Chapter 01

STRUCTURE OF AN ATOM

1. INTRODUCTION 2. They are deflected towards the positive plate in an electric

field suggesting that they are negatively charged. They were
In this chapter, we explore the inside world of atoms which is full of named as electrons by Stoney.
mystery and surprises. Whole chemistry is based on atoms and 3. They can make a light paddle wheel to rotate placed in their
their structures. We will also study the behaviour exhibited by the path. This means they possess kinetic energy and are material
electrons and their consequences. particles.
1.1 Discovery of fundamental particles 4. They ionise gases through which they travel.
Dalton’s atomic theory was able to explain the law of conservation 5. They produce X-rays when they strike a metallic target.
of mass, law of constant composition and law of multiple 6. The characterstics of cathode rays (electrons) do not depend
proportion very successfully but it failed to explain the results of on the material of electrodes and nature of the gas present in
many experiments like it was known that substances like glass or the cathode ray tube.
ebonite when rubbed with silk or fur generate electricity
7. The negatively charged material particles constituting the
1.1.1 Discovery of electron cathode rays are called electrons.
William Crookes in 1879 studied the electrical discharge in Thus, we can conclude that electrons are basic constituents
partially evacuated tubes known as cathode ray discharge tubes. of all matter.
A discharge tube is made of glass, about 60cm long containing 1.1.3 Charge to mass ratio of electron
two thin pieces of metals called electrodes, sealed in it. This is
known as Crookes Tube. The negative electrode is known as
cathode and positive electrode is known as anode.
When a gas enclosed at low pressure(  10-4 atm) in discharge
tube is subjected to a high voltage (  10,000V), invisible rays
originating from the cathode and producing a greenish glow
behind the perforated anode on the glass wall coated with
phosphorescent material ZnS is observed. These rays were called
cathode rays.

Apparatus of charge to mass ratio of electron

In 1897, J.J. Thomson measured e/m ratio of electron by using
cathode ray tube and applying electric and magnetic field
perpendicular to each other as well as to the path of electrons.
The extent of deviation of electrons from their path in the presence
of electric and magnetic field depends on:
(a) Charge on the electron
Production of Cathode rays (b) Mass of the particle
1.1.2 Properties of Cathode rays (c) The strength of electric or magnetic field

1. They produce sharp shadow of the solid object in their path When only electric field is applied, the electrons are deflected to
suggesting that they travel in straight line. the point A. When only magnetic field is applied, the electrons
are deflected to the point C. By balancing the strengths of electric
or magnetic fields, the electrons are allowed to hit the screen at cathode and then from the gas inside the discharge tube due to
point B i.e. the point where electrons hit in the absence of electric bombardment of the gas molecules by the high speed electrons
and magnetic field. By measuring the amount of deflections, emitted first from the cathode.
Thomson was able to calculate the value of e/m as
1.2.1 Discovery of proton
1.758820 × 1011C/kg.
Since the atom as a whole is electrically neutral and the presence
1.1.4 Charge on the electron
of negatively charged particles in it was established, therefore it
was thought that some positively charged particles must also be
present in the atom. So, during the experiments with cathode
rays, the scientist Goldstein designed a special type of discharge
tube. He discovered new rays called Canal rays. The name canal
rays is derived from the fact that the rays travelled in straight line
through a vacuum tube in the opposite direction to cathode rays,
pass through and emerge from a canal or hole in the cathode.
They are also known as anode rays.

Millikan oil drop apparatus

R.A Millikan devised a method known as oil drop experiment

to determine the charge on the electrons.
In this method, oil droplets in the form of mist, produced by the Production of Anode rays
atomiser, were allowed to enter through a tiny hole in the upper
plate of electrical condenser. The downward motion of these 1.2.2 Properties of anode rays
droplets was viewed through the telescope, equipped with a 1. They travel in straight lines.
micrometer eye piece. By measuring the rate of fall of these
2. They carry a positive charge.
droplets, Millikan was able to measure the mass of oil droplets.
The air inside the chamber was ionized by passing a beam of X- 3. They are made up of material particles.
rays through it. The electrical charge on these oil droplets was 4. The value of the charge on the particles constituting the
acquired by collisions with gaseous ions. The fall of these charged anode rays is found to depend on the nature of gas taken.
oil droplets can be retarded, accelerated or made stationary
5. The mass of the particles constituting the anode rays is
depending upon the charge on the droplets and the polarity and
found to depend on the nature of gas taken.
strength of the voltage applied to the plate. By carefully
measuring the effects of electrical field strength on the motion of 6. The charge to mass ratio(e/m) of the particles is also found
oil droplets, Millikan concluded that the magnitude of electrical to depend on the gas taken.
charge, q, on the droplets is always an integral multiple of the 7. Their behaviour in electric and magnetic field is opposite to
electrical charge e, that is, q = n e, where n = 1, 2, 3... . etc. that observed for electron.
Charge on the electron is found to be –1.6022 × 10–19C. 1.2.3 Origin of anode rays
The mass of electron was thus calculated as These rays are believed to be produced as a result of the knock
out of the electrons from the gaseous atoms by the bombardment
e 1.6022 1019 C of high speed electrons of the cathode rays on them. These
m   9.1094  1031 kg.
e / m 1.758820  1011 C / kg anode rays are not emitted from the anode but are produced in
the space between the anode and the cathode.
1.1.5 Origin of cathode rays
The lightest positively charged particles were obtained when
The cathode rays are first produced from the material of the the gas taken in the discharge tube was hydrogen. The e/m value
of these particles were maximum. They had minimum mass and
of  particles ( 42 He) is incident on Beryllium (Be), a new type of
unit positive charge. The particle was called a proton.
particle was ejected. It had mass almost equal to that of a proton
Charge on a proton = + 1.6022 × 10–19C
( 1.672 ×10–27kg) and carried no charge.
Mass of a proton = 1.672 × 10–27kg
11
5 B + 24 He  14 1
7 N + 0n
1.3.1 Discovery of neutron

The theoretical requirement for the existence of a neutron particle

9
4 Be + 42 He  12
6 C + 10 n
in the atomic nucleus was put forward by Rutherford in 1920.It
was proposed to be a particle with no charge and having mass
almost equal to that of a proton. He named it as neutron. In 1932,
Chadwick proved its existence. He observed that, when a beam
Properties of Fundamental Particles

Name Discoverer Symbol Charge Relative Mass

Charge

Electron J.J. Thomson e –1.6022 × 10–19 C –1 9.1094×10–31 Kg

Proton Goldstein p +1.6022 × 10–19C +1 1.6726 × 10–27 kg

Neutron Chadwick n 0 0 1.6749 × 10–27 kg

2. DEFINITIONS OF SUB-ATOMIC PARTICLES The charge distribution was such, that it gave the most stable
arrangement. This model of the atom was often called the plum –
2.1 Electron pudding model. Also the raisin pudding model or watermelon
A fundamental particle which carries one unit negative charge model.
and has a mass nearly equal to 1/1837th of that of hydrogen atom.
2.2 Proton
A fundamental particle which carries one unit positive charge
and has a mass nearly equal to that of hydrogen atom.
2.3 Neutron
A fundamental particle which carries no charge but has a mass
nearly equal to that of hydrogen atom.

3. THOMSON MODEL Thomson’s proposed model of atom

3.1 Drawbacks
Sir J. J. Thomson, who discovered the electron, was the first to Though the model was able to explain the overall neutrality of
suggest a model of atomic structure. the atom, it could not satisfactorily explain the results of scattering
(i) All atoms contain electrons. experiments carried out by Rutherford.
(ii) The atom as a whole is neutral. The total positive charge
and total negative charge must be equal. 4. RUTHERFORD’S -SCATTERING EXPERIMENT
He visualised all the positive charge of the atom as being spread
Rutherford conducted - particles scattering experiments in 1909.
out uniformly throughout a sphere of atomic dimensions (i.e.
In this experiment, a very thin foil of gold (0.004nm) is bombarded
approx. 10–10 m in diameter). The electrons were smaller particles
by a fine stream of alpha particles. A fluorescent screen (ZnS) is
together carrying a negative charge, equal to the positive charge
placed behind the gold foil, where points were recorded which
in the atom. They were studded in the atom like plums in a pudding.
were emerging from -particles. Polonium was used as the source
of -particles.

Scattering of   particles by thin gold foil

It has been found that radius of atom is of the order of
10 –10m while the radius of the nucleus is of the order of
10–15m.
Thus if a cricket ball represents a nucleus, then the radius of
atom would be about 5 km.
Rutherford’s Scattering Experiment
4.3 Rutherford’s nuclear atomic model
4.1 Observations
Rutherford carried out a number of experiments, involving the
scattering of   particles by a very thin foil of gold. Observations
were:
(i) Most of the   particles (99%) passes through it, without
any deviation or deflection.
(ii) Some of the   particles were deflected through small angles.
(iii) Very few   particles were deflected by large angles and
occasionally an   particle got deflected by 180o
4.2 Conclusions
(i) An atom must be extremely hollow and must consist of Rutherford’s Nuclear atomic model
mostly empty space because most of the particles passed (i) An atom consists of tiny positively charged nucleus at the
through it without any deflection. centre and it is surrounded by hollow portion called extra
(ii) Very few particles were deflected to a large extent. This nuclear part.
indicates that: (ii) The positive charge of the nucleus is due to nucleons which
(a) Electrons because of their negative charge and very low consist of protons and neutrons while the electrons, present
mass cannot deflect heavy and positively charged  particles in extra nuclear portion has negligible mass and carry a
(b) There must be a very heavy and positively charged body in negative charge.
the atom i.e. nucleus which does not permit the passage of (iii) The atom is electrically neutral, as the number of electrons is
positively charged  particles. equal to number of protons in it. Thus, total positive charge
(c) Because, the number of  particles which undergo deflection of the nucleus is balanced by the total negative charge of
of 180º, is very small, therefore the volume of positively electrons.
charged body must be extremely small fraction of the total (iv) The electrons in the extra nuclear part are revolving around
volume of the atom. This positively charged body must be the nucleus in circular paths called orbits. Thus, an atom
at the centre of the atom which is called nucleus. resembles the solar system in which the sun plays the role
of nucleus and the planets that of revolving electrons and
the model is known as planetary model.
(v) Electrons and nucleus are held together by the electrostatic NOTE
force of attraction.
The general notation that is used to represent the mass
(vi) Forces of attraction operating on the electron are exactly
A
balanced by centrifugal forces. number and atomic number of a given atoms is ZX
4.4 Drawbacks Where, X – symbol of element
(i) According to classical mechanics, any charged body in A – Mass number
motion under the influence of attractive forces should radiate
Z – atomic number
energy continuously. If this is so, the electron will follow a
spiral path and finally fall into the nucleus and the structure 5.3 Isotopes, Isobars, isotones and Isoelectronic
would collapse. This behaviour is never observed. 5.3.1 Isotopes:
Isotopes are the atoms of the same element having identicalatomic
number but different mass number. The difference is due
to the difference in number of neutrons.
The chemical properties of atoms are controlled by the number
of electrons. Thus, isotopes of an element show same chemical
behaviour.
Isotopes of Hydrogen
Isotope Formula Mass No. of No. of
number protons neutrons
Drawbacks of Rutherford’s model of atom 1
Protium 1H (H) 1 1 0
(ii) It says nothing about the electronic structure of atoms i.e.
how the electrons are distributed around the nucleus and Deuterium 2
1H (D) 2 1 1
what are the energies of these electrons.
3
5. ATOMIC NUMBER AND MASS NUMBER Tritium 1H (T ) 3 1 2

Isotopes of Oxygen
5.1 Atomic number (Z)
Isotope Mass number No. of No. of
Atomic number of an element is equal to the number of unit
protons neutrons
positive charges or number of protons present in the nucleus of
the atom of the element. It also represents the number of electrons 16
8 O
16 8 8
in the neutral atom. Eg. Number of protons in
Na = 11 , thus atomic number of Na = 11 17
8 O
17 8 9
5.2 Mass number (A)
18
8 O
The elementary particles (protons and neutrons) present in the 18 8 10
nucleus of an atom are collectively known as nucleons. Isotopes of some common elements
The mass number (A) of an atom is equal to the sum of protons Element Isotopes
and neutrons. It is always a whole number. Thus,
12 13 14
Mass number (A) = Number of protons(Z) + Number of Carbon (C) 6 C, 6 C, 6 C
neutrons(n)
14 15
Nitrogen (N) 7 N, 7 N
Therefore, number of neutrons (n) = Mass Number (A) – Number
of protons (Z) 233 235 238
Uranium 92 U, 92 U, 92 U
n =A– Z
32 33 34 36
Sulphur 16 S, 16 S, 16 S, 16 S
5.3.2 Relative Abundance 5.3.6 Isodiaphers

Isotopes of an element occur in different percentages in nature, Isodiaphers are the atoms of different elements which have the
which is termed as relative abundance. same diference of the number of neutrons & protons.

Using this relative abundance the average atomic mass of the Example
element can be calculated. For Example, 11 12
5
B 6
C
the average atomic mass of Cl is 35.5 due to existence of two p  5 p  6
isotopes 35Cl and 37Cl in 75% and 25% abundance respectively. n  6 n  p  1 n  7 n  p  1
e  5  e  6 
5.3.3 Isobars
Atoms of different elements having different atomic numbers 5.3.7 Isosteres
but same mass numbers are called isobars. Eg
Isosteres are the molecules which have the same number of atoms
Isobar Atomic Mass No. of No. of No. of & electrons.
number number electrons protons neutrons Example
40
18 Ar 18 40 18 18 22
CO 2 N2O
Atoms  1 2 Atoms  2 1
40
19 K 19 40 19 19 21 3 3
Electrons  6  8  2 Electrons  7  2  8
40  22e   22e 
20 Ca 20 40 20 20 20

5.3.4 Isotones 6. ELECTROMAGNETIC RADIATIONS

Atoms of different elements which contain the same number of
neutrons are called isotones. Eg Electromagnetic Radiations are waves which are formed as a result
of oscillating magnetic and electric fields which are perpendicular
Isotones Atomic Mass number No. of neutrons
to each other and both are perpendicular to direction of motion.
number
36
16 S 16 36 20

37
17 Cl 17 37 20

38
18 Ar 18 38 20

39
19 K 19 39 20

40
20 Ca 20 40 20
Electric & magnetic field components
5.3.5 Isoelectronic
They do not require any medium and can move in vacuum
The species (atoms or ions) containing the same number of unlike sound waves.
electrons are called isoelectronic. Eg.
Light is a form of radiation and has wave characterstics. The
O2–, F–, Na+, Mg2+, Al3+, Ne etc various characterstics of a wave are:
To go further into the atomic mysteries, we will have to
understand the nature of electromagnetic radiations and study
“Maxwell’s Electromagnetic Wave theory”.
James Maxwell was the first to give a comprehensive explanation
about the interaction between the charged bodies and the
behaviour of electric and magnetic fields.

Propagation of a Wave
1) Amplitude : It is height of the crest or trough (depth) of a 7.1 Electromagnetic Wave Theory
wave. Units : metre (m) The main points of this theory are:
2) Frequency (  ) : The number of waves passing through a (1) A source (like the heated rod) emits energy continuously
point in one second. Units : Hertz (Hz) or s –1 in the form of radiations (i.e. no change in wavelength or
frequency of the emitted radiations even on increasing the
3) Time Period : The time taken by a wave to complete one energy radiated).
vibration is called time period. Units : s
(2) These radiations are Electromagnetic in nature.
4) Velocity : The distance travelled by a wave in one second
7.2 Failure of EM wave theory
is called velocity. Units : m/s
In vacuum, all types of electromagnetic radiations travel at The theory failed because of two experiments:
the same speed i.e. 3 × 108 m/s. This is called speed of light.
(1) Black Body Radiation :
5) Wavelength(  ) : The distance between two adjacent crests
According to Maxwell’s theory on heating a body the
or troughs is called wavelength. Units : Angstrom(Å) intensity should increase, that is, energy radiated per unit
[1 Å=10–10m] area should increase without having any effect on the
wavelength or frequency.
6) Wave Number ( ) : It is the number of wavelengths per
But we observe that when we heat an iron rod, it first turns
centimetre of length. Units : m-1
to red then white and then becomes blue at very high
  1/  temperatures. This means that frequency of emitted
radiations is changing.
6.1 Relationship between velocity, frequency & wavelength
An ideal body, which emits and absorbs radiations of all
c   frequencies is called black body and radiation emitted by a
black body is called black body radiation
where c : speed of light i.e. 3 × 108 m/s in vaccum
The variation of intensity with wavelength at different
v : frequency; : wavelength temperatures for a black body is shown below:
7. ELECTROMAGNETIC SPECTRUM

When all the electromagnetic radiations are arranged in increasing

order of wavelength or decreasing frequency the band of
radiations obtained is termed as electromagnetic spectrum.

Wavelength-intensity relationship
Electromagnetic radiation & Visible spectrum
So it is observed that with increasing temperature,the
The visible spectrum is a subset of this spectrum (VIBGYOR) dominant wavelength in the emitted radiations decreases
whose range of wavelength is 380-760nm. and the frequency increases.
The wavelengths increase in the order: That is at higher temperatures, though the intensity rises as
Gamma Rays < X-rays < Ultra-violet rays < Visible< Infrared < predicted by Maxwell’s theory but the wavelength
Micro-waves <Radio waves. decreases. If T1>T2>T3 then 1< 2< 3.
(2) Photoelectric Effect :
Thus, these findings were contradictory to the Maxwell’s
When radiations with certain minimum frequency (0) strike theory. The number of electrons ejected and kinetic energy
the surface of a metal, the electrons are ejected from the associated with them should depend on the intensity of
surface of the metal. This phenomena is called photoelectric light. It has been observed that though the number of
effect. The electrons emitted are called photoelectrons. electrons ejected does depend upon the brightness of light,
the kinetic energy of the ejected electrons does not.
Light photons
To justify these findings Max Von Planck gave his Quantum
Electrons ejected
from the surface theory.

8. PLANCK’S QUANTUM THEORY

The main points of this theory are:

(i) The energy is emitted or absorbed not continuously but
discontinuously in the form of small discrete packets of
energy. Each such packet of energy is called a ‘quantum’.
Sodium metal In case of light this quantum of energy is called a photon.

Photoelectric effect

According to Maxwell’s Theory, the ejection of electrons

should depend on intensity of radiation that is if electrons
are not being ejected, then on increasing the intensity they
can be ejected.

Apparatus for studying the photoelectric effect

The following observations are made:

(i) The electrons are ejected from the metal surface as soon as
(ii) At a time only one photon can be supplied to one electron
the beam of light strikes the surface, i.e., there is no time lag
or any other particle.
between the striking of light beam and the ejection of
electrons from the metal surface. (iii) One quantum cannot be divided or distributed.
(ii) The number of electrons ejected is proportional to the
(iv) The energy of each quantum is directly proportional to the
intensity or brightness of light.
frequency of radiation.
(iii) For each metal, there is a characteristic minimum frequency,
0 (also known as threshold frequency) below which hc
E   or E  h 
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 h = Planck’s constant = 6.626 × 10-34Js
the increase of frequency of the light used. (v) The total energy emitted or absorbed by a body will be in
whole number quanta.
 nhc NOTE
Hence E = nh 
 Energy can also be expressed in Electron Volt(eV).
The energy acquired by an electron when it is accelerated
This is also called “Quantisation of energy”.
through a potential difference of one Volt.
8.1 Explanation of Black body radiation 1eV = 1.602 × 10–19J
As the temperature is increased the energy emitted increases Conclusion :
thereby increasing the frequency of the emitted radiations. As
Light has both the Wave nature (shows the phenomena of
the frequency increases the wavelength shifts to lower values.
diffraction and interference) and Particle nature (could explain
8.2 Explanation of Photoelectric effect the black body radiation and photoelectric effect) . Thus, light
(i) When light of some particular frequency falls on the surface has dual nature.
of metal, the photon gives its entire energy to the electron Bohr’s Model is based on “Atomic Spectra”, therefore before
of the metal atom. The electron will be ejected from the moving further we will study:
metal only if the energy of the photon is sufficient to
overcome the force of attraction of the electron by the 9. WHAT IS SPECTRUM?
nucleus. So, photoelectrons are ejected only when the
incident light has a certain minimum frequency (threshold A spectrum is a group or band of wavelengths/colours and the
frequency 0). The Threshold energy required for emission study of emission or absorption spectra is known as
spectroscopy.
is called “Work Function” that is “h0” . 9.1 Types of spectrum
(ii) If the frequency of the incident light () is more than the There are two types of spectrum
threshold frequency (0), the excess energy is imparted to 1) Emission Spectrum
the electron as kinetic energy. Hence, 2) Absorption Spectrum
Energy of one quantum = Threshold Energy + Kinetic 9.2 Emission Spectrum
Energy
When radiations emitted from a source are incident on a prism
h  h 0  (1/ 2) me v 2 and are separated into different wavelengths and obtained on a
photographic plate.
(iii) When  > 0, then on increasing the intensity, the number
(a) Continuous Emission Spectra:
of quanta incident increases thereby increasing the number
of photoelectrons ejected. There are no gaps between various wavelengths, one
wavelength merges into another.
(iv) When  > 0, then on further increasing the frequency, the
energy of each photon increases and thus kinetic energy of
each ejected electron increases.

Continuous spectrum of white light

Plot of kinetic energy versus frequency of absorbed photons

(b) Discontinuous Emission Spectra: They are numbered as 1,2,3,...... These numbers are known as
It is also known as Line Spectra or atomic spectra. Principal quantum Numbers.
In this, certain wavelengths go missing from a group and (a) Energy of an electron is given by:
that leaves dark spaces in between giving discontinuity to En= –RH (Z2/n2) n = 1,2,3.......
the spectrum. It is also known as fingerprint of an element.
where RH is Rydberg’s constant and its value is
2.18 × 10–18 J.
Z = atomic number

Z2
E n  2.18  1018 J / atom
n2

Z2
E n  13.6 eV / atom
n2

9.3 Absorption Spectra Z2

E n  1312 kJ / mol
When light from any source is first passed through the solution n2
of a chemical substance and then analysed, it is observed that
there are some dark lines in the continuous spectra. Thus, energies of various levels are in the order:

K < L < M < N...... and so on.

Energy of the lowest state(n=1) is called ground state.
(b) Radii of the stationary states:

52.9n 2
rn  pm
Z
For H-atom (Z = 1), the radius of first stationary state is
called Bohr orbit (52.9 pm)

(c) Velocities of the electron in different orbits:

Bohr studied the atomic spectra of hydrogen and based on that
he proposed his model. 2.188 106 Z
vn  m/s
n
10. BOHR’S MODEL
3) Since the electrons revolve only in those orbits which have
NOTE fixed values of energy, hence electrons in an atom can have
only certain definite values of energy and not any of their
This model is applicable to H-atom or H-like species like
own. Thus, energy of an electron is quantised.
He+,Li2+,Be3+. 4) Like energy, the angular momentum of an electron in an
10.1 Postulates atom can have certain definite values and not any value of
their own.
1) An atom consists of a small, heavy, positively charged
nucleus in the centre and the electrons revolve around it in nh
circular orbits. mr 
2
2) Electrons revolve only in those orbits which have a fixed
value of energy. Hence, these orbits are called energy levels Where n=1,2,3...... and so on.
or stationary states. 5) An electron does not lose or gain energy when it is present
in the same shell.
6) When an electron gains energy, it gets excited to higher
energy levels and when it de-excites, it loses energy in the
NOTE
form of electromagnetic radiations and comes to lower 1.09677 × 107 m-1 is also known as Rydberg’s constant.
energy values.
10.4 Line Spectrum of Hydrogen
When an electric discharge is passed through gaseous hydrogen,
the H 2 molecules dissociate and the energetically excited
hydrogen atoms produced emit electromagnetic radiations of
discrete frequency. The hydrogen spectra consists of several
lines named after their discoverer.
We get discrete lines and not a continuous spectra because the
energy of an electron cannot change continuously but can have
only definite values. Thus we can say that energy of an electron
is quantised.

