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ATOMIC STRUCTURE
SESSION -1

AIM
1) To introduce Fundamental particles
2) To introduce Thomson’s and Rutherford Atomic model
3) To introduce terms like atomic number, mass number and isotopes, isobars,
isotones
John Dalton coined the term atom. The atom is the fundamental particle of matter and
considered to be indivisible and indestructible.
In fact, the atom as the whole is electrically neutral as number of protons in it is equal to
number of electrons.
The electron, proton and neutron are the main fundamental particles of an atom.

Discovery of electron – study of Cathode rays:

J.J. Thomson observed that, when a high voltage is applied between the electrodes fitted
in discharge tube,at a very low pressure,some invisible radiations are emitted from the
cathode. At this stage wall of the discharge tube near cathode starts glowing.
Gas at low
Discharge tube
Pressure

Faint green glow


Cathode rays

To vacuum pump

Discharge tube experiment – production of cathode rays

Glowing is due to the bombardment of glass wall by the cathode rays. It may be noted that
when the gas pressure in the tube is 1 atm, no electric current flows through the tube.
This is because the gases are poor conductor of electricity.

Origin of Cathode rays:

Cathode rays are first produced in cathode due to bombardment of the gas molecules by the
high-speed electrons emitted first from the cathode.

Properties of Cathode rays

i. Cathode rays travel in straight lines with high speed.
ii. Cathode rays are made up of material particles.
iii.Cathode rays carry negative charge,the negatively charged material particles constituting
the cathode rays are called electrons.
iv. Cathode rays produce heating effect.
v. They cause ionization of the gas through which they pass.
vi. They produce X-rays when they strike against the surface of hard metals like tungsten,
molybdenum etc.
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vii.They produce green fluorescence on the glass walls of the discharge tube exp : zinc
sulphide.
viii. They affect the photographic plates.
ix. They possess penetrating effect (i.e., they can easily pass through thin foils of metals).
x. The nature of the cathode rays does not depend upon the nature of the gas, taken in
the discharge tube and the nature of cathode material.
xi. For each cathode rays, the ratio of charge (e) to mass (m) is constant

Discovery of proton – study of Anode rays:

Goldstein discovered the presence of positive rays.
He performed discharge tube experiment in which he took perforated cathode and a gas at
low pressure was kept inside a discharge tube.
On applying high voltage between electrodes, new rays were coming from the side of anode
and passing through the hole in the cathode gives fluorescence on the opposite glass wall
coated with zinc sulphide.
These rays were called anode rays or canal rays or positive rays.
Perforated cathode
H2 gas at low pressure
Anode rays

 ZnS coating

To vacuum pump
H.V
. Production of anode rays

Origin of anode or positive rays:

In the discharge tube the atoms of gas lose negatively charged electrons. These atoms,
thus, acquire a positive charge. The positively charged particle produced from hydrogen
gas was called the proton.
H→ H+(proton) + e-

Properties of Anode rays:

i) They travel in straight lines. However, their speed is much less than that of the
cathode rays.
ii) They are made up of material particles.
iii) They are positively charged, hence they called as canal rays or anode rays.’
iv) The nature of anode rays depends on the gas taken in the discharge tube.
v) For different gases taken in discharge tube the charge to mass ratio (e/m) of the
positive particles constituting the positive rays is different.
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Fundamental particles:
1) Electron: Electron is a universal constituent discovered by the J.J. Thomson.
* Charge: It was determined by Mullikan by oil drop experiment as -1.602x10-19coulombs
or 4.803x10-10 e.s.u.
* Mass:9.11x10-28g (nearly equal to 1/1837th of mass of hydrogen atom).
* Specific charge:e/m ratio is called specific charge & is equal to 1.76x108 coulombs/gm.
* Mass of one mole of electrons: It is 0.55 mg.
* Charge on one mole of electron is 96500 coulombs or 1 faraday.
* Density: 2.17x1017 g/cc.

2. Proton: (+1p0 or 1H1)

* It was discovered by Goldstein.
* Charge:It carries positive charge i.e.1.602 x 10-19coulombs or 4.803x10-10 esu.
* Mass:1.672x10-24g or 1.672x10-27kg.It is 1837 times heavier than an electron.
* Specific charge (e/m):9.58x104coulomb/gm.

3. Neutron (0n1)
* It was discovered by Chadwick by bombarding Be atom with high speed -particles.
𝟒𝐁𝐞𝟗 +𝟐 𝐇𝐞𝟒 → 𝟔𝐂
𝟏𝟐
+ 𝟎 𝐧𝟏
* Charge: Charge less or neutral particle.
* Mass:1.675x10-24 g or 1.675x10-27 kg.
* Density:1.5x1014 g/cm3 and is heavier than proton by 0.18%.
* Specific charge: It is zero.
* Among all the elementary particles neutron is the heaviest and least stable.

Properties of Electron, Proton and Neutron

Properties Electron Proton Neutron
Discovery J.J.Thomson Goldstein Chadwick
-19 -19
Charge -1.6022x10 C 1.6022x10 C Zero
-31 -27
Mass 9.109x10 kg 1.672x10 kg 1.675x10-27 kg
Spin ½ ½ ½
Charge -1 +1 0
Location Outside the nucleus In the nucleus In the nucleus
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Classical Models of Atom:

1) Thomson’s Atomic Model
According to Thomson, an atom is a sphere of positive charge having a number of
embedded electrons in it and sufficient enough to neutralize the positive charge. This
model is compared with a water melon in which seeds are embedded or pudding in which
raisins are embedded. Therefore, this model, sometime called watermelon model or
raisin or plum pudding model.

Thomson’s model of an atom

Limitation:
This model failed explain the results of scattering experiment of Rutherford and the
stability of atom.

2) Rutherford’s Atomic Model:

Rutherford, performed -ray scattering experiment in which he bombarded thin foils of
metals like gold, silver, platinum or copper with a beam of fast moving radioactive particles
originated from a lead block. The presence of 𝛼 particles at any point around the thin foil of
gold after striking it was detected with the help of a circular zinc sulphide screen. The
point at which a𝛼 particle strikes this screen, a flash of light is given out.

Observations and Conclusions

 particles ZnS screen

Beam of 
particles + Nucleus

Gold foil (100 nm thickness)

i. Most of the -particles passed through the gold foil without any deflection from
their original path.
Bcz atom has largely empty space as most of the -particles passed through the
foil undeflected.
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ii. A few of the alpha particles are deflected fairly at large angles while some are
deflected through small angles.
Bcz there is heavy positive charge at the center of the atom which causes
repulsions.
The entire mass of the atom is concentrated in the nucleus.
iii. A very few -particles are deflected back along their path.

According to Rutherford,
1. Atom is spherical and mostly hollow with a lot of empty space in it.
2. It has a small positively charged part at its center known as nucleus.
3. The nucleus is surrounded by electrons. The electrons revolve round the nucleus with
very high speeds in circular paths called orbits.
4. The number of extra nuclear electrons is equal to the number of units of positive charge
in the nucleus. Therefore the atom is electrically neutral. Electrons and the nucleus are
held together by electrostatic forces of attraction.
5. Rutherford’s model has resemblances with solar system. Hence it’s also known as
planetary model of the atom.
6. There is an empty space around the nucleus called extra nuclear part. In this part
electrons are present. As the nucleus of the atom is responsible for the mass of the
atom, the extra nuclear part is responsible for its volume.

Drawbacks:
1. According to the electromagnetic theory of Maxwell, when a charged particle moves
under the influence of attractive force it loses energy continuously in the form of
electromagnetic radiation. Therefore, an electron in an orbit will emit radiation.
As a result of this, the electron should lose energy at every turn and move closer
and closer to the nucleus following a spiral path.
The ultimate result is that it will fall into the nucleus thereby making the atom
unstable.
i.e., Rutherford’s model cannot explain the stability of the atom.

2. If the electrons lose energy continuously, the spectrum is expected to be continuous

but the actual observed spectrum consists of well-defined lines of definite
frequencies. Here the loss of energy by the electrons is not continuous in an atom.
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Atomic number (Z): Atomic number denotes the number of protons or the number of
electrons in the neutral atom.
Atomic number (Z) = Number of protons in the nucleus of an atom or ion
= Number of electrons in a neutral atom.

Mass number (A): The mass number is the total number of protons and neutrons present
in the nucleus of an atom of an element and indicated as A.
Protons and neutrons present in the nucleus of an atom are collectively known as nucleons.
Therefore, the mass number is also known as nucleon number.
Mass number (A) = Number of protons (Z) + Number of neutrons (n)
The number of neutrons (n) in an atom is equal to the difference between the mass
number and the atomic number.
n = A – Z

Mass Number A
X OR XA
Atomic Number Z Z

Symbol of Element

where X is the symbol for the element with superscript A and subscript Z, both on the left
hand side.

Isotopes, Isobars and Isotones:

Isotopes: The atoms of the same element which have the same atomic number but
different mass numbers are called isotopes.
Exp- 6 C12 , 6C13 , 6C14 1 2
1H , 1H , 1H
3

8 O16 , 8 O17 , 8 O18 17 Cl35 , 17 Cl37

Isotopes of an element differ in the number of neutrons present in the nucleus. But they
have the same number of protons and electrons.
Bcz of same number of electrons they show same chemical properties.They have
different number of neutrons, so they will have different masses and hence different
physical properties.
Isobars:The atoms of different elements which have the same mass number but
different atomic numbers are called isobars.
Exp: 18 Ar 40 , 19 K 40 , 20 Ca40 40
20𝐶𝑎

They have same number of nucleons. But they are differ chemically because the chemical
characteristics depend upon the number of electrons which is determined by the atomic
number.
Isotones:Isotones are the atoms of different elements which have the same number of
neutrons.
Eg: i. 6 C14 , 7 N15 , 8 O16 (n = 8) ii. 14 Si30 , 15P31, 16 S32 (n = 16)
Isotones show different physical and chemical properties.
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SESSION -1 CLASS EXERCISE

39
1. The number of neutrons present in 19
K is:
a) 39 b) 19 c) 20 d) None of these
2. The nucleus of the atom (Z > 1) consists of:
a) Proton and neutron b) Proton and electron
c) Neutron and electron d) Proton, neutron, electrons
3. The no, of electrons in a neutral atom is equal to it’s:
a) Atomic weight b) Atomic number
c) Equivalent weight d) Electron affinity
4. The specific charge of the canal rays:
a) Is not constant but changes with gas filled in discharge tube
b) Remains constant irrespective of the nature of gas in discharge tube
c) Is maximum when gas present in discharge tube is hydrogen
d) Is 9.58 x 104 coulombs/g
5. Proton is:
a) Nucleus of deuterium b) Ionized hydrogen molecule
c) Ionized hydrogen atom d) An α-particle
6. According to the Rutherford which statement is correct?
a) Electron revolves in fixed circular path around the nucleus
b) Electron revolves around the nucleus
c) Electron does not decrease its energy at the time of revolution
d)Electron obeys law of conservation of momentum at the time of revolution.
7. Rutherford's scattering experiments led to the discovery of
a) Nucleus
b) Presence of neutrons in the nucleus
c) Both a and b
d) Revolving nature of electrons around the nucleus
8. Deflection back of a few particles on hitting thin foil of gold shows that:
a) Nucleus is heavy b) Nucleus is small c) Both a and b
d) Electrons create hindrances in the movement of α –particles
9. α-particles are represented by
a) Lithium atoms b) Helium nuclei
c) Hydrogen nucleus d) None of the above

HOMEEXERCISE:
1. The species in which one of the fundamental particles is missing is
a) Helium b) Protium c) Deuterium d) Tritium
2.The discovery of neutron is late because neutron has
a) +ve charge b) –ve charge
c) neutral charge d) lightest particle
3. Which of the following statements are correct?
a) Isotopes have same number of protons
b) Isobars have same nucleon number.
c) Isobars have same number of protons
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d) Both a and b
4. The charge on electron is calculated by
a) Mullikan b) J J Thomson c) Ruther ford d) Newton
5. J J Thomson Model could able to explain the following?
a) Stability of Atom b) electrical neutrality of atom
c) Stability of nucleus d) all of these
6. The thickness of the gold foil used in Ruther Ford α ray scattering experiment
a) 0.0004 cm b) 0.0004 m c) 0.0004 mm d) 0.004 cm
7. What is the size of atom predicted by Ruther Ford?
a) 10-13cm b) 10-14cm c) 10-12cm d) 10-8cm
8. If Thomson Model is correct what should be the observation in α-ray scattering
experiment
a) All the α-rays should pass through the gold foil
b) Only few α-rays should pass through the gold foil
c) 98% of α-rays should get reflected back
d) Both b & c
9. Which part of atom is responsible for volume of atom?
a) Nucleus b) extra nuclear part c) protons d)
unknown particles
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SESSION – 2 AND 3
AIM-To understand nature of Electromagnetic Radiation

Nature of Light (Electromagnetic Radiation): Electromagnetic radiation do not need any

medium for propagation e.g visible, ultra violet, infrared, x-rays, -rays, radio waves, radiant
energy etc.
Two theories were proposed to explain the nature and the propagation of light
i. Corpuscular theory: This theory was proposed by Newton. According to this
theory light is propagated in the form of invisible small particles. i.e. light has
particle nature.
The particle nature of light explained some of the experimental facts such as
reflection and refraction of light but it failed to explain the phenomenon of
interference and diffraction. Therefore, was discarded and ignored.
ii. Wave theory of light (electromagnetic wave theory): was explained by James Clark
Maxwell in 1864 to explain and understand the nature of electromagnetic radiation.
Features of this theory are:
a. The light is a form of electromagnetic radiations.
b. The light radiations
consist of electric
and magnetic fields
oscillating
perpendicular to
each other.

Components of radiation
iii) The vertical component of wave, ‘E’ indicates the change in the strength of the electric
field and the horizontal component of the wave ‘H’ indicates the change in the strength of
the magnetic field.
iv) These radiations do not require any medium for propagation.
v) The radiations possess wave character and travel with the velocity of light i.e. 3x108
m/sec because of the above characteristics, the radiation is called electromagnetic
radiations or waves.

