : Section - C ( 19 - 24)
Part - II : Section - C ( 19 - 23)

21. Lecture : 7
12. HYDROGEN SPECTRUM :
Study of Emission and Absorption Spectra :
An instrument used to separate the radiation of different wavelengths (or frequencies) is called spectroscope
or a spectrograph. Photograph (or the pattern) of the emergent radiation recorded on the film is called a
spectrogram or simply a spectrum of the given radiation The branch or science dealing with the study of
spectra is called spectroscopy.
Spectrum
Based on
Nature
Based on
origin
Continuous Discrete Absorption
spectrum
Emission
spectrum
Band
Spectrum
Line
Spectrum
Emission spectra :
When the radiation emitted from some source e.g. from the sun or by passing electric discharge through a
gas at low pressure or by heating some substance to high temperature etc, is passed directly through the
prism and then received on the photographic plate, the spectrum obtained is called ‘Emission spectrum’.
Depending upon the source of radiation, the emission spectra are mainly of two type :
(a) Continuous spectra :
When white light from any source such as sun, a bulb or any hot glowing body is analysed by passing
through a prism it is observed that it splits up into seven different wide band of colours from violet to red.
These colours are so continuous that each of them merges into the next. Hence the spectrum is called
continuous spectrum.
(b) Discrete spectra : It is of two type
(i) Band spectrum
Band
Dark space
Band spectrum contains colourful continuous bands sepearted by some dark space.
Generally molecular spectrum are band spectrum

22. (2) Line Spectrum :
This is the ordered arrangement of lines of particular wavelength seperated by dark space eg. hydrogen
spectrum.
Line spectrum can be obtained from atoms.
2. Absorption spectra :
When white light from any source isfirst passed through the solution or vapours of a chemical substance and
then analysed by the spectroscope, it isobserved that some dark linesare obtained in the otherwisecontinuous
spectrum. These dark lines are supposed to result from the fact that when white light (containing radiations
of many wavelengths) is passed through the chemical substance, radiations of certain wavelengths are
absorbed, depending upon the nature of the element.
EMISSION SPECTRUM OF HYDROGEN :
When hydrogen gas at low pressure is taken in the discharge tube and the light emitted on passing electric
discharge is examined with a spectroscope, the spectrum obtained is called the emission spectrum of
hydrogen.
Line Spectrum of Hydrogen :
Line spectrum of hydrogen is observed due to excitation or de-excitation of electron from one stationary orbit
to another stationary orbit
Let electron make transition from n2
to n1
(n2
> n1
) in a H-like sample
photon
n1
n2
eV
– 13.6 Z
2
n2
2
– 13.6 Z2
n1
2 eV
Energy of emitted photon = (E)n2  n1

23. = 2
2
2
n
Z
6
.
13

– 






 
2
1
2
n
Z
6
.
13
= 13.6Z2








 2
2
2
1 n
1
n
1
Wavelength of emitted photon
 =
1
2 n
n
)
E
(
hc


 =








 2
2
2
1
2
n
1
n
1
Z
6
.
13
hc

1
= 







 2
2
2
1
2
n
1
n
1
hc
z
)
6
.
13
(
Wave number,

1
=  =








 2
2
2
1
2
n
1
n
1
RZ
R = Rydberg constant = 1.09678 × 107
m–1
; R ~1.1 × 107
m–1
; R =
hc
eV
6
.
13
; R ch = 13.6 eV
Ex.1 Calculate the wavelength of a photon emitted when an electron in H- atom maker a transition from n = 2 to n = 1
Sol.

1
= RZ2








 2
2
2
1 n
1
n
1


1
= R(1)2






 2
2
2
1
1
1


1
=
4
R
3
or
R
3
4



24. Lecture : 8
13. SPECTRA LINES OF HYDROGEN ATOM :
LYMAN SERIES
* It is first spectral series of H.
* It was found out in ultraviolet region in 1898 by Lyman.
* It’s value of n1 = 1 and n2 = 2,3,4 where ‘n1’ is ground state and ‘n2’ is called excited state of electron present
in a H - atom.
*