10.2 What does negative energy for Hydrogen atom means?

This negative sign means that the energy of the electron in the
atom is lower than the energy of a free electron at rest. A free
electron at rest is an electron that is infinitely far away from the
nucleus (n  ) and is assigned the energy value of zero. As
the electron gets closer to the nucleus (as n decreases), E n
becomes more and more negative. The most negative energy
Electronic Transitions in the hydrogen spectrum
value is given by n=1 which corresponds to the most stable
orbit. LYMAN SERIES :
10.3 Transition of Electron When an electron jumps from any of the higher states to the
ground state or first state (n = 1) ,the series of spectral lines
We know that energy is absorbed or emitted when electron excites
emitted lies in the ultra violet region and are called as Lyman
or de-excites respectively. The energy gap between the two orbits
series.
is
Therefore , in Rydberg’s formula n1= 1, n2 = 2,3,4,5...
E  E f  E i
BALMER SERIES :
When an electron jumps from any of the higher states to the
 R   R 
E    2H     2H  state with n=2,the series of spectral lines emitted lies in the visible
 nf   ni  region and are called as Balmer series.
Therefore , in Rydberg’s formula n1 = 2, n2 = 3,4,5,6....
 1 1   1 1  PASCHEN SERIES :
E  R H  2  2   2.18  1018  2  2  J / atom
 ni nf   ni nf  When an electron jumps from any of the higher states to the
The wavelength associated with the absorption or emission of state with n=3 ,the series of spectral lines emitted lies in the
the photon is: infrared region and are called as Paschen series.
Therefore , in Rydberg’s formula n1 = 3, n2 = 4,5,6...
1 E R H  1 1   1 1  1
      1.09677 10
7
 2  2m
 hc hc  n i2 n f2   ni nf 
This is known as Rydberg’s formula.
BRACKETT SERIES : 3) Inability to explain splitting of lines in the magnetic field
When an electron jumps from any of the higher states to the (Zeeman Effect) and in the electric field (Stark Effect). If the
state with n = 4,the series of spectral lines emitted lies in the source emitting the radiation is placed in magnetic or electric
infrared region and are called as Brackett series. field, it is observed that each spectral line splits up into a
Therefore , in Rydberg’s formula n1 = 4, n2 = 5,6,7... number of lines. Splitting of spectral lines in magnetic field
is known as Zeeman Effect while splitting of spectral lines
PFUND SERIES : in electric field is known as Stark Effect.
When an electron jumps from any of the higher states to the 4) It could not explain the ability of atoms to form molecules
state with n = 4,the series of spectral lines emitted lies in the by covalent bonds.
infrared region and are called as Pfund series.
5) It could not explain dual behaviour of matter and also
Therefore , in Rydberg’s formula n1 = 5, n2 = 6,7... contradicts Heisenberg uncertainty principle.
10.5 Ionisation Energy
11. TOWARDS QUANTUM MECHANICAL MODEL
It is the energy required to remove the electron completely from
the atom so as to convert it into a positive ion. This model was based on two concepts:
Thus, n1 = 1 and n2 =  (1) de Broglie Concept of dual nature of matter
10.6 Limiting Line (2) Heisenberg uncertainty Principle
It is the line of shortest wavelength i.e. n2 =  11.1 Dual behaviour of matter
NOTE de Broglie proposed that matter should also exhibit dual behaviour
i.e. both particle and wave like properties.
Although a hydrogen atom has only one electron, yet its
spectra contains large number of lines. This is because a
h h h h
sample of hydrogen gas contains large number of    
molecules. When such a sample is heated to a high mv p 2m(KE) 2mqV
temperature, the hydrogen molecules split into hydrogen
Where p is linear momentum of a particle.
atoms. The electrons in different hydrogen atoms absorb
different amounts of energies and are excited to different According to de Broglie, every object in motion has a wave
levels. When these electrons return, different electrons character. The wavelengths associated with ordinary objects are
adopt different routes to return to the ground state. Thus, so short (because of their large masses) that their wave properties
they emit different amounts of energies and thus large cannot be detected. The wavelengths associated with electrons
number of lines in atomic spectra in hydrogen. and other subatomic particles (with very small mass) can however
be detected experimentally.
Maximum no. of lines that can be emitted when an electron
in an excited state n2 de-excites to n1 (n2>n1) : 11.2 Heisenberg’s Uncertainty Principle
It is impossible to measure simultaneously the position and
(n 2  n1  1) (n 2  n1 ) momentum of a small particle with absolute accuracy. If an attempt
2 is made to measure any of these two quantities with higher
accuracy, the other becomes less accurate. The product of the
10.7 Limitations of Bohr’s Model
uncertainty in the position (x) and the uncertainty in momentum
1) Inability to explain line spectra of multi-electron atoms. (p) is always a constant and is equal to or greater than h/4.
2) It fails to account for the finer details (doublet-two closely (x). (p)  h/4
spaced lines) of the hydrogen spectra. or (x). (mv)  h/4
or (x). (v)  h/4m
11.2.1 Explanation
12.1 Probability Density
|  |2 is the probability of finding the electron at a point within an
atom.
12.2 Concept of Orbital
It is a three dimensional space around the nucleus within which
the probability of finding an electron of given energy is maximum
(say upto 90%).
12.3 Quantum Numbers
They may be defined as a set of four numbers with the help of
which we can get complete information about all the electrons in
an atom i.e. location, energy, type of orbital occupied, shape and
orientation of that orbital etc.
The three quantum numbers called as Principal, Azimuthal and
Magnetic quantum number are derived from Schrodinger wave
equation. The fourth quantum number i.e. the Spin quantum
number was proposed later on.
Suppose we attempt to measure both the position and momentum 1) Principal Quantum Number (n):
of an electron. To pin point the position of the electron we have It tells about the shell to which an electron belongs.
to use light so that the photon of light strikes the electron and
n = 1,2,3,4,5..... and so on.
the reflected photon is seen in the microscope. As a result of the
hitting, the position as well as the velocity of the electron are This number helps to explain the main lines of the spectrum
disturbed. on the basis of electronic jumps between these shells.
11.2.2 Significance of Uncertainity Principle (a) It gives the average distance of the electron from the
nucleus. Larger the value of n, larger is the distance from
It rules out the existence of definite paths or trajectories of
the nucleus.
electrons as stated in Bohr’s Model.
(b) It completely determines the energy in hydrogen atom or
NOTE hydrogen like species.
The effect of Heisenberg Uncertainty Principle is The energy of H-atom or H-like species depends only on
significant only for motion of microscopic objects, and is the value of n.
negligible for that of macroscopic objects.
Order of energy : 1 < 2 < 3 < 4 < 5....... and so on.
12. QUANTUM MECHANICAL MODEL OF ATOM For multi-electron species, energy depends on both principal
and azimuthal quantum number.
Quantum mechanics is a theoretical science that deals with the
study of the motion of microscopic objects which have both The maximum number of electrons present in any shell = 2n2
particle like and wave like properties. The fundamental equation
of quantum mechanics was developed by Schrodinger. 2) Azimuthal Quantum Number (l) :

This equation describes a function called electron wave function Also known as Orbital Angular momentum or Subsidiary
quantum number.Within the same shell, there are number
(  ). This wave function stores all the information about an
of sub-shells, so number of electronic jumps increases and
electron like energy, position, orbital etc. As such it does not
this explains the presence of fine lines in the spectrum. This
have any physical significance. The information stored in  about quantum number tells about :
an electron can be extracted in terms of Quantum Numbers.
(a) The number of subshells present in a shell.
(b) Angular momentum of an electron present in subshell.
(c) Shapes of various subshells present within the same shell. Under the influence of external magnetic field, electrons of a
(d) Relative energies of various subshells. subshell can orient themselves in a certain preferred regions of
Value of l varies from 0 to n – 1 space around the nucleus called orbitals.

For 1st shell (n = 1): l = 0 The magnetic quantum number determines the number of
preferred orientations of the electrons present in a subshell. Since
For 2nd shell (n = 2): l = 0,1 each orientation corresponds to an orbital, thus magnetic
For 3rd shell (n = 3): l = 0,1,2 quantum number determines the number of orbitals present in
For 4th shell (n = 4): l = 0,1,2,3 any subshell.
Value of m ranges from – l to +l including zero.
Value of l Designation of subshell
Subshell Orbitals (m) Number of orbitals
0 s
s-subshell (l=0) m= 0 1
1 p p-subshell (l=1) m = -1, 0, 1 3
2 d d-subshell (l=2) m = –2, –1, 0, 1 ,2 5
3 f f-subshell (l=3) m = –3, –2, –1, 0, 1, 2, 3 7
4 g
5 h
The notations s,p,d,f represent the initial letters of the word sharp, Orbital Value of m
principal, diffused and fundamental. In continuation l = 4 is called px m= 0
g subshell and l = 5 is called h subshell and so on.
py m=+1
Principal shell Subshells
pz m=–1
1st shell l = 0 (s-subshell)
2nd shell l = 0,1 (s & p subshell)
Orbital Value of m
3 shell
rd
l = 0,1,2 (s,p & d subshell)
4th shell l = 0,1,2,3 (s,p,d & f subshell) d z2 m=0

dxz m = +1
d yz m = –1
NOTE
The number of subshells present in any principal shell is d x2  y2 m = +2
equal to the number of the principal shell.
d xy m = –2
Energies of various subshell present within the same shell
is: s < p < d < f These orbitals of the same subshell having equal energy are
Angular momentum of an electron in orbital : .
called degenerate orbitals Eg.
h The three p-orbitals of a particular principal shall have the
l (l  1)   l (l  1)
2 same energy in the absence of magnetic field.
Similarly, all five orbitals of d-subshell of a particular shell have
(3) Magnetic Quantum Number(m): the same energy.
This quantum number is required to explain the fact that when Thus, for H-atom order of energy is:
the source producing the line spectrum is placed in a magnetic 1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f < ..........
field, each spectral line splits up into a number of line (Zeeman For multi electron atoms, the energy of the orbitals decreases
effect). with increase in effective nuclear charge.
Eg. E2s (H) > E2s (Li) > E2s (Na) > E2s (K) (2) Shape of p-orbitals:
The total possible values of m in a given subshell = 2l + 1
(a) It consists of two lobes present on either side of the
Total no. of orbitals in a given shell = n2 plane that passes through the nucleus. The p-orbital is
dumb-bell shaped.
4) Spin Quantum Number(s):
The electron in an atom not only moves around the nucleus (b) There are three possible orientations of electron cloud in
but also spins about its own axis. Since the electron in an p-orbitals. Therefore, the lobes of p-orbital may be
orbital can spin either in clockwise or anti-clockwise considered to be along x,y and z axis. Hence they are
direction. Thus s can have only two values designated as px, py, pz. The three p-orbitals are oriented
at right angles to one another.
1 1
 or  (c) First main energy level( Principal quantum number n = 1)
2 2
does not contain any p-orbital.
This quantum number helps to explain the magnetic
properties of substances. (d) The three p-orbitals of a particular energy level have same
energy in absence of an external electric and magnetic
NOTE field and are called degenerate orbitals.
An orbital cannot have more than two electrons and (e) Like s orbitals, p-orbitals increase in size with increase in
these electrons should be of opposite spin. the energy of main shell of an atom. Thus, value of
Thus, maximum number of electrons in s-subshell = 2 azimuthal quantum number is one (l=1) and magnetic
quantum number has three values (m= –1, 0, +1)
Maximum number of electrons in p-subshell = 6
Maximum number of electrons in d-subshell = 10
Maximum number of electrons in f-subshell = 14
12.4 Shapes of atomic orbitals
(1) Shape of s-orbitals:
(a) They are non-directional and spherically symmetric i.e.
probability of finding the electron at a given distance is
equal in all directions.
(b) 1s orbital and 2s orbital have same shape but size of 2s is
larger.
(c) There is a spherical shell within 2s orbital where electron Boundary surface diagrams of three 2p orbitals
density is zero and is called a node.
(3) Shapes of d-orbitals:
(d) The value of azimuthal quantum number(l) is zero (l=0)
and magnetic quantum number can have only one value (a) They are designated as dxy, dyz, dzx and d x 2 y2 . They
i.e. m = 0
have a shape like a four leaf clover. The fifth d orbital
designated as d z2 looks like a doughnut.

(b) All five d orbitals have same energy in the absence of

magnetic field.
(c) The d orbitals have azimuthal quantum number l = 2 and
magnetic quantum number values –2, –1,0,+ 1,+ 2.

Boundary surface diagram of the s orbital

(d) For principal shell number 1 and 2, there are no d orbitals.

12.6 Filling of orbitals in an atom

(1) Aufbau Principle:
In the ground state of the atoms, the orbitals are filled in
order of their increasing energies. In other words, electrons
Boundary surface diagrams of five 3d orbitals first occupy the lowest energy orbital available to them and
enter into higher energy orbitals only when the lower energy
We have drawn boundary surface diagrams i.e the surface is orbitals are filled.
drawn in space for an orbital on which probability density Unlike H-atom where energy of orbitals depend only on n,
()2 is constant. It encloses the region where probability of energy of orbitals of multi-electron orbitals depend on both
finding the electron is very high. We do not draw a boundary n and l. Their order of energy can be determined using (n+l)
surface diagram that encloses 100% probability of finding rule.
the electron because probability density has some value, According to this rule, lower the value of (n+l) for an orbital,
howsoever small it may be, at any finite distance from the lower is its energy. If two different types of orbitals have
nucleus. Thus it is not possible to draw a boundary surface the same value of (n+l), the orbital with lower value of n has
diagram of a rigid size in which the probability of finding the lower energy.
electron is 100%.
12.5 Nodes and nodal planes
NOTE
It is a region of zero probability.
There are two types of nodes:
(1) Radial or Spherical nodes: Three dimensional regions in
an orbital where probability of finding the electron
becomes zero.
Number of radial/ spherical nodes = n – l – 1

Order of filling of orbitals

1s22s22p63s23p64s23d104p65s24d105p66s2 ......
(2) Pauli Exclusion Principle: An orbital can have maximum
two electrons and these must have opposite spin.

Nodes in s-orbital
(2) Planar or Angular Nodes: They are the planes cutting
through the nucleus on which probability of finding the
(3) Hund’s rule of maximum multiplicity: Electron pairing in
electron is zero.
p,d and f orbitals cannot occur until each orbital of a given
Number of Planar/Angular Nodes = l subshell contains one electron each. Also all the singly
Total Number of nodes = n - 1 occupied orbitals will have parallel spin.
 Electronic configurations of elements in the ground state

Atomic No. Element Electronic Configuration

1 H 1s1
2 He 1s2
1
3 Li [He] 2s
2
4 Be [He] 2s
2 1
5 B [He] 2s 2p
2 2
6 C [He] 2s 2p
2 3
7 N [He] 2s 2p
2 4
8 O [He] 2s 2p
2 5
9 F [He] 2s 2p
2 6
10 Ne [He] 2s 2p
1
11 Na [Ne] 3s
2
12 Mg [Ne] 3s
2 1
13 Al [Ne] 3s 3p
2 2
14 Si [Ne] 3s 3p
2 3
15 P [Ne] 3s 3p
2 4
16 S [Ne] 3s 3p
2 5
17 Cl [Ne] 3s 3p
2 6
18 Ar [Ne] 3s 3p
1
19 K [Ar] 4s
2
20 Ca [Ar] 4s
1 2
21 Sc [Ar] 3d 4s
2 2
22 Ti [Ar] 3d 4s
3 2
23 V [Ar] 3d 4s
5 1
24 Cr [Ar] 3d 4s
5 2
25 Mn [Ar] 3d 4s
6 2
26 Fe [Ar] 3d 4s
7 2
27 Co [Ar] 3d 4s
8 2
28 Ni [Ar] 3d 4s
10 1
29 Cu [Ar] 3d 4s
10 2
30 Zn [Ar] 3d 4s
12.7 Exceptional Configuration of Cr & Cu 12.8 Electronic Configuration of Ions
The completely filled and completely half filled sub-shells are 12.8.1 Cations:
stable due to the following reasons: They are formed when outermost electrons are removed from
1. Symmetrical distribution of electrons: It is well known an atom. While removing the electrons, we must remove the
that symmetry leads to stability. The completely filled or electrons from the highest principal quantum number.
half filled subshells have symmetrical distribution of 12.8.2 Anions:
electrons in them and are therefore more stable. Electrons
They are formed when electrons are added to the innermost
in the same subshell (here 3d) have equal energy but empty shell.
different spatial distribution. Consequently, their shielding
12.8.3 Magnetic moment :
of one another is relatively small and the electrons are
more strongly attracted by the nucleus. n(n  2) B.M .
2. Exchange Energy : The stabilizing effect arises whenever
two or more electrons with the same spin are present in the B.M.  Bohr Magneton
degenerate orbitals of a subshell. These electrons tend to Where n is number of unpaired electrons.
exchange their positions and the energy released due to Species with unpaired electrons are called paramagnetic and
this exchange is called exchange energy. The number of the species with no unpaired electrons are called diamagnetic
exchanges that can take place is maximum when the
subshell is either half filled or completely filled. As a result
13. IMPORTANT RELATIONS
the exchange energy is maximum and so is the stability.
1  1 1 
eg. for Cr : [Ar] 4s1 3d5 Rydberg equation :   R H  2 2
n n
 1 2 

(RH = 109678 cm–1 and n2 > n1)

1
c     and  


hc
E  h or 

Bohr’s Model

1312Z2
En  kJ mol1
n2

2.178  10 18 Z 2 13.6Z2

 J / atom  eV / atom
n2 n2

2.165  106 Z
Velocity of electron, n  m/s
n

0.529 n 2
Radius of orbit  Å
Z

1
Photoelectric effect = h  h0  m
2

2
( 0  Threshold frequency)

h
de-Broglie equation :  
m
Possible exchange for a d5 configuration
h
Thus, total number of exchanges=10 Heisenberg’s uncetainity principle : x  p 
4
 Solve (iii) and (iv) to get :
14. MATHEMATICAL MODELLING OF BOHR’S
POSTULATES V
2KZe 2
and r  2
n 2h 2
nh 4 Kme 2 Z
Consider an ion of atomic number (Z) containing single electron
revolving its nucleus at a distance of ‘r’ as shown in the figure. put K = 9 × 109 Nm2C–2, e = 1.6 × 10–19C and h = 6.63 × 10–34 Js in
the above expressions to get :
Velocity of an electron in nth Bohr orbit 

V Z –1
Vn = 2.165 × 106 ms
n

n2
e– and Radius of the nth Bohr orbit  rn = 0.53 Å
+Ze r Z
Now, the Total Energy of the electron moving in nth orbit  K.E.n
+ P.E.n

1 K(Ze)(e)  K q1 q 2 
T.E.n  mVn2 
2 r  E.P.E.  r 
NOTE
Atomic number = Number of protons on the nucleus = Z 1  K Ze 2  K(Ze)(e)
  T.En= 2  r   rn [Using (iii)]
 n 
 Charge on the nucleus = + Ze
Electrostatic force of attraction (F) between the nucleus of charge -KZe 2
 En  T. En =
+ Ze and electron (–e) is given by : 2rn

K | q1 | | q 2 | 1 It can be shown from the above expressions that :

F where K  4
r2 0
1 K Ze 2 K Ze 2  K Ze2
K.E.n  , P.E.n  and E n 
= 9 × 109 Nm2C–2 2 rn rn 2rn

K | Ze | | e | KZe 2 or K.E.n = –En and P.E.n = 2En

F  2 ........(i)
r2 r Using the value of rn in the expression of En, we get :

mv 2 22 K 2 me 4 Z 2
The centrifugal forces acting on the electron is ........(ii) En 
r n2h2

This centrifugal force must be provided by the electrostatic force Z2

of attraction (F). En  -2.178×10-18 J / atom
n2
 From (i) and (ii), we have :
Z2
KZe 2
mV 2 or En  -13.6 eV / atom
 ...........(iii) n2
r2 r
Angular momentum of electron about the nucleus = 1eV  1.6  1019 J 

nh Z2
mVr  ........ (iv)  – 2.178 × 10–18 × 6.02 × 1023 J/mole
2 n2
where ‘n’ is a positive integer
Z2
(n = 1, 2, 3, .........  )  – 1312 kJ/mole
n2
15. PROBABILITY DISTRIBUTION DIAGRAMS FOR 1s AND 2s
 SUMMARY
 Constituents of atom: Atom is no longer considered as  Energy is emitted continuously from any source in the
indivisible. It is made up of electrons, protons and form of radiations travelling in the form of waves and
neutrons called fundamental particles. associated with electric and magnetic fields, oscillating
perpendicular to each other and to the direction of radiation
 Electron: A fundamental particle which carries one unit
. All of them travel with the velocity of light.
negative charge and has a mass nearly equal to 1/1837th of
that of hydrogen atom.  Relationship between velocity, frequency & wavelength:
c = 
 Proton: A fundamental particle which carries one unit
where c : speed of light i.e. 3 × 108 m/s in vacuum
positive charge and has a mass nearly equal to that of
 : frequency;  : wavelength
hydrogen atom.
 Electromagnetic spectrum: When all the electromagnetic
 Neutron: A fundamental particle which carries no charge radiations are arranged in increasing order of wavelength
but has a mass nearly equal to that of hydrogen atom. or decreasing frequency the band of radiations obtained is
 Thomson’s model of atom: An atom is a sphere of termed as electromagnetic spectrum.
positive electricity in which sufficient number of electrons  Black body radiation: If the substance being heated is a
were embedded to neutralize the positive charge just as black body (which is a perfect absorber and perfect radiator
seeds in a melon or raisins in pudding. It could not of energy) the radiation emitted is called black body radiation.
explain results of Rutherford’s scattering experiments.  Photoelectric effect: When radiation of certain minimum
 Rutherford’s model of atom: A thin foil of gold was frequency () strike the surface of a metal, electrons are
bombarded with  -particles. Most of the  -particles ejected. This minimum energy (h0) is called wave function
passed through the foil undeflected, a few were deflected (W0).
through small angle while very few were deflected back. It  Planck’s quantum theory: This theory was put forward to
was therefore, concluded that there was sufficient empty explain the limitations of electromagnetic wave theory. It
space within the atom and small heavy positively charged suggests that radiant energy is emitted or absorbed
body at the center called nucleus. Thus, atom consists of discontinuously in the form of small packets of energy called
a heavy positively charged nucleus in the centre quanta (called photons in case of light). Energy of each
containing all protons and neutrons and the electrons quantum (E) = hv where ‘h’ is Planck’s constant (= 6.626 ×
were revolving around the nucleus so that the centrifugal 10-34 Js). Total energy emitted or absorbed = nhv where n is
an interger. If n = N0, energy is called one einstein.
force balances the force of attraction.
 Emission and Absorption Spectra: When light emitted from
 Atomic number and mass number: The general notation
any source is directly passed on to prism and resolved, the
that is used to represent the mass number and atomic
spectrum obtained is called emission spectrum. In case of
A
number of a given atoms is ZX white light, e.g., from sun, it is resolved into seven colours
(VIBGYOR). The spectrum obtained is called contiuous
Where, X – symbol of element
spectrum. If light emitted from a discharge tube is resolved,
A – Mass number some coloured lines are obtained. The spectrum obtained
Z – atomic number is called line spectrum. It white light is first passed through
the solution of a compound or vapour of a substance and
 Isotopes: Isotopes are the atoms of the same element having then resolved, the spectrum obtained is called absorption
identical atomic number but different mass number. The spectrum. It has dark lines in the continuous spectrum.
difference is due to the difference in number of neutrons.
 Absorption spectrum of hydrogen: When H2 gas is taken in
 Isobars: Atoms of different elements having different atomic the dischange tube, series of lines obtained and the regions
numbers but same mass numbers are called isobars. in which they lie are as under:
 Isotones: Atoms of different elements which contain the
same number of neutrons are called isotones.
Series: Lyman Balmer
 Paschen Brackett Pfund
     
 Isoelectronic: The species (atoms or ions) containing
the same number of electrons are called isoelectronic. Region: UV Visible Infrared
 Rydberg formula: This formula is used to calculate wave  Azimuthal quantum number (l): It determines the number
number of different series of lines of the spectrum of of subshells present in any main shell (n) and the shape of
hydrogen or hydrogen like particles as : the subshell. For a given value of n, l = 0 to n - 1. Thus, for
n = 1, l = 0 (one subshell), for n = 2 , l = 0, 1, (2 subshell), for
 1 1  n = 3, l = 0, 1, 2 (3 subshells), for n = 4, l = 0, 1, 2, 3 (4
v  R  2  2  Z 2 (Z = 1 for hydrogen)
n 
 i nf  subshells). For l = 0, 1, 2 and 3. designation are s, p, d and f
respectively. Thus, subshells present are : n = 1 (1s), n = 2
where R = Rydberg constant = 109677 cm-1 or 1.097×107 m-1 (2s, 2p), n = 3 (3s, 3p, 3d), n = 4 (4s, 4p, 4d, 4f).
 Bohr’s Model:  Magnetic quantum number (m): It determines the number
of orbitals present in any subshell and the orientation of
1312Z 2
En  kJ mol1 each orbital. For a given value of l, m = - l to + l including ‘0’.
n2
 Spin quantum number (s): It tells about the spinning motion
2.178  10 18
Z 2
13.6Z2 of the electron, i.e., clockwise or anti-clockwise. For a given
 J / atom  eV / atom
n2 n2 1 1
value of m, s   and  . It helps to explain magnetic
2 2
2.165 106 Z
Velocity of electron, v n  m/s properties of the substances.
n

0.529 n 2  Shapes of atomic orbitals: The shape of an orbital is found

Radius of orbit  Å
Z by finding the probability    of the electron in that orbital
2

 Dual behaviour of particle: According to de Broglie, every at different points around the nucleus and representing by
object in motion has a wave character. The wavelengths the densiy of points. The shape of the electron cloud thus
associated with ordinary objects are so short (because of obtained gives the shape of the orbital. Some orbitals are
their large masses) that their wave properties cannot be found to have a region of space within it where probability
detected. The wavelengths associated with electrons and is zero. This is called a node. It may be spherical/radial or
other subatomic particles (with very small mass) can planar/angular.
however be detected experimentally  Rules for filling of electrons in orbitals:
 Heisenberg’s Uncertainty Principle: It is impossible to  Aufbau principle: Orbitals are filled in order of their
measure simultaneously the position and momentum of a increasing energy. The order of energy and hence that of
small particle with absolute accuracy. If an attempt is made filling orbitals is found by (n + l) rule. It states “lower the
to measure any of these two quantities with higher accuracy, (n + l) value, lower is the energy. If two orbitals have same
the other becomes less accurate. The product of the (n + l) value, orbital with lower value of n has lower energy.”
uncertainty in the position (x) and the uncertainty in Thus, the order is:
momentum (p) is always a constant and is equal to or 1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d....
greater than h/4. (n + l) 1 2 3 3 4 4 5 5 5 6 6 6 7 7
 Quantum mechanical Model of atom: Quantum mechanics  Hund’s rule of maximum multiplicity: Pairing of electrons
is a theoretical science that deals with the study of the does not occur in orbitals of the same subshell (degenerate
motion of microscopic objects which have both particle like orbitals) until each of them is first singly occupied.
and wave like properties. The fundamental equation of  Pauli exclusion principle: No two electrons in an atom can
quantum mechanics was developed by Schrodinger. have the same set of four quantum numbers or an orbital
can have maximum two electrons and these must have
 Quantum number: It is a set of four numbers which give
opposite spin.
complete information about any electron in an atom. These
 Electronic configuration of elements : Distribution of
are:
electrons of an atom into different shells, subshells and
 Principal quantum number (n): It determines the size of the
orbitals is called its electronic configuration. Complete
orbital. Its values are 1, 2, 3, etc. or K, L, M, etc. It also
electronic configuration is obtained by following the above
determines the energy of the main shell in which the elecron
rules, e.g.,
is present and maximum number of electrons present in the
Cl = 1s22s22p63s23p2x3p2y3p1z
nth shell (= 2 n2). 17
 SOLVED EXAMPLES
Example - 1

Distinguish between Proton, Neutron & Electron.

Sol.

Parameter Proton Neutron Electron

1. Charge unit positive no charge unit negative

(+1.6 × 10–19 C) (–1.6 × 10–19 C)

2. Mass 1.672 × 10–27 Kg 1.674 × 10–27 Kg 9.1094 × 10–31 Kg

3. Denoted 1
p1 0
n1 –1
e0

4. Location nucleus nucleus extra-nuclear region

5. Effected by Electric field & magnetic Remain undeflected Electric field & magnetic

field (towards negative in electric and field (towards positive plate)

plate) magnetic fields.

6. Discovered by Goldstein Chadwick Thomson

Example - 2

Give the difference between Isotopes, isobars & Isotones.

Sol.

Isotopes Isobars Isotones

1. They are atoms having the same They are atoms having the same They are atoms having the same

number of protons but differ in sum of protons and neutrons. number of neutrons but differ in

the number of neutrons. the number of protons.