Electromagnetic radiation is explained by following characteristics:

1. Wave length:
The distance between two successive crests, troughs or between any two consecutive
identical points in the same phase of a wave is called wave length. It is denoted by the
letter (lambda).
The wave length is measured in terms of meters (m), centimeters (cm), angstrom units
(A0) nanometers (nm), picometers (pm) and also in millimicrons (m).
The S.I. unit of wavelength is meter, m
1A0 = 10–10 m or 10–8 cm
1nm = 10–9 m or 10–7 cm = 10A0
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1pm = 10–12m or 10–10 cm =10−2 A0

2.Frequency:
The number of waves that pass-through a given point in one second is known as
frequency of radiation. It is denoted by the ‘v’ (nue).
Crest  Crest
a
a 
Trough Trough
Wave motion of the radiation

SI unit of frequency is per second(s–1) or Hertz (Hz). A cycle is said to be completed when
a wave consisting of a crest and a trough passes through a point.

3.Velocity:
The distance travelled by the wave in one second is called velocity or speed of the wave
(C).
SI unit is meters per second (ms–1).
C of electromagnetic radiation in vaccum is a constant commonly called the speed of light
and is denoted by ‘c’.It is equal to 3 × 108ms–1.
4.Wave number:
The number of waves that can be present at any time in unit length is called wave
number.
It is denoted by  (nue bar).
It is the reciprocal of wave length.
1
Wave number =  =


It is expressed in per centimeter (cm–1) or per meter (m–1).

The SI unit of wave number is m–1.
Wave length, wave number𝝂̅ , frequency 𝝂 and velocity c are related as follows

c = 
5.Amplitude:
The height of the crest or the depth of the trough of the wave is called amplitude of
the wave. It is denoted by A.
The amplitude determines the strength or intensity or brightness of radiation.
6.Time period:
It is the time taken by the wave for one complete cycle or vibrations. It is denoted by
T. It is expressed in second per cycle.
1 1
T= ( where  = frequency)
𝑉 
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Electromagnetic spectrum:
The arrangement of different types of electromagnetic radiations in the order of
increasing wavelengths or decreasing frequencies is known as electromagnetic spectrum.
 increases
 10-16 10-12 10-10 10-8 10-7 10-6 10-4 10-2 101 106
 decreases
Rays Cosmic - x- Ultra Visible Near Far Micro Radio Long E decreases
rays rays rays violet IR IR wave waves RW

V I B G Y O R

Violet Indigo Blue Green Yellow Orange Red

3800 Å 4300 4800 5300 5800 6300 6900 7600 (in Å)

Limitations of Electromagnetic Wave Theory :

Electromagnetic wave theory was successful in explaining the properties of light such as
interference, diffraction etc.
But it could not explain the following:
(i) The phenomenon of black body radiation.
(ii) The photoelectric effect.
(iii) The change heat capacity of solids as a function of T.
(iv) The line spectra of atoms with special reference to hydrogen.
These phenomena could be explained only if electromagnetic waves are supposed to have
particle nature.

Black body radiation:

When a radiant energy falls on the surface of a body, a part of it is absorbed, a part of it is
reflected and the remaining energy is transmitted.
An ideal body is expected to absorb completely the radiant energy falling on it is known as a
black body. A black body is not only a perfect absorber but also a perfect emitter of
radiant energy.
A hollow sphere coated inside with a platinum black, which has a small hole in its wall can act
as a near black body.
The radiation emitted by a black body kept at high temperature is called black body
radiation.A black body radiation is the visible glow that the solid object gives off when
heated.
A graph is obtained by plotting the intensity of radiation against wave length gives the
following details.
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1. The nature of radiation depends upon the T of the black body.

2. If the energy emitted is continuous the curve should be as shown by the dotted lines.
3. At a given temperature the intensity of radiation increases with the wave length,
reaches maximum and then decreases.
4. The intensity of radiation is greatest at the medium wave lengths and least at highest
and lowest wave lengths.
5. As the temperature increases the peak of maximum intensity shifts towards the shorter
wave lengths.

Planck’s quantum theory:

In order to explain black body radiation, Max Planck proposed quantum theory of radiation.
Postulates
1. The emission of radiation from a body is due to the vibrations of the charged particles in
the body.
2. The energy is emitted or absorbed by a body discontinuously in the form of small
packets of energy called quanta.
3. The energy of each quantum of light is directly proportional to the frequency of the
radiation.
E  or, E = h
Where ‘h’ is known as Planck’s constant.

The value of ‘h’, 6.6256 × 10–34 Jsec- or 6.6256 × 10–27ergs sec-

4. In case of light, the quantum of energy is called a photon.
The total amount of energy emitted or absorbed by a body will be some whole number
multiple of quantum,
E = nh  , where n is an integer such as 1,2,3 . . . . .
This means that a body can emit or absorb energy equal to hv, 2hv, 3hv . . . . . Or any
other integral multiple of h. This is called quantization of energy.
5. The emitted radiant energy is propagated in the form of waves.
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PhotoElectric Effect:
When radiations with certain minimum frequency (ν0 ) strike the surface of a metal, the
electrons are ejected from the surface of the metal. It is called photoelectric
effect,electrons emitted are called photoelectron.
Incident light

Detector
Metal Surface
Evacuated glass tube
Electrons
− +

− +

Photoelectric effect
For each metal a certain minimum frequency is needed to eject the electrons called as

threshold frequency (  ) which differs from metal to metal.

o
K.E. of photoelectron

K.E. of photoelectron
K. E. constant

o

Frequency of absorbed Intensity of Incident

photon radiation
K.E. as a function of frequency K.E. as a function of intensity

It was explained by Einstein. When light of suitable frequency falls on a metal surface, the
light photon gives its energy to the electron of metal atom and the electron is ejected from
metal surface by absorbing this energy. The minimum energy of a photon required to
eject an electron from a metal is called work function () of the metal. The remaining part
of the energy (h  - ) of photon is used to increase the kinetic energy of the ejected
electron. If o is the threshold frequency and  , the frequency of incident light then
Work function,  = h o .
 According to Einstein, E = h 
 Kinetic energy of photo electron Ek = E -  = h − ho

CLASS EXERCISE
1. The frequency of a radiation whose wave length is 600 nm is

a) 3 x 1014 sec-1 b) 4 x 1014 sec-1

c) 5 x 1014 sec-1 d) 3 x 1015 sec-1

2. The wavelength of light having wave number 4000 cm-1 is

a) 2.5 μm b) 250 μm c) 25 μm d) 25nm μm
-1
3. What is the energy of photons that corresponds to a wave number of 2.5 × 10-5 cm ?

a) 2.5 × 10-20 erg b) 5.1 × 10-23 erg

c) 5.0 × 10-22 erg d) 8.5 × 10-22 erg

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4. The frequency of radiation having wave number 10m-1 is:

a) 10s-1 b) 3×107s-1
c) 3×1010s-1 d) 3×109s-1
5. The wavelengths of two photons are 2000Å and 4000Å respectively. What is the ratio of
their energies?
a) 1/4 b) 4 c) 1/2 d) 2
6. In photo electric effect the number of photo electrons emitted is proportional to
a) Intensity of incident beam b) Frequency of incident beam
c) Velocity of incident beam d) Work function of photo cathode
7. The kinetic energy of the photo electrons does not depend upon
a) Intensity of incident radiation b) frequency of incident radiation
c) Wavelength of incident radiation d)wave number of incident radiation

8. The work function of a metal is 3.1x10-19 J. Which frequency of photons will not cause
the ejection of electrons?
-1
a) 5 x 1014 s b) 5 x 1015 kHz
-1
c) 6 x 1014 s d) 5 x 1012 Hz
9. The work function of a metal is 4.2 eV. If radiation of 2000 falls on the metal, then the
kinetic energy of the fastest photo electrons is

a) 1.6 × 10-19 J b) 16 × 1010 J

c) 3.2 × 10-19 J d) 6.4 × 10-10 J

10. A photo electric emitter has a threshold frequency v0. When light of frequency 2v0 is

incident, the speed of photo electrons is V. When light of frequency 5v0 is incident, the

speed of photo electrons will be

a) 4V b) 2V c) 2.5V d)√2.5V

HOME EXERCISE
1. Wave theory failed to explain the following properties
a) diffraction b) interference
c) black body radiation d) all the above
2. Plank’s quantum theory is explained which of the following properties
a) quantization b) black body radiation
c) diffraction d) both a & b
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3. The electromagnetic radiation with high energy

a) radio waves b) X-rays c) Infra-red radiation d) visible light
4. The atomic transition gives rise to radiation of frequency 104 Hz. The change in energy
per mole of atoms taking place would be:
a) 3.99 × 10–6J b) 3.99J c) 6.62×10––24J d) 6.62× 10–30J
5. Two electromagnetic radiations having energy ratio 3:2 is falling on metal surface and
producing metallic luster what is the ratio of wave numbers of those radiation?
a) 1:2 b) 2:3 c) 3:2 d) 9:4
6. The energy of the photons which corresponds to light of frequency

3×1015 sec-1 is

a) 1.9876×10-15 ergs b) 2.9876×10-8 ergs

c) 1.9876×10-10 ergs d) 1.9876×10-11 ergs

7. Find the frequency of light that correspond to photons energy

5.0 x 10-5 erg

a) 2.2x1011 sec-1 b) 7.5x1021 sec-1

c) 4.0 x 10-5 sec-1 d) 4.0 x 104 sec-1

8. Photoelectric effect shows:
a) Particle-like behavior of light
b) Wave like behavior of light
c) Both wave like and particle-like behavior of light behavior of light
d) none
9. When the frequency of light incident on a metallic plate is doubled, the KE of the emitted
photoelectrons will be:
a) Doubled b) Halved
c) Increased but more than doubles of previous KE d) Unchanged
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SESSION - 4
AIM - To introduce Atomic Spectra

ATOMIC SPECTRA
Spectrum is the impression produced on a screen when radiations of a particular
wavelengths areanalyzed through a prism or diffraction grating. Spectra are broadly
classified into two.
(i) Emission Spectrum.
(ii) Absorption Spectrum.
1. Emission Spectrum:When the radiation emitted from some source, e.g., from the sun
or by passing electric discharge through a gas at low pressure or by heating some
substance to high temperature etc. is passed directly through the prism and then
received on the photographic plate, the spectrum obtained is called ‘Emission
spectrum’.
The spectrum of a radiation emitted by a substance in its excited state is an emission
spectrum.
Emission Spectrum is of two types:
a) Continuous Spectrum and b) Discontinuous Spectrum
a. Continuous Spectrum: When white light from any source such as sun, a bulb or any hot
glowing body is analyzed by passing through a prism, it is observed that it splits up into
seven different colours from violet to red,(like rainbow), as shown in fig .

7-colours
White light

Beam
•
VIBGYOR

Slit Prism
Photographic
plate

These colors are so continuous that each of them merges into the next. Hence, the spectrum
is called continuous spectrum.
It may be noted that on passing through the prism, red colour with the longest wavelength is
dedicated least while violet colour with shortest wavelength is deviated the most.
b. Discontinuous Spectrum: When gases or vapours of a chemical substance are heated
in an electric Arc or in a Bunsen flame, light is emitted. If the ray of this light is
passed through a prism, a line spectrum is produced.
• A discontinuous spectrum consisting of distinct and well-defined lines with dark areas
in between is called line spectrum. It is also called atomic spectrum.
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• The emission spectrum consisting of a series of very closely spaced lines is called band
spectrum.
Band spectrum is the characteristic of molecules. Hence it is also known as molecular
spectrum. The band spectrum is due to vibrations and rotations of atoms present in a
molecule.
Differences between line and band spectrum
Line spectrum Band spectrum
1. The line spectrum has sharp, 1. The band spectrum has many closed
distinct well defined lines. lines.

2. The line spectrum is the 2. The band spectrum is

characteristic of atoms and is characteristic of molecules and is
also called atomic spectrum. also called molecular spectrum.
3. The line spectrum is due to 3. The band spectrum is due to
transition of electrons in an vibrations and rotations of atoms in
atom. a molecule
rotations of atoms in a
4. The line spectrum is given by 4. The band spectrum is given by hot
inert gases, metal vapors and metals and molecular nonmetals.
atomized nonmetals.

2. Absorption spectra: When white light from any source is first passed through the
solution or vapours of a chemical substance and then analysed by the spectroscope, it is
observed that some dark lines are obtained.Further, it is observed that the dark lines are at
the same place where coloured lines are obtained in the emission spectra for the same
substance.
Difference between emission spectra and absorption spectra
EMISSION SPECTRA ABSORPTION SPECTRA
1. Emission spectrum is 1. Absorption spectrum is obtained when
obtained when the radiation the white light is first passed through
from the source are directly the substance and the transmitted light
analyses in the is analyzed in the spectroscope.
spectroscope.
2. It consists of bright 2. It consists of dark lines in the
coloured lines separated by otherwise continuous spectrum.
dark spaces.
3. Emission spectrum can be 3. Absorption spectrum is always
continuous spectrum (if discontinuous spectrum of dark lines.
source emits white light) or
discontinuous, i.e., line
spectrum if source emits
some coloured radiation.
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Emission Spectrum of Hydrogen:

When hydrogen gas at low pressure is taken in the discharge tube and the light emitted on
passing electric discharge is examined with a spectroscope,the spectrum obtained is
called the emission spectrum of hydrogen which contain large number of lines which
are grouped into different 5 different series,
• Lyman series,
• Balmer series
• Paschen series
• Brackett series
• Pfund series.
• Humpry series
The wave numbers of all the lines in all the series can be calculated by the Rydberg equation.
1 1 1
ν̅ = = RZ2 ( 2 − 2 )
λ n1 n2
Where n1 and n2 are whole numbers, n2> n1.

For one electron species like He+, Li2+ and Be3+, the value of R is 109677 cm–1× Z2, where Z
is the atomic number of the species.