1
= RH








 2
2
2
n
1
1
1
where n2 > 1 always.
* The wavelength of marginal line =
H
2
1
R
n
for all series. So for lyman series  =
H
R
1
.
* st
line of lyman series  2  1
nd
line of lyman series = 3  1
Last line of lyman series =   1
[10.2 eV  (E)lyman  13.6 eV]
6
.
13
12400
 lyman  2
.
10
12400
Aº
Aº
* Longest line : longest wavelength line  longest
or max.
=
min
)
E
(
12400

* Shortest line : shortest wavelength line  shortest
or min
=
max
)
E
(
12400

* First line of any spectral series is the longest ( max
) line.
* Last line of any spectral series is the shortest ( min
) line.
Series limit :
t is the last line of any spectral series.
Wave no of st
line of Lyman series
=

1
=  = R × 12 





 2
2
2
1
1
1
 = R × 12 




 
4
1
4
 =
4
3
R 
=
4
R
3
 







R
3
4
Wave no of last line of Lyman series
 = R × 12 






 2
2
1
1
1
 = R
For Lyman series,
 longest
=
1
2
)
E
(
12400


, shortest =
  1
E
12400




25. BALMER SERIES :
* It is the second series of H-spectrum.
* It was found out in 1892 in visible region by Balmer.
* It’s value of n1 = 2 and n2 = 3,4,5,.............
* The wavelength of marginal line of Balmer series =
H
2
1
R
n
=
H
2
R
2
=
H
R
4
*

1
= RH 







 2
2
2
n
1
2
1
where n2 > 2 always.
1.9  (E)balmer  3.4 eV.
.
All the lines of balmer series in H spectrum are not in the visible range. nfact only st
4
lines belongs to visible range.
4
.
3
12400
Aº   balmer  Å
9
.
1
12400
3648 Å   balmer  6536 Å
Lines of balmer series (for H atom) lies in the visible range.
Ist
line of balmer series = 3  2
last line of balmer series =   2
(  ) 1st
line = R ×1 





 2
2
3
1
2
1
=
36
R
5
(  ) last line = R 






 2
2
1
2
1
=
4
R
PASCHEN SERIES :
(a) It is the third series of H - spectrum.
(b) It was found out in infrared region by Paschen.
(c) It’s value of n1 = 3 and n2 = 4,5,6 ........
(d) The wavelength of marginal line of Paschen series =
H
2
1
R
n
=
H
2
R
3
=
H
R
9
.
(e)

1
= RH








 2
2
2
n
1
3
1
where n2 > 3 always.
BRACKETT SERIES :
(a) It is fourth series of H - spectrum.
(b) It was found out in infrared region by Brackett.
(c) It’s value of n1 = 4 and n2 = 5,6,7 ..............
(d) The wavelength of marginal line of brackett series =
H
2
1
R
n
=
H
2
R
4
=
H
R
16
(e)

1
= RH








 2
2
2
n
1
4
1
where n2 > 4 always.
PFUND SERIES :
(a) It is fifth series of H- spectrum.
(b) It was found out in infrared region by Pfund.
(c) It’s value of n1 = 5 and n2 = 6,7,8 ............... where n1 is ground state and n2 is excited state.
(d) The wavelength of marginal line of Pfund series =
H
2
1
R
n
=
H
2
R
5
=
H
R
25
(e)

1
= RH








 2
2
2
n
1
5
1
where n2 > 5 always.

26. HUMPHRY SERIES :
(a) It is the sixth series of H - spectrum.
(b) It was found out in infrared region by Humphry.
(c) It’s value of n1 = 6 and n2 = 7 , 8 , 9 ...................
(d) The wavelength of marginal line of Humphry series =
H
2
1
R
n
=
H
2
R
6
=
H
R
36
(e)

1
= RH








 2
2
2
n
1
6
1
where n2 > 6.
Ex. Calculate wavelength for 2nd
line of Balmer series of He+
ion
Sol.