2. They have the same atomic They have the same mass They have different atomic number

number but differ in their mass number but differ in their atomic and mass number

number. number

3. They are atoms of the same They are atoms of different They are atoms of different elements

element hence they have elements hence there is no hence there is no similarity in their

identical chemical properties. similarity in their chemical chemical properties.

properties.
Example - 3 Example - 6
Why electrons are called Planetary according to An atom of an element contains 29 electrons and 35
Rutherford model? neutrons. Deduce
Sol. This model was analgous to the solar system, where the (a) the number of protons and
nucleus may be compared to the sun and the electrons to
(b) the electronic configuration of the element
the planets. The Coulombic force between the nucleus and
(c) Mass Number
kq1q 2
the electron is where q1 and q2 are the charges, r is Sol. Number of electrons = 29
r2
the distance of separation between the molecules and the Number of neutrons = 35
electron and k is the proportionality constant. This is similar For neutral atom :
to gravitational force between two masses m1 and m2 as Number of protons = Number of electrons
mm (a) Therefore, number of protons = 29
G. 1 2 2 where G is gravitational constant and r is the
r (b) Atomic Number = 29
distance of separation between planet and the sun.
Element is Cu.
Example - 4
1s2 2s2 2p6 3s2 3p6 4s1 3d10
(a) Why e/m ratio of anode rays is different for different
(c) Mass Number = p + n
gases.
= 29 + 35 = 64
(b) Why was pressure of air inside the tube reduced to
Example - 7
10–2 atm in cathode ray experiment ?
Sol. (a) The charge to mass (e/m) ratio of anode rays is (i) Which model of an atom is called apple pie model?
dependent upon the nature of the gas taken in the (ii) Why are the atomic masses of most element is
discharge tube. This is because positively charged fractional?
particles are produced by the loss of one or more Sol. (i) Thomson Model
electrons from the neutral atoms of the gas contained
(ii) It is due to the existence of various isotopes of an
in the discharge tube. Therefore, the mass of the
element in various percentage abundance.
positively charged particles will depend upon the
nature of the gas. In case of hydrogen, the charge to Example - 8
mass (e/m) ratio was maximum. Explain the Rutherford’s scattering experiment and also
(b) This is because at very low pressure gas gets ionized give its drawbacks.
and become conducting. Sol. A very thin foil of gold (0.004nm) is bombarded by a fine
Example - 5 stream of alpha particles. A fluorescent screen (ZnS) is
placed behind the gold foil, where points were recorded
Calculate the number of neutrons in
which were emerging from -particles. Polonium was used
(a) dipositive zinc ion (b) Mg+2 as the source of -particles.
Sol. (a) Number of protons = 30 Observations
Mass Number = 65 Rutherford carried out a number of experiments, involving
 p + n = 65 the scattering of -particles by a very thin foil of gold.
n = 65 – 30 = 35 Observations were:
(b) Number of protons = 12
(i) Most of the -particles (99%) passes through it, without
Mass Number = 24 any deviation or deflection.
p + n = 24
(ii) Some of the -particles were deflected through small angles.
n = 24 – 12 = 12.
(iii) Very few -particles were deflected by large angles and
occasionally an -particle got deflected by 180o
Conclusions
Example - 9
(i) An atom must be extremely hollow and must consist of The nuclear radius is of the order of 10-13cm while atomic
mostly empty space because most of the particles passed radius is of the order 10-8cm. Assuming nucleus to be
through it without any deflection. spherical, what fraction of atomic volume is occupied by
(ii) Very few particles were deflected to a large extent. This nucleus?
indicates that:
4 3
(a) Electrons because of their negative charge and very low Sol. Volume of a sphere = r
3
mass cannot deflect heavy and positively charged 

particles 4
Therefore, volume of nucleus =  (1013 )3
3
(b) There must be a very heavy and positively charged body
in the atom i.e. nucleus which does not permit the passage 4
Volume of atom =  (108 )3
of positively charged  particles. 3
Fraction of atomic volume occupied by
(c) Because, the number of  particles which undergo
deflection of 180º, is very small, therefore the volume of
4
positively charged body must be extremely small fraction  (1013 )3
of the total volume of the atom. This positively charged nucleus = 3  1015
4 8 3
body must be at the centre of the atom which is called  (10 )
3
nucleus.
Example - 10
Drawbacks
(i) Calculate the total number of electrons in 1.6g of
methane.
(ii) An ion with mass number 56 contains 3 units of
positive charge and 30.4% more neutrons than
electrons. Assign symbol to this ion.
Sol. (i) 1 molecule of CH4 has = 10 electrons
16g CH4 has = 10NA electrons
1.6g CH4 has = 1.0 NA electrons
= 6.022 × 1023 electrons
(i) According to classical mechanics, any charged body in
motion under the influence of attractive forces should (ii) Suppose number of electrons in the ion, M3+ = x
radiate energy continuously. If this is so, the electron will
follow a spiral path and finally fall into the nucleus and the 30.4
 No. of neutrons = x  x  1.304x
structure would collapse. This behaviour is never 100
observed. No. of electrons in the neutral atom = x + 3
(ii) It says nothing about the electronic structure of atoms i.e.  No. of protons = x + 3
how the electrons are distributed around the nucleus and
Mass no. No. of protons + No. of neutrons
what are the energies of these electrons.
56 = x + 3 + 1.304 x or 2.304x = 53 or x = 23
 No. of protons = Atomic no. = x + 3 = 23 + 3 = 26
Hence, the symbol of the ion will be 56
26 Fe3 .
Example - 11 (4) So it is observed that with increasing temperature the
dominant wavelength in the emitted radiations
(i) Arrange X-rays, cosmic rays and radiowaves
decreases and the frequency increases.
according to frequency.
(ii) Calculate the wavenumber of yellow radiation (c) When radiations with certain minimum frequency ( 0 )
having wavelength of 5800Å. strike the surface of a metal, the electrons are ejected
(iii) Define threshold frequency. from the surface of the metal. This phenomena is called
photoelectric effect. The electrons emitted are called
Sol. (i) Cosmic rays > x-rays > Radio waves
photoelectrons.
1 1 If the frequency of the incident light (  ) is more than
(ii)    = 1.72 × 106 m–1
 5800 1010
the threshold frequency (  0 ), the excess energy is
(iii) The minimum frequency required to eject an electron imparted to the electron as kinetic energy. Hence,
from the surface of metal.
Energy of one quantum = Threshold Energy + Kinetic
Example - 12
Energy
Define
h  h0  (1/2) me v2
(a) Electromagnetic spectrum
(b) Black Body and Black body radiation. When    0 , then on increasing the intensity the
(c) Photoelectric Effect number of quanta incident increases thereby
Sol. increasing the number of photoelectrons ejected.
(a) When all the electromagnetic radiations are arranged When    0 , then on further increasing the
in increasing order of wavelength or decreasing
frequency, the energy of each photon increases and
frequency the band of radiations obtained is termed thus kinetic energy of each ejected electron increases.
as electromagnetic spectrum.
The visible spectrum is a subset of this spectrum
(VIBGYOR) whose range of wavelength is 380-760nm.
The wavelengths increase in the order:
Gamma Rays < X-rays < Ultra-violet rays < Visible<
Infrared < Micro-waves <Radio waves.
(b)
(1) An ideal body, which emits and absorbs radiations of
all frequencies is called black body and radiation
emitted by a black body is called black body radiation. Example - 13
(2) According to Maxwell’s theory on heating a body the Differentiate between absorption spectrum and emission
intensity should increase, that is, energy radiated per spectrum?
unit area should increase without having any effect
Sol. The main points of difference between absorption &
on the wavelength or frequency.
emission spectra are summed up in the below :
Absorption Spectrum :
1. Absorption spectrum is obtained when the white light
is first passed through the substance and the
transmitted light is analysed in the spectroscopy.
(3) 2. It consists of dark lines in the otherwise continous
spectrum
3. Absorption spectrum is always discontinuous
spectrum consisting of dark lines.
 Emission Spectrum : Example - 15
1. Emission spectrum is obtained when the radiation from A 100 watt bulb emits electromagnetic light of wavelength
the source are directly analysed in the spectroscope. 400nm. Calculate the number of photons emitted per
2. It consists of bright coloured lines separated by dark second by the bulb.
spaces. Sol. Power of the bulb = 100 watt
3. Emission spectrum can be continuous spectrum (if = 100 J s–1
source emits white light) or discontinuous, i.e., line
spectrum if source emits some coloured radiation. hc
Energy of one photon, E = h 
Example - 14 

(a) Calculate the wavelength, frequency and wave

(6.626  10 34 Js)  (3  108 m s 1 )
number of light wave whose time period is 
400  109 m
2 × 10–10sec?
(b) Yellow light emitted from sodium lamp has = 4.969 × 10–19 J
wavelength of 580 nm. Calculate the frequency and  Number of photons emitted
wave number of yellow light.
100 J s 1
(c) How long will it take for a radio wave of frequency =
4.969  10 19 J
6 × 1013 Hz, sent by a path finder to travel from Mars
to earth over a distance of 8 × 107 km = 2.012 × 1020 s–1.

1 1 Example - 16
Sol. (a) Frequency () =   5 109 s 1.
Period 2.0  1010 s The threshold frequency  0 for a metal is 7 × 1014s-1.
Calculate the kinetic energy of an electron emitted when
c 3.0 108 m s 1 radiation of frequency 1 × 1015s-1 hits the metal.
Wavelength,     6.0 × 10–2 m
 5  109 s 1
Sol. 0 = 7 × 1014 s–1 ;  = 1015 s–1

1 1 According to photo electric effect

Wavenumber,     16.66 m1
 6  102 m h = h0 + K.E.
(b) C   K.E. = h ( – 0)
= 6.626 × 10–34 (1015 – 7 × 1014)
3  108
  = 1.9878 × 10–19 J.
580  109
Example - 17
= 5.17 × 1014 Hz
If photon of wavelength 150pm strikes an atom and one
1 1 of its inner bound electron ejected out with a velocity of
   1.7  106 m1
 580  109 1.5 ×107m/s. Calculate the energy with which it is bound
to the nucleus.
(c) All radiations in vacuum travel with the same speed,
Sol. Energy of incident photon
C = 3 × 108 m/s
Distance to be travelled = 8 × 107 ×103 m hc
=

= 8 × 1010 m

8  1010 (6.626 1034 Js)  (3.0 108 ms 1 )

 2.66  102 sec 
Time taken = (150  1012 m)
3  108
= 4 min. 44 sec. = 1.325 × 10–15 J
 Energy of ejected electron
1
1 K.E. = mv2
 m 2 2
2
2  K.E. 2  0.93  10 19
1  v2   = 0.204 × 1012
  (9.11 1031 kg)  (1.5  107 ms 1 ) 2 m 9.11 10 31
2
= 1.025 × 10–16 J v  0.204  1012 = 0.452 × 106
Energy with which the electron was bound to the nucleus  v = 4.52 × 105 ms–1
–16 –16
= 13.25 × 10 – 1.025 × 10
= 12.225 × 10 –16
J Example - 19

Electrons are emitted with zero velocity from a metal

12.225  1016
or  = 7.63 × 103 eV.. surface when it is exposed to radiations of wavelength
1.602  1019
6800Å. Calculate the threshold frequency and work
Example - 18 function of the metal.
The work function for caesium atom is 1.9eV. Calculate Sol. K.E. = h – h0
(a) threshold frequency (b) threshold wavelength of
If the velocity is zero, K.E. = 0
the radiation (c) If the cesium atom is irradiated with a
wavelength of 500nm, calculate the kinetic energy and  0 = h – h0
the velocity of ejected electron. or h = h0 or  = 0
Sol. (a) Work function (w0) = h 0
c 3.0  108 ms 1
v = 
 6800  1010 m

W0 1.9  1.602  1019 J = 4.41 × 1014 s–1

 0   (1 eV  1.602  10 19 )
h 6.626  1034 J s Threshold frequency = 4.41 × 1014 s–1
= 4.59 × 1014 s–1 Work function = h0 = 6.626 × 10–34 × 4.41 × 1014
= 2.92 × 10–19 J.
c 3.0  108 m s 1
(b)  0    6.54  107 m Example - 20
 0 4.59  1014 s 1
Why electronic energy is negative?
= 654 × 10–9 m = 654 nm Sol. The negative sign of energy means that the energy of the
(c) Work function, hv0 = 1.9 × 1.602 × 10–19J electron in the atom is lower than the energy of a free
electron at rest. A free electron at rest is an electron that is
= 3.04 × 10–19 J
sufficiently far away from the nucleus and its energy is
hc assumed to be zero. Mathematically, it corresponds to
Energy of incident radiation, hv =
 setting n equal to infinity in the equation so that E = 0. As
the electron moves closer to the nucleus due to electrostatic
6.626  1034  3  108 attraction, work is done by the electron itself and hence
 = 3.97 × 10–19 J
500  109 energy is released. Consequently, its energy decreases and
it takes energy values less than zero, which means negative
hv = hv0 + K.E.
values. The negative sign also indicates that the electron
 K.E. = hv – hv0 is bound to the nucleus and a hydrogen atom is in a stable
= 3.97 × 10–19 – 3.04 × 10–19 state in comparison to a state where electron is sufficiently
far away from the nucleus.
K.E. = 0.93 × 10–19J
Example - 21 52.9n 2
rn = pm
Describe postulates of Bohr’s model. Z
Sol. Postulates: For H-atom (Z = 1), the radius of first stationary state
1) An atom consists of a small, heavy positively charged
is called Bohr orbit (52.9 pm)
nucleus in the centre and the electrons revolve around
it in circular orbits. (c) Velocities of the electron in different orbits:
2) Electrons revolve only in those orbits which have a
2.188 ×106 Z
fixed value of energy. Hence, these orbits are called Vn = m/s
energy levels or stationary states. n
They are numbered as 1,2,3,...... These numbers are 3) Since the electrons revolve only in those orbits which
known as Principal quantum Numbers. have fixed values of energy, hence electrons in an
atom can have only certain definite values of energy
and not any of their own. Thus, energy of an electron
is quantised.
4) Like energy, the angular momentum of an electron in
an atom can have certain definite values and not any
value of their own.

nh
mνr =
2π
Where n=1,2,3...... and so on.
5) An electron does not lose or gain energy when it is
(a) Energy of an electron is given by: present in the same shell.
En= –RH(Z2/n2) n = 1,2,3....... 6) When an electron gains energy, it gets excited to higher
where RH is Rydberg’s constant and its value is energy levels and when it de-excites, it loses energy
in the form of electromagnetic radiations and comes
2.18 × 10–18 J.
to lower energy values.
Z = atomic number
Example - 22
Z 2 What is meaning of “Quantisation of angular
E n = -2.18 ×10-18 J/atom momentum”?
n2
Sol. According to Bohr’s Model, angular momentum of an
Z2 h
E n = -13.6 2 eV/atom electron, moving in an orbit is a fixed multiple of .
n 2

Z2 nh
E n = -1312 kJ/mol mr  where n = 1, 2, 3, 4........
n2 2
It means that when an electron gains or loses energy, it
Thus, energies of various levels are in the order: does so in such a way that n has a value which is a whole
number. In other words, electron does not gain or lose
K < L < M < N...... and so on.
energy in a continuous manner but in jumps (or bursts).
Energy of the lowest state(n=1) is called ground state. This led to the concept of quantisation of energy which
(b) Radii of the stationary states: means that radiant energy is emitted or absorbed in bursts
(or jumps) rather than as continuous flow.
Example - 23
R H . 22
What transition of Li+2 spectrum will have the same I.E. for He+ =
12
wavelength as that of second line of Balmer series of
He+ spectrum. = 2.18 × 10–18 × 4
Sol. Using Rydberg’s formula : = 8.72 × 10–18 J
Example - 25
1  1 1 
 R H  2  2  Z2 What transition in the hydrogen spectrum would have
  n1 n 2  the same wavelength as the Balmer transition,n=4 to n=2
of He+ spectrum?
1  1 1  Sol. For He+ ion,
For Li 2 :  RH  2  2  (3)
2
....... (1)
 n
 1 n 2 

1 1 1
 RZ2  2  2
1 1 1 2  n
 1 n 2
For He+ :  R H  2  2  (2) ........ (2)
 2 4 
Now, n1 = 2, n2 = 4 and Z = 2
Compairing (1) and (2)
1 1 1 3
 R(2) 2  2  2   R ...... (i)
 1 1  2  1 1 2  2 4  4
 2  2  (3)   2  2  (2)
 n1 n 2  2 4 
1 1 1
2 For H atom   R  2  2  ........ (ii)
 1 1   1 1 2 n
 1 n 2
 2  2  2  2  
n
 1 n 2  2 4 3
Equating equations (i) and (ii) ( is the same)

 1 1  1 1  1 1 3
 2  2     
 n1 n 2   9 36  n12 n 22 4

Now, if n1 = 1 and n2 = 2
 1 1  1 1 
 2  2  2  2  Therefore, the transition n = 2 to n = 1 in H atom will have the
 n1 n 2   3 6  same wavelength as the transition from n = 4 to n = 2 in He+.
Hence, n1 = 3 and n2 = 6 Example - 26
Example - 24 The wavelength of the first line in the Balmer series is
6561Å. Calculate the wavelength of second line and
Calculate the energy required for the process
limiting line in Balmer series.
He+ He+2 +e– . the ionisation energy for the hydrogen
atom in the ground state is 2.18 × 10–18J/atom–1. Sol. According to Rydberg equation,

Sol. For ionisation, n2 =  1  1 1 

R 2  2
 n
 1 n 2 
2  1 1 
Ionisation Energy, I.E. = R H .Z  n 2  n 2 
 1 2  For first line in Balmer series, n1 = 2, n2 = 3

 1
1  1 1  5 
1  R . Z2  6561  R  22  32   R  36  ........ (i)
 R H . Z2  2  2   H 2    
 n1   n1
For second line in Balmer series, n1 = 2, n2 = 4
2
R H . (1)
I.E. for H in ground state = = RH 1  1 1  3
12    R  22  42   R  16  ....... (ii)
   
–18 –1
= 2.18 × 10 J atom
 Dividing eq. (i) by (ii),
1 1
 
 5 16 8R H 8  1.067 107
 
6561 36 3
  1.17 108 m
6561 5 16
  = 4860 Å  = 11.7 nm.
36  3
Example - 29
For limiting line in Balmer series, n1 = 2 & n2 = 
What is the wavelength of light emitted when electron in
H-atom undergoes transition from energy level with n =
1  1 1  R
  R 2  2   (iii) 4 to an energy level n = 2?
  2   4
Dividing Eq. (i) by (iii) 1  1 1 
Sol.  RZ 2  2  2 
 n
 1 n 2 
 45

6561 36
1  1 1 
 1.097 107  12  2  2 
 '  3645Å  2 4 
Example - 27
1 1 
Electromagnetic radiation of wavelength 242nm is just  1.097 107    
 4 16 
sufficient to ionize the sodium atom. Calculate the
ionisation energy of sodium in kJ/mol.
3
 1.097 107 
c 16
Sol. E  N h  N h

16

1 34
(6.02  10 mol )  (6.626 10 J s)  (3  10 m s )
23 8 1 1.097  107  3

242  10 9 m
 4.8617  1017 m
= 4.945 × 105 J mol–1 = 494.5 kJ mol–1.
 4861.7  10 10 m
Example - 28
The atomic spectrum of Li+2 arises due to transition of an  4861.7 Å
electron from n2 to n1 level if n1+n2=4 and n2–n1= 2.
Calculate the wavelength (in nm)of transition. Example - 30
Sol. Solving, For Hydrogen atom calculate the energy required to
remove the electron completely from n=2 orbit. What is
n1 + n2 = 4
the longest wavelength of light in cm that can be used to
n2 – n1 = 2 cause this transition.
n2 = 3 and n1 = 1
 1 1 
Using Rydberg’s formula : Sol. E = 2.18 × 10–18 Z2  n 2  n 2  J / atom
 1 2 

1 1 1 n1 = 2 n2 = 
 RH  2  2  Z2
  n1 n2 
 1 1
E = 2.18 × 10–18 (1)2  2  
2 
1 1 1
 RH  2  2 (3)2
 1 3  = 5.45 × 10–19 J

hc
1 E 
 8R H 

 hc 6.626  1034  3  108 5
   109,677  cm 1
E 5.45  10 19 36

= 3.647 × 10–7 m = 15232.9 cm–1 = 1.523 × 106 m–1.

= 3.647 × 10–5cm. Example - 34
Example - 31 Which state of triply ionised Be+3 has same orbital radius
as that of ground state of H-atom.
The energy associated with first orbit in hydrogen atom
is – 2.17 × 10–18J/atom. What is the energy associated (1) 2
with fifth orbit? Sol. For hydrogen atom, r = 0.529 Å
(1)
Sol. Energy associated with nth orbit in hydrogen atom is
= 0.529 Å
18
2.17  10 J
En   atom 1 0.529n 2
n2 For Be+3, r =
(4)
 Energy associated with 5th orbit,
0.529 n 2
2.17  1018 J Thus, 0.529 
E5   4
52
= – 8.68 × 10–20 J  n = 2.
Example - 32 Example - 35
(a) Which transition between Bohr orbits corresponds Give the difference between Electromagnetic waves and
to third line in the Balmer series of the hydrogen matter waves.
spectrum.
Sol :
(b) Calculate the radius of Bohr’s fifth orbit for hydrogen
Electromagnetic Waves :
atom.
1. The electromagnetic waves are associated with electric
Sol. (a) n1 = 2 n2 = 5
and magnetic fields, perpendicular to each other and to
Transition : 5  2 the direction of propagation.
(b) Radius of Bohr’s nth orbit is given as : 2. They do not require any medium for propagation, i.e., they
rn = 0.0529 n2 nm can pass through vacuum.
 For n = 5 3. They travel with the same speed as that of light.
r5 = 0.0529 × (5)2 nm = 1.3225 nm 4. They leave the source, i.e., they are emitted by the source.

Example - 33 c
5. Their wavelength is given by  
Calculate the wavenumber for the longest wavelength 
transition in the Balmer series of atomic hydrogen. Matter waves :
Sol. For Balmer series, 1. Matter waves are not associated with electric and magnetic
fields.
 1 1 
  109, 677  2  2  cm 2. They require medium for their propagation, i.e., they cannot
2 n 
pass through vaccum.
 1 3. They travel with lower speeds. Moreover, it is not constant
 will be maximum of  is minimum     for all matter waves.
 
4. They do not leave the moving particle, i.e., they are not
for n = 3. The value will be
emitted by the particle.
 1 1
 = 109, 677 cm–1  22  32 
 
 h Or (x). (mv)  h/4
5. Their wavelength is given by  
m
Example - 36 Or (x). (v)  h/4m
Give the difference between particle and a wave. Explanation
Sol.
Particle :
1. A particle occupies a well-defined position in space, i.e., a
particle is localized in space, e.g., a grain of sand, cricket
ball, etc.
2. When a particular space is occupied by one particle, the
same space cannot be occupied simultaneously by any
other particle. In other words, particles do not interfere.
3. When a number of particles are present in a given region
of space, their total value is equal to their sum, i.e., it is
neither less nor more.
Wave :
Change of momentum and position of
1. A wave is spread out in space, e.g., on throwing a stone in
electron on impact with a photon
a pond of water, the waves start moving out in the form of
concentric circles. Similarly, the sound of the speaker Suppose we attempt to measure both the position and
reaches everybody in the audience. Thus, a wave is momentum of an electron. To pin point the position of
delocalized in space. the electron we have to use light so that the photon of
light strikes the electron and the reflected photon is
2. Two or more waves can coexist in the same region of space
seen in the microscope. As a result of the hitting, the
and hence interfere.
position as well as the velocity of the electron are
3. When a number of waves are present in a given region of disturbed.
space, due to interference, the resultant wave can be larger
It rules out the existence of definite paths or trajectories
or smaller than the individual waves, i.e., inter-ference may
of electrons as stated in Bohr’s Model.
be constructive or destructive.
(ii) On the basis of Heisenberg’s uncertainty principle, it
Example - 37 can be shown why electrons cannot exist within the
(i) State and illustrate Heisenberg’s uncertainty atomic nucleus. This is because the diameter of the
principle atomic nucleus is of the order of 10–15m. Hence, if the
(ii) Why electron cannot exist in the nucleus? electron were to exist within the nucleus, the maximum
Sol. (i) It is impossible to measure simultaneously the position uncertainty in its position would have been 10–15m
and momentum of a small particle with absolute (i.e., x = 10–15 m) Taking the mass of electron as 9.1 ×
accuracy. If an attempt is made to measure any of these 10–31 kg, the minimum uncertainty in velocity can be
two quantities with higher accuracy, the other calculated by applying uncertainty principle as
becomes less accurate. The product of the uncertainty follows :
in the position (x) and the uncertainty in momentum h h
(p) is always a constant and is equal to or greater x. p  or x. (m  ) 
4 4
than h/4.
h
or  
(x). (p)  h/4 4 x  m
6.6  1034 kg m 2 s 1

4  3.14  (10 15 m)  (9.1 1031 kg)
= 5.77 × 1010 m s–1
 This value is much higher than the velocity of light (viz, Example - 41
3 × 108 m s–1) and hence is not possible. Calculate the mass of a photon with wavelength 3.6Å
Example - 38 Sol. Here,  = 3.6 Å = 3.6 × 10–10 m. As photon travels with the
Show that the circumference of the Bohr orbit for velocity of light,
hydrogen atoms is an integral multiple of the de-
 = 3.0 × 108 m s–1
Broglie’s wavelength associated with the electron
moving around the orbit. h
By de Broglie equation,  =
Sol. According to Bohr postulate of angular momentum, m

h h h
mr  n or 2  r  n ...... (i) or m 
2 m 

h 6.626  1034 J s
According to de Broglie equation,   .... (ii) 
m (3.6  1010 m) (3.0  108 m s 1 )
Substituting this value in eqn. (i), we get 2  r = n
= 6.135 × 10–29 kg.
Thus, the circumference (2  r) of the Bohr orbit for
hydrogen atom is an integral multiple of de Broglie Example - 42
wavelength. The kinetic energy of an electron is 4.55 × 10-25 J.
Example - 39 Calculate the wavelength of the electron.
Sol. Here, we are given
Find velocity of electron for H-atom in its first Bohr orbit
of radius a0. Also, find the de-broglie wavelength. kinetic energy
Sol. According to Bohr’s model, angular momentum is 1
quantised. i.e., m 2  4.55  1025 J
2
nh m = 9.1 × 10–31 kg
mr 
2 h = 6.6 × 10–34 kg m2 s–1
nh 1
   (9.1 1031 )  2  4.55 1025
2 ma 0 2

de-Broglie wavelength,
4.55  10 25  2
or    106
2

h h (2 ma 0 ) 9.1  10 31

 
m m (nh)
1 h
or   10 m sec   
3

m
2 a 0

n
6.6  1034 kg m 2s 1
Example - 40 
(9.1 10 31 kg)  103 m s 1
Does Bohr model satisfy Heisenberg principle ?
= 7.25 × 10–7 m.
Sol. No, it rules out the well defined circular paths (orbits) or
trajectories proposed by Bohr. Since for a subatomic particle Example - 43
like an electron, it is not possible to simultaneously
determine the position and velocity at any moment with (a) Calculate the de-Broglie wavelength of an electron
good degree of precision, therefore, it is not possible to moving with 1% speed of light.
talk about the trajectory of an electron or well defined
(b) A molecule of O2 and that of SO2 travel with the same
circular orbits.
velocity. What is the ratio of their wavelengths ?
 h 1
Sol. (a) We know that   Sol. Kinetic Energy = m 2  qV
m 2

m = 9.1 × 10–31kg, h = 6.63 × 10–34kg m2 s–1 mv 2

V ....... (1)
2q
1 3.0  108
 = 1% of speed of light = ms 1
100
h
According to de-Broglie,  
= 3.0 × 106m s–1 (  speed of light = 3.0 × 108m s–1) m

6.63  10 34 kg m 2s 1 h
   
(9.1 1031 kg)  (3.0  106 m s 1 ) m

= 2.43 × 10–10 m. Putting in (1)

2
h m  h 
(b)   V  
m 2q  m 

 O2 mSO2 64 h2
  V
SO2 mO 2 32 2mq 2

 O2 2 (6.626  10 34 ) 2
 V
SO2 1 (2  1.67  10 27 )  (1.6  1019 ) (0.005 10 9 ) 2

Example - 44 V = 32.8 Volts

An electron is moving with KE of 3 × 10-25 J. Calculate its Example - 46

wavelength and frequency. What will happen to the wavelength associated with a
moving particle if its velocity is doubled?
1
Sol. K.E.  m 2 Sol. Wavelength becomes half of the original value.
2
h 1
(because   , i.e.,   )
2 K.E. 2  3.0  1025 J m 
  
m 9.1 1031 kg
Example - 47
= 812ms –1 2 –2
(1J = 1 kg m s ) The mass of an electron is 9.1 × 10–31kg. If its kinetic
energy is 10MeV, calculate its wavelength.
h
By de Broglie equation,   Sol. 1eV = 1.6 × 10–19 J
m
KE = 10 × 106 eV
34
6.626 10 J s K.E. = 10 × 106 × 1.6 × 10–19J
 = 8.967 × 10–7 m
(9.1 1031 kg) (812 m s 1 )
K.E = 1.6 × 10–12 J
= 8967 Å.
1 2
K.E = m
c 3  10 8 2
   3.34  1014 Hz
 8.967  10 7
2 (K.E.)
Example - 45 
m
Calculate the accelerating potential that must be applied
on a proton beam to give it an effective wavelength of h h
 
0.005 nm. m 2m (KE)
 Sol. m = 10–27g = 10–30 kg
6.626  1034
 h
2  9.1 1031 (1.6  1012 ) x . =
4m
  3.88 1013 m
h
 
Example - 48 4 mx
A golf ball has a mass of 40g and a speed of 45m/s. If the
speed can be measured within accuracy of 2%. Calculate 6.626  1034
uncertainty in the position.  
4  3.14  10 30  10 11
Sol. Uncertainty in speed = 2% of 40 m s–1 = 5.25 × 106 m/s.
2 Example - 51
i.e.,    45  0.9 m s1
100 An electron has a speed of 500m/s with an uncertainty
Applying uncertainty principle of 0.02% what is the uncertainty in locating its position.

h 0.02
x (m  )  Sol. Uncertainity in speed =  500  0.1 m / s
4 100

h h
or x  x .  
4m 4m

h 6.626  1034
6.626 1034 kg m 2 s 1 x  
 4 m 4  3.14  9.1 10 31  0.1
4  3.14  (40  10 3 kg) (0.9 m s 1 )

= 1.46 × 10–33 m x  5.77  10 4 m.