Different series of spectral lines in hydrogen emission spectrum

Name of the n1 n2 Spectral region
series
Lyman series 1 2,3,4,5,6,7….. Ultraviolet
Balmer series 2 3,4,5,6,7… Visible
Paschen series 3 4,5,6,7…… Near infrared
Brackett series 4 5,6,7…. Infrared
Pfund series 5 6,7…. Far infrared
The wave number for any single electron species like He+, Li2+ and Be3+ can be calculated
1 1
from the equation ν̅ = Z 2 R H (n2 − n2 )
1 2
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CLASS EXERCISE
1. Number of spectral lines possible when an electron falls from fifth orbit to ground state
in hydrogen atom is
a) 4 b) 15 c) 10 d) 21
2. Which of the following electronic transitions require the largest amount of energy?
a) n = 1 to n = 2 b) n = 2 to n = 3
c) n = 3 to n = 4 d) n = 4 to n = 5
3. Which of the following spectral line is associated with a minimum wavelength?
a) n = 5 to n = 1 b) n = 4 to n = 1
c) n = 3 to n = 1 d) n = 2 to n = 1
4. Of the following transitions in hydrogen atom the one which gives an absorption line is
lowest frequency is
a) n =1 to n= 2 b) n = 3 to n = 5
c) n = 2 to n = 1 d) n = 5 to n = 3
5. The first emission line of Balmer series in H spectrum has wave number equal to
9R 7R 3R 5R
a) 400H cm−1 b) 144H cm−1 c) 4H cm−1 d) 36H cm−1
0
6. If the series limit of wave length of the Lyman series for hydrogen atoms is 912A. then
the series limit of wave length for the Balmer series of hydrogen atom is
912
a) 912Ao b) 2 × 912A° c) 4×912A° d) 2 A°

HOMEEXERCISE
1. There are three energy levels in an atom. How many spectral lines are possible in its
emission spectra?
a) One b) Two c) Three d) Four
2. The wave length of second line in the Balmer series of hydrogen spectrum is equal to
(R=Rydberg constant)
a) 36/5R b) 5R/36 c) 3R/16 d) 16/3R
3. When an electron falls from higher orbit to third orbit in hydrogen atom, the spectral
time observed
a) Balmer series b) Lyman series
c) Brackett series d) Paschen series
4. Which of the following electronic transitions require the largest amount of energy?
a) n = 1→n =2 b) n = 2→n = 3 c) n = 3→n = 4 d) n = 4 →n = 5
5. The wave number of the series limiting line for the Lyman series for hydrogen atom is (R =
109678 cm-1).
a) 82259 cm-1 b) 109678 cm-1
c) 1.2157 x 10-5 cm d) 9.1176 x 10-6 cm
20 ACTIVE SITE EDUTECH - 9844532971

SESSION – 5 AND 6

AIM-To introduce Bohr’s and Sommerfeld’s Atomic models

To overcome the objections of Rutherford model and to explain the hydrogen spectrum,Bohr
proposed a quantum mechanical model.

POSTULATES OF BOHR’S THEORY

• The electrons revolve round the nucleus with definite velocity in certain fixed closed
circular paths called orbits (or) shells (or) stationary state. These shells are
numbered as 1, 2, 3, 4 or termed as K, L, M, N from the nucleus.

• Each orbit is associated with a definite amount of energy. As long as an electron is

revolving in an orbit it neither loses nor gains energy. Hence these orbits are called
stationary states or stable orbits+
• The centrifugal force of the revolving electron in a stationary orbit is balanced by
the electrostatic attraction between the electron and the nucleus.
• Electron can revolve only in orbits whose angular momentum are an integral multiple of
the factor h/2 π.
nh
mvr = 2π
Where m = mass of electron,
v = velocity of electron,
r = radius of the orbit and
‘n’ is the integral number like, 1, 2, 3, 4 . . . , is called principal quantum number and h =
Planck’s constant
• The energy of an electron changes only when it moves from one orbit to another.
Outer orbits have higher energies while inner orbits have lower energies.
The energy is absorbed when an electron moves from inner orbit to outer orbit. The
energy is emitted when the electron jumps from outer orbit to inner orbit.
• The energy emitted or absorbed in a transition is equal to the difference between
the energies of the two orbits (E2 – E1). Energy emitted or absorbed is in the form
of quanta.
E=E2 – E1 = hv
Here E1 and E2 are the lower and higher allowed energ states.

• Expressions for radius of orbit:

Consider an electron of mass ‘m’ and charge ‘e–’ revolving round the nucleus of charge ‘Ze’ in a
circular orbit of radius ‘r’.
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Let ‘v’ be the tangential velocity of the electron. As per coulomb’s law, the electrostatic
force of attraction between the moving electron andthenucleus is –Ze2/r.

For the atom to be stable an equal centrifugal force must act away from the nucleus. This
centrifugal force is equal to –mv2/r, where ‘m’ is the mass of electron and ‘r’ is the radius of
the orbit.
In a stationary orbit
–Ze2 −mv2 Ze2
= or = mv2
r2 r r
𝑛ℎ
As per Bohr’s quantum condition, mvr = 2𝜋
nh 2 𝑛2 ℎ 2
∴v= 2πmror v = 4𝜋2𝑚2 𝑟 2

Substituting the value of v2, we get

Ze2 mn2 h2 n2 h2
or = 4π2 m2r2 or r =
r 4π2 mZe2

Radius for ‘nth’ orbit, rn =

n2 h2
4π2 mZe2

Substituting the standard values, of h, , m and e, we get radius of nth orbit rn =

0.529×n2
A°
Z
For hydrogen, Z=l and n=1 for first orbit,
The radius of the first orbit of hydrogen is 0.529 A0 or 0.0529 nm or 52.9 pm. This value
is known as Bohr’s radius. As the value of n increases, the radius of the orbit will increase.
n2 h2
In S.I units, rn = 4π2mKZe2
1
Where,K = 4π∈ (ϵ0 = permitivity of air = 8.854 × 10−12 Farad Metre)
0

• Expression for Energy of electron:

The total energy of electron is the sum of kinetic and potential energies.Kinetic energy due
to motion of electron is mv2, where m is the mass of electron and v is its velocity.
1
2
Ze2 Ze2
K.E = 2mv2 =
1
∵ mv 2 =
2r 2r
−Ze2
P.E of electron = 𝑟
Total energy of electron, En = K.E + P.E
Ze2 Ze2 1 Ze2
En= 2r − = −2
r r

Substituting the value of r, we get energy of electron in nth orbit,

−Ze2 4π2 mZe2 −2π2 mZ2 e4
En = or En=
2n2 h2 n2 h2
Substituting the values of m, e, h and𝜋 in the equation, we get
−13.6 ×𝑍2
En = eV per atom
𝑛2
−313.6×𝑍 2
or En = k cal mol–1
𝑛2
−1312×Z2 –1
or En = kJ mol
n2
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−2.18×10−11 𝑍 2
or En = erg per atom
𝑛2
−2.18×10−18 𝑍 2
orEn = j/atom
𝑛2
2π2 mK2 Z2 e4
In S.I units: En = − n 2 h2
–1
WhereK = 4𝜋𝜖 and 𝜀𝑜 beingpermittivity of air and is equal to 8.854 × 10–12 Farad metre
1
0

• Derivation of Rydberg equation:

When a gas is subjected to electric discharge or heated by a flame, the electrons in the
ground state of the atom absorb energy and they are promoted to higher quantum states.
As theyare not stable in these states, they emit energy and return to ground state or any
other lower energy states.If E2 is the energy of the higher energy state, E1 is the energy of
the lower energy state and 𝜈 is the frequency of emitted radiation E2 – E1 = h𝜈
If the numbers of the higher and lower energy states are n2 and n1 respectively, En2 =
−2π2 mZe4 1
. n2
h2 2
−2π2 mZe4 1
En1 = . n2
h2 1
−2π2 mZe4 1 1
En2 − E n1 = [n2 − n2 ]
h2 1 2
En2 − En1
But En2 − En1 = hcν̅ and ν̅ = ch
1 −2π2 mZe4 1 1
ν̅ = λ = [n2 − n2 ]
ch2 1 2

This equation is similar to Rydberg equation.

1 1 1
[ν̅ = = R × ( 2 − 2 )]
λ n1 n2
2π2 mZ e4
Rydberg constant R should be equal to R = ch3

Substituting the values, we get RH= 1,09,681cm-1.This value is almost equal to Rydberg’s
constant 1,09,677 cm–1.
The frequencies of the spectral lines in the hydrogen spectrum calculated by using Bohr’s
equation are in excellent agreement with the experimental values. This is a concrete
proof of the validity of Bohr’s theory of hydrogen atom.
• Expression for velocity of electron:
As per Bohr’s quantum conditions,
nh nh
mvr = 2π or v = 2πmr
n2 h2
∵ r = 4π2 m Ze2
nh 4π2 m Ze2
∴ v = 2πm × n2 h2

2πZe2
v= cms −1
n
Substituting the values of 𝜋, e and h in the above expression
2.18×108 ×𝑍 –1
vn= cm s
𝑛
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8
Thus the velocity of electron in the first orbit of hydrogen atom is 2.18 ×10 cm s–1. As
the number of the orbit increases, the velocity of the electron decreases.

Explanation of Hydrogen Spectrum by Bohr’s Theory:

Bohr’s theory successfully explains the origin of lines in hydrogen emission spectrum.
Hydrogen atom has only one electron. It is present in K shell of the atom (n = 1). When
hydrogen gas is subjected to electric discharge, energy is supplied. The molecules absorb
energy and split into atoms. The electrons in different atoms absorb different amounts of
energies. By the absorption of energy the electrons are excited to different higher energy
levels.

Atoms in the excited state are unstable. Therefore the electrons jump back into different
lower energy states in one or several steps. In each step the energy is emitted in the form
of radiation and is indicated by a line.
Each line has a definite frequency and thus the emission spectrum of hydrogen has many
spectral lines.
• Lyman series are obtained in UV region, when electron returns to the ground state
from higher energy levels 2, 3, 4, 5 ......... and so on.
• Balmer series are obtained in visible region when electron returns to second energy
level from higher energy levels 3, 4, 5, 6 and so on.
• Paschen series are obtained in near infrared region, when electron returns to third
energy level from higher energy levels 4, 5, 6.... And so on.
• Brackett series are obtained in mid infrared region when electron returns to fourth
energy level from higher energy levels 5, 6, 7 . . . and so on.
• Pfund series are obtained in far infrared region when electron returns to the fifth
energy level from higher energy levels 6, 7…….
The maximum number of lines produced when electrons jumps from nth level to ground level
𝑛(𝑛−1)
is equal to, Or ∑(𝑛2 − 𝑛1 )
2

Where, n2 = higher energy level.

n1 = lower energy level.
n = difference in the two energy levels.

Merits and demerits of Bohr’s Atomic model:

1. Bohr’s model explains the stability of the atom. The electron revolving in a stationary
orbit does not lose energy and hence it remains in the orbit forever.
2. Bohr’s theory successfully explains the atomic spectrum of hydrogen.
3. This theory not only explains hydrogen spectrum but also explains the spectra of one
2+ 3+
electron species such as He+, Li and Be etc.
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4. The experimentally determined frequencies of spectral lines are in close agreement with
those calculated by Bohr’s theory.
5. The value of Rydberg constant for hydrogen calculated from Bohr’s equation tallies with
the value determined experimentally.
Limitations of Bohr’s model:
1. Bohr’s theory fails to explain the spectra of multielectron atoms.
2. It could not explain the fine structure of atomic spectrum.
3. It does not explain the splitting of spectral lines into a group of finer lines under the
influence of magnetic field (Zeeman Effect) and electric field (Stark effect).
4. Bohr’s theory predicts definite orbits for revolving electron. It is against the wave
nature of electron.
5. Bohr’s theory is not in agreement with Heisenberg’s uncertainty principle.

Sommerfeld’s Atomic Model:

It is an extension of Bohr’s model. In this model, the electrons in an atom revolve around
the nuclei in elliptical orbit. The circular path is a special case of ellipse. Association of
elliptical orbits with circular orbits explains the fine line spectrum of atoms.
Radial Velocity
Tar velocity
•
Avg Velocity
• major axis
focus

Minor axis
n=4,k=4
n=4,k=3
n=4,k=2

• n=4, k=1, k  0

Sommerfeld’s orbits in hydrogen atom

The main postulates are:
i) The motion of electron in closed circular orbits is influenced by its own nucleus and is set
up into closed elliptical paths of definite energy levels.
ii) The nucleus is one of the foci for all these orbits.
iii) The angular momentum of electron in closed elliptical paths is also quantized i.e. k (h/2),
where k is another integer except zero.
25 ACTIVE SITE EDUTECH - 9844532971

n length of major axis

iv) The ratio = length of suggests for the possible number of subshells in a shell.
k min or axis