 2
2
2
1
2
n
1
n
1
)
2
(
R
1
n1
= 2 n2
= 4








 2
2
2
4
1
2
1
)
2
(
R
1
4
R
3
1


 =
3
R
4
Ans.
14. No. of photons emitted by a sample of H atom :
If an electron is in any higher state n = n and makes a transition to ground state, then total no. of different
photons emitted is equal to
2
)
1
n
(
n 

.
If an electron is in any higher state n = n2
and makes a transition to another excited state n = n1
, then total
no. of different photons emitted is equal to
2
)
1
n
(
n 


, where n = n2
– n1
Note In case of single isolated atom if electron make transition from nth
state to the ground state then max.
number of spectral lines observed = (n-1)
Ex. If electron make transition from 7th
excited state to 2nd
in H atom sample find the max. number of spectral
lines observed.
n = 8 – 2 = 6
spectral lines = 6 




 
2
1
6
= 6×
2
7
= 21
Ex. An electron in isolated hydrogen atom is in 4th
excited state, then, upon de excitation:
(A) the maximum number of possible photons will be 10
(B) the maximum number of possible photons will be 6
(C) it can emit two photons in ultraviolet region
(D*) if an infrared photon is generated, then a visible photon may follow this infrared photon.
Ex. In a hydrogen like sample, ions are in a particular excited state, if electrons make transition upto 1st
excited state, then it produces maximum 15 different types of spectral lines then electrons were in
(A) 5th state (B) 6th state (C*) 7th state (D) 8th state
Sol.
 
2
1
n
n 


= 15
 n
 = 5
 n – 2 = 5
 n = 7

27. Ex. In a Hydrogen like sample, all the ions are in a particular excited state. When electron make transition up to
ground state, then 6 diff. typesof photons are observed. If second excitation potential of the sample is 193.64
V, then :
(a) dentify the sample
(b) Find the excited state of sample
(c) Find ionisation energy
(d) Find binding energyof 3rd
state
(e) Shortest and longest wavelength belonging to above transition
(f) Series limit of Brackett series (wavelength) (in terms of R)
(g) Find the diff between the wave number of 2nd
line of Lyman series and st
line of Balmer series
(h) If 2nd
line of Balmer series of this sample is used to excite the He+
ion already in ground state, then find
the final excited state of He+
(i) Find the no of lines not in visible range.
(j) If single isolated atom is considered in the above sample then find the max. number of spectral lines
observed in the above transition
Sol. (a) Second excitation energy = 193.6 eV.
12.1 Z2
= 193.6 eV
Z2
=
1
.
12
6
.
193
= 16
Z2
= 16
Z = 16 = 4
Hence sample is Be3+
(b)
2
)
1
n
(
n 
=
1
6
n = 4
Hence sample is in 3rd
excited state or 4th
state
(c) .E. = 13.6 Z2
=
6
.
218
16
6
.
13 
eV
(d) B.E. of 3rd
state = 1.5 × 42
= 1.5 × 16 = 24.0 eV
(e) longest
=
3
4
)
E
(
12400


= 2
4
)
85
.
0
5
.
1
(
12400


 shortest
=
1
4
)
E
(
12400


(f) 
1
= R × 42 






 2
2
1
4
1

1
= R,  =
R
1
(g) Lyman series 1
 = R × 42 





 2
2
3
1
1
1
=
9
R
8
×16 =
9
R
128
( 2
 ) balmer = R × 42 





 2
2
3
1
2
1
=
36
80
R 
=
9
R
20
2
1 

 =
9
R
128
–
9
R
20
=
9
R
20
R
128 
=
9
R
108

28. (h)
3
2
n = 1
-3.4 eV
-13.6 eV
H
5
4
3
2
n = 1
6
-3.4 eV
-13.6 eV
Li
++
Note : Let electron make transition from n2
to n1
in a Hydrogen like sample having atomic number Z1
and n3  n4
transition in a sample having atomic no z2
then,
(E)
1
2 n
n  = (E)
3
4 n
n 
13.6 Z1
2








 2
2
2
1 n
1
n
1
= 13.6 Z2
2








 2
4
2
3 n
1
n
1
2
2
2
1
2
1
2
1
n
Z
n
Z
 = 2
4
2
2
2
3
2
2
n
Z
n
Z

On comparing,
2
3
2
2
2
1
2
1
n
Z
n
Z

3
2
1
1
n
Z
n
Z

 
2
1
3
1
Z
Z
n
n

& 2
4
2
2
2
2
2
1
n
Z
n
Z
 
2
1
4
2
Z
Z
n
n


3
1
n
n
=
4
2
n
n
=
2
1
Z
Z
H4  2
use for He+
sample n2
= 4, n1
= 2, Z1
= 1
3
n
2
=
4
n
4
=
2
Z
1
n3
= 2Z2
, n4
= 4Z2
nd
line of Balmer series = 4  2
n1
= 2 , n2
= 4,
Z1
= 4, Z2
= 2.