Example - 52
Example - 49 (a) What is the physical significance of  and 2
Calculate the uncertainty in position of dust particle with (b) What is quantum mechanics ?
mass equal to 1mg if the uncertainity in velocity is 5.5 × Sol. (a) In the phsyical sense,  gives the amplitude of the
10–20m/s. wave associated with the electron. We know that in
Sol. m = 10–3 g = 10–6 kg the case of light waves, the square of the amplitude of
the wave at a point is proportional to the intensity of
= 5.5 × 10–20 m/s.
light. Extending the same cocept to electron wave
h motion, the square of the wave function, 2 may be
x ×  = taken as intensity of electron at any point. In other
4m
words, 2 determines the probability of finding the
h moving electron in a given region i.e. it gives the
x = 4 m probability density. Thus, 2 has been called the
probability density and  the probability amplitude.
Large value of 2 means a high probability of finding
6.626  10 34
x = the electron at that place and a small value of 2 means
4  3.14  106  5.5  10 20
low probability. If 2 is almost zero at a particular point,
x  9.59 × 10–10m it means that the probability of finding the electron at
that point is negligible.
Example - 50
(b) A branch of science that takes dual nature of matter
The approximate mass of an electron is 10–27g. Calculate
into consideration is known as quantum mechanics.
the uncertainty in its velocity if the uncertainty in its
position were of the order 10–11m.
 Example - 53
Distinguish between Orbit and Orbital (vi) Outline the shapes of: (a) 3s (b) 3p z (c) 3d xz
Sol : Orbit (d) 3d x 2 -y2
(i) An orbit is a well defined circular path around the nucleus Sol. :
in which the electron revolves. (i) An orbital may be defined as a region in space around
the nucleus where the probability of finding the
(ii) An orbit represents the planar motion of an electron around electron is maximum.
the nucleus.

(iii) All the orbits are circular.

(iv) The concept of an orbit is not in accordance with the wave

(ii)
character of electrons (de Broglie’s hypothesis and

Heisenberg’s uncertainity principle)

(v) The orbits do not have any directional characteristics. (iii) Similarities : (i) Both have spherical shape (ii) Both
h
(vi) The maximum number of electrons in any orbit is given by 2n2, have same angular momentum as it is = l (l  1) .
2
where n is the number of orbit. Differences :
Orbital (i) 1s has no node while 2s has one node.
(i) An orbital is the three dimensional space around the (ii) Energy of 2s is greater than that of 1s.
nucleus within which the probability of finding an electron (iii) Size of 2s is larger than that of 1s.
is maximum.
(iv) Shape of s-orbital is spherically symmetric because
(ii) An orbital represents a three dimensional region of space the probability of finding the electron is same in all
around the nucleus. the directions at a particular distance from the nucleus.
(iii) Different orbitals have different shapes. e.g. (v) (1) Shapes of s-orbitals:
s-orbitals are spherically symmetric, p-orbitals are dumb-
bell shaped and so on. (a) They are non-directional and spherically symmetric
i.e. probability of finding the electron at a given
(iv) The concept of an orbital is in accordance with the wave distance is equal in all directions.
character of electrons and uncertainty principle.
(b) 1s orbital and 2s orbital have same shape but size of
(v) All the orbitals, except s-orbitals, have directional 2s is larger.
characteristics.
(c) There is a spherical shell within 2s orbital where
(vi) The maximum number of electrons present in any orbital is electron density is zero and is called a node.
two.
(d) The value of azimuthal quantum number (l ) is zero
Example - 54
(l =0) and magnetic quantum number can have only
(i) What is an orbital? one value i.e. m=0
(ii) Which d-orbital does not have four lobes? Draw
its shape?
(iii) Compare the shapes of 1s and 2s orbital.
(iv) Why the shape of s-orbital is spherically
symmetric?
(v) Explain the shapes of s,p,d orbitals
(2) Shapes of p-orbitals:
(a) It consists of two lobes present on either side of the
plane that passes through the nucleus. The p-orbital
is dumb-bell shaped.
(b) There are three possible orientations of electron cloud (vi)
in p-orbitals. Therefore, the lobes of p-orbital may be
considered to be along x,y and z axis. Hence they are
designated as px,py,pz. The three p-orbitals are oriented
at right angles to one another.
(c) First main energy level( Principal quantum number=1)
does not contain any p-orbital.
(d) The three p-orbitals of a particular energy level have
same energy in absence of an external electric and
magnetic field and are called degenerate orbitals.
(e) Like s orbitals, p-orbitals increase in size with increase
in the energy of main shell of an atom. Thus, value of
azimuthal quantum number is one (l =1) and magnetic
quantum number has three values (m= –1, 0, +1)
Example - 55
(a) What is the lowest value of n which allows a g-orbital
to exist?
(b) What are degenerate orbitals ?
Sol. (a) For g-subshell, l = 4. As l = 0 to n–1, hence to have
l = 4, minimum value of n = 5, i.e., 5th shell.
For l = 4, m = –4, –3, – 2, –1, 0, +1, +2, +3, + 4, i.e., 9 values
which means 9 orbitals.
(b) The orbitals of same shell and sub-shell having equal
energy are called degenerate orbitals.
eg. 3px, 3py, 3pz
Example - 56
(i) Designate the orbital with n = 4, l = 2 and m = 0
(3) Shapes of d-orbitals:
(ii) List the quantum numbers of electrons for 3d
(a) They are designated as dxy, dyz, dzx and dx2-y2. They orbital
have a shape like a four leaf clover. The fifth d
(iii) An atomic orbital has n=3. What are the possible
orbital designated as d 2z looks like a doughnut. values of l and m.
(b) All five d orbitals have same energy in the absence (iv) Which of the following orbitals are possible?
of magnetic field. 1p, 2s,2p,3f
(c) The d orbitals have azimuthal quantum number l = 2 (v) Using s,p,d notations describe the following
and magnetic quantum number values –2, quantum numbers:
–1,0,+ 1,+ 2. (a) n=1, l = 0 (b) n=3, l = 2
(d) For principal shell number 1 and 2, there are no d (c) n=3, l = 1 (d) n=4, l = 3
orbitals. (e) n=2, l = 1
(vi) Write the values of n,l,m,s for 4p
(vii) What is the total number of orbitals in the 4f
sub-shell?
 (viii) What is the maximum number of electrons that Example - 57
can occupy the 4d sub-shell What is the angular momentum of an electron in
(ix) How many electrons will be present in possible (i) 2s orbital (ii) 4f orbital (iii) 2p angular momentum
orbital having n = 3, l = 1, m = –1
h
(x) Calculate the number of electrons in  (l  1)
2
(a) 3pz orbital (b) 3d subshell (c) 7s subshell.
Sol. (i) For 2s, l = 0, thus angular momentum = 0
(xi) How many electrons in an atom may have the
following quantum numbers h
(ii) For 4f orbital, l = 3, angular momentum = 12
2
1
(a) n = 4, ms = - (b) n = 3, l = 0
2 2h
(iii) For 2p orbital, l = 1, angular momentum 
Sol. (i) 4d 2z 2
Example - 58
1 1
(ii) n = 3 l = 2 m = –2, –1, 0, +1, +2 s   , (i) How many electrons will be present in all the possible
2 2
orbital having (a) n + l = 4 (b) n + l = 5
(iii) n=3 l = 0, 1, 2 (ii) How many electrons in sulphur (Z=16) can have
when l=0 m= 0 n+l=3
l=1 m = –1, 0, +1 (iii) How many electrons in Cl have n + l = 3
l=2 –2, –1, 0, +1, +2 Sol
(iv) 1p is not possible because when n = 1, l = 0 only (i) (a) Subshells with n + l = 4 are 4s, 3p
(for p, l = 1) Hence, electrons present = 2 + 6 = 8
2s is possible becuase when n = 2, l = 0 (for s, l = 0) (b) Subshells with n + l = 5 are 5s, 4p, 3d. Hence,
2p is possible because when n = 2, l = 0, 1 (for p, l = 1) electrons present = 2 + 6 + 10 = 18.

3f is not possible because when n = 3, l = 0, 1, 2 (ii) For 1 s2, n + l = 1 + 0 = 1

(for f, l = 3) For 2 s2, n + l = 2 + 0 = 2
(v) (a) 1s, (b) 3d, (c) 3p, (d) 4f, (e) 2p For 2 p6, n + l = 2 + 1 = 3

1 1 For 3 s2, n + l = 3 + 0 = 3
(vi) n = 4 ; l = 1 ; m = –1, 0, + 1 s   ,
2 2 For 3 p4, n + l = 3 + 1 = 4
(vii) Number of orbitals in f-subshell = 7 Thus, n + l = 3 for 2 p6 and 3 s2 electrons, i.e., for 8
(viii) Maximum number of electrons in d-subshell - 10 electrons.

(ix) m = –1 represents an orbital and orbital can have (iii) Cl  1s22s22p63s23p5
maximum of two electrons. For 1s2, n + l = 1 + 0 = 1
(x) (a) An orbital can have maximum of 2 electrons For 2s2, n + l = 2 + 0 = 2
(b) d-subshell has maximum of 10 electrons For 2p6, n + l = 2 + 1 = 3
(c) s-subshell has maximum of two electrons
For 3s2, n + l = 3 + 0 = 3
2 2
(xi) (a) Total electrons in n = 4 are 2n , i.e., 2 × 4 = 32.
For 3p5, n + l = 3 + 1 = 4
1 Thus, (n + l) = 3 for 2p6 & 3s2 electrons, i.e. for 8 electrons.
Half of them, i.e., 16 electrons have ms = – .
2
(b) n = 3, l = 0 means 3s orbital which can have
2 electrons.
Example - 59 Example - 63
What are quantum number of the valence electrons in
Why half filled and fully filled orbitals are stable?
potassium atom[Z =19] in ground state?
Sol. The completely filled and completely half filled sub-shells
Sol. K[19] : 1s2 2s2 2p6 3s2 3p6 4s1
are stable due to the following reasons:
1 1 1. Symmetrical distribution of electrons: It is well known
Therefore, n = 4, l = 0, m = 0, s = + or 
2 2 that symmetry leads to stability. The completely filled or
half filled subshells have symmetrical distribution of
Example - 60
electrons in them and are therefore more stable. Electrons
What information do you get from the principal quantum in the same subshell (here 3d) have equal energy but
number about an atom ? different spatial distribution. Consequently, their shielding
Sol. (i) It gives us the average distance of the electron from of one another is relatively small and the electrons are
the nucleus.
more strongly attracted by the nucleus.
(ii) It determines the energy of the electron in H-atom
and hydrogen like particles. 2. Exchange Energy : The stabilizing effect arises whenever
two or more electrons with the same spin are present in the
(iii) The maximum number of electrons present in any
degenerate orbitals of a subshell. These electrons tend to
shell is given by 2n2 where ‘n’ is the number of exchange their positions and the energy released due to
principal shell. this exchange is called exchange energy. The number of
Example - 61 exchanges that can take place is maximum when the
Explain pauli exclusion principle & Why Pauli exclusion subshell is either half filled or completely filled. As a result
principle is called exclusion principle? the exchange energy is maximum and so is the stability.
Sol. No two electrons in an atom can have the same set of four eg. Cr (24) : [Ar] 4s1 3d5
quantum numbers.
If one electron in an atom has some particular values for
the four quantum numbers, then all the other electrons in
that atom are excluded from having the same set of values.
It is because of this reason that this principle is called
exclusion principle.
Example - 62
Write short note on Hund’s rule of maximum multiplicity.
Why it is called multiplicity rule?
Sol. Electron pairing in p, d and f orbitals cannot occur until
each orbital of a given subshell contains one electron each
or is singly occupied.
This is due to the fact that electrons being identical in
charge, repel each other when present in the same orbital.
This repulsion can, however, be minimised if two electrons
move as far apart as possible by occupying different
degenerate orbitals.
Further, all the singly occupied will have parallel spins, i.e.,
in the same direction, viz., either clockwise or anticlockwise.
This is due to the fact that two electrons with parallel spins
(of course in different orbitals) will encounter less inter-
electronic repulsions in space than when they have opposite Thus, total number of exchanges=10
spins and total spin of unpaired electrons is maximum.
 Example - 64 Example - 66
Why the three electrons present in 2p subshell of (a) Indicate the number of unpaired electron in
nitrogen remain unpaired?
(i) P (ii) Cr(iii) Si (iv) Kr (v)Fe+2
Sol. According to Hund’s rule, electron pairing in p, d and f
orbitals cannot occur until each orbital of a given subshell (b) Which is more paramagnetic:
contains one electron each. so that total spin of unpaired Fe+2 or Fe+3
electrons is maximum.
(c)Which is more stable Fe+2 or Fe+3
For the element nitrogen, which contains 7 electrons, the
Sol.
following configurations can be written :
(a) (i) 15
P = 1s2 2s2 2p6 3s2 3p1x 3p1y 3p1z .

No. of unpaired electron = 3

(ii) 24
Cr = 1s2 2s2 2p6 3s2 3p6 3d5 4s1.
Total spin of unpaired electrons No. of unpaired electrons = 6

1 1 1 1 (iii) Si = 1s2 2s2 2p6 3s2 3p1x 3p1y .

   1 14
2 2 2 2
No. of unpaired electrons =2
Example - 65
(iv) 36
Kr = Noble gas. All orbitals are filled. Unpaired
Write the electronic configuration of electrons = 0.
(i) Cu (ii) Cu+ (iii) Cu+2 (iv) Cr (v) Cr+3 (vi) Co+3 (v) Fe = 1s2 2s2 2p6 3s2 3p6 3d6. No. of unpaired
26
–
(vii) O (viii) Fe (ix) Fe (x) Zn (xi) H (xii) Na
–2 +3 +2 +2 +
electrons = 4 (in 3 d)
(xiii) O–2 (xiv) F– (xv) Al+3 (xvi) Sc (xvii)Cl– (b) As Fe3+ contains 5 unpaired electrons while Fe2+
Sol. (i) 1s2 2s2 2p6 3s2 3p6 4s1 3d10 contains only 4 unpaired electrons, Fe3+ is more
paramagnetic.
(ii) 1s2 2s2 2p6 3s2 3p6 3d10
(c) Fe+2 : [Ar] 4s0 3d6
(iii) 1s2 2s2 2p6 3s2 3p6 3d9.
Fe+3 : [Ar] 4s0 3d5
(iv) 1s2 2s2 2p6 3s2 3p6 3d5 4s1.
(v) 1s2 2s2 2p6 3s2 3p6 3d3 Since d-subshell is half filled in Fe+3, hence it is more stable

(vi) 1s2 2s2 2p6 3s2 3p6 3d6 Example - 67

(vii) 1s2 2s2 2p6 (i) What are the atomic numbers of the elements whose
outermost electrons are represented by (a)3s1 (b) 2p3
(viii) 1s2 2s2 2p6 3s2 3p6 3d5
(c) 3d5
(ix) 1s2 2s2 2p6 3s2 3p6 3d6
(x) 1s2 2s2 2p6 3s2 3p6 3d10 (ii) Which atoms are indicated by the following
configurations? (a) [He]2s 1 (b) [Ne]3s 2 3p 3
(xi) 1s 2 (c) [Ar]4s23d1
(xii) 1s2 2s2 2p6 Sol.
2 2 6
(xiii) 1s 2s 2p (i) (a) Total electrons : 2 + 2 + 6 + 1 = 11
(xiv) 9
F = 1s2 2s2 2p5  F– = 1s2 2s2 2p6 Atomic Number = 11
(xv) 1s2 2s2 2p6 (b) Total electrons : 2 + 2 + 3 = 7
(xvi) 1s2 2s2 2p6 3s2 3p6 4s2 3d1 Atomic Number = 7

(xvii) 1s2 2s2 2p6 3s2 3p6

 (c) There can be two cases : Example - 70
2 2 6 2
For Cr : 1s 2s 2p 3s 3p 4s 3d6 1 5 How many radial nodes & angular nodes are present in
Thus atomic number = 24 (a) 5d (b) 4p (c) 6d
Sol. (a) Radial nodes : n – l – 1
For Mn : 1s2 2s2 2p6 3s2 3p6 4s2 3d6
=5–2–1=2
Thus atomic Number = 25
Angular nodes = l = 2
(ii) (a) 1s2 2s1, Thus 3Li
(b) Radial nodes = n – l – 1
(b) 1s2 2s2 2p6 3s2 3p3, Thus 15P
=4–1–1=2
2 2 6 2 6 2 1
(c) 1s 2s 2p 3s 3p 4s 3d , Thus 21Sc Angular nodes = l = 1
Example - 68 (c) Radial nodes = n – l – 1
(i) Which orbital in the following pair is lower in energy =6–2–1=3
in a many electron atom : 3p and 3d Angular nodes = l = 2
(ii) Why 4s orbital is filled before 3d? Example - 71
Sol. (i) Orbital having lower value of (n + l) Estimate the highest velocity of the electron being
ejected by light of  = 2.4×1015Hz for a metal with a work
will have lower energy.
function of 10 eV.
For 3p : n + l = 4 (n = 3, l = 1)
Sol. Energy of incident photon = h
3d : n + l = 5 (n = 3, l = 2) = 6.63 × 10–34 × 2.4 × 1015
Thus 3p has lower energy. = 1.63 × 10–18 J
(ii) Orbitals are filled with electrons starting with the orbital Threshold Energy = 10eV × 1.6×10–18J
of lowest energy.
Therefore, KE = 1.63 ×10–18 – 1.6×10–19 J
Orbital having lower value of (n + l) will have lower = 1.47 × 10–18J
energy.
½ mev2 = 1.47 × 10–18
For 4s : n + l = 4 (n = 4, l = 0)
For 3d : n + l = 5 (n = 3, l = 2) v= 3.23  10  m /s
12
= 1.79×106 m/s Ans.
Thus 4s is filled before 3d. Example - 72
Example - 69 A H-like species is in their excited state
(i) What is the total number of electrons in 2p subshell (A) and absorbs a photon of energy 3.868 eV and get
of a chlorine atom in the ground state. excited to a new state (B). The electron from B on returning
(ii) Which sub-shells are occupied in the outermost to a lower orbit, can give a maximum of ten different
principal energy level of Argon atom in the ground emissions. Some of the radiations have energies greater
state. than it and some equal to 3.868 eV. Exactly 2 radiations
have energies less than 3.868 eV. Determine the orbit
(iii) How many electrons are there in the valence quantum numbers of states A and B and also identify the species.
level of copper (Atomic number=29)?
Sol. Total number of emissions from state B = 10.
Sol. (i) Cl (17) : 1s2 2s2 2p6 3s2 3p5
Thus,nB (nB – 1)/2 = 10  nB = 5 Ans.
Total electrons in 2p = 6
Also, number of spectral lines with energies less than 3.868
(ii) Ar (18) : 1s2 2s2 2p6 3s2 3p6 eV = 2. So, there are two transitions possible between nB to
p-subshell is occupied nA excluding the direct transition from B to A. Therefore, nA
= 3 Ans.
(iii) Cu (29) : 1s2 2s2 2p6 3s2 3p6 4s1 3d10
Also I.P = 13.6Z2 eV and from the given information 3.868
Valence quantum level of Cu = 4 =13.6Z2 ([1/32) – [1/52]) Thus, Z = 2 Ans.
It has 1 electron
Example - 73
v
Estimate the De-broglie’s wavelength of Hence frequency =
2r
(a) An electron moving with a velocity of 7.28×107m/s.
Calculate velocity (v2) and radius (r2) for electron in 2nd
(b) A 100 kg motorbike moving at 6.63 m/s Bohr orbit in H-atom (Z = 1)
Sol. (a) mass of electron = 9.1×10-31kg and given velocity Z = 1 for H-atom
= 7.28×107m/s
n2
Thus, momentum of electron Using rn  0.529 Å
Z
= m × v = 9.1×7.28×10–24 kgm/s = 6.625×10–23
 2
2
Therefore, wavelength = h/mv = 10–8m
r2  0.529  1010 m  2.12  1010 m
= 10–11m Ans. 1
(b) Momentum = 100 × 6.63 kgm/s = 663 kgm/s vn = 2.165 × 106 (1/n) m/s
Therefore, wavelength = h/mv = 10 m Ans.
-36
v2 = 2.165 × 106 (1/2) = 1.09 × 106 m/s
Example - 74
A electron is moving with a velocity of 108 ± 105 m/s. v2 1.09  106
Hence frequency  
Calculate the uncertainty in its position. 2r2 2( ) (2.12  10 10 )
Sol. v = 2×105. Therefore p = mv = 1.82×10-25 v = 8.18 × 1014 Hz Ans.
Thus, x = h/4p = 2.9 × 10 m Ans.
–10
Example - 76
Example - 75 Determine the maximum number of lines that can be
Determine the frequency of revolution of the electron in emitted when an electron in H atom in
2nd Bohr’s orbit in hydrogen atom. n = 6 state drops to the ground state. Also find the
Sol. The frequency of revolution of electron is given by : transitions corresponding to the lines emitted.
Sol. The maximum number of lines can be calculated by using
1
Frequency = (n 2  n1  1) (n 2  n1 )
time period the formula where n2 = 6 and n1 = 1
2
are 15
v The distinct transition corresponding to these line are
61
r 6  2, 21
6  3, 3  2, 31
+Ze
6  4, 4  3, 4  2, 41
6  5, 5  4, 5  3, 5  2, 51
Example - 77
Find the wavelength of radiation required to excite the
Time period
electron in ground level of Li++ (Z = 3) to third energy
Total distance covered in 1 revolution 2r level. Also find the ionisation energy of Li2+. (R = 109,
= 
velocity v 677 cm–1)
Sol. Ground level : n = 1
Each of dxy, dxz, dyz, d (x 2  y2 ) dz2 has 2 nodal planes,
1  1 1 
Use :  RZ 2  2  2  which means a total of 10 nodal planes.
 n
 1 n 2 

Hence for n = 3, a total of 13 nodal planes are there.

Putting the values : n1 = 1, n2 = 3, Z = 3
Example - 79
1 1 1
We get :  1.09  10  3   2  2 
7 2
Write down the electronic configuration of following
 1 3  species. Also find the number of unpaired electrons in
each.
1 1
 8.776  107 m 1   = 113.97Å Ans. (a) Fe, Fe2+, Fe3+ (Z of Fe = 26)
 v
Ionisation energy is the energy required to remove the (b) Br, Br– (Z of Br = 35)
electron from ground state to infinity i.e. corresponding Sol. Follow the order of increasing energy (Aufbau Rule) :
transition responsible is 1   . 1s, 2s, 2p, 3s, 3p, 4s, 3p, 4d, 5s, 4d ......
(a) Fe (Z = 26) : 1s2 2s2 2p6 3s2 3p6 4s2 3d6
Note that 3d orbital are not fully filled.

3d6 
Orbitals filled as per Hund’s Rule.
Clearly the number of unpaired electrons is 4.