Possible values of k for n = 4 are 1, 2, 3, 4 respectively. For any given value of n, k cannot
be zero as in that case, the ellipse would degenerate into a straight line passing through
the nucleus. When n = k, path becomes circular.
CLASS EXERCISE
1. The ratio of radius of 2nd and 3rd Bohr orbit is
a) 3 : 2 b) 9 : 4 c) 2 : 3 d) 4 : 9
2. According to Bohr’s model, the angular momentum of an electron in 4th orbit is
a) h/3 b)h/2 c) 2h/ d) 3h/2
3. The radius of Bohr’s first orbit in hydrogen atom is 0.053nm. The radius of second orbit

of He+ would be
a) 0.0265 nm b) 0.053 nm c) 0.116 nm d) 0.212 nm
4. The minimum energy required to excite a hydrogen atom from its ground state
a) 13.6 eV b) -13.eV c) 3.4 eV d) 10.2 eV
5. The ratio of kinetic energy and potential energy of an electron in a Bohr orbit of a
hydrogen atom is
a) 1:2 b) -1:2 c) 1:1 d) -1:1
6. The ratio of potential energy and total energy of an electron in a Bohr orbit of a hydrogen
atom is
a) 2:1 b)-1:2 c) 1:1 d) -1:2
7. The ratio of kinetic energy and total energy of an electron in a Bohr orbit of a hydrogen
atom is
a) 1 : -1 b) -2:1 c) 1:1 d) -1:2
8. In a certain Bohr orbit the total energy is -4.9 eV for this orbit, the kinetic energy and
potential energy are respectively.
a) 9.8 eV, - 4.9 eV b) 4.9 eV, - 98 eV
c) 4.9 eV, - 4.9 eV d) 9.8 eV, - 9.8 eV
9. If speed of electron in first Bohr orbit of hydrogen be ‘x’, then speed of the electron in
second orbit of He+ is:
a) x/2 b) 2x c) x d)4x
10. The ratio of the difference in energy between the first and second Bohr orbits to that
between the second and third Bohr orbit is
a) 1/2 b) 1/3 c) 4/9 d) 27/5
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HOME EXERCISE
+2
1. Calculate the ratio of the radius of in 3rd energy level of Li ion of 2nd energy level of He+
ion
a)3:2 b)1:2 c)2:3 d)1:1
2. Of the following, which of the statement(s) regarding Bohr’s theory is wrong?
a) Kinetic energy of an electron is half of the magnitude of its potential energy
b) Kinetic energy of an electron is negative of total energy of electron
c) Energy of electron decreases with increase in the value of the principal quantum
number
d) The ionization energy of H-atom in the first excited state is negative of one fourth of
the energy of an electron in the ground state.
3. If first ionization energy of hydrogen is E, then the ionization energy of He+ would be:
a) E b) 2E c) 0.5E d) 4E
4. The ratio of radii of first orbits of H, He+ and Li2 is:
a) 1:2:3 b) 6:3:2 c) 1:4:9 d) 9:4:1
5. The angular momentum of an electron in the M shell of H- atom is
a) 3h /2  b) h/ 2  c) h /  d) 2h / 

6. If ionization potential of H-atom is 13.6 eV, the ionization potential of He+ is

a) 54.4 eV b) 6.8eV c)13.6eV d) 27.2eV
18
7. The ionization energy of H-atom is its ground state is 2.17x10- J. The ionization energy
+2
of Li in the ground state will be

a) 1.953 x 10-15 J b) 1.953 x 10-16 J

c) 1.953 x 10-17J d) 1.953 x 10-18 J

8. If the value of E = - 78.5 K.cal /mole. The order of the orbit in hydrogen atom is
a) 4 b) 3 c) 2 d) 1
9. The ionization potential of hydrogen atom is 13.6 eV. The energy required to remove an
electron in the n = 2 state of the hydrogen atom is
a) 3.4 eV b) 6.8 eV c) 13.6 eV d) 27.2 eV
10. The minimum energy (numerical value) required to be supplied to H-atom to push its
electron from 2nd orbit to the 3rd orbit
a) 1.9 eV b) 2.2 eV c) 2.7 eV d) 7.0 eV
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SESSION- 7
AIM
1) To introduce de Broglie’s theory
2) To introduce Heisenberg’s Uncertainty principle.

DUALNATURE OF MATTER(DE BROGLIE’S WAVE THEORY)

Light exhibits different properties such as diffraction, interference, photoelectric
effect, Compton effect, reflection and refraction. The phenomenon of diffraction and
interference can be explained by the wave nature of the light. But the phenomenon of
photoelectric effect and Compton Effect can be explained by the particle nature of the
light. Thus light has dual nature. De-Broglie proposed that matter like radiation, should also
exhibit dual behavior.
hc
Einstein’s generalization of Planck’s theory is given as, E = hν = λ
2
Einstein’s mass energy relationship is E = mc
Equating above two equations, we get
hc 2 h h
= mc or = mc or λ = mc
λ λ
h
Where ‘c’ is the velocity of light. If the velocity of micro particle is ‘v’ then, λ = mV
This is de Broglie’s equation,
Where ‘λ’ is the de Broglie’s wave length, ‘m’ is the mass of the moving particle and ‘h’ is
Planck’s constant.
h
P = mv or λ = P .
Here 𝜆 signifies wave nature and P signifies particle nature.
This is applicable to microparticles like electron, proton, etc., and not applicable for
macrobodies like cricket ball, bullet etc.
The electron moving with high speed possesses both the particle nature and the wave nature.
The waves associated with material particles are known as matter waves or particle waves.

The Heisenberg’s uncertainty principle:

“It is impossible to determine simultaneously and accurately the exact position and
momentum or velocity of a sub-atomic particle like electron in an atom”.
One can determine the position of a particle very accurately, and then the determination of
its velocity becomes less accurate. Similarly, one can determine the velocity of a particle
28 ACTIVE SITE EDUTECH - 9844532971

very accurately, and then the determination of its position becomes less accurate. The
certainty in one factor introduces the uncertainty in another factor.
If the uncertainty in the determination of the position of a small particle is given by Δx and
uncertainty in its momentum is Δp, then
ℎ
(Δx) (Δp) ≥ 𝑛𝜋
Where n = 1,2,3,4.........
For an electron revolving around the nucleus in an atom the value of n is nearly 4.
Thus Heisenberg’s principle can also be stated as the product of uncertainty in position and
momentum of an electron like micro particle moving with high speed cannot be less than h/4.
Heisenberg’s equation can also be written as,
ℎ
(Δx) (Δv) ≥ 4𝜋𝑚
Where m is the mass of the particle and Δv is uncertainty in velocity.
If the position of the particle is known exactly (Δx = 0), Δv becomes infinity (∞) and vice
versa. Heisenberg's uncertainty principle is not applicable to those objects which cannot
change their position by themselves when a light falls on them. It is applicable for micro
particles like electrons.
Significance of Heisenberg’s uncertainty principle:
Like de Broglie equation, although Heisenberg’s uncertainty principle holds good for all
objects but it is significance only for microscopic particles. The reason for this is quite
obvious. The energy of the photon is insufficient to change the position and velocity of
bigger bodies when it collides with them. For example, the light from a torch falling on a
running rat in a dark room, neither change the speed of the rat nor its direction, i.e.,
position.
This may be further illustrated with the following examples:
For a particle of mass 1 mg, we have
ℎ 6.625×10−34 𝑘𝑔𝑚2 𝑠−1
Δx.Δ𝜐 = = = 10−28 𝑚2 𝑠 −1
4𝜋𝑚 4×3.1416×(10−6 𝑘𝑔)
Thus, the product of Δx and Δ𝜐 is extremely small. For particles of mass greater than 1 mg,
the product will still smaller. Hence, these values are negligible.
For a microscopic particle like an electron, we have
ℎ 6.625×10−34 𝑘𝑔𝑚2 𝑠−1
Δx.Δ𝜐 = 4𝜋𝑚 = 4×3.1416×(9×10−31 𝑘𝑔) ≈ 10−4 𝑚2 𝑠 −1

CLASS EXERCISE
–1
1. A ball of 100g mass is thrown with a velocity of 100ms . The wavelength of the de Broglie
wave associated with the ball is about
a) 6.63 × 10–35 m b) 6.63 × 10–30 m
c) 6.63 × 10–35 cm d) 6.63 × 10–33 m
2. If kinetic energy of a proton is increased nine times the wavelength of the de-Broglie
wave associated with it would become
a) 3 times b) 9 times c) 1/3 times d) 1/9 times
29 ACTIVE SITE EDUTECH - 9844532971

3. Number of waves made by a Bohr electron in one complete revolution in the 3rd orbit
a) 1 b) 2 c) 3 d) 4
-10
4. The uncertainty in position and velocity of a particle are 10 m and
5.27x10-24ms-1 respectively. Calculate the mass of mass of the particle.(h=6.625 10-34 J-s)
5. Calculate the uncertainty in velocity a cricket ball of mass 150g. if the uncertainty in its
0
positionis the order of 1A (h=6.6x10-34kg m2 s-1)
6. In an atom, an electron is moving with a speed of 600 m sec-1 with an accuracy of 0.005%
certainty with the position of the electron can be located is:(h=6.6x10-34kg m2 s-1, mass
of electron=9.1 x 10-31kg)
a) 1.52x10-4 m b)5.1x10-3 m c)1.92 x10-3 m d) 3.84 x 10-3

HOME EXERCISE
1. The de Broglie wavelength of 1mg grain of sand blown by a 20ms-1 wind is:
a) 3.3x10-29 b)3.3x10-21 m c) 3.3  10-49M d) 3.3  10-42 m
2. If the kinetic energy of an electron is increased 4 times, the wavelength of the Broglie
wave associated with it would become:
1 1
a) 4times b) 2times c) times d) times
2 4
3. The momentum of the particle having the wave length of 1Å is
a) 6.6x 10-19 gram cm/sec b) 6.6 x 1019 gram cm/sec
c) 6.6 x 1034 gram cm/sec d) 6.6 x 10-34 gram cm/sec

4. If the uncertainty in the position of an electron is 10-8cm, the uncertainty in its velocity
is
a) 3×108 cm/sec b) 5.8×107 cm/sec
c) 6.625×109 cm/sec d) 7.35 × 10-8 cm/sec
5. The uncertainty in momentum of an electron is 1x10-5 kg-m/s. The uncertainty in its
position will be (h = 6.6x10-34 Joule-sec)
a) 1.05 x 10-28m b) 1.05 x 10-26 m
c) 5.27 x 10-30 m d) 5.25 x 10-28 m
6. The uncertainty in the momentum of an electron is 10-5kg.m/sec. The uncertainty in its
position will be
a) 1.05 x 10-28 m b) 1.05 x 10-26m
c) 5.27 x 10-30m d) 5.25 x 10-25m
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SESSION –8 AND 9
AIM
1) To introduce wave mechanical model of atom.
2) To introduce Quantum numbers.
3) To understand shapes of orbitals and Probability distribution.
THEORY
Classical mechanics, based on Newton’s laws of motion, was successful in explaining the
motion of macroscopic bodies like falling stones or motion of planets around the sun etc. But
it failed when applied to microscopic particles like electrons, atoms, molecules etc.Hence
new branch introduced called as ‘Quantum mechanics’.
Schrodinger Wave Equation:
Quantum mechanics, as developed by Erwin Schrodinger is based on the wave
motionassociated with the particles. The Schrodinger differential wave equation is given by
∂2 ψ ∂2 ψ ∂2 ψ 8π2 m
+ + + (E − V)ψ
∂x2 ∂z2 ∂y2 h2

Here x, y, z are Cartesian coordinates of the electron

m = mass of electron
h = Planck’s constant
E = total energy of the electron (KE + PE)
V = potential energy of the electron (PE)
ψ= wave function of the electron.
Significance of ψ: ψ is the wave function. It gives the amplitude of the electron wave.

The intensity of light is proportional to the square of amplitude (ψ2). Just as 𝛙2 indicates
the density of photons in space, 𝛙2 in case of electron wave denotes the probability of
finding an electron in the space or probability of finding the electron is also maximum.
Quantum numbers:
The behavior of an electron in an atom is described mathematically by a wave function or
orbital. They are principal quantum number, azimuthal quantum number, magnetic quantum
number and spin quantum number.
‘Set of numbers used to describe energy, size, shape of orbitals in an atom’ called as
quantum numbers.
1.Principal quantum number(n):
• ‘n’ can be any whole number value such as 1,2,3,4, etc. The energy shells corresponding
to these numbers are K, L, M, N, etc.
• Principal Quantum no. indicates the main energy level to which the electron belongs.
It also indicates the average distance of an electron from nucleus and also the
speed of the atomic electron.
• As the ‘n’ value increases the distance of electron from the nucleus increases and its
energy also increases.
31 ACTIVE SITE EDUTECH - 9844532971

• The maximum no. of electrons that can be present in an orbit is given by 2𝑛2 . The
maximum no. of electron in K, L, M, and N shells are 2,8,18 and 32 respectively.
0.529×𝑛2
• The radius of the orbit is given by the expression: rn = Ao.
𝑍
• The energy of the electron/orbit is given by the expression.
−13.6×𝑍 2
En= cm/sec
𝑛
2.18×108 ×𝑍
• The velocity of the electron is given by the expression. Vn= cm /sec.
𝑛

2. Azimuthal Quantum Number:

• Azimuthal Quantum number was introduced by Sommerfeld’s to explain the fine
spectrum.
• It is also called as secondary quantum no. or orbital angular momentum quantum
number or subsidiary quantum number.
• It is denoted by l.
• ‘l’ can have the values from 0 to (n-1), a total of ‘n’ values. ‘l’ values 0,1,2,3 indicates
s,p,d,f. s,p,d and f are spectroscope terms which indicates sharp. Principle, diffuse and
fundamental respectively.
• Azimuthal Quantum number indicates the sub-shell to which the electron belongs. It
also determines the shapes of the orbital in which the electron is present.
• Each main energy shell can have ‘n’ number of sub-shells.
n l

1 0 (1s)

2 0 (2s), 1 (2p)

3 0 (3s), 1 (3p), 2(3d)

4 0 (4s), 1(4p), 2(4d), 3(4f)

• The orbital angular momentum (L) of an electron is given by the expression: L=

ℎ
√𝑙(𝑙 + 1)2𝜋

3. Magnetic Quantum number:

• Magnetic quantum number was introduced by Lande to explain Zeeman Effect.
• It is denoted by m or ml.
• This quantum number refers to different orientations of electron could in a particular
subshell. These orientations are called the orbitals.
• An electron due to its orbital motion around the nucleus generates an electric .This
electric field in turn produces a magnetic field which can interact with the external
magnetic field. Thus, under the influence of the external magnetic field, the electrons
of a subshell can orient themselves in certain preferred regions of space around the
32 ACTIVE SITE EDUTECH - 9844532971

nucleus called orbitals. The magnetic quantum number determines the number of
preferred orientations of the electron present in a subshell.
Since each orientation corresponds to an orbital, therefore, the magnetic quantum
number determines the number of orbitals present in any subshell.
• ‘m’ can have values from – 𝑙 to +𝑙 including zero, a total (2 𝑙+1) values.
Subshell 𝒍 m values No. of orientations (Orbitals)
s 0 0 1
p 1 -1, 0, +1 3
d 2 -2, -1, 0, +1, +2 5
F 3 -3, -2, -1, 0, +1, +2, +3 7
• When l = 0, m has only one value, m = 0. The sub-level‘s’ has one orbital called s orbital.
• When l =1, m can have 3 values m = –1, 0, +1. The sub-level ‘p’ has three space
orientations or three orbitals. The three orbitals are designated as px, py and pz.
• When l = 2, m can have 5 values m = –2,–1, 0, +1, +2. The sub-level ‘d’ has five space
orientations or five orbitals. The five orbitals are designated as dxy, dyz, dzx,
dx2 −y2 and dz2 .
• When l = 3, m can have 7 values m = –3,–2,–1,0,+1,+2,+3. The sub-level ‘f’ has seven
space orientations or seven orbitals.
The magnetic quantum number gives orientation of orbitals in space. All the orbitals
present in a sublevel have same energy and shape. They are called ‘degenerate
orbitals’, which differ in their spatial orientation.
• Each value of ‘m’ constitutes an orbital in the sublevel.
• Maximum no. of electrons in subshell : 2(2𝑙+1) or (4 𝑙+2).