29. 3
n
2
=
4
n
4
=
2
4








2
n
1
n
4
3
(i)
(E)4–3
= (1.5 – 0.85) × 42
eV
40
.
10
16
65
.
0 
 visible
range 4000   visible  8000 Aº
Aº
8000
12400
 E 
400
12400
1.55 eV  E  3.1 eV
There are no lines in visible range all the lines are lie in the ultraviolet region.
(j)
4
3
2
1
Be
3+
Maximum number of spectral lines observed = 3 (4 3, 3  2, 2  1)
Home - Work
NCERT (Reading) 2.3.3, 2.4.1
NCERT (Exercise) 2.17, 2.33, 2.55, 2.56
Sheet (Ex - 1) Part - I : Section - D (25 - 31)
Part - II : Section - D (24 - 31)

30. Lecture : 9
15. Dual nature of electron (de-Broglie Hypothesis):
(a) Einstein had suggested that light can behave as a wave as well as like a particle i.e. it has dual character.
(b) In 1924, de-Broglie proposed that an electron behaves both as a material particle and as a wave.
(c) This proposed a new theory wave mechanical theory of matter. According to this theory, the electrons
protons and even atom when in motion possess wave properties.
(d) According to de-Broglie, the wavelength associated with a particle of mass m, moving with velocity v is
given by the relation,
 =
mv
h
where h is Planck’s constant
(e) This can be derived as follows according to Planck’s equation.
E = h  =

c
.
h
Energy of photon on the basis of Einstein’s mass energy relationship
E = mc2 or  =
mc
h
Equating both we get

c
.
h
= mc2
or  =
mc
h
Which is same as de - Broglie relation.
This was experimentally verified by Davisson and Germer by observing diffraction effects with an electron
beam.
Let the electron is accelerated with a potential of V then the K.E. is
2
1
mv2 =eV
m2v2 = 2emV
mv = emV
2 = p (momentum)
 =
emV
2
h
If we associate Bohr’s theory with de - Broglie equation then
2r = n or  =
n
r
2
From de-Broglie equation
 =
mv
h
therefore
mv
h
=
n
r
2
so, mvr =

2
nh
m = dynamic mass = 2
0
c
v
1
m







m0
= rest mass of particle
depended on velocity
c = velocity of light

31. If velocity of particle is zero then :
dynamic mass = rest mass
Rest mass of photon is zero that means photon is never at rest
* K.E. =
2
1
mv2
m (K.E.) =
2
1
m2
v2
multiplied by mass on both side
 m.v. = .)
E
.
K
(
m
2
 = .)
E
.
K
(
m
2
h
If a charge q is accelerated through a potential difference of ‘V’ volt from rest then K.E. of the charge is equal
to : “ q.V”
 = )
V
.
q
(
m
2
h
* If an electron is accelerated through a potential difference of ‘V’ volt from rest then :
  =
)
eV
(
m
2
h
e
  =
2
1
V
150






Å (on putting values of h, me
and e)
  =
V
3
.
12
Å (V in volt)
* mvr = n ×

2
h
 =
mv
h
mv =

h
putting this in mvr =

2
nh


h
r =

2
nh
 




 


n
r
2
de Broglie wavelength
Ex. (NCERT)
What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10 m s–1
?
Sol. According to de Broglie equation
=
mv
h
=
)
s
m
10
(
)
kg
1
.
0
(
)
Js
10
626
.
6
(
1
–
34
–

= 6.626 × 10–34
m (J = kg m2
s–2
)
Ex. (NCERT)
The mass of an electron is 9.1 × 10–31
kg. If its K.E. is 3.0 × 10–25
J, calculate its wavelength.
Sol. Since K.E. =
2
1
mv2
v =
2
1
m
.
E
.
K
2






=
2
1
31
–
2
–
2
25
–
kg
10
1
.
9
s
m
kg
10
0
.
3
2











= 812 m s–1

32.  =
v
m
h
=
)
s
m
812
(
)
kg
10
1
.
9
(
Js
10
626
.
6
1
–
31
–
34
–


= 8967 × 10–10
m = 896.7 nm
Ex. Find the wavelength associated with a dust particle of mass 5 mg. moving with velocity 200 m/s.
 =
mv
h
=
200
10
5
10
62
.
6
6
34