 Fe2+ : (Z = 24)
While Writing electronic configuration (e.c.) of
cations, first write e.c. of neutral atom and then
“remove desired number of electrons from outermost
orbital”.
In Fe2+, remove 2e– from 4s2 since 4s orbital (through
1 1  lower in energy then 3d) is the outermost. Hence e.c.
i.e. E (1 )  13.6  32  2  2  eV
1   of Fe2+ is : 1s2 2s2 2p 3p6 3d6 4s0
Note that number of unpaired electrons remains same
Ionisation energy = E (1 )  122.4 eV
as that in Fe, i.e. 4.
= 1.95 × 10–17J Ans. [1eV  1.6  10 19 J]  Fe3+ (Z = 23)
Example - 78 Now remove 2e– from 4s2 and 1e– from 3d6 to get e.c.
In all, how many nodal planes are there in the atomic as : 1s2 2s2 2p6 3s2 3p6 3d5 4s0
orbitals for the principal quantum number n = 3.
Note that, now all ‘d’ orbits have an odd electron (i.e.
Sol. Shell with n = 3 has 1 ‘s’ (3s), 3 ‘p’ (px, py, pz) and 3 ‘d’ are half filled).
2
(dxy, dxz,dyz, d (x 2  y2 ) and d z ) orbitals.

 ‘s’ has no nodal plane. Hence number of unpaired electrons in Fe3+ is 5.

(b) Br (Z = 35)
 Each of px, py, pz has one nodal plane, which means a
Following Aufbau rule, e.c. is :
total of 3 nodal planes.
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5
 Clearly one of 4p5 orbitals contains unpaired electrons :
  n(n  2) B.M
4p5 
(n = number of unpaired electrons ; BM : unit of magnetic
Orbitals filled as per Hund’s Rule. moment in Bohr’s Magneton)
Hence Br has only one unpaired electron.
 1.73  n(n  2)
 Br– (Z = 36)
On solving for n, we get n = 1. This means that vanadium
Since anion (s) is (are) formed by adding electron (s), so ion (Z = 23) in the compound has one unpaired electron.
simply write e.c. as per total number of electrons finally.
3d :
For Z = 36, e.c. is : 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6
So its electronic configuration (e.c.) must be :
Clearly there are no unpaired electrons.
Example - 80 1s2 2s2 2p6 3s2 3p6 3d1

A compound of Vanadium has a magnetic moment of i.e vanadium exists as V4+ ion in the compound since the
1.73 B.M. Work out the electronic configuration of ground state e.c. of 23V is :
vanadium in the compound. 1s2 2s2 2p6 3s2 3p6 3d3 4s2
Sol. The magnitude of magnetic moment () of a compound/
species/ion is given by : 3d : 4s :
EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS
Subatomic Particles & Special Terms 9. Which of the following atoms and ions are isoelectronic
i.e. have the same number of electrons with the neon atom
1. Which is not basic postulate of Dalton’s atomic theory ?
–
(a) F (b) Oxygen atom
(a) Atoms are neither created nor destroyed in a chemical
–
reaction (c) Mg (d) N
(b) In a given compound, the relative number and kinds 10. The triad of nuclei that is isotonic is :
of atoms are constant. 14 15 17 12 14 19
(a) 6 C, 7 N, 9 F
(b) 6 C, 7 N, 9 F
(c) Atoms of all elements are alike, including their masses.
(d) Each element is composed of extremely small particles 14 14 17 14 14 19
(c) 6 C, 7 N, 9 F
(d) 6 C, 7 N, 9 F
called atoms.
11. An isotone of 32Ge76 is
2. The number of electrons in a neutral atom of an element is
equal to its : (i) 32Ge77 (ii) 33As77

(a) atomic weight (b) atomic number (iii) 34Se77 (iv) 34Se78

(c) equivalent weight (d) electron affinity (a) Only (i) and (ii) (b) Only (ii) and (iii)

3. The e/m for positive rays in comparison to cathode rays (c) Only (ii) and (iv) (d) (ii), (iii), and (iv)
is : Rutherford's Nuclear Model of an Atom and
(a) very low (b) high Electromagnetic Spectrum
(c) same (d) none 12. When a gold sheet is bombarded by a beam of –particles,
4. Which has highest e/m ratio ? only a few of them get deflected whereas most go straight,
(a) He2+ (b) H+ undeflected. This is because
(c) He+ (d) H (a) The force of attraction exerted on the –particles by
the oppositely charged electrons is not sufficient.
5. Cathode rays have :
(b) A nucleus has a much smaller volume than that of an
(a) mass only (b) charge only
atom.
(c) neither mass nor charge
(c) The force of repulsion acting on the fast moving
(d) mass and charge both –particles is very small.
6. Mass of neutron is .............. times the mass of electron. (d) The neutrons in the nucleus do not have any effect
(a) 1840 (b) 1480 on the –particles.
(c) 2000 (d) None of these 13. Discovery of the nucleus of an atom was due to the
7. Positive rays or canal rays are : experiment carried out by

(a) electromagnetic waves (a) Bohr (b) Mosley

(b) a steam of positively charged gaseous ions (c) Rutherford (d) Thomson

(c) a stream of electrons 14. Rutherford’s scattering experiment is related to the size of
the
(d) neutrons
(a) Nucleus (b) Atom
8. The ion that is isoelectronic with CO is :
(c) Electron (d) Neutron
(a) CN (b) O 2

(c) O2 (d) N 2
15. Rank the following types of radiations from the highest 24. Which of the following is not a characteristic of Planck’s
energy to the lowest. quantum theory of radiation ?
ultraviolet/visible/X-ray/microwave/infrared (a) Energy is not absorbed or emitted in whole number
(a) X-ray, ultraviolet, microwave, infrared, visible multiples of quantum.
(b) ultraviolet, X-ray, visible, infrared, microwave (b) Radiation is associated with energy.
(c) infrared, microwave, ultraviolet, visible, X-ray (c) Radiation is associated with energy emitted or absorbed
(d) X-ray, ultraviolet, visible, infrared, microwave continuously but in the form of small packets called quanta.

16. The velocity of light is 3.0 × 108 ms–1. Which value is (d) The magnitude of energy associated with quantum is
closest to the wavelength in nanometres of a quantum of proportional to frequency.
light with frequency of 8 × 1015 s–1 25. The number of photons emitted in 10 hours by a 60 W
(a) 3 × 107 (b) 2 × 10–25 sodium lamp ( of photon = 6000 Å)
(c) 5 × 10–18 (d) 3.7 × 101 (a) 6.50 × 1024 (b) 6.40 × 1023
17. The frequency of a wave of light is 12 × 1014 s–1. The (c) 8.40 × 1023 (d) 3.40 × 1023
wave number associated with this light is Bohr's Model of an Atom
(a) 5 × 10–7 m (b) 4 × 10–8 cm–1
26. The Bohr orbit radius for the hydrogen atom (n = 1) is
(c) 2 × 10–7 m–1 (d) 4 × 104 cm–1 approximately 0.530 Å. The radius for the first excited state
18. The frequency of a green light is 6 × 1014 Hz. Its wavelength (n = 2) orbit is
is : (a) 0.13 Å (b) 1.06 Å
(a) 500 nm (b) 5 nm (c) 4.77 Å (d) 2.12 Å
(c) 50,000 nm (d) None of these 27. The energy of electron in 3rd orbit of hydrogen atom is
Planck's Quantum Theory (a) –1311.8 kJ mol–1 (b) –82.0 kJ mol–1
19. Which wave property is directly proportional to energy of (c) –145.7 kJ mol–1 (d) –327.9 kJ mol–1
electromagnetic radiation : 28. The ionization energy of H atom is 13.6 eV. The ionization
(a) velocity (b) frequency energy of Li2+ ion will be
(c) wave number (d) all of these (a) 54.4 eV (b) 40.8 eV
20. The energy E corresponding to intense yellow line of (c) 27.2 eV (d) 122.4 eV
sodium of  589 nm is : 29. The ratio of the difference in energy between the first and
the second Bohr orbit to that between second and third
(a) 2.10 eV (b) 43.37 eV
Bohr orbit is
(c) 47.12 eV (d) 2.11 kcal
21. The relation between energy of a radiation and its 1 1
(a) (b)
frequency was given by : 2 3
(a) de Broglie (b) Einstein 27 4
(c) Planck (d) Bohr (c) (d)
5 9
22. The number of photons of light of  = 2.5 × 106 m–1 necessary 30. Energy of electron of hydrogen atom in second Bohr
to provide 1 J of energy are orbit is
(a) 2 × 1018 (b) 2 × 1017 (a) –5.44 × 10–19 J (b) –5.44 × 10–19 kJ
(c) 2 × 1020 (d) 2 × 1019 (c) –5.44 × 10–19 cal (d) –5.44 × 10–19 eV
23. Minimum number of photons of light of wavelength 4000Å, 31. The energy of second Bohr orbit in the hydrogen atom is
which provide 1J energy : –3.4 eV. The energy of fourth orbit of He+ ion would be
(a) 2 × 1018 (b) 2 × 109 (a) –3.4 eV (b) –0.85 eV
(c) 2 × 1020 (d) 2 × 1010 (c) –13.64 eV (d) +3.4 eV
32. The energy of an electron in the first Bohr orbit of H atom (d) There are interelectronic repulsions.
is –13.6 eV. The possible energy value(s) of the excited 40. Ratio of frequency of revolution of electron in the second
state(s) for electrons in Bohr orbits to hydrogen is (are) excited state of He+ and second state of hydrogen is
(a) –3.4 eV (b) –4.2 eV
32 27
(c) –6.8 eV (d) +6.8 eV (a) (b)
27 32
33. The ionization energy of hydrogen atom is 13.6 eV. The
energy required to excite the electron in a hydrogen atom 1 27
(c) (d)
from the ground state to the first excited state is 54 2
(a) 1.69 × 10–18 J (b) 1.69 × 10–23 J
Hydrogen Spectra
(c) 1.69 × 1023 J (d) 1.69 × 1025 J
41. The line spectrum of two elements is not identical because
34. In a Bohr’s model of atom when an electron jumps from
(a) They do not have same number of neutrons
n = 1 to n = 3, how much energy will be emitted or absorbed
(1erg = 10–7 J) (b) They have dissimilar mass number

(a) 2.15 × 10–11 erg (b) 0.1911 × 10–10 erg (c) They have different energy level schemes

(c) 2.389 × 10–12 erg (d) 0.239 × 10–10 erg (d) They have different number of valence electrons

35. An electron in H-atom is moving with a kinetic energy of 42. The line spectrum observed when electron jumps from
5.45×10–19J. What will be energy level for this electron ? higher level to M level is known as

(a) 1 (b) 2 (a) Balmer series (b) Lyman series

(c) 3 (d) None of these (c) Paschen series (d) Brackett series

36. The energy required to dislodge electron from excited 43. In hydrogen spectrum, the series of lines appearing in
isolated H-atom, IE1 = 13.6 eV is ultra violet region of electromagnetic spectrum are called

(a) = 13.6 eV (b) > 13.6 eV (a) Balmer lines (b) Lyman lines
(c) Pfund lines (d) Brackett lines
(c) < 13.6 and > 3.4 eV (d)  3.4 eV
44. How many spectral lines are produced in the spectrum of
37. The radius of first Bohr’s orbit for hydrogen is 0.53 Å. hydrogen atom from 5th energy level ?
The radius of third Bohr’s orbit would be
(a) 5 (b) 10
(a) 0.79 Å (b) 1.59 Å
(c) 15 (d) 4
(c) 3.18 Å (d) 4.77 Å
45. An electron jumps from 6th energy level to 3rd energy
38. According to Bohr model, angular momentum of an electron level in H-atom, how many lines belong to visible region ?
in the 3rd orbit is :
(a) 1 (b) 2
3h 1.5h (c) 3 (d) Zero
(a) (b)
  46. The wavenumber for the shortest wavelength transition
3 9h in the Balmer series of atomic hydrogen is
(c) (d)
h  (a) 27420 cm–1 (b) 28420 cm–1
39. Electronic energy is a negative energy because (c) 29420 cm–1 (d) 12186 cm–1
(a) Electron carries negative charge. 47. When electrons in N shell of excited hydrogen atom return
to ground state, the number of possible lines spectrum is :
(b) Energy is zero near the nucleus and decreases as the
distance from the nucleus increases. (a) 6 (b) 4
(c) Energy is zero at an infinite distance from the nucleus (c) 2 (d) 3
and decreases as the electron comes closer to the
nucleus.
48. What transition in He+ ion shall have the same wave number 56. Photoelectric effect shows
as the first line in Balmar series of H atom ?
(a) particle-like behaviour of light
(a) 7  5 (b) 5  3
(b) wave-like behaviour of light
(c) 6  4 (d) 4  2
(c) both wave-like and particle-like behaviour of light
49. The difference in wavelength of second and third lines of
(d) neither wave-like nor particle-like behaviour of light
Balmer series in the atomic spectrum is
57. Increase in the frequency of the incident radiations
(a) 131 Å (b) 524 Å
increase the :
(c) 324 Å (d) 262 Å
(a) rate of emission of photo-electrons
50. The third line in Balmer series corresponds to an electronic
(b) work function
transition between which Bohr’s orbits in hydrogen atom
(c) kinetic energy of photo-electrons
(a) 5  3 (b) 5  2
(d) threshold frequency
(c) 4  3 (d) 4  2
58. Ultraviolet light of 6.2 eV falls on aluminium surface (work
51. The wave number of the first line of Balmer series of H
function = 4.2 eV). The kinetic energy (in joule) of the
atom is 15200 cm–1. What is the wave number of the first
fastest electron emitted is approximately :
line of Balmer series of Li2+ ion ?
(a) 3 × 10–21 (b) 3 × 10–19
(a) 15200 cm–1 (b) 6080 cm–1
(c) 3 × 10–17 (d) 3 × 10–15
(c) 76000 cm–1 (d) 136800 cm–1
59. The kinetic energy of the photoelectrons does not depend
52. A certain transition in H spectrum from an excited state to
upon
the ground state in one or more steps gives rise to a total of
10 lines. How many of these belong to the UV spectrum ? (a) Intensity of incident radiation

(a) 3 (b) 4 (b) Frequency of incident radiation

(c) 6 (d) 5 (c) Wavelength of incident radiation

(d) Wave number of incident radiation.
Particle Nature of Electromagnetic Radiations
60. The threshold wavelength for photoelectric effect on
53. Einstein’s theory of photoelectric effect is based on :
sodium is 5000 Å. Its work function is :
(a) Newtons corpuscular theory of light
(a) 4 × 10–19 J (b) 1J
(b) Huygen’s wave theory of light
(c) 2 × 10–19J (d) 3 × 10–10 J
(c) Maxwell’s electromagnetic theory of light
61. If the threshold frequency of a metal for photoelectric effect
(d) Planck’s quantum theory of light is 0, then which of the following will not happen ?
54. In photoelectric effect the number of photo-electrons (a) If the frequency of the incident radiation is 0, the
emitted is proportional to : kinetic energy of the electrons ejected is zero.
(a) intensity of incident beam (b) If the frequency of the incident radiation is , the kinetic
(b) frequency of incident beam energy of the electrons ejected will be h– h0.
(c) velocity of incident beam (c) If the frequency is kept same at  but intensity is
(d) work function of photo cathode increased, the number of electrons ejected will increase.

55. Threshold wavelength depends upon : (d) If the frequency of incident radiation is further
increased, the number of photoelectrons ejected will
(a) frequency of incident radiation
increase.
(b) velocity of electrons
(c) work function
(d) None of the above
Dual Nature of Particle 69. The Heisenbergs Uncertainty Principle states that ........... .
62. The de–Broglie wavelength associated with a material (a) no two electrons in the same atom can have the same
particle is set of four quantum numbers
(a) Directly proportional to its energy (b) two atoms of the same element must have the same
(b) Directly proportional to momentum number of protons
(c) Inversely proportional to its energy (c) it is impossible to determine accurately both the
(d) Inversely proportional to momentum position and momentum of an electron simultaneously

63. The de Broglie wavelength of a tennis ball of mass 66 g (d) electrons of atoms in their ground states enter
moving with the velocity of 10 metres per second is energetically equivalent sets of orbitals singly before
approximately they pair up in any orbital of the set

(a) 10–35 metres (b) 10–33 metres 70. For an electron, if the uncertainty in velocity is v, the
(c) 10–31 metres (d) 10–36 metres uncertainty in its position (x) is given by :
64. The wavelength of a cricket ball weighing 100 g and
h 
travelling with a velocity of 50 m/s is (a) m (b)
2 hm
(a) 1.3 × 10–28 m (b) 1.3 × 10–37 m
(c) 1.3 × 10–34 m (d) 1.3 × 10–30 m h 2m
(c) (d)
65. A cricket ball of 0.5 kg is moving with a velocity of 100 ms–1. 4m h
The wavelength associated with its motion is 71. A ball of mass 200g is moving with a velocity of 10m sec–1.
(a) 1/100 cm (b) 66 × 10–34 m If the error in measurement of velocity is 0.1%, the
uncertainty in its position is :
(c) 1.32 × 10–35 m (d) 6.6 × 10–28 m
(a) 3.3 × 10–31 m (b) 3.3 × 10–27 m
66. An electron with velocity v is found to have a certain
value of de Broglie wavelength. The velocity that the (c) 5.3 × 10–25 m (d) 2.64 × 10–32 m
neutron should possess to have the same de Broglie 72. If uncertainty in the measurement of position and
wavelength is momentum of an electron are equal then uncertainly in the
(a) v (b) v/1840 measurement of its velocity is approximately :
(c) 1840v (d) 1840/v (a) 8 × 1012 m s–1 (b) 6 × 1012 m s–1
67. An electron has kinetic energy 2.8 × 10–23 J de–Broglie (c) 4 × 1012 m s–1 (d) 2 × 1012 m s–1
wavelength will be nearly(me = 9.1 × 10–31 kg) 73. In the Schrodinger’s wave equation  represents
(a) 9.24 × 10–4 m (b) 9.24 × 10–7 m (a) Orbit (b) Wave function
(c) 9.24 × 10–8 m (d) 9.24 × 10–10 m (c) Wave (d) Radial probability
Heisenberg's Uncertainty Principle and 74. The quantum number not obtained from Schrodinger
Schrodinger's Wave Equation equation is

68. If uncertainty in the position of an electron is zero, the (a) n (b) l

uncertainty in its momentum would be (c) m (d) s

h Quantum Numbers
(a) zero (b) 
4 75. For each value of , the number fo ms values are
(a) 2 (b) n
h
(c)  (d) infinite (c) 2 + 1 (d) n – 
4
76. A subshell with n = 6,  = 2 can accommodate a maximum (c) V3+ (d) Fe3+
of 85. Azimuthal quantum number for the last electron in Na atom
(a) 10 electrons (b) 12 electrons is
(c) 36 electrons (d) 72 electrons (a) 1 (b) 0
77. The correct designation of an electron with n = 4, l = 3, (c) 2 (d) 3
m = 2, and s = 1/2 is :
86. Presence of three unpaired electrons in phosphorus atom
(a) 3d (b) 4f can be explained by
(c) 5p (d) 6s (a) Pauli’s rule
78. A 3d-electron having s = +1/2 can have a magnetic quantum (b) Uncertainty principle
no :
(c) Aufbau’s rule
(a) +2 (b) +3 (d) Hund’s rule
(c) –3 (d) +4
87. The quantum numbers for the outermost electron of an
79. Which of the following sets of quantum number is correct element are given below
for an electron in 4f orbital ?
1
(a) n = 3,  = 2, m = –2, s = +1/2 n  2,   0, m  0, ms  
2
(b) n = 4,  = 4, m = –4, s = –1/2
The atom is
(c) n = 4,  = 3, m = +1, s = +1/2
(a) hydrogen (b) lithium
(d) n = 4,  = 3, m = +4, s = +1/2
(c) beryllium (d) boron
80. For a d-electron, the orbital angular momentum is
88. Which electronic configuration does not follow the Pauli’s
(a) 6  h / 2  (b) 2  h / 2  exclusion principle ?

(c) h/2 (d) zero (a) 1s2, 2s2 2p4 (b) 1s2, 2s2 2p4, 3s2
(c) 1s2, 2p4 (d) 1s2, 2s2 2p6, 3s3
Electronic Configuration
89. Magnetic quantum number for the last electron in sodium
81. According to (n + l) rule after completing ‘np’ level the
is :
electron enters to :
(a) 3 (b) 1
(a) (n – 1) d (b) (n + 1) s
(c) 2 (d) zero
(c) nd (d) (n + 1) p
90. How many spherical nodes are present in 4s orbital in a
82. The correct ground state electronic configuration of
hydrogen atom ?
chromium atom (Z = 24) is
(a) 0 (b) 2
(a) [Ar] 3d5 4s1 (b) [Ar] 3d4 4s2
(c) 3 (d) 4
(c) [Ar] 3d6 4s0 (d) [Ar] 4s1 4p5
91. The number of nodes possible in radial wave function of
83. In manganese atom, Mn (Z = 25), the total number of
3d orbital is
orbitals populated by one or more electrons (in ground
state) is (a) 1 (b) 2

(a) 15 (b) 14 (c) 0 (d) 3

(c) 12 (d) 10 92. The d-orbital with the orientation along X and Y axes is
called :
84. Which of the following has maximum number of unpaired
electrons ? (a) d z2 (b) d zy
(a) Mg 2+
(b) Ti 3+

(c) d yz (d) d x 2 y2

93. Which d-orbital does not have four lobes ? 100. For which one of the following sets of four quantum
numbers an electron will have the highest energy
(a) d x 2 y2 (b) d xy
n  m s
(c) d z2 (d) d xz (a) 3 2 1 1/2
(b) 4 1 0 –1/2
94. The number of angular nodes in a 3s atomic orbital is
(c) 4 2 –1 1/2
(a) 0 (b) 1
(d) 5 0 0 –1/2
(c) 2 (d) 3
95. The number of radial nodes in a 3s atomic orbital is 101. Nitrogen has the electronic configuration 1s 2 ,2s2 2p1x
(a) 0 (b) 1
2p1y 2p1z and not 1s 2 , 2s 2 2p x2 2p1y 2p 0z . It was proposed by :
(c) 2 (d) 3
(a) Aufbau principle
96. The electrons identified by n and l
(b) Pauli’s exclusion principle
(i) n = 4, l = 1 (ii) n = 4, l = 0
(c) Hund’s rule
(iii) n = 3, l = 2 (iv) n = 3, l = 1
(d) Uncertainty principle
can be placed in order of increasing energy, from lowest to
highest 102. Which of the following has maximum number of unpaired
electron (atomic number of Fe 26)
(a) (iv) < (ii) < (iii) < (i) (b) (ii) < (iv) < (i) < (iii)
(a) Fe (b) Fe (II)
(c) (i) < (iii) < (ii) < (iv) (d) (iii) < (i) < (iv) < (ii)
(c) Fe (III) (d) Fe (IV)
97. The correct set of quantum numbers for the unpaired
electron of chlorine atom is 103. The total spin and magnetic moment for the atom with
atomic number 7 are
n  m
(a) 2 1 0 (a)  3, 3 BM (b)  1, 8 BM
(b) 2 1 1
3
(c) 3 1 1 (c)  , 15 BM (d) 0, 8 BM
2
(d) 3 0 0
104. 3py orbital has .......... nodal plane
98. The maximum number of 4d-electrons having spin quantum
(a) XY (b) YZ
1
number s   are (c) ZX (d) Any
2
105. A compound of vanadium has magnetic moment of 1.73
(a) 10 (b) 7 B.M. The electronic configuration of the vanadium ion in
(c) 1 (d) 5 the compound is
99. Consider the following ions (a) [Ar] 4s03d1 (b) [Ar] 4s13d0
1. Ni2+ 2. Co3+ 3. Cr2+ 4. Fe3+ (c) [Ar] 4s23d0 (d) [Ar] 4s03d3
(Atomic number : Cr = 24, Fe = 26, Co = 27 and Ni = 28)
The correct sequence of increasing number of unpaired
electrons in these ions is
(a) 1, 2, 3, 4 (b) 4, 2, 3, 1
(c) 1, 3, 2, 4 (d) 3, 4, 2, 1
EXERCISE - 2 : PREVIOUS YEAR JEE MAINS QUESTIONS
1. In Bohr series of lines of hydrogen spectrum, the third line 7. Of the following sets which one does not contain
from the red end corresponds to which one of the following isoelectronic species ? (2005)
inner-orbit jumps of the electron for Bohr orbits in an atom
(a) BO33 , CO32  , NO3 (b) SO32  , CO 32  , NO 3
of hydrogen ? (2003)
(a) 3  2 (b) 5  2 (c) CN  , N 2 , C 22  (d) PO 34 , SO 42  , ClO 4
(c) 4  1 (d) 2  5
8. Uncertainty in the position of an electron
2 The wavelength of the radiation emitted, when in a (mass = 9.1 × 10–31 kg) moving with a velocity 300 ms–1,
hydrogen atom electron falls from infinity to stationary Accurate upto 0.001%, will be (h = 6.63 × 10–34 Js) (2006)
state 1, would be (Rydberg constant = 1.097 × 107 m–1)
(a) 19.2 × 10–2 m (b) 5.76 × 10–2 m
(2004)
(c) 1.92 × 10–2 m (d) 3.84 × 10–2 m
(a) 91 nm (b) 192 nm
9. According to Bohr’s theory, the angular momentum of an
(c) 406 nm (d) 9.1 × 10–8 nm
electron in 5th orbit is (2006)
3. Which of the following sets of quantum numbers is correct
for an electron in 4f orbital ? (2004) h h
(a) 25 (b) 1.0
(a) n = 4, l = 3, m = + 4, s = + 1/2  

(b) n = 4, l = 4, m = – 4, s = – 1/2
h h
(c) n = 4, l = 3, m = + 1, s = + 1/2 (c) 10 (d) 2.5
 