4. Spin Quantum Number:

• Spin Quantum number was proposed by Uhlenbeck and Goudsmith.
• It is denoted by ‘s’ or ‘ms’.
• It indicates the direction of spinning of electron present in any orbital.
• Since the electron in an orbital can spin either in the clockwise direction or in anti-
clockwise direction, hence for a given value of m, s can have only two values, i.e., +1/2
and -1/2 or these are very often represented by two arrows pointing in the opposite
direction, i.e.,↑and ↓.
If an orbital contains two electrons, the two magnetic moments oppose and cancel
each other.
Thus, in an atom, if all the orbitals are fully filled, net magnetic moment is zero and
the substance is diamagnetic (i.e., repelled by the external magnetic field). However,
33 ACTIVE SITE EDUTECH - 9844532971

if some half-filled orbitals are present, the substance has a net magnetic moment and
is paramagnetic (i.e., attracted by the external magnetic field).
• The spin angular momentum (𝜇 s) of an electron is given by
h
μs = √s(s + 1) 2π

Atomic Orbital:
The three dimensional space around the nucleus where the probability of finding the
electron is maximum is called an atomic orbital.
Differences between orbit and orbital:
Orbit Orbital
1. n orbit is a well-defined circular 1. An orbital is the region of space
path 1.aroundthe
An orbital isaround
the region
theof space
nucleus
around the
around the nucleus in which the where theprobability of finding the
electronrevolves. electron is maximum (95%)

2. An orbit represents the movement 2. An orbital represents the movement

of electron in one plane. of electron in three dimensional
spaces.
3. An orbit means the position as well 3. In an orbital it is not possible to
as thevelocity of the electron can find theposition as well as velocity
be known with Certainty. of the electroncan be known with
certainty.
4. Orbits are circular or elliptical 4. They have different shapes like
shaped. spherical,dumbbell etc Orbitals ha

5. Orbits do not have directional 5. Except ‘s’ orbitals, all other

characteristics. orbitals have directional
characteristics
6. An orbit can have a maximum 6. An orbital can accommodate a
number of2n2 electrons. maximumof only two electrons.

Node- The three dimensional space around the nucleus where the probability of finding
the electron is minimum or zero.
y
z
Nucleus
node
x

(2s)

(1s)
Types of Nodes:
Nodes are of two types: a) Radial Node b) Angular Node
34 ACTIVE SITE EDUTECH - 9844532971

A radial node is the spherical region around then nucleus, where the probability if finding
the electron is zero (Ψ2 = 0).
Similarly,nodal plane(angular plane) have zero probability of finding electron.
Calculation of no. of nodes:
No. of Radial nodes = n−𝑙 − 1
No. of angular nodes = 𝑙
Total no. of nodes = n-1
Ex: In a 3p -orbital
No. of Radial nodes = 3-1-1 = 1
No. of angular nodes = 1
Total no. of nodes = 2.
Shapes of Orbitals:
• s –Orbitals: s- Orbital can accommodate electrons with l = 0 and these orbitals are
present in every orbit starting from 1st orbit.

Orbital in which e-s with n=1 , l = 0 are present is called 1s - orbital.

All s-orbitals are spherical in shape and the size of sphere increases with ‘n’ value. s -
Orbitals are spherically symmetrical because the probability of finding the electron around
the nucleus is same in all directions.
• p – Orbitals:
p- Sublevel begins from 2nd orbit. For p - sublevel l = 1, indicates that each p - sub level
contains three orbitals with ‘m’ values –1, 0, +1. These are designated as px, py and pz,
depending on the axis in which electron density is present.

In px-orbital, electron density is concentrated along the x-axis.

p-Orbitals have dumb-bell shape. Each p -orbital has two lobes separated by one nodal plane.
The probability density function is zero on the plane where the two lobes touch each other.
The nodal planes for px, py and pz - orbitals are YZ, ZX and XY - planes respectively.
The three orbitals present in a given p - sublevel will have same shape, size and energy but
different orientations (differ in m value). These three orbitals are perpendicular to each
other and the angle between any two p - orbitals is 90o.
• d - Orbitals:begins from 3rd orbit (n = 3). For d- sub level l= 2, indicates that each d
- sublevel contains five orbitals with ‘m’ values –2, –1, 0, +1, +2. These are designated as
dxy,dyz,dzx, 𝑑𝑥 2 −𝑦 2 and d𝑧 2 .
35 ACTIVE SITE EDUTECH - 9844532971

All the d-orbitals (except d𝑧 2 ) have double dumb-bell shape. Each d-orbital has four lobes
separated by two nodal planes.
In case of dxy, dyz and dzxorbitals, lobes are present in between the corresponding axes.
i.e.,between x and yaxis in case of dxy orbital. Whereasin d𝑥 2 −𝑦 2 and d𝑧 2 orbitals lobes are
present along the axes. dxy Orbital contains yz and zx as nodal planes. dyz and dzx contain
(xy,zx) and (xy,yz) planes respectively. d𝑥 2 −𝑦 2 orbitalcontains two nodal planes perpendicular
to each other and which make an angle of 45o with respect to x and y axes. 𝑑 2 orbital does
𝑧
not contain nodal planes.
5 dorbitals present in a given d- sublevel will have same energy in the ground state.

CLASS EXERCISE
1. If the above radial probability curve indicates ‘2s’ orbital, the distance between the
peak points X.Y is:

a) 2.07Å b) 1.59Å c) 0.53Å d) 1.1Å

2. The wave function curve which crosses ‘x’ axis maximum number of times in the graph
drawn between distance from nucleus r(on x axis) and radial wave function R(ψr).
a)4d b)4p c)4s d)4f

3. The number of nodal planes is greatest for the orbital:

a) 4s b) 2p c) 3d d) 2s
4. The radial distribution curve of the orbital with double dumbbell shape in the 4th principle
shell consists of ‘n’ nodes, n is
a) 2 b) 0 c)1 d) 3
5. Which one of the following sets of quantum numbers represents as impossible
arrangements?
n l m s
a) 3 2 –2 ½
b) 4 0 0 ½
c) 3 2 –3 ½
d) 5 3 0 –1/2.
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6. Correct set of four quantum numbers for the valence (outermost) electron of rubidium (Z
= 37) is
1 1 1 1
a) 5, 0, 0, +b) 5, 1, 0, + c) 5, 1, 1, + d) 6, 0, 0, +
2 2 2 2
7. The maximum number of electrons in an orbital having same spin quantum number will be:
a) l + 2 b) 2l + 1 c) l(l + 1) d) √l(l + 1)
1
8. The four quantum number of last electron of an atom are 4, 0, 0, + then atomic number
2
of that element could be
a) 19 b) 55 c) 36 d) 37
9. The number of atomic orbitals with quantum numbers n = 3, l = 1, m = 0
a) 1 b) 6 c) 3 d) 5
HOME EXERCISE
1. Which of the following can be negative?
a) 4πr2ψ2 b) 4πr2ψ2dr c) ψ d) ψ2

2. The quantum number not obtained from the Schrodinger’s wave equation is
a) n b) l c) m d) s
3. Maxima’s in Radial probability distribution curve of 2s is
a) One b) Two c) Three d) Four
4. In which of following case would the probability of finding an electron in dxyorbital be
zero?
a) Xy and yz plane b) xy and planes
c) xz and yz planes d) z-direction, yz and xz planes
5. The principal quantum number of an atom is related to the
a) Size of the orbital b) spin angular momentum
c) Orbital angular momentum d) orientation of the orbital in space
6. The orbital angular momentum of an electron in 2s orbital is:
1 h h h
a) . b) zero c) d) 2.
2 2 2 2
7. What will be all 4-Sets of Quantum Number for last electron of sodium?
a) n = 3 l=0 m=0 s = +1/2
b) n = 3 l=1 m=1 s = +1/2
c) n = 2 l=0 m=0 s = +1/2
d) n = 2 l = 1 m=1 s = +1/2
8. p - orbitals of an atom in presence of magnetic field are:
a) Threefold degenerate b) Two fold degenerate
c) Non-degenerate d) none of these
9. The quantum number that is no way related to an orbital
a) principal b) azimuthal c) magnetic d) spin
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SESSION – 10
AIM-To introduce Electronic Configuration
ENERGY OF ORBITALS
The energy of an electron in a hydrogen atom is determined only by the principal quantum
number. Within a shell, all hydrogen orbitals have the same energy, independent of the other
quantum numbers.
1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f
Although the shapes of 2s and 2p orbitals are different, an electron has the same energy
when it is in 2s orbital or 2p orbital. The energy of an electron in a multielectron atom
depends, not only on its principal quantum number, but also on its azimuthal quantum number.
The s, p, d and f orbitals within a given shell have slightly different energies in a multi
electron atom.

Electronic configuration of multi electron atoms:

The distribution and arrangement of electrons in the main shells, subshells and orbitals
of an atom is called electronic configuration of the element.
• Aufbau Principle:
“In the ground state of the atoms, the orbitals are filled in order of their increasing
energies”.
In other words electrons first occupy the lowest energy orbital available to them and enter
into higher energy orbitals only after the lower energy orbitals are filled.
The relative energy of an orbital is given by
(n +l )rule. As(n+l) value increases, the energy of orbital increases.
• The orbital with the lowest (n + l) value is filled first.
• When two or more orbitals have the same (n +l) value, the one with the lowest ‘n’
value (or) highest ‘l ’ value is preferred in filling.
Exp-Consider two orbitals 3d and 4s.
n+l value of 3d = 3 + 2 = 5 and of 4s = 4 + 0 = 4. Since 4s has lowest(n +l) value, it is filled
first before filling taking place in 3d.
Consider the orbitals 3d, 4p and 5s
The (n + l) value of 3d = 3 + 2 = 5
The (n +l) value of 4p = 4 + 1 = 5
The (n +l) value of 5s = 5 + 0 = 5
These three values are same. Since the ‘n’ value is lower to 3d orbitals, the electrons prefer
to enter in 3d, then 4p and 5s.
The order of increasing energy of atomic orbitals is:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s and so on.
The sequence in which the electrons occupy various orbitals can be easily remembered with
the help of Moeller’s diagram as shown in Fig
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• Pauli’s Exclusion principle:stated as “No two electrons in an atom can have the
same set of values for all the four quantum numbers”.This means that two
electrons in an orbital may have the same n, same l and same m but differ in spin
quantum number. In an orbital if one electron has clockwise spin, the other has
anticlockwise spin. It follows that an orbital can hold a maximum of two electrons with
opposite spins.
Exp- helium atom has two electrons in its 1s orbital. Quantum numbers for first electron are
n =1, l = 0, m = 0 and s = +1/2. Quantum numbers for second electron are:
n =1, l = 0, m =0, s = –1/2.
The two electrons have the same value for n, same value for l and same value for m but
differ in s.

• Hund’s rule of maximum multiplicity:

According to this rule, when electrons are filled in degenerate orbitals of a subshell,
pairing of an electron takes place only when each orbital of the subshell is filled with
one electron each.It can be also stated that, in ground state of an atom, the
configuration which has more number of unpaired electrons is most stable.
Thus in s, p, d and f subshells, pairing starts from 2nd, 4th, 6th and 8th electrons respectively.
Ex: Electronic configuration of N (7) is 1s2 2s2 2p3.
The electrons in 2p subshell are occupied sing ally. i.e., 1s2 2s2 2𝑝𝑥1 2𝑝𝑦1 2𝑝𝑧1
39 ACTIVE SITE EDUTECH - 9844532971

Electronic configuration of elements from 1 to 30

1 H 1s1 1s1
2 He 1s2 1s2
3 Li 1s2 2s1 [He] 2s1
4 Be 1s2 2s2 [He] 2s2
5 B 1s2 2s2 2p1 [He] 2s2 2p1
6 C 1s2 2s2 2p2 [He] 2s2 2p2
7 N 1s2 2s2 2p3 [He] 2s2 2p3
8 O 1s2 2s2 2p4 [He] 2s2 2p4
9 F 1s2 2s2 2p5 [He] 2s2 2p5
10 Ne 1s2 2s2 2p6 [He] 2s2 2p6
11 Na 1s2 2s2 2p6 3s1 [Ne] 3s1
12 Mg 1s2 2s2 2p6 3s2 [Ne] 3s2
13 Al 1s2 2s2 2p6 3s2 3p1 [Ne] 3s2 3p1
14 Si 1s2 2s2 2p6 3s2 3p2 [Ne] 3s2 3p2
15 P 1s2 2s2 2p6 3s2 3p3 [Ne] 3s2 3p3
16 S 1s2 2s2 2p6 3s2 3p4 [Ne]3s2 3p4
17 Cl 1s2 2s2 2p6 3s2 3p5 [Ne] 3s2 3p5
18 Ar 1s2 2s2 2p6 3s2 3p6 [Ne] 3s2 3p6
19 K 1s2 2s2 2p6 3s2 3p6 4s1 [Ar] 4s1
20 Ca 1s2 2s2 2p6 3s2 3p6 4s2 [Ar] 4s2
21 Sc 1s2 2s2 2p6 3s2 3p6 3d1 4s2 [Ar] 3d1 4s2
22 Ti 1s2 2s2 2p6 3s2 3p6 3d2 4s2 [Ar] 3d2 4s2
23 V 1s2 2s2 2p6 3s2 3p6 3d3 4s2 [Ar] 3d3 4s2
24 Cr 1s2 2s2 2p6 3s2 3p6 3d5 4s1 [Ar] 3d5 4s1
25 Mn 1s2 2s2 2p6 3s2 3p6 3d5 4s2 [Ar] 3d5 4s2
26 Fe 1s2 2s2 2p6 3s2 3p6 3d6 4s2 [Ar] 3d6 4s2
27 Co 1s2 2s2 2p6 3s2 3p6 3d7 4s2 [Ar] 3d7 4s2
28 Ni 1s2 2s2 2p6 3s2 3p6 3d8 4s2 [Ar] 3d8 4s2
29 Cu 1s2 2s2 2p6 3s2 3p6 3d10 4s1 [Ar] 3d10
4s1
30 Zn 1s2 2s2 2p6 3s2 3p6 3d10 4s2 [Ar] 3d10
4s2
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Stability of atoms
Extra stability is associated with atoms in which degenerate orbitals are either half-filled
or completely filled due to
(1) Symmetrical distribution of electrons
(2) Exchange energy. Greater the exchange energy greater is the stability.
The presence of half-filled and completely filled degenerate orbitals gives greater
stability to atoms.
1
It is for this reason the electronic configurations of Cr and Cu are represented as [Ar] 4s
5 1 10
3d and [Ar] 4s 3d respectively.