= 6
34
10
1000
10
62
.
6




= 6.62 ×10–31
m
Note : Since order of the wavelength is extremely small, therefore de-Broglie wavelength calculation for daily life
particle has no physical significance
Ex. Find the wavelength associated with the electron moving with the velocity 106
m/s.
 =
mv
h
= 6
31
34
10
10
1
.
9
10
62
.
6





= 
1
.
9
10
62
.
6 9


0.7 × 10–9
m = 7× 10–10
m = 7 Aº
Aº
Note Since wavelength associated with the electron has considerable value therefore we can conclude that
electron has more of wave nature and less of particle nature.
Ex. Calculate ratio of wavelength for an  particle & proton accelerated through same potential difference.
Sol.
V
m
q
2
h


 
 , V
m
q
2
h
p
p
p 

, m
= 4mp
, q
= 2qp
2
2
1
8
1
4
1
.
2
1
m
.
q
m
q p
p
p









Ans.
Ex. (a) Hydrogen sample is in st
excited state A photon of energy 8eV is used to excite the hydrogen sample.
Find the de-Broglie wavelength of the electron.
(a) n = 2
K.E. of electron = 8 – 3.4 = 4.6 eV
 =
6
.
4
150
Aº = 5.71 Å
OR
2
1
mv2
= 4.6 × 1.6 × 10–19
(m = 9.1 × 10– 31
kg)
v = 1.27 × 106
m/sec
Putting the value of v in  =
mv
h
, we get :
 = 6
31
–
34
–
10
27
.
1
10
1
.
9
10
626
.
6




 5.7 Å
Ex. An electron having Initial K.E. of 6eV is accelerated through a P.D. of 4V find the wavelength of associated
with electron
Ans. Total K.E. = 6eV + 4eV = 10eV
 =
2
1
10
150






=
10
150
= 15 = 3.87 Å

33. Ex. De-Broglie wavelength of electron in second orbit of Li2+
ion will be equal to de-Broglie of wavelength of
electron in
(A) n = 3 of H-atom (B*) n = 4 of C5+
ion (C) n = 6 of Be3+
ion (D) n = 3 of He+
ion
Sol.  =
mv
h
Now v  





n
z
so  





z
n
for second orbit of Li2+
 





3
2
for n = 4 of C5+
ion,  





6
4
=
3
2
Hence the result.
Ex. An electron, practically at rest, is initially accelerated through a potential difference of 100 volts. It then has
a de Broglie wavelength = 1
Å. It then get retarded through 19 volts and then has a wavelength 2
Å.Afurther
retardation through 32 volts changes the wavelength to 3
, What is
1
2
3




?
(A)
41
20
(B)
63
10
(C*)
63
20
(D)
41
10
Sol. 1
=
)
100
(
mq
2
h
=
10
k
, 2
=
)
81
(
mq
2
h
=
9
k
3
=
)
49
(
mq
2
h
=
7
k
1
2
3 –



=
10
k
9
k
–
7
k
=












10
k
63
k
2
=
63
20
so the required answer.
16. Heisenberg’s Uncertainty Principle :
The exact position and momentum of a fast moving particle cannot be calculated precisely at the same
moment of time. If x is the error in the measurement of position of the particle and if p is the error in the
measurement of momentum of the particle, then:
x . p  
4
h
or x . (mv) 

4
h
where, x = uncertainty in position
p = uncertainty in momentum
h = Planck’s constant
m = mass of the particle
v = uncertaily in velocity
If the position of a particle is measured precisely, i.e. x  0 then p   .
If the momentum of the particle is measured precisely. i.e. p  0 then x   .
This is because of a principle of optics that if a light of wavelength ‘’ is used to locate the position of a
particle then maximum error in the position measurement will be  
i.e. x =  
If x  0 ;   0
But, p =

h
 p  
So, to make x  0 ,   0 a photon of very high energy is used to locate it.
 When this photon will collide with the electron then momentum of electron will get changed by a
large amount.