(d) n = 3, l = 2, m = – 2, s = + 1/2
10. Which of the following sets of quantum numbers represents
4. Consider the ground state of Cr atom (Z = 24). The numbers the highest energy of an atom ? (2007)
of electrons with the azimuthal quantum numbers, l = 1 and
(a) n = 3, l = 1, m = 1, s = + 1/2
2 are, respectively (2004)
(b) n = 3, l = 2, m = 1, s = + 1/2
(a) 12 and 4 (b) 12 and 5
(c) n = 4, l = 0, m = 0, s = + 1/2
(c) 16 and 4 (d) 16 and 5
(d) n = 3, l = 0, m = 0, s = + 1/2
5. Which of the following statements in relation to the
hydrogen atom is correct ? (2005) 11. The ionization enthalpy of hydrogen atom is 1.312 × 106 J
mol–1. The energy required to excite the electron in the
(a) 3s, 3p and 3d orbitals all have the same energy
atom from n = 1 to n = 2 is (2008)
(b) 3s and 3p orbitals are of lower energy than 3d orbital
(a) 8.51 × 105 J mol–1 (b) 6.56 × 105 J mol–1
(c) 3p orbital is lower in energy than 3d orbital
(c)7.56 × 105 J mol–1 (d) 9.84 × 105 J mol–1
(d) 3s orbital is lower in energy than 3p orbital
12. In an atom, an electron is moving with a speed of 600 m/s
6. In a multi electron atom, which of the following orbitals with an uncertainity of 0.005%, the position of the electron
described by the three quantum numbers will have the can be (h = 6.6 × 10 –34 kg m2s –1, mass of electron
same energy in the absence of magnetic and electric fields ? em = 9.1 × 10–31 kg) (2009)
(A) n = 1, l = 0 m = 0 (a) 1.52 × 10–4 m (b) 5.01 × 10–3 m
(B) n = 2, l = 0, m = 0 (c) 1.92 × 10–3 m (d) 3.84 × 10–3 m
(C) n = 2, l = 1, m = 1 13. Calculate the wavelength (in nanometer) associated with a
(D) n = 3, l = 2, m = 1 proton moving at 1.0 × 103 ms–1 (Mass of proton =
1.67 × 10–27 kg and h = 6.63 × 10–34 Js) (2009)
(E) n = 3, l = 2, m = 0 (2005)
(a) 0.032 nm (b) 0.40 nm
(a) (D) and (E) (b) (C) and (D)
(c) 2.5 nm (d) 14.0 nm
(c) (B) and (C) (d) (A) and (B)
14. The energy required to break one mole of Cl–Cl bonds in 21. The energy of an electron in first Bohr orbit of H-atom is
Cl2 is 242 kJ mol–1. The longest wavelength of light capable –13.6 eV. The energy value of electron in the excited state
of breaking a single Cl–Cl bond is (2010) Li2+ is: (Online 2014 SET-1)
(a) 594 nm (b) 640 nm (a) –27.2 eV (b) 30.6 eV
(c) 700 nm (d) 494 nm (c) – 30.6 eV (d) 27.2 eV
15. Ionisation energy of He+ is 19.6 × 10–18 J atom–1. The energy 22. Chloro compound of vanadium has only spin magnetic
of the first stationary state (n = 1) of Li2+ is (2010) moment of 1.73 BM. This vanadium chloride has the
(a) 4.41 × 10–16 J atom–1 (b) – 4.41 × 10–17 J atom–1 formula: (at. no. of V = 23) (Online 2014 SET-1)
(c) – 2.2 × 10 J atom
–15 –1
(d) 8.82 × 10–17 J atom–1 (a) VCl2 (b) VCl4
16. A gas absorbs photon of 355 nm and emits at two wave- (c) VCl3 (d) VCl5
lengths. If one of the emission is at 680 nm, the other is at 23. Based on the equation:
(a) 1035 nm (b) 325 nm
1 1
(c) 743 nm (d) 518 nm
E = – 2.0 ×10–18J  2
 2  the wavelength of the light
17. The frequency of light emitted for the transition n = 4 to  n2 n1 
n = 2 of He+ is equal to the transition in H atom corresponding that must be absorbed to excite hydrogen electron from
to which of the following ? (2011) level n= 1 to level n = 2 will be (h = 6.625 × 10–34 Js,
(a) n = 3 to n = 1 (b) n = 2 to n = 1 c = 3 × 108 ms–1) (Online 2014 SET-2)
(c) n = 3 to n = 2 (d) n = 4 to n = 3 (a) 2.650 × 10–7 m (b) 1.325 × 10–7 m
18. The electrons identified by quantum numbers n and l (c) 5.300 × 10 m–10
(d) 1.325 × 10–10 m
(2012) 24. If 0 and  be the threshold wavelength and wavelength of
incident light, the velocity of photoelectron ejected from
(1) n = 4, l = 1 (2) n = 4, l = 0
the metal surface is: (Online 2014 SET-2)
(3) n = 3, l = 2 (4) n = 3, l = 1
can be placed in order of increasing energy as 2h  1 1  2h
(a)    (b)  0   
(a) (3) < (4) < (2) < (1) (b) (4) < (2) < (3) < (1) m  0   m
(c) (2) < (4) < (1) < (3) (d) (1) < (3) < (2) < (4)
19. Energy of an electron is given by 2hc 2hc   0   
(c) 0    (d) 
m  0 

m
 Z2 
E  2.178  10 18 J  2 
n  25. The de-Broglie wavelength of a particle of mass 6.63 g
moving with a velocity of 100 ms–1 is:
Wavelength of light required to excite an electron in an
hydrogen atom from level n = 1 to n = 2 will be (2013) (Online 2014 SET-3)
(h = 6.62 × 10 –34
Js and c = 3.0 × 10 ms )
8 –1 (a) 10–33
m (b) 10–35
m
(a) 1.214 × 10 m
–7
(b) 2.816 × 10 m –7 (c) 10–25
m (d) 10–31
m
(c) 6.500 × 10 m
–7
(d) 8.500 × 10 m –7 26. Excited hydrogen atom emits light in the ultraviolet
region at 2.47 × 10 15 Hz. With this frequency, the
20 The correct set of four quantum numbers for the valence
energy of a single photon is:
electrons of rubidium atom (Z=37) is : (2014)
(h = 6.63 × 10–34 Js) (Online 2014 SET-3)
1 1
(a) 5,1,0,  (b) 5,1,1,  (a) 8.041 × 10–40 J (b) 6.111 × 10–17 J
2 2
(c) 2.680 × 10–19 J (d) 1.640 × 10–18 J
1 1
(c) 5,0,1,  (d) 5,0,0, 
2 2
27. If m and e are the mass and charge of the revolving 32. The total number of orbitals associated with the principal
electron in the orbit of radius r for hydrogen atom, the quantum number 5 is (Online 2016 SET-1)
total energy of the revolving electron will be: (a) 5 (b) 10
(Online 2014 SET-3) (c) 20 (d) 25
me2 e2 33. Aqueous solution of which salt will not contain ions
(a) (b)  with the electronic configuration 1s22s22p63s23p6 ?
r r
(Online 2016 SET-2)
1 e2 1 e2 (a) NaF (b) NaCl
(c) (d) 
2 r 2 r
(c) KBr (d) CaI2
28. Ionization energy of gaseous Na atoms is 495.5 kJ mol–1. 34. The radius of the second Bohr orbit for hydrogen atom is:
The lowest possible frequency of light that (2017)
ionizes a sodium atom is (h = 6.626 × 10 –34 Js,
NA = 6.022 × 1023 mol–1) (Online 2014 SET-4) (Planck’s Const. h = 6.6262 × 10–34 Js;
Mass of electron = 9.1091 × 10–31 kg;
(a) 7.50 × 104 s–1 (b) 4.76 × 1014 s–1
Charge of electron e = 1.60210 × 10–19 C;
(c) 1.24 × 1015 s–1 (d) 3.15 × 1015 s–1
29. Which of the following is the energy of a possible excited Permittivity of vacuum
state of hydrogen ? (2015) 0  8.854185  1012 kg 1m3 A 2 )
(a) – 3.4 eV (b) + 6.8 eV
(a) 4.76 Å (b) 0.529 Å
(c) + 13.6 eV (d) – 6.8 eV
(c) 2.12 Å (d) 1.65 Å
30. At temperature T, the average kinetic energy of any
35. If the shortest wavelength in Lyman series of hydrogen
3 atom is A, then the longest wavelength in Paschen series
particle kT. The de Broglie wavelength follows the
2 of He+ is (Online 2017 SET-1)
order : (Online 2015 SET-1)
5A 9A
(a) Visible photon > Thermal neutron > Thermal (a) (b)
9 5
electron
(b) Thermal proton > Thermal electron > Visible 36A 36A
photon (c) (d)
5 7
(c) Thermal proton > Visible photon > Thermal
36. The electron in the hydrogen atom undergoes transition
electron
from higher orbitals to orbital of radius 211.6 pm. This
(d) Visible photon > Thermal electron > Thermal transition is associated with : (Online 2017 SET-2)
neutron
(a) Lyman series
31. A stream of electrons from a heated filament was passed
(b) Balmer series
between two charged plates kept at a potential difference
V esu. If e and m are charge and mass of an electron, (c) Paschen series
respectively, then the value of h/ (where  is wavelength (d) Brackett series
associated with electron wave) is given by: 37. Ejection of the photoelectron from metal in the
(2016) photoelectric effect experiment can be stopped by
applying 0.5 V when the radiation of 250 nm is used.
(a) 2meV (b) meV The work function of the metal is :
(c) 2meV (d) meV (Online 2018 SET-1)
(a) 4 eV (b) 4.5 eV
(c) 5 eV (d) 5.5 eV
38. The de-Broglie’s wavelength of electron present in first (c) 1, 3 (d) 3, 4
Bohr orbit of ‘H’ atom is : (Online 2018 SET-2)
42. Which of the graphs shown below does not represent
(a) 0.529 Å the relationship between incident light and the electron
ejected from metal surface?
(b) 2  0.529Å
(JEE Main 10-01-2019 Shift-I)
0.529
(c) Å
2

(d) 4  0.529Å
39. Which of the following statements is false ? (a)

(Online 2018 SET-3)

(a) Photon has momentum as well as wavelength.
(b) Splitting of spectral lines in electrical field is called
Stark effect.
(c) Rydberg constant has unit of energy.
(b)
(d) Frequency of emitted radiation from a black body
goes from a lower wavelength to higher wavelength as
the temperature increases.
40. For emission line of atomic hydrogen from ni = 8 to nf = n,

 1 
to the plot of wave number    against  2  will be
n 
(c)
(The Rydberg constant, RH is in wave number unit)
(JEE Main 09-01-2019 Shift-I)
(a) Linear with intercept -RH
(b) Non linear
(c) Linear with slope RH
(d) Linear with slope -RH
(d)
41. Which of the following combination of statements is true
regarding the interpretation of the atomic orbitals?
1. An electron in an orbital of high angular momentum 43. The ground state energy of hydrogen atom is –13.6 eV.
stays away from the nucleus than an electron in the The energy of second excited state of He+ ion in eV is:
orbital of lower angular momentum. (JEE Main 10-01-2019 Shift-II)
2. For a given value of the principal quantum number, (a) –54.4 (b) –3.4
the size of the orbit is inversely proportional to the
(c) –6.04 (d) –27.2
azimuthal quantum number.
44. Heat treatment of muscular pain involves radiation of wave-
3. According to wave mechanics, the ground state an-
length of about 900 nm. Which spectral line of H-atom is
h suitable for this purpose?
gular momentum is equal to
2 (JEE Main 11-01-2019 Shift-I)
4. The plot of  vs r for various azimuthal quantum [RH = 1 × 10 cm , h = 6.6 × 10–34 Js,
5 –1

numbers, shows peak shifting towards higher r value. c = 3 × 108 ms–1]

(JEE Main 09-01-2019 Shift-I) (a) Paschen,   3 (b) Paschen, 5  3
(a) 1, 4 (b) 1, 2
(c) Balmer,   2 (d)   1
 50. If p is the momentum of the fastest electron ejected from
45. The de Broglie wavelength () associated with a photo-
a metal surface after the irradiation of light having wave-
electron varies with the frequency (v) of the incident ra- length  , then for 1.5p momentum of the photoelectron,
diation as, [v0 is threshold frequency]
the wavelength of the light should be:
(JEE Main 11-01-2019 Shift-II)
(Assume kinetic energy of ejected photoelectron to be
1 1 very high in comparison to work function):
  (JEE Main 08-04-2019 shift-II)
(a) 5
(b) 1

(v v 0 ) 2
(v v 0 ) 4
3 1
(a)  (b) 
1 1 4 2
 
(c) 3
(d) 1

(v v 0 ) 2 (v v 0 ) 2 3 4
(c)  (d) 
2 9
46. What is the work function of the metal if the light of wave-
length 4000Å generates photoelectrons of velocity 6 × 51. For any given series of spectral lines of atomic hydrogen,
105 ms–1 from it?
let v  vmax  v min be the difference in maximum and mini-
(Mass of electron = 9 × 10–31 kg
mum frequencies in cm–1. The ratio v Lyman / vBalmer is:
Velocity of light = 3 × 108 ms–1
(JEE Main 09-04-2019 Shift-I)
Planck’s constant = 6.626 × 10–34 Js
(a) 4: 1 (b) 9: 4
Charge of electron = 1.6 × 10–19 Je V–1)
(JEE Main 12-01-2019 Shift-I) (c) 5: 4 (d) 27: 5

(a) 0.9 eV (b) 3.1 eV 52. Which one of the following about an electron occupying
the 1s orbital in a hydrogen atom is incorrect? (The Bohr
(c) 2.1 eV (d) 4.0 eV radius is represented by a0).
47. If the de Broglie wavelength of the electron in nth Bohr (JEE Main 09-04-2019 Shift-II)
orbit in a hydrogenic atom is equal to 1.5  a0 (a0 is Bohr (a) The probability density of finding the electron is maxi-
radius), then the value of n/z is mum at the nucleus.
(JEE Main 12-01-2019 Shift-II) (b) The electron can be found at a distance 2a0 from the
(a) 0.40 (b) 1.50 nucleus.
(c) 1.0 (d) 0.75 (c) The magnitude of the potential energy is double that
48. The size of the isoelectronic species Cl , Ar and Ca is
– 2+ of its kinetic energy on an
affected by: (JEE Main 08-04-2019 Shift-I) (d) The total energy of the electron is maximum when it is
(a) azimuthal quantum number of valence shell at a distance from the nucleus.
(b) electron-electron interaction in the outer orbitals 53.
2
The graph between  and r (radial distance) is shown
(c) principal quantum number of valence shell below. This represents: (JEE Main 10-04-2019 Shift-I)
(d) nuclear charge
49. The quantum number of four electrons are given below:
I. n = 4, l = 2 , m1 = –2, ms = –1/2
II. n = 3, l = 2, m1 = –2, ms = +1/2
III. n = 4, l = 1 , m1 = 0, ms = +1/2
IV. n = 3, l = 1 , m1 = 1, ms = –1/2
The correct order of their increasing energies will be:
(JEE Main 08-04-2019 Shift-I)
(a) IV < III < II < I (b) I < II < III < IV (a) 3s orbital (b) 2s orbital
(c) IV < II < III < I (d) I < III < II < IV (c) 1s orbital (d) 2p orbital
54. The isoelectronic set of ions is : A. The integer n1 = 2.
(JEE Main 10-04-2019 Shift-I) B. The ionization energy of hydrogen can be calculated
(a) N3–, O2–, F– and Na+ from the wave number of these lines.
(b) N3–, Li+, Mg2+ and O2– C. The lines of longest wavelength corresponds to n =2
(c) F–, Li+, Na+ and Mg2+ to n = 3.
(d) Li+, Na+, O2– and F– D. As wavelength decreases, the lines of the series con-
verge. (JEE Main 07-01-2020 Shift-II)
55. The ratio of the shortest wavelength of two spectral se-
ries of hydrogen spectrum is found to be about 9. The (a) B, C, D (b) A, B, D
spectral series are: (JEE Main 10-04-2019 Shift-II) (c) A, C, D (d) A, B, C
(a) Lyman and Paschen 59. The radius of second Bohr orbit, in terms of the Bohr
(b) Balmer and Brackett radius, a0, in Li2+ is: (JEE Main 08-01-2020 Shift-II)
(c) Brackett and Pfund 2a 0 4a 0
(d) Paschen and Pfund (a) (b)
3 9
56. The electrons are more likely to be found:
4a 0 2a 0
(c) (d)
3 9
60. The de Broglie wavelength of an electron in the 4th Bohr
orbit is: (JEE Main 09-01-2020Shift-I)
(JEE Main 12-04-2019 Shift-I)
(a) 4a 0 (b) 2a 0

(a) in the region a and c (c) 8a 0 (d) 6a 0

(b) in the region a and b 61. The figure that is not a direct manifestation of the quan-
(c) only in the region a tum nature of atoms is : (JEE Main 02-09-2020 Shift-I)

(d) only in the region c

57. Amongst the following statements, that which was not
proposed by Dalton was: (a)
(JEE Main 07-01-2020 Shift-I)
(a) Matter consists of indivisible atoms.
(b) When gases combine or reproduced in a chemical
reaction they do so in a simple ratio by volume provided
(b)
all gases are at the same T & P.
(c) Chemical reactions involve reorganisation of atoms.
These are neither created nor destroyed in a chemical
reaction.
(d) All the atoms of a given element have identical prop-
erties including identical mass. Atoms of different ele- (c)
ments differ in mass.
58. For the Balmer series in the spectrum of H-atom,
The correct statements among (A) to (D) are:

1 1 (d)
v  RH  2  2 
n
 1 n 2 
62. The number of subshells associated with n = 4 and
(c) 6s, 6p, 6d, 6f
m= –2 quantum numbers is :
(d) 6s, 5f, 6d, 6p
(JEE Main 02-09-2020 Shift-II)
68. The difference between the radii of 3rd and 4th orbits of
(a) 4 (b) 8
Li2+ is R 1 . The difference between the radii of 3rd and
(c) 2 (d) 16
4th orbits of He+ is R 2 Ratio R 1 : R 2 is :
63. The work function of sodium metal is 4.41 × 10–19J. If pho-
tons of wavelength 300 nm are incident on the metal, the (JEE Main 05-09-2020 shift-I)
kinetic energy of the ejected electrons will be (h = 6.63 × (a) 8 : 3 (b) 3 : 8
10–34 J s; c = 3 × 108 m/s) …………. × 10–21 J. (c) 3 : 2 (d) 2 : 3
(JEE Main 02-09-2020 Shift-II)
69. The correct statement about probability density (except
64. Consider the hypothetical situation where the azimuthal
at infine distance from nucleus) is :
quantum number, l, takes values 0, 1, 2, ………. n + 1, (JEE Main 05-09-2020 Shift-II)
where n is the principal quantum number. Then, the ele-
(a) It can be zero for 3p orbital
ment with atomic number :
(JEE Main 03-09-2020 Shift-II) (b) It can be zero for 1s orbital
(a) 13 has a half-filled valence subshell (c) It can never be zero for 2s orbital
(b) 9 is the first alkali metal (d) It can negative for 2p orbital
(c) 8 is the first noble gas 70. According. Bohr's atomic theory (2021-02-24-Shift-II)
(d) 6 has a 2p-valence subshell
Z2
65. The region in the electromagnetic spectrum where the (A) Kinetic energy of electron is 
n2
Balmer series lines appear is:
(JEE Main 04-09-2020 Shift-I) (B) The product of velocity (v) of electron and principal
(a) Microwave quantum number (n), ' vn '  Z 2 .
(b) Infrared
Z3
(c) Ultraviolet (C) Frequency of revolution of electron in an orbit is  .
n3
(d) Visible
66. The shortest wavelength of H atom in the Lyman series is Z3
 . The longest wavelength in the Balmer series of He + (D) Coulombic force of attraction on the electron is .
n4
is : (JEE Main 04-09-2020 Shift-II) Choose the most appropriate answer from the options given
below
91 21
(a) (b) (a) (A) only
5 5
(b) (C) only
361 51 (c) (A), (C) and (D) only
(c) (d)
4 9
(d) (A) and (D) only
67. In the sixth period, the orbitals that are filled are :
(JEE Main 05-09-2020 Shift-I)
(a) 6s, 5d, 5f, 6p
(b) 6s, 4f, 5d, 6p
71. The plots radial distribution functions for various orbitals splitting of spectral lines in the presence of a magnetic
of hydrogen atom against ‘r’ are given below field.
(2021-02-22-Shift-I) In the light of the above statements, choose the most
appropriate answer from the options given below
(2021-03-18-Shift-II)
(a) Both statement I and statement II are false.

(A) (b) Both statement I and statement II are true.

(c) Statement I is false but statement II is true.
(d) Statement I is true but statement II is false.
75. If the Thomson model of the atom was correct, then the
result of Rutherford's gold foil experiment would have been:
(2021-07-27-Shift-II)

(B) (a) All of the  particles pass through the gold foil without
decrease in speed
(b)  particles are deflected over a wide range of angles.
(c) All particles get bounced back by 180°
(d)  particles pass through the gold foil deflected by
small angles and with reduced speed
76. Given below are two statements: (2021-07-27-Shift-I)
(C)
Statement I : Rutherford's gold foil experiment cannot
explain the line spectrum of hydrogen atom.
Statement II: Bohr's model of hydrogen atom contradicts
Heisenberg's uncertainty principles.
In the light of the above statements, choose the most
appropriate answer from the options given below
(D)
(a) Statement I is false, but Statement II is true.
(b) Statement I is true, but Statement II is false.

The correct plot for 3s orbital is (c) Both Statement I and Statement II are false

(a) (B) (b) (A) (d) Both Statement I and Statement II are true

(c) (D) (d) (C) 77. Given below are two statements. (2021-08-26-Shift-I)

72. The orbital having two radial as well as two angular nodes Statement I: According to Bohr's model of an atom,
is: (2021-02-26-Shift-I) qualitatively the magnitude of velocity of electron increase
with decrease in positive charges on the nucleus as there
(a) 3p (b) 5d
is no strong hold on the electron by the nucleus.
(c) 4f (d) 4d
Statement II: According to Bohr's model of an atom,
73. A certain orbital has no angular nodes and two radial nodes. qualitatively the magnitude of velocity of electron increases
The orbital is: (2021-03-18-Shift-I) with decrease in principal quantum number.
(a) 2s (b) 3s In the light of the above statements, choose the most
(c) 2p (d) 3p appropriate answer from the options given below:
74. Given below are two statements: (a) Both Statement I and Statement II are false
Statement I: Bohr’s theory accounts for the stability and (b) Both Statement I and Statement II are true.
line spectrum of Li + ion. (c) Statement I is false, but Statement II is true.
Statement II: Bohr’s theory was unable to explain the (d) Statement I is true, but Statement II is true.
78. Identify the element for which electronic configuration in 88. A source of monochromatic radiation of wavelength 400
+3 oxidation state is [Ar]3d5: (2021-09-01-Shift-II) nm provides 1000 J of energy in 10 seconds. When this
(a) Ru (b) Mn radiation falls on the surface of sodium, x × 1020 electrons
(c) Co (d) Fe are ejected per second. Assume that wavelength 400 nm is
79. The azimuthal quantum number for the valence electrons sufficient for ejection of electron from the surface of sodium
of Ga+ ion is______. (Atomic number of Ga = 31) metal. The value of x is _____. (Nearest interger)
(2021-07-20-Shift-I) (2021-07-25-Shift-I)
80. A proton and a Li3+ nucleus is accelerated by the same
 h  6.626 10 –34
Js 
potential. If Li and  p denote the de Broglie wavelengths
89. An accelerated electron has a speed of 5 × 106 m s-1 with an
 Li uncertainty of 0.02%. The uncertainty in finding its location
of Li3+ and proton respectively, then the value  is while in motion is x × 10-9 m. The value of x is __.
p
(2021-07-25-Shift-II)
x 10 1 . The value of x is _________. (Rounded off to [Nearest integer]
the nearest integer) [Use mass of electron = 9.1 × 10-31 kg, h = 6.63 ×10-34Js, =
[Mass of Li = 8.3 mass of proton] [2021-02-24-Shift-I]
3+ 3.14]
81. The spin only magnetic moment of a divalent ion in an 90. A metal surface is exposed to 500 nm radiation. The
aqueous solution (atomic number 29) is (round-off)___ threshold frequency of the metal for photoelectric current
BM. (2021-02-25-Shift-II) is 4.3 × 1014 Hz. The velocity of ejected electron is ___ x 105
82. A ball weighing 10g is moving with a velocity of 90 ms-1. If ms-1 (Nearest integer) (2021-08-26-Shift-II)
the uncertainty in its velocity is 5%, then the uncertainty  Use : h = 6.63×10 –34 Js, me =9.0×10 –31 kg 
33
 
in its position is ________ 10 m .(Rounded off to the
91. The kinetic energy of an electron in the second Bohr orbit
nearest integer) (2021-02-26-Shift-II)
h2
Given : h  6.63  1034 Js  of a hydrogen atom is equal to . The value of 10x is
  xma02
83. When light of wavelength 248 nm falls on a metal of ____. (Nearest integer) (a0 is radius of Bohr’s orbit)
threshold energy 3.0 eV, the de-Broglie wavelength of (2021-08-27-Shift-I)

emitted electrons is ______ A . [2021-03-16-Shift-I]  Given : π  3.14
Use : 3  1.73, h  6.626  1034 js  92. The number of photons emitted by a monochromatic (single
  frequency) infrared range finder of power 1 mW and
m e  9.1 1031 kg; c  3.0  108 ms 1 ;1eV  1.6  1019 J  wavelength of 1000 nm, in 0.1 second is x × 1013. The value
of x is _____ . (Nearest integer) (2021-08-27-Shift-II)
84. The number of orbitals with n = 5, m1 = +2 is (Round off to (h = 6.63 × 10-34 Js, c = 3.00 × 108 ms-1)
the Nearest interger). [2021-03-16-Shift-II] 93. Ge (Z =32) in its ground state electronic configuration has
85. A certain orbital has n = 4 and ml = –3. The number of radial x completely filled orbitals with m1 = 0. The value of x is__.
nodes in this orbital is ........... (Round off to the Nearest (2021-08-31-Shift-I)
Integer). (2021-03-17-Shift-I) 94. The value of magnetic quantum number of the outermost
86. The wavelength of electrons accelerated from rest through electron of Zn+ ion is _____. (2021-08-31-Shift-II)
a potential difference of 40 kV is x × 10-12 m. The value of x 95. A 50 watt bulb emits monochromatic red light of wavelength
is______. (Nearest integer) (2021-07-20-Shift-II) of 795 nm. The number of photons emitted per second by
Given : Mass of electron = 9.1 × 10–31 kg the bulb is x × 1020. The value of x is _____. (nearest integer)
Charge on an electron = 1.6 × 10–19 C (2021-09-01-Shift-II)
Planck’s constant = 6.63 ×10–34 Js
Given : h  6.63  1034 Js and c = 3.0  108 ms -1 
87. Number of electrons that Vanadium (Z = 23) has in p-orbitals  
is equal to _____ (2021-07-22-Shift-II)
EXERCISE - 3 : ADVANCED OBJECTIVE QUESTIONS
Objective Questions I [Only one correct option] 9. In any subshell, the maximum number of electrons having
1. The minimum real charge on of any particle, which can exist same value of spin quantum number is
is :
(a)     1 (b)  + 2
(a) 1.6 × 10–19 coulomb (b) 1.6 × 10–10 coulomb
(c) 4.8 × 10–10 coulomb (c) zero (c) 2 + 1 (d) 4 + 2