CLASS EXERCISE
1. When 3d-orbital is complete, the newly entering electron goes into:
a) 4f b) 4s c) 4p d) 4d
2. An electron will have the highest energy in the set:
a)3, 2, 1, ½ b)4, 2, –1, 1/2 c)4, 1, 0, –1/2 d) 5, 0, 0, ½
3. Which has minimum number of unpaired d-electrons?
3+ 3+ 2+ 3+
a) Fe b) Co c) Co d) Mn
4. Which of the following is violation of Pauli’s exclusion principle?
2s 2p 2s 2p 2s 2p 2s 2p

a) b) c) d)
5. The number of electrons in M shell of an element with atomic number 24 is
a) 24 b) 12 c) 13 d) 8

HOME EXERCISE
1. The maximum number of unpaired electrons present in 4f -energy level is:
a) 5 b) 7 c) 10 d) 6
2. The number of unpaired electrons in fluorine atom is:
a) 7 b) 5 c) 1 d) 2
3. Which set has the same number of unpaired electrons in their ground state?
– 3+ 3+
a) N, P, V b) Na, P, Cl c)Na + , Mg 2+ , Al d)Cl ,Fe ,Cr
4. In which of the following electron distributions in ground state, only the Hund’s rule is
violated
2s 2p 2s 2p 2s 2p 2s 2p

a) b) c) d)
8 2
5. Electronic configuration of Ni is [Ar] 3d , 4s . The electronic configuration of
next element is:
10 1 9 2
a) [Ar] 3d , 4s b) [Ar] 3d , 4s
8 2 1
c) Ar] 3d , 4s , 4p d) none of these
 School/Board Exam. Type Questions
Very Short Answer Type Questions:
1. What is the SI unit of frequency?
Sol. hertz or s–1.
2. Which quantum number determines the orientation of the orbital?
Sol. Magnetic quantum number (ml).
3. What are isoelectronic species?
Sol. These are the chemical species which have same number of electrons.
4. Out of manganese and iron, which has higher number of unpaired electrons?
Sol. 25Mn  [Ar]3d 54s 2

26Fe  [Ar]3d 6 2
4s
Manganese has highest number of unpaired electrons (5) as compared to iron (4).
5. What are the possible value of ‘ml’ for 2p-orbitals?
Sol. For 2p-orbital, n = 2; l = 1

Therefore, ‘ml’ has values = –l to +l including 0. The possible values of ‘ml’ are = 1,1,0 .
6. Atomic spectra is a line spectra. What does this mean?
Sol. Atomic spectra is a line spectra or discontinuous spectra. This means that the spectrum of atoms consists
of a series of bright lines separated from each other by a dark space.
7. How many angular nodes are present in 3dxy orbital?
Sol. Angular nodes = ‘l’
For 3dxy orbital, l = 2
Therefore angular nodes are 2.
8. Name the series of lines obtained when electrons in hydrogen atoms fall from higher energy level to first energy
level.
Sol. Lyman series.
9. Which scientist discovered nucleus?
Sol. Rutherford.
24 Structure of Atom Solution of Assignment (Set-1)

10. What is the formula for calculating the orbit angular momentum of an electron in a stationary state?

h
Sol. Angular momentum, mevr = n ; where n = 1, 2, 3, ...
2
Short Answer Type Questions:
11. Define atomic number. Give the atomic number of zinc.
Sol. Atomic number is defined as the total unit positive charges on the nucleus i.e., the equal to the number of
protons. Atomic number designated as ‘Z’.
Atomic number of zinc is 30 (Z = 30).
12. Write the three main observations made by Rutherford from his -scattering experiment.
Sol. (i) Most of the -particles striking the thin gold foil passed undeflected.
(ii) A small fraction of -particles were deflected by small angles.
(iii) A very few -particles (1 in 20,000) bounced back, i.e., they were deflected by nearly 180°.
13. Write down the important feature of plum pudding model of atom.
Sol. Plum Pudding model is the Thomson’s model of atom. An important feature of this model is that the mass of
the atom is assumed to be uniformly distributed over the atom.
3+ 2+ion?
14. Why Fe ion is more stable than Fe
Sol. The electronic configuration of iron

 26Fe = [Ar]

3d6 4s2

 Fe2+= [Ar]

3d6 4s0
[In this case (Fe2+), two electrons gets removed from the 4s-orbital leaving four unpaired electron in
3d-orbitals]

Fe3+ = [Ar]

3d 5 4s0
[In this case (Fe3+), two 4s electrons and one 3d electron gets removed to form Fe3+ion]
In Fe3+, the d-orbitals are half-filled and symmetrical. So they are very stable.
3+ 2+
Therefore, Fe ion is more stable than Fe ion.
15. If the atomic number of an element ‘X’ is 27, then deduce the possible values of four quantum numbers for
thelectron of element ‘X’.
the 20
Sol. Element ‘X’ has Z = 27
Then the electronic configuration of the element ‘X’ is [Ar]4s 23d 7
19, 20 21, 26 22, 27 23 24 25
Electronic configuration = 18 [Ar]
4s2 3d7
thelectron is 4s 2
The 20
So, the n value is 4
1
l = 0, ml= 0, ms 
2
Solution of Assignment (Set-1) Structure of Atom 25
16. What are orbits? Why Bohr’s orbits are called stationary states?
Sol. Orbits : Orbits are the fixed circular paths in which the electrons revolve around the nucleus.
Bohr’s orbits are called stationary states because the radius and energy of orbits in which the electrons revolve
around the nucleus is fixed.
17. Give the explanation behind following the Hund’s Rule of maximum multiplicity along with an example.
Sol. Hund’s rule of maximum multiplicity is followed so as to minimize the electronic repulsions in an atom. The
electron-electron repulsions are minimum when the electrons are as far apart as possible with parallel spins
i.e., when all the orbitals of the subshells are half-filled with parallel spins.
For example :
1, 2 3, 4 5, 6 7

Nitrogen : 7N (Wrong way)

2 2 3
1s 2s 2p
The above is the wrong way of filling the orbitals, as pairing cannot occur unless all the degenerate orbitals
are singly filled.

1, 2 3, 4 5 6 7

So, Nitrogen 7N ( Correct way)

2 2 3
1s 2s 2p

The above is the correct way of assigning the electrons in which p-orbitals are singly filled.
18. 2+
Calculate the energy associated with the first orbit of Li ion.
+2
Sol. Atomic number of Lithium, Li (Z = 3)
Energy of hydrogen-like species is given as

⎛ Z2 ⎞
En  2.18  10 18 ⎜ 2 ⎟ J
⎜n ⎟
⎝ ⎠
‘n’ here is 1, as 1storbit

⎛ 32 ⎞
So, E1= –2.18 × 10 –18
⎜⎜ 2 ⎟⎟ J
⎝1 ⎠
17
1 J101.962 E
19. Write two points of difference between cathode rays and canal rays.

Sol.
Cathode rays Canal rays
(i) These consist of negatively charged particles (i) Canal rays is the other name of anode rays
called electrons which consist of positively charged particles
(ii) The mass of negatively charged particles does (ii) The mass of the positive particles depend upon
not depend upon the nature of gas taken in the the nature of gas taken in the discharge tube
discharge tube

20. Who discovered neutron and how?

Sol. Neutron was discovered by Chadwick. It was discovered when he bombarded a thin foil of Beryllium with fast
moving -particles.
4  9Be 
2 He 0n
12
 4
1
6 
(particle) (Neutron)

C
26 Structure of Atom Solution of Assignment (Set-1)

21. Find the number of protons, electrons and neutrons in Sc3+ion having mass number 45.
Sol. Mass number (A) for scandium = 45. The number of protons is equal to the atomic number of scandium which
is 21 i.e., Z = 21
 Number of protons = 21
Number of electrons = 21 – 3 = 18
3+ion is formed by the loss of three electrons from Sc atom)
(Sc
Number of neutrons = A – Z
= 45 – 21 = 24
22. Write a short note on Planck’s quantum theory.
Sol. Planck’s quantum theory was given by Max Planck. He suggested that the energy is radiated or absorbed
by atoms and molecules not continuously but discontinuously in the form of small discrete quantities or
packets.
He gave the name quantum to the small quantity of energy absorbed or radiated in the form of electromagnetic
radiations. In case of light, the quantum is called ‘photon’. The energy of quantum of radiation is proportional
to its frequency

E h
23. Write down the main postulates of Bohr’s model for hydrogen atom.
Sol. The main postulates of Bohr’s model for hydrogen atom are as follows :
(i) The electrons in the hydrogen atom revolve around the nucleus in a circular paths of fixed radius and energy
called orbits or allowed energy states.
(ii) An electron do not radiate energy if it stayed in one orbit and therefore do not fall into the nucleus.
(iii) An electron moves from a lower energy state to higher energy state after absorbing energy but energy is
emitted when electron moves from the higher energy state to lower energy state.
24. Calculate the wavelength of moving object ‘X’ so that its momentum is four times that of the moving object
–10m.
‘Z’ having wavelength 2.6 × 10
–10
Sol. Wavelength of object ‘Z’ = 2.6 × 10 m
Momentum (p) of object ‘X’ is four times that of object ‘Z’ i.e., x  z (Given)
p 4p
Wavelength of object X = ?
Applying de-Broglie equation
h

p
For object ‘Z’;  = 2.6 × 10–10m, we get

–10m = h
...(i)
2.6 × 10 pz
For object X;  = ?

h
x  ...(ii)
px
Now, we know p = 4p
x z

Putting in equation (ii), we get

h
x 
4p z
Solution of Assignment (Set-1) Structure of Atom 27
Now taking ratio
x h p
  z
 z 4p z h
x 1

z 4
2.6  10 10 m
Therefore,  x=
4
11
x m106.5 
25. What is meant by
(i) Stark effect?
(ii) Zeeman effect?
Sol. (i) Stark effect : The splitting of spectral lines in electric field is called stark effect.
(ii) Zeeman effect : The splitting of spectral line in magnetic field is called zeeman effect.
26. Show how the wavelength of a moving particle is related to its kinetic energy(KE).
Sol. According to de-Broglie;

h
Wavelength of a moving particle () = ...(i)
mv
mv = momentum

1 2
Kinetic energy (E) = mv
2
2E
v= ...(ii)
m
Substituting the value of (ii) in (i) we get
h

2E
m
m
h

(2mE) 1/2
27. Write the complete symbol of the element with the given information
(i) Z = 24, A = 52, (ii) Z = 20, A = 40, (iii) Z = 4, A = 9
Sol. (i) Z = 24, A = 52
Chromium (Cr) is the element with atomic number 24.
52
So, 24 Cr

(ii) Z = 20, A = 40
Calcium (Ca) is the element with atomic number 20.
40
20 Ca

(iii) Z = 4, A = 9
Beryllium (Be) is the element with atomic number 4.
9
4 Be
28 Structure of Atom Solution of Assignment (Set-1)

28. Write a short note on Heisenberg’s uncertainty principle. Is it applicable to macroscopic objects?
Sol. According to Heisenberg’s uncertainty principle : “It is impossible to measure simultaneously both the position
and velocity (or momentum) of a microscopic moving particle”.

h
x v 
4m
where x = uncertainty in position and v = uncertainty in velocity
This uncertainty principle led to the concept of probability of finding the electron around the nucleus.
Uncertainty principle is not applicable to macroscopic objects as from the equation it is clear
1
x v 
m
m = Mass of the object
As, the mass of the macroscopic objects is large, therefore the uncertainties become insignificant.
29. Which of the following sets of quantum numbers are not possible and why?
1
(i) n = 3, l = 2, ml= 0, m s= 
2
1
(ii) n = 4, l = 4, ml= +3, m s= 
2
1
(iii) n = 3, l = 1, ml= –2, m s= 
2
1
(iv) n = 2, l = 0, m l= 0, m s= 
2
Sol. (i) Possible.
(ii) Not possible, as ‘l’ can never have value equal to ‘n’.
‘l’ has values = 0 to (n – 1)
(iii) Not possible
l = 1, m = –2
l
has value = –l to +l including 0.
Magnetic quantum number = –2 is not possible here as ml
So, the possible values of mlcan be = –1, 0, +1
(iv) Possible.
30. State (n + l) rule. Illustrate with the help of example.
Sol. The energy of an orbital is determined by (n + l) rule as follows :
(i) The orbital with the lower (n + l) value have lower energy. For example, 4s and 3d-orbitals.
4s-orbital 3d-orbital
n=4 n=3
l=0 l=2

nl  4 nl 5
Thus 4s-orbital has a lower energy than 3d-orbital, hence filled first.
(ii) When the two orbitals have the same (n + l) value, then the orbital having lower value of n has lower energy.
For example : 2p and 3s-orbitals
2p-orbital 3s-orbital
n=2 n=3
l=1 l=0

nl 3 nl 3
Thus 2p-orbital has lower energy than 3s-orbital, hence filled first.
Solution of Assignment (Set-1) Structure of Atom 29
Long Answer Type Questions