34. * p.x  
4
h
(multiplied & divided by t)
x
.
t
t
P






4
h
(
t
P


= rate of change in momentum = F)
F.x.t 

4
h
E . t 

4
h
E 
 uncertainty in energy
t 
 uncertainty in time
 In terms of uncertainty in energy E, and uncertainty in time t, this principle is written as,
E.t 

4
h
.
 Heisenberg replaced the concept of definite orbits by the concept of probability.
Ex. (NCERT)
A golf ball has a mass of 40 g, and a speed of 45 m/s. If the speed can be measured within accuracy of 2%,
calculate the uncertainty in the position.
Sol. The uncertainty in the speed is 2%, i.e., 45 ×
100
2
= 0.9 m s–1.
Using the equation
x =
v
m
4
h


=
)
s
m
9
.
0
(
10
40
14
.
3
4
10
626
.
6
1
–
3
–
34
–




= 1.46 × 10–33 m
This is nearly ~ 1018 times smaller than the diameter of a typical atomic nucleus. As mentioned earlier for
large particles, the uncertainty principle sets no meaningful limit to the precision of measurements.
Ex. A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1Å:
What is the uncertainty involved in the measurement of its velocity?
[Mass of electron = 9.1 x 1031
kg and Planck’s constant (h) = 6.626 x 1034
Js)
Sol. x × p =

4
h
or x × mv =

4
h
v =
m
x
4
h




=
kg
10
1
.
9
m
10
1
.
0
14
.
3
4
s
J
10
626
.
6
31
10
34









= 0.579 × 107
ms–1
(1J = 1 kgm2
s–2
)
= 5.79 × 106
ms–1
Ex. Uncertainty in position is twice the Uncertainty in momentum Uncertainty in velocity is -
(A)

h
m
1
(B)

h
m
2
1
(C*)

2
h
m
2
1
(D) m
4
h

Sol. x = 2p
Now x. p


4
h
2p2


4
h
; 2(mv)2


4
h
; v


2
h
m
2
1
Home - Work
NCERT (Reading) 2.5, 2.5.1, 2.5.2
NCERT (Exercise) 2.20, 2.21, 2.32, 2.57- 2.61
Sheet (Ex - 1) Part - I : Section - E (32 - 37)
Part - II : Section - E (32 - 38)

35. Lecture : 10
17. Orbital :
An orbital may be defined as the region of space around the nucleus where the probability of finding an
electron is maximum (90% to 95%)
Orbitals do not define a definite path for the electron, rather they define only the probability of the electron
being in various regions of space around the nucles.
Difference between orbit and orbitals
Orbit Orbitals
1. It is well defined circular path followed 1. It is the region of space around the nucleus
by revolving electrons around the nucleus where electron is most likely to be found
2. It represents planar motion of electron 2. It represents 3 dimensional motion of
an electron around the nucleus.
3. The maximum no. of electron in an 3. Orbitals can not accomodate more than
orbits is 2n2
where n stands for no. 2 electrons.
of orbit.
4. Orbits are circular in shape. 4. Orbitals have different shape e.g. s-orbital is
spherical, p - orbital is dumb- bell shaped.
5. Orbit are non directional in character. 5. Orbitals (except s-orbital) have directional character.
Hence, they cannot explain shape of Hence, they can account for the shape of molecules.
molecules
6. Concept of well defined orbit is against 6. Concept of orbitalsis in accordance with Heisenberg’s
Heisenberg’s uncentainty principle. principle
18. Shape of the orbitals :
Shape of the orbitals are related to the solutions of Schrodinger wave equation, and gives the space in which
the probability of finding an electron is maximum.
s- orbital : Shape  spherical
s- orbital is non directional and it is closest to the nucleus, having lowest energy.
s-orbital can accomodate maximum no. of two electrons.

36. p-orbital : Shape  dumb bell
Dumb bell shape consists of two loops which are separated by a region of zero probability called node.
p - orbital can accomodate maximum no. of six electrons.
d - Orbital : Shape  double dumb bell
d - orbital can accomodate maximum no. of 10 electrons.
f - orbital : Shape  leaf like
f - orbital can accomodate maximum no. of 14 electrons.