2. The ratio of e/m, i.e., specific charge for a cathode ray : 10. The ratio of specific charge (e/m) of an electron to that of a
hydrogen ion is :
(a) has the smallest value when the discharge tube is filled
with H2 (a) 1 : 1 (b) 1840 : 1

(b) is constant (c) 1 : 1840 (d) 2 : 1

(c) varies with the atomic number of gas in the discharge 11. The nucleus of an atom can be assumed to be spherical.
tube The radius of the nucleus of mass number A is given by 1.25
× 10–13 × A1/3 cm Radius of atom is one Å. If the mass number
(d) varies with the atomic number of an element forming the
is 64, then the fraction of the atomic volume that is occupied
cathode
by the nucleus is
3. Which are isoelectronic with each other ?
(a) 1.0 × 10–3 (b) 5.0 × 10–5
(a) Na+ and Ne (b) K+ and O
(c) 2.5 × 10–2 (d) 1.25 × 10–13
(c) Ne and O (d) Na+ and K+
12. Li2+ and Be3+ are :
4. Naturally occurring elements are mixtures of :
(a) isotopes (b) isomers
(a) isotone (b) isobars
(c) isobars (d) isoelectronic
(c) isotopes (d) isomers
13. The mass number of three isotopes of an element are 11, 12,
5. Suppose 10–17J of light energy is needed by the interior of and 13 units. Their percentage abundances 80, 15, and 5,
human eye to see an object. The photons of green light (= respectively. What is the atomic weight of the element ?
550 nm) needed to see the object are :
(a) 11.25 (b) 20
(a) 27 (b) 28
(c) 16 (d) 10
(c) 29 (d) 30
14. The given diagram indicates the energy levels of a certain
6. The approximate quantum number of a circular orbit of atom. When the system moves from 2E level to E level, a
diameter, 20.6 nm of the hydrogen atom according to Bohr’s photon of wavelength  is emitted. The wavelength of
theory is :
4E
(a) 10 (b) 14 photon produced during the transition from to E level
3
(c) 12 (d) 16 is :
7. As electron moves away from the nucleus, its potential
energy
(a) Increases (b) Decreases
(c) Remains constant (d) None of these
8. In a sample of H-atoms, electrons make transitions from n =
5 to n = 1. If all the spectral lines are observed, then the line  3
(a) (b)
having the 3rd highest energy will correspond to 3 4
(a) 5  3 (b) 4  1
4
(c) 3  1 (d) 5  4 (c) (d) 3
3
15. A photon of 300 nm is absorbed by a gas and then re-emits 21. Photoelectron emission is observed for three different metals
two photons. One re-emitted photon has wavelength 496 A, B and C. The kinetic energy of the fastest photoelectrons
nm, the wavelength of second re-emitted photon is : versus frequency ‘’ is plotted for each metal. Which of the
(a) 759 (b) 857 following graph shows the phenomenon correctly ?
(c) 957 (d) 657
16. 4000 Å photon is used to break the iodine molecule,then the
% of energy converted to the K.E. of iodine atoms if bond
dissociation energy of I 2 molecule is 246.5 kJ/mol (a) (b)

(a) 8% (b) 12%

(c) 17% (d) 25%
17. Consider a 20 W light source that emits monochromatic
light of wavelength 600 nm. The number of photons ejected
per second in the form of Avogadro’s constant NAV is (c) (d)
approximately :
(a) NAV (b) 10–2NAV
(c) 10–4 NAV (d) 10–6 NAV 22. Which of the following electronic transition in a hydrogen
atom will require the largest amount of energy
18. The work function of a metal is 4.2 eV. If radiations of 2000 Å
fall on the metal. then the kinetic energy of the fastest (a) From n = 1 to n =2 (b) From n = 2 to n=3
photoelectron is (c) From n   to n  1 (d) From n = 3 to n = 5
(a) 1.6 × 10 –19
J (b) 16 × 10 J
10
23. If the total energy of an electron in the 1st shell of H
(c) 3.2 × 10 –19
J (d) 6.4 × 10 –10
J atom = 0.0 eV then its potential energy in the 1st excited state
19. If a certain metal was irradiated by using two different light would be
radiations of frequency ‘x’ and ‘2x’, the maximum kinetic (a) + 6.8 eV (b) + 20.4 eV
energy of the ejected electrons are ‘y’ and ‘3y’ respectively. (c) – 6.8 eV (d) + 3.4 eV
The threshold frequency of the metal will be : 24. Ratio of frequency of revolution of electron in the 2nd excited
(a) x/3 (b) x/2 state of He+ and 2nd state of hydrogen is
(c) 3x/2 (d) 2x/3 (a) 32/27 (b) 27/32
20. Which of the correct graphical representation based on (c) 1/54 (d) 27/2
photoelectric effect (assuming v > v0) 25. Which of the following curves represent the speed of the
electron of hydrogen atom as a function of principal quantum
number ‘n’ ?

(a) 1 (b) 2
(c) 3 (d) 4
(a) I and II (b) II and III
(c) III and IV (d) II and IV
26. For a hypothetical H like atom which follows Bohr’s model, 33. What is the de–Broglie wavelength associated with the
some spectral lines were observed as shown. hydrogen electron in its third orbit
If it is known that line ‘E’ belongs to the visible region, (a) 9.96 × 10–10 cm (b) 9.96 × 10–8 cm
then the lines possibly belonging to ultra violet region will (c) 9.96 × 104 cm (d) 9.96 × 108 cm
be ( n1 is necessarily ground state) 34. De-Broglie wavelength for an electron is related to applied
n5 voltage as
n4
C B 12.3 12.3
n3 (a)   Å (b)   Å
D A
n2
h V
E
n1
12.3 12.3
[Assume for this atom, no spectral series shows overlaps (c)   Å (d)   Å
E m
with other series in the emmission spectrum]
35. If E1, E2 and E3 represent respectively the kinetic energies of
(a) B and D (b) D only
an electron, an alpha particle and a proton each having same
(c) C only (d) A only
de Broglie wavelength then :
27. If each hydrogen atom in the ground state of 1.0 mole of H-
(a) E1 > E3 > E2 (b) E2 > E3 > E1
atoms is excited by absorbing photons of energy 8.4eV, 12.09
(c) E1 < E3 < E2 (d) E1 = E2 = E3
eV and 15.0 eV of energy, then the number of spectral lines
emitted is equal to : 36. If the value of (n + l) is more than 3 and less than 6, then
what will be the possible number of orbitals ?
(a) None (b) Two
(c) Three (d) Four (a) 6 (b) 9
28. A certain transition in H-spectrum from an excited state to (c) 10 (d) 13
ground state in one or more steps gives rise to a total of ten 37. Which of the following pairs of ions have the same electronic
lines. How many of these belong to visible spectrum ? configuration ?
(a) 3 (b) 4 (a) Cr3+, Fe3+ (b) Fe3+, Mn2+
(c) 5 (d) 6 (c) Fe3+, Co3+ (d) Sc3+, Cr3+
29. The series limit for Balmer series of H-spectra is 38. How many of the following ions have the same magnetic
(a) 3800 Å (b) 4200 Å moments ?
(c) 3646 Å (d) 4000 Å Fe2+ Mn2+ Cr2+ Ni2+
30. What is the total number of pairs of electrons at least three (a) 1 (b) 2
same quantum numbers for Be ?
(c) 3 (d) 4
(a) 2 (b) 4
39. In following two plots,  2 is plotted against the distance ‘r’
(c) 3 (d) 8
from nucleus.
31. The accelerating potential that must be imparted to proton
beam to give an wavelength 5 pm.
(a) 32.8 V (b) 3.28 V
(c) 328 V (d) 0.328 V
32. A proton accelerated from rest through a potential difference
of ‘V’ volts has a wavelength  associated with it. An alpha
particle in order to have the same wavelength must be Select the correct statement :
accelerated from rest through a potential difference of
(a) ‘A’ is for 1s and ‘B’ for 2s
(a) V volt (b) 4V volt
(b) ‘A’ is for 2s and ‘B’ for 1s
(c) 2V volt (d) V/8 volt
(c) ‘A’ is for 2s and ‘B’ for 2p
(d) ‘A’ is for 2p and ‘B’ for 2s
40. In the following two figures, ( 2 ) is plotted against (r) the 45. Which of the following parameters are not same for all
hydrogen like atoms and ions in their ground state ?
distance from nucleus :
(a) Radius of orbit (b) Speed of electron
(c) Energy of the atom
(d) Orbital angular momentum of electron
46. The angular momentum of electron can have the value (s) :

h h
(a) 0.5 (b)
(a) Both (A) and (B) are for 1s  
(b) Both (A) and (B) are for 2s h h
(c) (d) 2.5
(c) (A) is for 1s and (B) is for 2s 0.5 2
(d) (A) is for 2s and (B) is for 1s 47. Which of the following are the limitations of Bohr’s model?
41. Let us consider following graph for radial distribution (a) It could not explain the intensities or the fine spectrum
function. Which of the following has correct matching of of the spectral lines.
curve and orbital ?
(b) No justification was given for the principle of the
quantization of angular momentum.
(c) It could not explain why atoms should combine to form
bond.
(d) It could not be applied to multi-electron atoms.
48. ‘g’ orbital is possible if
(a) n = 5, l = 4 (b) it will have 18 electrons
(a) A (3p) B (3d) C (3s) (b) A (3s) B (3p) C (3d) (c) it will have 9 types of orbitals
(c) A (3d) B (3p) C (3s) (d) A (3s) B (3d) C (3p) (d) it will have 22 electrons.
Objective Questions II 49. Which of the following statement about quantum number is
correct ?
[One or more than one correct option]
(a) If the value of l = 0, the electron distribution is spherical.
42. The number of orbitals in nth Bohrs orbit of an atom
equal to (b) The shape of orbital is given by subsidiary quantum
number.
(a) n2 (b) 2n2
(c) (2l + 1) (d) possible values of ‘m’ (c) The Zeeman’s effect is explained by magnetic quantum
number.
43. The number of electrons in Na(11) having l = 0 are
equal to (d) The spin quantum number gives the orientations of
electron cloud.
(a) there are two electrons in 1s orbital.
(b) there are two electrons in 2s orbital. 50. Which of the following configuration is/are correct in the
first excited state ?
(c) there are two electrons in 2p orbital.
(d) there are one electrons in 3s orbital. (a) Cr [Ar] 3d5 4s1 (b) Fe2+ [Ar] 3d5 4s1
(c) Mn2+ [Ar] 4s0 3d5 (d) Co3+ [Ar] 3d5 4s1
44. The ionisation energy of hydrogen atom is 13.6 eV. Hydrogen
atoms in the ground state are excited by monochromatic 51. The magnetic moment of Xn+ is 24 BM. Hence, the
light of energy 12.1 eV. The spectral lines emitted by
species can be
hydrogen atoms according to Bohr’s theory will be
(a) Fe2+ (b) Cr2+
(a) n = 3 to n = 1 (b) n = 3 to n = 2
(c) Mn3+ (d) Co3+
(c) n = 2 to n = 1 (d) n = 4 to n = 1
52. The spin only magnetic moment of V (Z = 23), Cr (Z = 24) (b) the shortest wavelength in Balmer series of He+ is x
and Mn (Z = 25) are x, y and z respectively. Which of the
4
following is are correct relationships ? (c) the longest wavelength in Lyman series of H atom is x
3
(a) x > y (b) x > z
(c) y > z (d) x < z 16x
(d) the longest wavelength in Paschen series of Li2+ is
53. For radial probability distribution curves, which of the 7
following is/are correct ? 58. The ratio of the de Broglie wavelength of a proton and  -
(a) The number of maxima in 2s orbital are two particles will be 1 : 2 if their :
(b) The number of spherical or radial nodes is equal to (a) velocity are in the ratio 1 : 8
n–l–1 (b) velocity are in the ratio 8 : 1
(c) The number of angular nodes are ‘l’ (c) kinetic energy are in the ratio 16 : 1
(d) 3d 2z has two angular nodes. (d) kinetic energy are in the ratio 1 : 16
59. The uncertainties in measurement of position and momentum
54. Pick out the orbitals with the maximum number of nodal
of an electron are equal. Choose the correct statement.
planes ?
(a) The uncertainty in measurement of speed = 8 × 1012.
(a) 3dxy (b) 4d z 2
(b) The uncertainty in measurement of kinetic energy =
(c) 4dxy (d) 2px
h
55. Select the correct plots for the photoelectric current. .
8m
(c) The uncertainty in measurement of time = 9.1 × 10–31.
(d) Increasing the wavelength of light used in the experiment
(a) (b) will decrease uncertainty in position and increase the
uncertainty in momentum.
60. Which of the following is/are correct ?
(a) The orbital angular momentum for a d-electron is
h
6 .
2
(c) (d)
(b) The number of orbitals in a shell with principal quantum
number n is 2n2.
56. If the radius of first Bohr’s orbit of H-atom is x, which of the (c) The correct set of quantum numbers for the last unpaired
following is the correct conclusion ?
 1
(a) The de-Broglie wavelength in the third Bohr orbit of H- electron of Cl atoms is 3,1,1,   .
 2
atom = 6x.
(d) The ratio of energy in the first Bohr orbit of H-atom to
(b) The fourth Bohr’s radius of He+ ion = 8x.
the electron in the first excited state of Be3+ is 1 : 4.
(c) The de-Broglie wavelength in third Bohr’s orbit of
Li2+ = 2x Numeric Value Type Questions
(d) The second Bohr’s radius of Be2+ = x 61. With what velocity should an alpha () particle travel towards
the nucleus of a copper atom so as to arrive at a distance
57. If the shortest wavelength of transition of H-atom in Lyman
10–13 m from the nucleus of the copper atom ? (in 106 m/s)
series is x, then the correct conclusion is (are)
62. How many of the following atoms/ions are isoelectronic
x with Ca2+ :
(a) the longest wavelength in Balmer series of He+ is
4 Ar, Na+, Mg2+, K+, Cl–, F–, S2–, N3–
63. Infrared lamps are used in restaurants to keep the food warm. (a) A (b) B
The infrared radiation is strongly absorbed by water, raising (c) C (d) D
its temperature and that of the food. If the wavelength of
70. Assertion (A) : When   rays hit a thin foil of gold, only
infrared radiation is assumed to be 1500 nm, then the number
of photons per second of infrared radiation produced by an a few   particles are deflected back.
infrared lamp that consumes energy at the rate of 100 W and Reason (R) : Within an atom, there is a very small positively
is 12% efficient only is x × 1019. Here, x is charged heavy body is present.
64. Calculate the velocity of electron (in 105 m/s) ejected from (a) A (b) B
platinum surface when radiation of 200 nm falls on it. Work (c) C (d) D
function of platinum is 5 eV. (1 eV = 1.6 × 10–19)
71. Assertion (A) : According to the wave theory, the radiation
65. A hydrogen atom in the ground state is hit by a photon emitted by the body being heated should have the same
exciting the electron to 3rd excited state. The electron then colour but its intensity varies as the heating is continued.
drops to 2nd Bohr orbit. What is the frequency of radiation
Reason (R) : Energy of any electromagnetic radiation
emitted in the process ? (in 1014 hz)
depends upon its frequency.
66. The energy of separation of an electron is 30.6 eV moving in (a) A (b) B
an orbit of Li+2. Find out the number of waves made by the (c) C (d) D
electron in one complete revolution in the orbit. 72. Assertion (A) : Emission spectrum of a pure atom is line
Assertion Reason spectrum, not the continuous.

(A) If both Assertion and Reason are correct and Reason (R) : Energy of the atoms are quantized.
Reason is the correct explanation of (a) A (b) B
Assertion. (c) C (d) D

(B) If both Assertion and Reason are true but 73. Assertion (A) : Accurate measurement of both positions
and momentum can be done simultaneously for a
Reason is not the correct explanation of
macroparticle.
Assertion.
Reason (R) : Heisenberg’s uncertainty principle is more
(C) If Assertion is true but Reason is false. significant to microparticles only.
(D) If Assertion is false but Reason is true. 74. Assertion (A) : Half-filled and fully-filled degenerate orbitals
are more stable.
Charge
67. Assertion (A) : ratio of anode rays is found Reason (R) : Extra stability is due to the symmetrical
Mass distribution of electrons and exchange energy.
different for different gases.
(a) A (b) B
Reason (R) : Proton is the fundamental particle present in (c) C (d) D
the gases.
75. Assertion (A) : The spin only magnetic moment of Zn2+ is
(a) A (b) B zero.
(c) C (d) D Reason (R) : Zn2+ had 3d10 4s0 configuration. If has no
68. Assertion (A) : Cathode rays are produced only when the unpaired electron.
pressure of the gas inside the discharge tube is very low. (a) A (b) B
Reason (R) : At high pressure, no electric current flows (c) C (d) D
through the tube as gases are poor conductor of electricity.
76. Assertion (A) : Wave number of visible light with wave-
(a) A (b) B length of 5000 Å is 2 × 106 m–1.
(c) C (d) D Reason (R) : Wave number is defined as the number of
69. Assertion (A) :   Particles are helium nuclei. waves present in 1 unit length.
Reason (R) : They are deflected slightly towards the negative (a) A (b) B
plate and hence carry positive change. (c) C (d) D
77. Assertion (A) : The photoelectrons produced by a 81. Matching Column Types
monochromatic light beam incident on a metal surface have Column - I Column - II
a spread in their kinetic energy.
(A) Configuration of Cr is (P) 1s2 2s2 2p6 3s2 3p6 3d10
Reason (R) : The work function of the metal varies as the
(B) Configuration of Cu is (Q) 5
function of depth from surface.
(C) Number of unpaired (R) 1s2 2s2 2p6 3s2 3p6
(a) A (b) B
electrons in Fe 3+
3d5 4s1
(c) C (d) D
(D) Electronic configuration (S) 1s2 2s2 2p6 3s2 3p6
78. Assertion (A) : In the emission spectrum of hydrogen atom,
of Zn2+ 3d10 4s1
lines are closely spaced in the region of large wavelengths.
(T) 1s2 2s2 2p6 3s2 3p6 3d9 4s2
Reason (R) : In the region of large wavelengths, electronic
(U) 1s2 2s2 2p6 3s2 3p6 3d4 4s2
transitions occur more frequently.
82. Matching Column Types
(a) A (b) B
Column - I Column - II
(c) C (d) D
Match the Following
Each question has two columns. Four options (A) Radial probability distribution (P)
are given representing matching of elements
graphs for 1s orbital
from Column-I and ColumnII. Only one of these
four options corresponds to a correct
matching.For each question, choose the op- (B) Radial probability distribution (Q)
tion corresponding to the correct matching.
79. Column - I Column - II
–19 (C) Radial probability distribution (R)
(A) Plum-Pudding model (P) 1.6022 × 10 C
(B) Planetory model of atom (Q) Thomson’s model
graph for 2p orbital
(C) Atoms are indivisible (R) Rutherford’s model

(D) Charge on electron (S)  1/2

(D) Electron cloud picture of 2s (S)
(E) The spin of electron is (T) Dalton theory
80. Matching Column Types orbital
Column - I Column - II 83. Pn = potential energy, En = total energy
f = frequency, Z = atomic number
(A) Orbit angular momentum (P) n(n  2)
vn = velocity of nth orbit Tn = time period in nth orbit
(B) Orbital angular momentum (Q) nh / 2 Column - I Column - II

(C) Spin angular momentum (R) s(s  1) h (A) En  r , y = ?

y
(P) 1/2
(B) En/Pn (Q) 1
(D) Magnetic moment (S) l (l  l) h
1
(C) f  x  z, x  ? (R) 2
(T) n (n  l) h n

(D)  v n  Tn   rn, t  ?
t
(S) –1
84. Match the entries in Column I with the correctly related
87. Which of the following is correct relation of wavelength
quantities in Column II.
among various radiations :
Column - I Column - II
(a) Cosmic < X-rays < Micro-waves < -rays
(A) Angular momentum (P) Increase by increasing n
(b) UV rays < Radio waves < visible < IR rays
(B) Kinetic energy (Q) Decreases by decreasing Z
(c) Cosmic < UV rays < IR rays < Radio wave
(C) Potential energy (R) Increases by decreasing Z
(d) -rays < Cosmic rays < IR rays < Micro waves
(D) Velocity (S) Decreases by decreasing n
88. Which of the following is incorrect w.r.t Maxwell wave
85. Matching Column Types theory?
Quantum number Orbitals (a) EM radiations travel with the speed of light.
(A) n = 2, l = 1, m = –1 (P) 2px, or 2py (b) The wave number of radiations having frequency of 4 ×
(B) n = 4, l = 2, m = 0 (Q) 4 dz 2 1014 Hz is 1.33 × 104 cm–1
(C) n = 3, l = 1, m =  1 (R) 3px or 3py (c) Electric field of radiations is perpendicular to magnetic
field but parallel to direction of propagation.
(D) n = 4, l = 0, m = 0 (S) 4s
(d) Radiant energy is a continuous form of energy.
(E) n = 3, l = 2, m =  2 (T) 3d x 2 y2 , 3d xy
Use the following passage, to anwers Q. 89 to
Paragraph Type Questions Q. 91
Use the following passage, to anwers Q. 86 to Passage
Q. 88
In a mixture of H – He+ gas (He+ is singly ionized He atom),
Passage H atoms and He+ ions are excited to their respective first
Electromagnetic wave theory was proposed by James Clark excited states. Subsequently, H atoms transfer their total
Maxwell in 1864. Acc. to this theory, the energy is emitted excitation energy to He+ ions (by collisions). Assuming that
from any energy source continuously in the form of the Bohr model of atom is applicable, answer the following
radiations & is called radiant energy. questions.

Some important characteristics of a wave are : 89. The ratio of the potential energy of the n = 2 electron for the H
atom to that of He+ ion is :
1. Wavelength () :- Distance between any two
consecutive crests or troughs (a) 1/4 (b) 1/2
(c) 1 (d) 2
2. Frequency () :- Number of waves passing through a
point in one second. 90. The quantum number n of the state finally populated in He+
ions is :
3. Amplitude :- Height of crest or depth of trough
(a) 2 (b) 3
1 (c) 4 (d) 5
4. Wave Number (ν) =
 91. The wavelength of light emitted in the visible region by He+
Relation between velocity, wavelength & frequency ions after collisions with H atoms is : (in Å)
c = × (a) 6.5 × 10–7 m (b) 5.6 × 10–7 m
86. How long would it take a radiowave of frequency, 6 × 103 (c) 4.8 × 10–7 (d) 4.0 × 10–7 m
sec–1 to travel from Mars to Earth, a distance of 8 × 107 km ?
In visible region Balmer series e has to come from n
(a) 266 sec (b) 246 sec
n  4 to n  2
(c) 280 sec (d) None of these
 Use the following passage, to anwers Q. 92 to 92. If the wavelength of series limit of Lyman series for He+ ion
Q. 94 is x Å, then what will be the wavelength of series limit of
Balmer series for Li2+ ion ?
Passage
One of the fundamental laws of physics is that matter is 9x 16x
(a) Å (b) Å
most stable with the lowest possible energy. Thus, the 4 9
electron in a hydrogen atom usually moves in the n = 1
orbit, the orbit in which it has the lowest energy. When the 5x 4x
electron is in this lowest energy orbit, the atom is said to be (c) A (d) A
4 7
in its ground electronic state. If the atom receives energy
from an outside source, it is possible for the electron to 93. The emission spectra is observed by the consequence of
move to an orbit with a higher n value, in which case the transition of electron from higher energy state to ground
atoms is in an excited with a higher energy. state of He+ ion. Six different photons are observed during
the emission spectra, then what will be the minimum
The law of conservation of energy says that we cannot
wavelength during the transition ?
create or destroy energy. Thus, if a certain amount of external
energy is required to excite an electron from one energy 4 4
level to another, then that same amount of energy will be (a) 27R (b) 15R
H H
liberated when the electron returns to its initial state.
Lyman series is formed when the electron returns to the 15 16
lowest orbit while Balmer series is formed when the electron (c) 16R (d) 15R
H H
returns to second orbit. Similarly Paschen, Brackett and
Pfund series are formed when electrons returns to the third, 94. What transition in the hydrogen spectrum would have the
fourth and fifth orbits from higher energy orbits respectively. same wavelength as Balmer transition, n = 4 to n = 2 in the
When an electron returns from n2 to n1 state, the number of He+ spectrum ?
lines in the spectrum will equal to (a) n = 3 to n = 1 (b) n = 3 to n = 2
(c) n = 4 to n = 1 (d) n = 2 to n = 1
(n 2  n1 ) (n 2  n1  1)
2 Use the following passage, to anwers Q. 95 to
If the electron comes back from energy level having energy
Q. 97
E2 to energy level having energy, E1, then the difference Passage
may be expressed in terms of energy of photon as :
Werner Heisenberg considered the limits of how precisely
hc we can measure the properties of an electron or other
E 2  E1  E, E  microscopic particle. He determined that there is a

fundamental limit to how closely we can measure both
Since, h and c are constants, E corresponds to definite position and momentum. The more accurately we measure
energy ; thus, each transition from one energy level to the momentum of a particle, the less accurately we can
another with produce a radiation of definite wavelength. determine its position. The converse is also true. This is
This is actually observed as a line in the spectrum of summed up in what we now call the Heisenberg uncertainty
hydrogen atom. principle.
Wave number of a spectral line is given by the formula
h
The equation is x.(mv) 
 1 1  4
R 2  2
 n1 n 2  The uncertainty in the position or in the momentum of a
macroscopic object like a baseball is too small to observe.
where R is a Rydberg’s constant (R = 1.1 × 107 m–1) However, the mass of microscopic object such as an electron
is small enough for the uncertainty to be relatively large and
significant.
95. If the uncertainties in position and momentum are equal, the (b) If (n + l) value is same for many sub-shells, priority of
uncertainty in the velocity is : electron filling is given to the sub-shell with lowest n
value.
h h
(a) (b) (c) (i) Fulfilled sub shell is most stable.
 2
(ii) "half filled orbitals are more stable than partially filled
orbital".
1 h
(c) (d) None of these 98. Which pair of sub-shell has same energy for above
2m 
described exceptional element under rule (a) ?
96. If the uncertainty in velocity and position is same, then the
(a) 1s, 2s (b) 2s, 2p
uncertainty in momentum will be :
(c) 3d, 4p (d) 5p, 4d
hm h 99. If Hunds rule is not obeyed by some elements given below
(a) (b) m
4 4 then which atom has maximum magnetic moment.
(a) Fe (b) Cu
h 1 h
(c) (d) (c) Cr (d) Mn
4m m 4
100. Which pair of element follow rule (c) (ii) ?
97. What would be the minimum uncertainty in de-Broglie (a) Cr, Mo (b) Mn, Fe
wavelength of a moving electron accelerated by potential
difference of 6 volt and whose uncertainty in position is (c) Cu, Ag (d) N, P