31. The wave number of a radiation is 500 cm–1. Find out its
(i) Wavelength
(ii) Frequency
(iii) Time period
(iv) Energy in terms of joule per photon
(v) Energy in terms of kcal per mol of photon
Sol. (i) Wavelength ()

1
Wavelength =
Wave number
1 1 1
  cm
 500
 = 2 × 10–3cm
(ii) Frequency ()

c


c = 3 × 108ms –1

= 3 × 1010cm s –1

3 1010 cms 1
 
  103 cm
2 

  10
1.5 131
s

(iii) Time period (T)

1
T


1
T 1
1.5  1013 s

T  6.6 10
14 s

(iv) Energy in J per photon
Ephoton = h
= 6.626 × 10–34Js × 1.5 × 10 13 s–1

E  9.939 10
21 J

(v) Energy in kcal per photon
–21
E = 9.939 × 10 J
photon –21J × 6.022 × 10 23mol –1

For 1 mol of photon, 3energy–1= 9.939 × 10

Energy = 5.985 × 10 J mol
(1 cal = 4.184 J)
–1
Energy = 1.430 kcal mol
30 Structure of Atom Solution of Assignment (Set-1)

32. Discuss photoelectric effect in detail.

Sol. When a beam of light of suitable frequency falls on a metal surface, electrons are ejected from the surface
of metal. This phenomenon of ejection of electrons from the surface of a metal by the action of light is known as
photoelectric effect. The electron so ejected are called photoelectrons. It explains the particle nature of light.
1 2
K.E. =
2 mv
Incident light Electron

Electron

Metal
Photoelectric effect
Generally the metals which show this effect are as Cs, K and Rb.
The three important facts about photoelectric effect are
(a) There is no time lag between the ejection of the electrons from metal and the striking of beam of light on
the surface.
(b) The number of electron ejected is directly proportional to the intensity of light.
(c) There is some characteristic minimum frequency for a metal called threshold frequency, below which the
photoelectric effect cannot be observed.
Photoelectric effect follows the law of conservation of energy principle.
1
h = h0+ me v 2
2
h = Energy of incident light
h0= Threshold energy or work function
1
me v 2 = K.E. of the photoelectrons
2
While the number of electrons ejected depends upon the intensity of incident light but their energies do not
depend upon the energy of the incident light.
e
33. What is the charge, mass and ratio of electron? Write few properties of cathode rays.
m e
Sol. Electron is a negatively charged particle. It has
–19
Charge = –1.6022 × 10 C
–31
Mass = 9.11 × 10 kg
e
= 1.758 × 1011C kg –1
me
Few properties of cathode rays are :
(i) The cathode rays start from cathode and move towards anode.
(ii) The rays travels in straight lines.
(iii) These rays are not visible but their behaviour can be observed with the help of certain kind of
phosphorescent materials like zinc sulphide (ZnS).
(iv) On applying electric field, these rays are deflected towards the positive plate. This shows that the cathode
rays constitute negatively charged particles called electrons.
(v) The mass of the particles do not depend on the nature of the gas taken in the discharge tube.
Solution of Assignment (Set-1) Structure of Atom 31
34. (i) Discuss the line spectrum of hydrogen.
(ii) Find out the longest wavelength of absorption line for hydrogen gas containing atoms in ground state.
Sol. (i) Line spectrum of hydrogen : When an electric discharge is passed through gaseous hydrogen, the H2
molecules dissociate and the energetically excited hydrogen atoms produced emit electromagnetic radiation
of discrete frequencies.
There are five series of lines called Lyman, Balmer, Paschen, Brackett and Pfund series.
All the series of lines in the hydrogen spectrum could be described by the following expression
⎛ 1 1⎞ 1
  109677 ⎜⎟2 cm
2
⎝ n1 n2⎜⎟⎠
Where n1for Lyman series = 1
Balmer series = 2
Paschen series = 3
Brackett series = 4
Pfund series = 5
n2= n 1 + 1, n1 +2, .....
(ii) For longest wavelength, energy has to be lowest, that means transition will occur to the first excited energy
level.
When the hydrogen atom is in the ground state n = 1, n will be = 2 (first excited energy level)
1 2
Putting the values in equation
⎡1 1⎤ 1
 RZ 2 2 2⎢⎥ cm
⎣⎢ n1 n2 ⎥⎦

1 ⎛1 1 ⎞ 1
109677  (1)2 2
cm
 ⎝ 12 2⎜⎟ ⎠

1 3
 = 109677 × 1 cm
 4
1
 = 82257.75 cm –1

= 1.215 × 10  –5cm
35. What is the ground state electronic configuration of copper and Cu+ ion? Why fully-filled orbitals are more stable?
Sol. The ground state electronic configuration of copper (29Cu) is

29Cu = 1s2 2s22p 6 3s 23p 63d 104s 1

29Cu = [Ar]3d 10 4s 1

= [Ar]

3 d 10 4s1
Cu+ = [Ar]3d 10 4s 0

= [Ar]
3 d 10 4s0
The stability of fully-filled orbitals are highly stable due to the following reasons :
(i) Symmetrical distribution of electrons : It is well-known fact that symmetry leads to stability. The
completely filled orbitals are symmetrical, therefore stable.
32 Structure of Atom Solution of Assignment (Set-1)

(ii) Exchange energy : The stabilizing effect arises whenever two or more electrons with the same spin are
present in the degenerate orbitals of a subshell. These electrons with same spin can exchange their
positions and the energy involved is called exchange energy. Greater the number of exchange, greater will
be the exchange energy and more will be stability. In case of fully filled orbitals, maximum number of
exchanges are possible, therefore, it has maximum stability.
36. Define the four properties used to characterize the electromagnetic radiations along with their SI units.
Sol. The following are the properties associated with the radiations which are used to characterize them.
(i) Wavelength () : The distance between two neighbouring troughs or crests is known as
wavelength. The SI unit of wavelength = metre (m)
(ii) Frequency () : It is defined as the number of waves that pass a given point in one second.
–1
SI unit = Hertz (Hz) or (s )
(iii) Wave number ( ) : It is defined as the number of waves per unit length.
–1
SI unit = m
(iv) Velocity (v) : The distance travelled by the wave in one second is called its velocity.
–1
SI unit = ms
37. (i) Define isotopes and isobars.
32
(ii) How many protons, electrons and neutrons are present in 2.4 g of 16 S?

Sol. (i) (a) Isotopes : These are the atoms of same element having different mass numbers but same atomic
number.
(b) Isobars : These are the atoms of different elements with same mass number but different atomic
numbers.
32
(ii) 16 S

Number of protons (Z) = 16

Number of electron = 16
Number of neutrons = A – Z
= 32 – 16 = 16
32 g of sulphur contains = 16 × 6.022 × 1023neutrons, protons and electrons
23electrons, protons and neutrons
= 96.352 × 10

96.352  10 23
1 g of sulphur contains = neutrons, protons and electrons
32

96.352  10 23
2.4 g of sulphur contains =  2.4
32

= 7.22 × 1023neutrons, protons and electrons

Therefore, the number of protons in 2.4 g of S = 7.22 × 1023protons

The number of electrons in 2.4 g of S = 7.22 × 1023 electrons
The number of neutrons in 2.4 g of S = 7.22 × 1023neutrons
Solution of Assignment (Set-1) Structure of Atom 33
38. (i) What is the wavelength associated with 200 eV electron?
(ii) What are the shapes of s and p orbitals?
Sol. (i) 1 eV = 1.6 × 10–19J
h
Wavelength () = ...(i)
mv
1
K.E. = mv 2
2
2 K.E.
v ...(ii)
m
From (i) and (ii), we get
h

2 m KE
Therefore,

6.626  10 34 Js

 19

10(2 9.11 1.6 
kg)(200 10 J)
3
1
= 0.867 × 10–10m

= 0.867 Å

(ii) Shapes of s, p and d-orbitals are

s-orbital = Spherical shape

p-orbital = Dumb-bell shape

39. If the sodium metal is irradiated with a wavelength 450 nm, calculate the kinetic energy and the velocity of
the ejected photoelectron. (Given w0 (work function) = 2.3 eV)
Sol. Given w0= 2.3 eV

1 eV = 1.6 × 10–19J

w0= 2.3 × 1.6 × 10 –19J

19
w 0  3.68 0 J

Wavelength 1of irradiated light () = 450 nm

= 450 × 10–9m

E = h

hc
h


c
= 6.626 × 10–34Js ×


6.626  10 34 Js 10 8

3  ms 
= 1
450  10 9 m

  4.417  10 19 J
h
34 Structure of Atom Solution of Assignment (Set-1)

We know,

h = h (w
0 0) + K.E.

K.E. = h – h
0–19
J – 3.68 × 10–19J
K.E. = 4.417 × 10
–19 J

K.E. = 0.737 × 10
1
K.E.  mv 2
2

2 K.E.
v
m

2  0.737  19 J

9.11  10131 kg
0
v = 4.01 × 105ms –1
40. State and explain Pauli’s exclusion principle along with its applications.
Sol. Pauli’s exclusion principle states that “no two electrons in an atom can have the same set of four quantum
numbers”. That means that the number of electrons to be filled in the various orbitals is restricted by the
exclusion principle. Two electrons in an orbital can have the same values of three quantum numbers n, l and
mlbut must have the opposite spin quantum numbers (m s).
For example, for K shell i.e., ‘n’ = 1
nl m s

1
1 0 0  (For first electron)
2
1
1 0 0  (For second electron)
2
Two points we can conclude from this principle
(i) An orbital cannot have
thatmore than 2 electrons.

(ii) If an orbital has two electrons, they must have opposite spin.
Application of the principle
The maximum number of electrons in different sub-shells and energy levels can be deduced from this principle.
For L shell ‘n’ = 2

n l m s Number of electrons
in subshell
2 0 0 ±1/2 2
–1 0 ±1/2 2
1 ±1/2 2
+1 ±1/2 2

There are 8 electrons in L shell

2 in s-subshell
6 in p-subshell.
Solution of Assignment (Set-1) Structure of Atom 35
41. A proton is accelerated to one-tenth of the velocity of light. Suppose velocity can be measured with a precision
of ±2%, what can be the uncertainty in its position?

Sol. Mass of proton = 1.672 × 10–27kg

8ms –1
Velocity of light = 3 × 10
1 1
Velocity of proton =  3 1 8 ms
10
0

= 3 × 107ms –1

Uncertainty in velocity (v) = 0.02 × 3 × 107

5
= 6 × 10 –1
ms

Applying, Heisenberg’s uncertainty principle

h
x × v 

We get, 4 m

6.626  10 34 Js
x   5 1
 10(4 3.1429.1 kg) 6( 10
3
1 ms )
x  9.655  10 11 m

42. (i) What are (a) Atomic orbital, (b) Quantum numbers, (c)
Orbits?

(ii) List the four quantum along with their functions.

Sol. (i) (a) Atomic orbital : It is defined as the 3-dimensional region of space around the nucleus where the
probability of finding an electron is maximum.
(b) Quantum numbers may be defined as a set of four numbers with help of which we can give the
complete information about the electron present in an atom.

(c) Orbits : It is a rigid circular path, in which the electron revolves around the nucleus.

(ii) Four quantum numbers are :

(a) Principal quantum number (n) : It gives the energy of the main shell.

(b) Azimuthal quantum number (l) : It gives the shape of the orbitals in the subshell.

(c) Magnetic orbital quantum number (ml) : It gives the orientation of degenerate orbitals in space.

(d) Spin quantum number (ms) : It gives the spin of the electrons.
43. Write down the conclusion made from -scattering experiment and the defects of the Rutherford model of atom.

Sol. Conclusions of -scattering experiment

(a) Most of the space in the atom is empty as most of the -particles passed undeflected.
(b) A few positively charged -particles were deflected. This shows that they are deflected by the small
positively charged body in the atom.

(c) The positively charged body called nucleus is very small of radius 10–15m as compared to the radius of
–10
atom which is about 10 m.
36 Structure of Atom Solution of Assignment (Set-1)

Defects
(i) Position of electrons : The exact position of the electrons from the nucleus was not mentioned.
(ii) Stability of atom : According to electromagnetic theory, when a charged body moves under the influence
of an attractive force, it loses energy continuously in the form of electromagnetic radiations. The electron
should therefore continuously emit radiation and lose energy. As a result of this a moving electron will come
closer to the nucleus through a spiral path, it should ultimately fall into the nucleus. So, Rutherford’s model
of atom could not account for the stability of atom.

–
+

44. (a) What are electromagnetic radiations?