37. Lecture : 11
19. QUANTUM NUMBERS :
The set of four numbers required to define an electron completely in an atom are called quantum numbers.
The first three have been derived from Schrodinger wave equation.
(i) Principal quantum number (n) : (Proposed by Bohr)
It describes the size of the electron wave and the total energy of the electron. It has integral values
1, 2, 3, 4 ...., etc., and is denoted by K, L, M, N. ..., etc.
* Number of subshell present in nth shell = n
n subshell
1 s
2 s, p
3 s, p, d
4 s, p, d, f
* Number of orbitals present in nth shell = n2 .
* The maximum number of electrons which can be present in a principal energy shell is equal to 2n2.
No energy shell in the atoms of known elements possesses more than 32 electrons.
* Angular momentum of any orbit =

2
nh
(ii) Azimuthal quantum number () : (Proposed by Sommerfield)
It describes the shape of electron cloud and the number of subshells in a shell.
* It can have values from 0 to (n – 1)
* value of  subshell
0 s
1 p
2 d
3 f
* Number of orbitals in a subshell = 2 + 1
* Maximum number of electrons in particular subshell = 2 × (2 + 1)
* Orbital angular momentum L =

2
h
)
1
( 

 =  )
1
( 

 







2
h

i.e. Orbital angular momentum of s orbital = 0, Orbital angular momentum of p orbital =

2
h
2 ,
Orbital angular momentum of d orbital =

2
h
3
(iii) Magnetic quantum number (m) : (Proposed by Linde)
It describes the orientations of the subshells. It can have values from – to +  including zero, i.e.,
total (2 + 1) values. Each value corresponds to an orbital. s-subshell has one orbital, p-subshell three
orbitals (px, py and pz), d-subshell five orbitals )
d
,
d
,
d
,
d
,
d
( 2
2
2
z
y
x
zx
yz
xy  and f-subshell has seven orbitals.
The total number of orbitals present in a main energy level is ‘n2’.
(iv) Spin quantum number (s) : (Proposed by Goldschmidt & Uhlenbeck)
It describes the spin of the electron. It has values +1/2 and –1/2. (+) signifies clockwise spinning and
(–) signifies anticlockwise spinning.
* Spin magnetic moment s = mc
2
eh
 )
1
s
(
s  or  = )
2
n
(
n  B.M. (n = no. of unpaired electrons)
* It represents the value of spin angular momentum which is equal to

2
h
)
1
s
(
s 
* Maximum spin of atom =
2
1
x No. of unpaired electron.

38. Ex. (NCERT)
What is the total number of orbitals associated with the principal quantum number n = 3 ?
Sol. For n = 3, the possible values of  are 0, 1 and 2. Tthere is one 3s orbital (n = 3,  = 0 and m = 0) ; there are
three 3p orbitals (n = 3,  = 1 and m= – 1, 0, + 1) ; there are five 3d orbitals
(n = 3,  = 2 and m= – 2, – 1, 0, + 1+, + 2).
Therefore, the total number of orbitals is 1 + 3 + 5 = 9
The same value can also be obtained byusing the relation; number of orbitals = n2, i.e. 32 = 9.
Ex. (NCERT)
Using s, p, d, f notations, describe the orbital with the following quantum numbers
(a) n = 2,  = 1, (b) n = 4, = 0, (c) n = 5,  = 3, (d) n = 3, = 2
Sol. n  orbital
(a) 2 1 2p
(b) 4 0 4s
(c) 5 3 5f
(d) 3 2 3d
Ex. Find orbital angular momentum of an electron in (a) 4s subshell and (b) 3p subshell
Ans. (a) 0
(b)

2
h
2
Ex. Orbital angular momentum of an electron in a particular subshell is

h
5 then find the maximum number
of electrons which may be present in this subshell.
Sol. Orbital angular momentum = )
1
( 



2
h
 )
1
( 



2
h
= 5 
h
)
1
( 

 = 5
2 = 20   = 4.
hence maximum number of electrons in this subshell = 2(2 + 1) = 18.
Ans. 18
Ex. Which of the following set of quantum numbers is not valid.
(A) n = 3, l = 2, m = 2, s = +
2
1
(B) n = 2, l = 0, m = 0, s = –
2
1
(C) n = 4, l = 2, m = –1, s = +
2
1
(D*) n = 4, l = 3, m = 4, s = –
2
1
( m >  is not possible)
20. Electronic configuration :
Pauli’s exclusion principle :
No two electrons in an atom can have the same set of all the four quantum numbers, i.e., an orbital cannot
have more than 2 electrons because three quantum numbers (principal, azimuthal and magnetic) at the most
may be same but the fourth must be different, i.e., spins must be in opposite directions.
Aufbau principle :
Aufbau is a German word meaning building up. The electrons are filled in various orbitals in order of their
increasing energies. An orbital of lowest energy is filled first. The sequence of orbitals in order of their
increasing energy is :
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, ....
The energy of the orbitals is governed by (n + ) rule.