7
nm?
22
(a) 6.25 Å (b) 6 Å
(c) 0.625 Å (d) 0.3125 Å
Use the following passage, to anwers Q. 98 to
Q. 100
Passage
The sequence of filling electrons in sub-shells of elements
with few exceptions in d-block and f-block elements is
governed by Aufbau principle followed by Hund’s rule and
Pauli’s exclusion principle.
(a) The electron prefers to enter into sub-shells with lower
(n + l) value.
The energy for any sub-shell of an element other than
hydrogen is proportional to the sum of principal quantum
number (n) and angular momentum quantum number
EXERCISE - 4 : PREVIOUS YEAR JEE ADVANCED QUESTIONS
8. The ratio of the energy of a photon of 200 Å wavelength
Objective Questions I
radiation to that of 4000 Å radiation is (1986)
[Only one correct option]
1
1. Any p-orbital can accommodate up to (1983) (a) (b) 4
4
(a) four electrons (b) six electrons
(c) two electrons with parallel spins 1
(c) (d) 20
(d) two electrons with opposite spins 2
2. Which electronic level would allow the hydrogen atom to 9. Which one of the following sets of quantum numbers
absorb a photon but not be emit a photon ? (1984) represents an impossible arrangement ? (1986)
(a) 3s (b) 2p n l m s
(c) 2s (d) 1s 1
3. Correct set of four quantum numbers for the valence (a) 3 2 –2
2
(outermost) electron of rubidium (Z = 37) is (1984)
1
1 1 (b) 4 0 0
(a) 5, 0, 0,  (b) 5, 1, 0,  2
2 2
1
1 1 (c) 3 2 –3
(c) 5, 1, 1,  (d) 6, 0, 0,  2
2 2
4. The increasing order (lowest first) for the values of e/m 1
(d) 5 3 0 
(charge/mass) for electron (e), proton (p), neutron (n) and 2
alpha particle () is (1984) 10. Rutherford’s alpha particle scattering experiment eventually
(a) e, p, n,  (b) n, p, e,  led to the conclusion that (1986)
(c) n, p, , e (d) n, , p, e (a) mass and energy are related
5. Electromagnetic radiation with maximum wavelength is (b) electrons occupy space around the nucleus
(1985) (c) neutrons are burried deep in the nucleus
(a) ultraviolet (b) radio wave (d) the point of impact with matter can be precisely
(c) X-ray (d) infrared determined
6. The radius of an atomic nucleus is of the order of 11. The sum of the number of neutrons and proton in the isotope
(1985) of hydrogen is (1986)
(a) 10–10 cm (b) 10–13 cm (a) 6 (b) 5
(c) 10 –15
cm (d) 10 cm–8
(c) 4 (d) 3
7. Bohr’s model can explain (1985) 12. The energy of an electron in the first Bohr orbit of H-atom
(a) the spectrum of hydrogen atom only is –13.6 eV. The possible energy value (s) of the excited
state (s) for electrons in Bohr orbits of hydrogen is (are)
(b) spectrum of an atom or ion containing one electron
only (1988)
(c) the spectrum of hydrogen molecule (a) – 3.4 eV (b) – 4.2 eV
(d) the solar spectrum (c) – 6.8 eV (d) + 6.8 eV
13. The orbital diagram in which the aufbau principle is violated 20. For a d-electron, the orbital angular momentum is
(1988) (1997)

(a) (b)  h   h 
(a) 6  (b) 2 
 2   2 
(c) (d)
 h   h 
(c)   (d) 2  
14. The wavelength of a spectral line for an electronic transition  2   2 
is inversely related to (1988)
21. The first use of quantum theory to explain the structure of
(a) the number of electrons undergoing the transition atom was made by (1997)
(b) the nuclear charge of the atom (a) Heisenberg (b) Bohr
(c) the difference in the energy of the energy levels (c) Planck (d) Einstein
involved in the transition
22. The energy of an electron in the first Bohr orbit of H-atom
(d) the velocity of the electron undergoing the transition is –13.6 eV. The possible energy value(s) of the excited
15. The correct set of quantum numbers for the unpaired state(s) for electrons in Bohr orbits of hydrogen is (are)
electron of chlorine atom is (1989) (1998)
n l m n l m (a) –3.4 eV (b) – 4.2 eV
(a) 2 1 0 (b) 2 1 1 (c) – 6.8 eV (d) + 6.8 eV
(c) 3 1 1 (d) 3 0 0 23. The electrons, identified by numbers n and l,
16. The correct ground state electronic configuration of (i) n = 4, l = 1, (ii) n = 4, l = 0, (iii) n = 3, l = 2, and (iv) n = 3,
chromium atom is (1989) l = 1 can be placed in order of increasing energy, from the
(a) [Ar] 3d5 4s1 (b) [Ar] 3d4 4s2 lowest to highest, as (1999)

(c) [Ar] 3d 4s
6 0
(d) [Ar] 4d5 4s1 (a) (iv) < (ii) < (iii) < (i) (b) (ii) < (iv) < (i) < (iii)

17. The orbital angular momentum of an electron in 2s orbital (c) (i) < (iii) < (ii) < (iv) (d) (iii) < (i) < (iv) < (ii)
is (1992) 24. The electronic configuration of an element is 1s2, 2s22p6,
3s23p63d5, 4s1. This represents its (2000)
1 h (a) excited state (b) ground state
(a)  . (b) zero
2 2
(c) cationic form (d) anionic form
h h 1 1
(c) (d) 2. 25. The quantum numbers  and  for the electron spin
2 2 2 2
18. Which of the following does not characterise X-rays ? represent (2001)
(1992) (a) rotation of the electron in clockwise and anticlockwise
(a) The radiation can ionise gases direction respectively
(b) It causes ZnS to fluoresce (b) rotation of the electron in anticlockwise and clockwise
(c) Deflected by electric and magnetic fields direction respectively

(d) Have wavelengths shorter than ultraviolet rays (c) magnetic moment of the electron pointing up and down
respectively
19. Which of the following has the maximum number of
unpaired electrons ? (1996) (d) two quantum mechanical spin states which have no
classical analogue
(a) Mg2+ (b) Ti3+
(c) V3+ (d) Fe2+
26. The wavelength associated with a golf ball weighing 200g
h2 h2
and moving at a speed of 5 m/h is of the order (c) (d)
32  2 ma 02 64 2 ma 02
(2001)
(a) 10–10 m (b) 10–20 m 33. P is the probability of finding the 1s electron of hydrogen
(c) 10 m–30
(d) 10 –40
m atom in a spherical shell of infinitesimal thickness, dr, at a
distance r from the nucleus. The volume of this shell is
27. Rutherford’s experiment, which established the nuclear
4r2dr. The qualitative sketch of the dependence of P on r
model of the atom, used a beam of (2002)
is (2016)
(a) -particles, which impinged on a metal foil and got
absorbed
(b) -rays, which impinged on a metal foil and ejected
electrons
(c) Helium atoms, which impinged on a metal foil and got (a) (b)
scattered
(d) Helium nuclei, which impinged on a metal foil and got
scattered
28. If the nitrogen atom had electronic configuration 1s7, it
would have energy lower than that of the normal ground
state configuration 1s2 2s2 2p3, because the electrons (c) (d)
would be closer to the nucleus, yet 1s7 is not observed
because it violates (2002)
(a) Heisenberg uncertainty principle 34. consider the Bohr’s model of a one-electron atom whre the
elctron moves around the nuceus. In the following List-I
(b) Hund’s rule contains some quantities for the nth orbit of the atom and
(c) Pauli exclusion principle List-II contains options showing how they depend on n.
(d) Bohr postulate of stationary orbits List-I List-II
29. Which hydrogen like species will have same radius as that (I) Radius of the n orbit
th
(P)  n–2
of Bohr orbit of hydrogen atom ? (2004) (II) Angular momentum of (Q)  n–1
(a) n = 2, Li2+ (b) n = 2, Be3+ the electron in the nth orbit
(c) n = 2, He+ (d) n = 3, Li2+ (III) Kinetic energy of the (R)  n0
30. The number of radial nodes in 3s and 2p respectively are electron in the nth orbit
(2005) (IV) Potential energy of the (S)  n1
(a) 2 and 0 (b) 0 and 2 electron in the nth orbit
(c) 1 and 2 (d) 2 and 1 (T)  n2
31. The number of nodal planes in a px orbital is (U)  n1/2
(2005)
Which of thle following has the correct combination con-
(a) One (b) Two sidering List-I and List-II? (2019)
(c) Three (d) Zero (a) (III), (S) (b) (IV), (Q)
32. The kinetic energy of an electron in the second Bohr orbit (c) (III), (P) (d) (IV), (U)
of a hydrogen atom is [a0 is Bohr radius] (2013)

h2 h2
(a) (b)
4 2 ma 02 162 ma 02
35. Consider the Bohr’s model of a one-electron atom whre the 40. Which of the following relates to photons both as wave
elctron moves around the nuceus. In the following List-I motion and as a stream of particles ?
contains some quantities for the nth orbit of the atom and (1992)
List-II contains options showing how they depend on n.
(a) Inference (b) E = mc2
List-I List-II
(c) Diffraction (d) E = h
(I) Radius of the nth orbit (P)  n–2
41. Which of the following statement (s) is (are) correct?
(II) Angular momentum of (Q)  n–1
(1998)
the electron in the nth orbit
(a) The electronic configuration of Cr is [Ar] 3d5 4s1.
(III) Kinetic energy of the (R)  n0 (Atomic number of Cr = 24).
electron in the nth orbit (b) The magnetic quantum number may have a negative
(IV) Potential energy of the (S)  n1 value.
electron in the nth orbit (c) In silver atom, 23 electrons have a spin of one type and
(T)  n2 24 of the opposite type. (Atomic number of Ag = 47).
(U)  n1/2 (d) The oxidation state of nitrogen in HN3 is –3.
Which of thle following has the correct combination con- 42. Ground state electronic configuration of nitrogen atom can
sidering List-I and List-II? (2019) be represented by (1999)
(a) (II), (R) (b) (II), (Q) (a)
(c) (I), (P) (d) (I), (T)
Objective Questions II (b)

[One or more than one correct option] (c)

36. The atomic nucleus contains (1984)
(a) protons (b) neutrons (d)
(c) electrons (d) photons 43. The ground state energy of hydrogen atom is –13.6eV. Con-
37. When alpha particles are sent through a thin metal foil, sider an elctronic state  of H3+ whose energy, azimuthal
most of them go straight through the foil because (1984) quantum number and magnetic quantum number are –3.4
(a) alpha particles are much heavier than electrons ev, 2 and 0, respectively. Which of the following statement(s)
is(are) true for the state  ? (2019)
(b) alpha particles are positively charged
(c) most part of the atom is empty space (a) It is a 4d state

(d) alpha particles move with high velocity (b) The nuclear charge experienced by the electron in this
state is less than 2e, where e is the magnitude of the
38. Many elements have non-integral masses because electronic charge
(1984) (c) It has 3 radial nodes
(a) they have isotopes (d) It has 2 angular nodes
(b) their isotopes have non-integral masses
Numeric Value Type Questions
(c) their isotopes have different masses
44. No considering the electronic spin, the degeneracy of the
(d) the constiuents, neutrons, protons and electrons, second excited state (n = 3) of H atom is 9, while the
combine to give fractional masses degeneracy of the second excited state of H– is (2014)
39. An isotone of 76
32 Ge is (1984) 45. In an atom, the total number of electrons having quantum
numbers n = 4, |ml| = 1 and ms = –1/2 is
76 77
(a) 32 Ge
(b) 33 As
(2015)

77 78
(c) 34 Se
(d) 34 Se
46. Consider a helium (He) atom that absorbs a photon of (B) A hydrogen-like one (Q) Azimuthal quantum
wavelength 330 nm. The change in the velocity (in cm s-1) of
-electron wave function number
He atom after the photon absorption is ______. (nearest
integer) (Assume : Momentum is conserved when photon obeying Pauli principle
(C) Shape, size and orientation (R) Magnetic quantum
is absorbed).
of hydrogen like atomic number
(Use : Planck constant = 6.6 × 10-34 Js, Avogadro number
orbitals
= 6 × 1023 mol-1, Molar mass of He = 4 mol-1). (2021)
(D) Probability density of (S) Electron spin quantum
Match the Following
electron at the nucleus number
Each question has two columns. Four in hydrogen -like atom
options are given representing matching of
Paragraph Type Questions
elements from Column-I and ColumnII. Only
Use the following passage, to anwers Q. 49 to
one of these four options corresponds to a
Q. 51
correct matching.For each question, choose
the option corresponding to the correct Passage
matching. The hydrogen-like species Li2+ is in a spherically symmetric
state S1 with one radial node. Upon absorbing light the ion
47. In the following, Vn, Kn and En represent potential energy,
undergoes transition to a state S2. The state S2 has one
kinetic energy and total energy of an electron in the nth
radial node and its energy is equal to the ground state
Bohr orbit (radius : r) of hydrogen like species (nuclear
energy of the hydrogen atom
charge : Z). Match the entries on the left with those given
on right. (2006) 49. The state S1 is (2010)
(A) Vn/Kn = ? (P) –1 (a) 1s (b) 2s
(B) Vn/En = ? (Q) – 2 (c) 2p (d) 3s
(C) Kn/En = ? (R) 1 50. Energy of the state S1 in units of the hydrogen atom ground
state energy is (2010)
(D) 1/r  (Z)x : x = ? (S) 2
(a) 0.75 (b) 1.50
48. Match the entries in Column I with the correctly related
quantum number(s) in Column II. Indicate your answer by (c) 2.25 (d) 4.50
darkening the appropriate bubbles of the 4 × 4 matrix given 51. The orbital angular momentum quantum number of the
in the ORS (2008) state S2 is (2010)
Column - I Column - II (a) 0 (b) 1
(A) Orbital angular momentum (P) Principal quantum (c) 2 (d) 3
of the electron in a number
hydrogen
-like atomic orbital
Paragraph Type Questions
Use the following passage, to anwers Q. 52 to Q. 54
Passage
Answer Q. 52, Q. 53 and Q. 54 appropriately matching the information given in the three columns of the following table.

The wave function  n ,l , m l

, is a mathematical function whose value depends upon spherical polar coordinates (r, , ) of the
electron and characterized by the quantum numbers n, l and ml. Here r is distance from nucleus, q is colatitude and f is azimuth. In the
mathematical functions given in the Table, Z is atomic number and a0 is Bohr radius. (2018)
Column 1 Column 2 Column 3

3
 Zr 
 Z  2  a 
(I) 1s-orbital (i)  n,l,ml   e  0 (P)
 a0 

1
(II) 2s-orbital (ii) One radial node (Q) Probability density at nucleus  
a 30

5
 Zr 
 Z  2  a 
(III) 2pz-orbital (iii)  n,l,ml    re  0  cos (R) Probability density is maximum at nucleus
 a0 

(IV) 3d 2z  orbital (iv) xy-plane is a nodal plane (S) Energy needed to excite electron from n = 2 state

27
to n = 4 state is times the energy needed to excite
32
electron from n = 2 state to n = 6 state

52. For He+ ion, the only INCORRECT combination is For Subjective Questions
2s orbital no. of radial nodes = n - l - 1 = 1 (2018) 55. The energy of the electron in the second and third Bohr’s
(a) (I) (i) (S) (b) (II) (ii) (Q) orbits of the hydrogen atom is – 5.42 × 10–12 erg and
–2.41 × 10–12 erg respectively. Calculate the wavelength of
(c) (I) (iii) (R) (d) (I) (i) (R)
the emitted light when the electron drops from the third to
53. For the given orbital in Column 1, the only the second orbit. (1981)
CORRECT combination of any hydrogen- like species is 56. Calculate the wavelength in Angstroms of the photon that
(2018) is emitted when an electron in the Bohr’s orbit, n = 2 returns
(a) (II) (ii) (P) (b) (I) (ii) (S) to the orbit, n = 1 in the hydrogen atom. The ionisation
potential of the ground state hydrogen atom is 2.17 × 10–11
(c) (IV) (iv) (R) (d) (III) (iii) (P)
erg per atom. (1982)
57. Give reason why the ground state outermost electronic
54. For hydrogen atom, the only CORRECT combination is configuration of silicon is : (1985)
(2018)
(a) (I) (i) (P) (b) (I) (iv) (R)
(c) (II) (i) (Q) (d) (I) (i) (S)
58. What is the maximum number of electrons that may be 68. With what velocity should an -particle travel towards the
present in all the atomic orbitals with principal quantum nucleus of a copper atom so as to arrive at a distance
number 3 and azimuthal quantum number 2 ? (1985) 10–13 m from the nucleus of the copper atom ? (1997)
59. The electron energy in hydrogen atom is given by 69. Calculate the energy required to excite 1 L of hydrogen gas
at 1 atm and 298 K to the first excited state of atomic
21.7  10 12
En  erg. Calculate the energy required to hydrogen. The energy for the dissociation of H—H bond
n2 is 436 kJ mol–1. (2000)
remove an electron completely from the n = 2 orbit. What is
70. The wavelength corresponding to maximum energy for
the longest wavelength (in cm) of light that can be used to
hydrogen is 91.2 nm. Find the corresponding wavelength
cause this transition ? (1984)
for He+ ion. (2003)
60. According to Bohr’s theory, the electronic energy of
71. (a) The Schrodinger wave equation for hydrogen atom is
hydrogen atom in the nth Bohr’s orbit is given by
(2004)
21.76  1019
En  J. Calculate the longest wavelength  1 
3/2
 r0   r /a 0
n2 1
 2s    2 e
of light that will be needed to remove an electron from the 4 2  a 0   a 0 
third Bohr orbit of the He+ ion.
where a0 is Bohr’s radius. If the
(1990)
radial node in 2s be at r0, then find r0 in terms of a0.
61. What transition in the hydrogen spectrum would have the
(b) A base ball having mass 100 g moves with velocity 100
same wavelength as the Balmer transition n = 4 to n = 2 of
m/s. Find out the value of wavelength of base ball.
He+ spectrum ? (1993)
72. (a) Calculate velocity of electron in first Bohr orbit of
62. Estimate the difference in energy between 1st and 2nd
hydrogen atom (Givern, r = a0)
Bohr’s orbit for a hydrogen atom. At what minimum atomic
number, a transition from n = 2 to n = 1 energy level would (b) Find de-Broglie wavelength of the electron in first Bohr
result in the emission of X-rays with l = 3.0 × 10–8 m ? orbit.
Which hydrogen atom-like species does this atomic number (c) Find the orbital angular momentum of 2p-orbital in terms
correspond to? (1993) of h/2 units. (2005)
63. Find out the number of waves made by a Bohr’s electron in Fill in the Blanks
one complete revolution in its 3rd orbit. (1994) 73. Elements of the same mass number but of different atomic
64. Iodine molecule dissociates into atoms after absorbing light numbers are known as ................ . (1983)
to 4500Å. If one quantum of radiation is absorbed by each 74. The uncertainty principle and the concept of wave nature
molecule, calculate the kinetic energy of iodine atoms. of matter were proposed by .............. and .............
(Bond energy of I2 = 240 kJ mol–1) (1995) respectively. (1988)
65. A bulb emits light of  = 4500Å. The bulb is rated as150 75. The 2px, 2py and 2pz orbitals of atom have identical shapes
watt and 8% of this energy is emitted as light. How many but differ in their ............ . (1993)
photons are emitted by the bulb per second ? 76. Wave functions of electrons in atoms and molecules are
(1995) called ................ . (1993)
66. Consider the hydrogen atom to be proton embedded in a 77. The light radiations with discrete quantities of energy are
cavity of radius a 0 (Bohr’s radius) whose charge is called ................... . (1993)
neutralised by the addition of an electron to the cavity in
True/False
vacuum, infinitely slowly. Estimate the average total energy
of an electron in its ground state in a hydrogen atom as the 78. The energy of the electron in the 3d-orbital is less than
work done in the above neutralisation process. Also, if the that in the 4s-orbital in the hydrogen atom. (1983)
magnitude of the average kinetic energy is half of average 79. Gamma rays are electromagnetic radiations of wavelengths
potential energy, find the average potential energy. of 10–6 cm to 10–5 cm. (1983)
(1996) 80. The electron density in the XY plane in 3d x 2  y2 orbital is
67. Calculate the wave number for the shortest wavelength zero. (1986)
transition in the Balmer series of atomic hydrogen.
(1996)
 Answer Key
CHAPTER -1 STRUCTURE OF AN ATOM
EXERCISE - 1 : EXERCISE - 2 :
BASIC OBJECTIVE QUESTIONS PREVIOUS YEAR JEE MAINS QUESTIONS

1. (c) 2. (b) 3. (a) 4. (b) 5. (d) 1. (b) 2. (a) 3. (c) 4. (b) 5. (a)
6. (a) 7. (b) 8. (a) 9. (a) 10. (a) 6. (a) 7. (b) 8. (c) 9. (d) 10. (b)
11. (c) 12. (b) 13. (c) 14. (a) 15. (d) 11. (d) 12. (c) 13. (b) 14. (d) 15. (b)
16. (d) 17. (d) 18. (a) 19. (d) 20. (a) 16. (c) 17. (b) 18. (b) 19. (a) 20. (d)
21. (c) 22. (a) 23. (a) 24. (a) 25. (a) 21. (c) 22. (b) 23. (b) 24. (d) 25. (a)
26. (d) 27. (c) 28. (d) 29. (c) 30. (a) 26. (d) 27. (d) 28. (c) 29. (a) 30. (d)
31. (a) 32. (a) 33. (a) 34. (b) 35. (b) 31. (c) 32. (d) 33. (a) 34. (c) 35. (d)
36. (d) 37. (d) 38. (b) 39. (c) 40. (a) 36. (b) 37. (b) 38. (b) 39. (d) 40. (d)
41. (c) 42. (c) 43. (b) 44. (b) 45. (d) 41. (d) 42. (d) 43. (d) 44. (d) 45. (d)
46. (a) 47. (a) 48. (c) 49. (b) 50. (b) 46. (c) 47. (d) 48. (d) 49. (c) 50. (d)
51. (d) 52. (b) 53. (d) 54. (a) 55. (c) 51. (b) 52. (d) 53. (b) 54. (a) 55. (a)
56. (a) 57. (c) 58. (b) 59. (a) 60. (a) 56. (a) 57. (b) 58. (c) 59. (c) 60. (c)
61. (d) 62. (d) 63. (b) 64. (c) 65. (c) 61. (d) 62. (c) 63. (222.00) 64. (a)
66. (b) 67. (c) 68. (d) 69. (c) 70. (c) 65. (d) 66. (a) 67. (b) 68. (d) 69. (a)
71. (d) 72. (a) 73. (b) 74. (d) 75. (c) 70. (d) 71. (c) 72. (b) 73. (b) 74. (c)
76. (a) 77. (b) 78. (a) 79. (c) 80. (a) 75. (d) 76. (d) 77. (c) 78. (d)
81. (b) 82. (a) 83. (a) 84. (d) 85. (b) 79. (0.00) 80. (2.00)
86. (d) 87. (b) 88. (d) 89. (d) 90. (c) 81. (2.00) 82. (1.00)
91. (c) 92. (d) 93. (c) 94. (a) 95. (c) 83. (8.68) 84. (3.00)
96. (a) 97. (c) 98. (d) 99. (a) 100. (c) 85. (0.00) 86. (6.00)
101. (c) 102. (c) 103. (c) 104. (c) 105. (a) 87. (12.00) 88. (2.00)
89. (58.00) 90. (5.00)
91. (3155.0) 92. (50.00)
93. (7.00) 94. (0.00)
95. (2.00)
EXERCISE - 3 :
ADVANCED OBJECTIVE QUESTIONS

1. (a) 2. (b) 3. (a) 4. (c) 79. (A-Q ; B- R; C-T; D-P ; E-S)

5. (b) 6. (b) 7. (a) 8. (c) 80. (A-Q ; B- S; C-R; D-P)
9. (c) 10. (b) 11. (d) 12. (d) 81. (A-R ; B- S; C-Q; D-P)
13. (a) 14. (d) 15. (a) 16. (c) 82. (A-P ; B- Q; C-R; D-S)
17. (c) 18. (c) 19. (b) 20. (d) 83. (A-S ; B- P; C-P; D-Q)
21. (c) 22. (a) 23. (a) 24. (a) 84. (A-P,S ; B- Q; C-P,R,S ; D-S)
25. (a) 26. (d) 27. (c) 28. (a) 85. (A-P ; B- Q; C-R ; D-S; E - T)
29. (c) 30. (b) 31. (a) 32. (d) 86. (a) 87. (c) 88. (c) 89. (a)
90. (c) 91. (c) 92. (b) 93. (b)
33. (b) 34. (b) 35. (a) 36. (d)
94. (d) 95. (c) 96. (a) 97. (c)
37. (b) 38. (b) 39. (a) 40. (c) 98. (c,d) 99. (c) 100. (a)
41. (a) 42. (a,d) 43. (a,b,d) 44. (a,b,c)
45. (a,b,c) 46. (a,b,c) 47. (a,b,c,d) 48. (a,b,c)
49. (a,b,c) 50. (b,d) 51. (a,b,c,d) 52. (c,d)
53.(a,b,c,d)54. (a,c) 55. (b,d) 56. (a,b)
57. (b,c,d) 58. (b,c) 59. (a,b,c) 60. (a,c,d)
61. (6.00) 62. (4.00) 63. (9.00) 64. (6.00)
65. (6.00) 66. (2.00) 67. (c) 68. (a)
69. (b) 70. (a) 71. (a) 72. (c)
73. (d) 74. (a) 75. (a) 76. (a)
77. (a) 78. (a)
EXERCISE - 4 :
PREVIOUS YEAR JEE ADVANCED QUESTIONS

1. (d) 2. (d) 3. (a) 4. (d) 5. (b)

6. (b) 7. (b) 8. (d) 9. (c) 10. (b)
11. (a) 12. (a) 13. (b) 14. (c) 15. (c)
16. (a) 17. (b) 18. (c) 19. (d) 20. (a)
21. (b) 22. (a) 23. (a) 24. (b) 25. (d)
26. (c) 27. (d) 28. (c) 29. (b) 30. (a)
31. (a) 32. (c) 33. (a) 34. (c) 35. (c)
36. (a,b) 37. (a,c) 38. (a,c) 39. (b,d) 40. (b,d)
41.(a,b,c) 42. (a,d) 43. (a,b) 44. (3.00) 45. (6.00)
46.(30.00)
47. (A-Q ; B- S; C-P; D-R)
48. (A-Q ; B- S; C-P,Q,R; D-P,Q,R)
49. (b) 50. (c) 51. (b) 52. (c) 53. (a)
54. (d)