39. ‘n + Rule :
The relative order of energies of various sub-shell in a multi electron atom can be predicated with the help
of ‘n + ’ rule
 The sub-shell with lower value of
(
n + 
)
has lower energy and it should be filled first.
eg. 3d 4s
(n +) = 3 + 2 (n +) = 4 + 0
= 5 = 4
Since, (n + ) value of 3d is more than 4s therefore, 4s will be filled before 3d.
 If two sub-shell has same value of
(
n + 
)
then the sub-shell with lower value of n has lower energy and it
should be filled first.
eg. 3d 4p
(n +) = 3 + 2 = 4 + 1
= 5 = 5
3d is filled before 4p.
MEMORY MAP :
1s
2s
3s
4s
5s
6s
2p
3p
4p
5p
6p
3d
4d 4f
5d 5f
1 – s
2
3 – s,p
4
– s,p
– s,d,p
5 – s,d,p
6 – s,f,d,p
6 – s,f,d,p
7 – s,f,d,p
Hund’s rule :
No electron pairing takes place in the orbitals in a sub - shell until each orbital is occupied by
one electron with parallel spin. Exactly half filled and fully filled orbitals make the atoms more stable, i.e., p3,
p6, d5, d10, f7 and f14 configuration are most stable.
Ex. Write the electronic configuration and find the no. of unpaired electrons as well as total spin for the following
atoms :
(1) 6
C
(2) 8
O
(3) 15
P
(4) 21
Sc
(5) 26
Fe
(6) 10
Ne
(i) 6
C  1s2
, 2s2
, 2p2
No. of unpaired electrons  2.
Total spin =
2
2

or
2
2


40. (ii) 8
O  1s2
, 2s2
, 2p4
1s 2s 2p
 No. of unpaired electrons = 2
Total spin =
2
2

or
2
2

(iii) 15
P 1s2
, 2s2
, 2p6
, 3s2
, 3p3
3s 3p
 No. of unpaired electrons = 3
Total spin =
2
3

or
2
3

(iv) 21
Sc  1s2
, 2s2
, 2p6
, 3s2
, 3p6
, 4s2
, 3d1
or [Ar] 4s2
3d1
3d 4s
[Ar] 3d1
4s2
 No. of unpaired electrons = 1
 Total spin =
2
1

or
2
1

(v) 26
Fe  1s2
, 2s2
2p6
, 3s2
, 3p6
, 4s2
3d6
or [Ar] 4s2
, 3d6
3d
4s
No. of unpaired electrons = 4
 Total spin =
2
4

or
2
4

(vi) 10
Ne  1s2
, 2s2
2p6
No. of unpaired electrons = 0
Total spin = 0
EXCEPTIONS :
(1) 24
Cr = [Ar] 4s2
, 3d4
(Not correct)
[Ar] 4s1
, 3d5
(correct : as d5
structure is more stable than d4
structure)
(2) 29
Cu = [Ar] 4s1
, 3d10
(correct : as d10
structure is more stable than d9
structure)
Ex. Find the electronic configuration of Fe2+
and Cr3+
and their magnetic moments.
Ans.

41. (i) Fe  [Ar] 3d6
, 4s2
Fe2+
[Ar] 3d6
3d
No. of unpaired electrons = 4
 Magnetic moment = )
2
n
(
n  B.M. = )
2
4
(
4  B.M. = 6
4 B.M. = 24
B.M. = 6
2 B.M.
(ii) Cr  [Ar] 3d5
, 4s1
Cr3+
 [Ar] 3d3
 No of unpaired electrons = 3
 Magnetic moment = )
2
n
(
n  B.M. = )
2
3
(
3  B.M. = 5
3 B.M. = 15 B.M.
Ex. Write all four quantum numbers for the last electrons of Na
1s2
, 2s2
,2p6
, 3s1
n = 3
 = 0
m = 0
s = ±
2
1
Home - Work
NCERT (Reading) 2.6.4, 2.6.5
NCERT (Exercise) 2.23 - 2.31, 2.62 - 2.67
Sheet (Ex - 1) Part - I : Section - F (38 - 43)
Part - II : Section - F (39 - 49)