(b) What is electromagnetic spectrum?
(c) Arrange the different regions of electromagnetic spectrum in the increasing order of wavelength.
(d) What does  and L designate with respect to the structure of atom?
Sol. (a) Electromagnetic radiations : The radiations which have both magnetic as well as electric field components
which oscillate in the phase perpendicular to each other as well as perpendicular to the direction of
propagation are called electromagnetic radiations. All of these move with the same speed in vacuum
8 –1
regardless of their wavelength with the speed of light of 3 × 10 ms .
(b) Electromagnetic spectrum : An arrangement of radiations of all kinds, in the order of decreasing frequency
or increasing wavelength is called electromagnetic spectrum.
(c) Gamma rays < X-rays < UV rays < IR < Microwaves < Radiowaves
(d) l stands for subshell.
ndshell.
L stands for 2
45. (a) Write the two limitations of Bohr’s model.
(b) Name the various subshells associated with (i) n = 4, l = 2, (ii) n = 3, l = 1.
(c) Arrange the subshells in 3rdshell in the order of increasing energy incase of (i) Hydrogen atom and
(ii) Iron atom.
Sol. (a) Two limitations of Bohr’s model of atom are
(i) Bohr’s theory was unable to explain the splitting of spectral lines in the presence of magnetic field
(Zeeman effect) and in an electric effect (Stark effect)
(ii) Only applicable to single electron system.
(b) (i) n = 4; l = 2
Subshell = 4d
(ii) n = 3; l = 1
Subshell = 3p
(c) (i) Hydrogen atom
3s = 3p = 3d
(ii) Iron atom
3d > 3p > 3s
Solution of Assignment (Set-1) Structure of Atom 37

SECTION - B

Model Test Paper

Very Short Answer Type Questions :
1. What is the approximate radius of nucleus?
Sol. The radius of nucleus is approximately 10–15m.
2. Define an orbital.
Sol. It is three-dimensional region of space around the nucleus where the probability of finding the electron is
maximum.
3. Who discovered neutron?
Sol. Chadwick discovered neutron.
4. What is the shape of p-orbitals?
Sol. p-orbitals are dumb-bell in shape.
5. Which series of hydrogen spectrum lies in the visible spectrum?
Sol. Balmer series.
6. What is the atomic number of element which as 7 protons and 8 neutrons in the nucleus of its atom?
Sol. The atomic number of element is 7 (Z = 7).
7. What are isobars?
Sol. Isobars are the atoms of different elements with same mass number but different atomic number.
8. Which quantum number signify the energy of the shell?
Sol. Principal quantum number gives the energy of the shell.

Short Answer Type Questions :

9. Write the central idea of Planck’s quantum theory.
Sol. Planck’s quantum theory states that the radiant energy is not emitted continuously but discontinuously in the
form of small packets of energy called quantum having energy equal to h.
10. What is work function?
Sol. The minimum amount of energy required to eject the photoelectron is called work function.
Work function (W0) = h0
0= Threshold frequency
h = Planck’s constant
11. Describe briefly about azimuthal quantum number.
Sol. Azimuthal quantum number is denoted as ‘l’. This quantum number determines the angular momentum of the
electron. It defines the three dimensional shape of the orbital. For a given value of ‘n’, ‘l’ can have values 0
to (n – 1).
12. Why the effect of Heisenberg’s uncertainty principle insignificant for macroscopic moving objects?
h
Sol. Heisenberg’s uncertainty principle is given by the formula x × v 

Macroscopic objects have large mass, so, the value of x × v becomes4 m extremely small and insignificant
and the associated uncertainties are hardly of any real consequence.
13. Describe Thomson’s model of atom.
Sol. Thomson’s atom is a sphere of positive charge in which the electrons are embedded into it. This model is known
by different names as plum-pudding model, raisin pudding or watermelon.
38 Structure of Atom Solution of Assignment (Set-1)

14. What is electromagnetic spectrum?

Sol. An arrangement of electromagnetic radiations which differ from one another in wavelength or frequency in the
increasing order of wavelength is called electromagnetic spectrum.
15. Name the rules which are followed during the filling of electrons in orbitals in atom.
Sol. The following are the rules/principles which are followed during the filling of orbitals in atom :
(a) Aufbau principle
(b) Pauli’s exclusion principle
(c) Hund’s rule of maximum multiplicity
Short Answer Type Questions :
16. Write down the drawbacks of Rutherford’s model of atom.
Sol. (i) According of Maxwell, the charged moving bodies radiate energy in the form of electromagnetic radiations.
As, electron is also a charged body revolving around the nucleus i.e., under acceleration must radiate
energy. After losing energy it must fall into the nucleus spirally. But this does not happen as atom is
stable. Therefore, Rutherford’s atomic model could not account for the stability.

–
+

(ii) Rutherford’s model gave no idea about how the electrons are distributed around the nucleus and what are
the energies of these electrons.
17. Write down the electronic configuration of Nickel, Copper and Chromium.
Sol. Nickel (Z = 28)
Electronic configuration : 28Ni = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 8 4s 2
Copper (Z = 29) Electronic
configuration : 29Cu = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 1
Chromium (Z = 24)
Electronic configuration of : 24Cr = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 1
18. Calculate the wavelength in nm associated with a radiation of frequency 3.4 × 1012Hz and also find out its
energy.

Velocity
Sol. Wavelength =
Frequency
Frequency = 3.4 × 1012Hz (s –1)

8ms –1
Velocity = 3 × 10
3  108 ms 1 –5m
Wavelength = = 8.82 × 10
3.4  1012 s1

m108.82  
5 (1 nm = 10–9m)

  8.82  10 4 nm
Energy of radiation = h
–34Js × 3.4 × 10 12 s–1
E = 6.626 × 10
E  2.25 10
21 J

Solution of Assignment (Set-1) Structure of Atom 39
19. Explain Hund’s rule of maximum spin multiplicity.
Sol. This rule says that no pairing of electron starts in any of the degenerate orbitals until all the orbitals of the
subshell contain one electron each with parallel spin.
For example, each of the three p-orbitals of the p subshell gets one electron of parallel spin before any one
of them receives second electron of opposite spin.
Example :

7 N=
1s2 2s2 2p3
This principle is very important in guiding the filling of p, d, f and g subshells which have more than one kind
of orbitals.
20. Write some properties of anode rays.
Sol. Anode rays are also known as canal rays
(i) These rays are positively charged.
(ii) These rays originate in the region between the two electrodes in the discharge tube.
(iii) The mass of positive particles which constitute these rays depend upon the nature of the gas in the tube.
(iv) The magnitude of charge on anode rays varies from particle to particle depending upon the number of
electrons lost by gaseous atom.
21. What do we mean by the dual nature of electromagnetic radiations?
Sol. Dual nature of electromagnetic radiations mean that the radiations possess both particle as well as wave
character. This is evident that particle nature of light can explain the black body radiation and photoelectric
effect but cannot explain the phenomenon of interference and diffraction which can only be explained by wave
nature of light. Therefore light (electromagnetic radiation) possess dual behaviour either as a wave or as a stream
of particles when radiation interacts with matter, or as a wave when it displays wave like properties like
diffraction and interference.
Long Answer Type Questions :
22. Discuss in detail the shape of s-orbital.
Sol. For s-orbital, l = 0 and ‘ml’ value is also 0. As the ‘ml’ value is 0. This means that the probability of finding
an electron in s-orbital is same in all directions at a given distance from the nucleus. The shape of s-orbitals
is spherical and non-directional.
The s-orbitals in all energy levels i.e., higher energy levels (2s, 3s etc.) are also spherically symmetrical and
non-directional. As the principal quantum number (n) increases the effective volume or size of the orbitals
increases. This means 2s-orbital is larger than 1s-orbital.
Nodes : 2s-orbital consists of two spherical regions of high electron probability. In between these two spheres,
there is a spherical shell or region where the probability of finding the electron is zero. This spherical region
is called node or nodal surface. More precisely this node is called radial node. The only way in which 2s-orbital
differs from 1s-orbital is its larger size and presence of node.
Number of nodes for s-orbital = n – 1
Therefore,
1s-orbital = 0 node
2s-orbital = 1 node

1s-Orbital 2s-Orbital
40 Structure of Atom Solution of Assignment (Set-1)

23. Define wavelength and wave number. Calculate the wave number of longest wavelength transition in the Paschen
series of hydrogen atom.
Sol. Wavelength () : It is the distance between centre of two adjacent crests and
troughs.
Wave number ( ) : It is defined as the number of waves per unit
1
length. For Paschen series, n = 3
⎛ 1 1 ⎞
 R
2 ⎜⎟2⎜⎟
⎝ 3 n 2 ⎠

1
We know that  

 will be longest when  is shortest. So, for  to be shortest then n2has to be minimum i.e., when, n 2= 4
⎛ 1 1 ⎞ 7
  109677 ⎟⎜2 2 109677 
⎝3 4 ⎠ 144
= 5331.52 cm–1
Hence, for this wave number, wavelength is maximum.
 Objective Type Questions
76
1. An isotone of 32 Ge is
77 77 77 78
(1) 32 Ge (2) 33 As (3) 34 Se (4) 74 Se
Sol. Answer (2)
Isotone means same number of neutrons

Number of neutrons  Mass number  Atomic number

Number of neutron in Ge = 76 – 32 = 44
Number of neutron in As = 77 – 33 = 44

2. The ratio of specific charge of an electron to that of a proton is

(1) 1 : 1 (2) 1837 : 1 (3) 1 : 1837 (4) 2 : 1
Sol. Answer (2)
e e
electron : proton electron Proton
m m
e
 e
Both e–  protons have same charge 1 :
1837 1
th
1
Mass e is the mass of protons  1837 : 1
1837

3. Atomic number and mass number of an element M are 25 and 52 respectively. The number of electrons,
protons and neutrons in M2+ ion are respectively
(1) 25, 25 and 27 (2) 25, 27 and 25 (3) 27, 25 and 27 (4) 23, 25 and 27
Sol. Answer (4)
Atomic number = 25
Mass number = 52
Number of proton = Atomic number = 25
Number of neutron = (Mass number – Atomic number)
52 – 25 = 27
26 Structure of Atom Solution of Assignment (Set-2)

Number of e– = Number of protons but M2+ ion means 2e– are removed
 Number of e– = Number of protons – 2
= 25 – 2 = 23

4. According to Bohr’s theory angular momentum of an electron in 6th orbit is

h h h 2.5h
(1) 2.5 (2) 6 (3) 3 (4)
   2
Sol. Answer (3)
nh
mvr  (n = number of shell)
2
Angular momentum

6h 3h
for 6th shell  
2 

5. If r1 is the radius of the first orbit of hydrogen atom, then the radii of second, third and fourth orbits in term
of r1 are
(1) r12, r13, r14 (2) 4r1, 9r1, 16r1 (3) 8r1, 27r1, 64r1 (4) 2r1, 6r1, 8r1
Sol. Answer (2)

r0  n2
rn 
z
rn = Radius of n orbit
r0 = Radius of Ist orbit
n = Number of orbit
z = Charge on nucleus
For 'H' atom z=1
(for 2nd orbit) (for 3rd orbit) (for 4th orbit)
 r 2 = r 1 × 22 r3 = r 1 × 3 2 r4 = r 1 × 4 2
= 4r1, 9r1, 16r1

6. Electronic energy is negative because

(1) Electron has negative charge
(2) Energy is zero near the nucleus and decreases as the distance from nucleus increases
(3) Energy is zero at infinite distance from the nucleus and decreases as the electron comes towards nucleus
(4) These are interelectronic repulsions
Sol. Answer (3)
At infinite distance energy of electron will be zero and it will decreases as the electron approaches towards
nucleus

7. An electron jumps from lower orbit to higher orbit, when

(1) Energy is released (2) Energy is absorbed (3) No change in energy (4) It radiates energy
Sol. Answer (2)
When e– jumps from lower to higher energy level absorbed.
When e– comes from higher to lower energy level released.
Solution of Assignment (Set-2) Structure of Atom 27
8. If the energy difference between the ground state and excited state of an atom is 4.4 × 10–19 J. The wavelength
of photon required to produce this transition is
(1) 4.5 × 10–7 m (2) 4.5 × 10–7 nm (3) 4.5 × 10–7 Å (4) 4.5 × 10–7 cm
Sol. Answer (1)

E = [Excited state  ground state]  4.4 × 10 19 J

nhc hc
According to Plank's quantum theory  E = =
 E

6.6  10 34 J-s  3×108 m/s

 19
 4.5  10 7 m
4.4  10 J

9. The number of photons of light of wavelength 7000 Å equivalent to 1 J are

(1) 3.52 × 10–18 (2) 3.52 × 1018 (3) 50,000 (4) 10,0000
Sol. Answer (2)
nhc  E 7000  1010 m  1J
E= n= 
 h  c 6.6  1034 J-s  3×108 m/s

10. The threshold energy is given as E0 and radiation of energy E falls on metal, then K.E. is given as

E  E0 E
(1) (2) E – E0 (3) E0 – E (4)
2 E0
Sol. Answer (2) E
E = Incident energy E0 K.E.
E0 = Threshold energy
e– e– e–
E = E0 + K.E.
e– e– e–
K.E. = E  E0 E = Incident energy

11. The frequency of a wave is 6 × 1015 s–1. Its wave number would be
(1) 105 cm–1 (2) 2 × 107 m–1 (3) 2 × 107 cm–1 (4) 2 × 105 cm–1
Sol. Answer (2)

1 Velocity c
Wave number   
Wavelength  Frequency 
c 
c       c 
 c

6  1015 s1 1
  8
 2  107 m
3.0  10 m/s

12. If threshold wavelength (°) for ejection of electron from metal is 330 nm, then work function for the
photoelectric emission is
(1) 6 × 10–10 J (2) 1.2 × 10–18 J (3) 3 × 10–19 J (4) 6 × 10–19 J
Sol. Answer (4)
hc  0  threshold frequency
Work function = h0 
0  0  threshold wavelength

6.6  10 34  3  108

0 = 330 × 10–9 m Work function = J
330  10 9
28 Structure of Atom Solution of Assignment (Set-2)

13. The ionization energy of the electron in the lowest orbit of hydrogen atom is 13.6 eV. The energies required
in eV to remove electron from three lowest orbits of hydrogen atom are
(1) 13.6, 6.8, 8.4 (2) 13.6, 10.2, 3.4 (3) 13.6, 27.2, 40.8 (4) 13.6, 3.4, 1.51
Sol. Answer (4)
I. E. = E  E1

I.E. = 13.6 eV given

E  0

13.6  0  E1

E1  13.6 eV
E1 13.6
 1  
2
E2  2
 3.4 eV
n 4
E1 13.6
E3    1.51 eV
3 2 9

 I  E1  E  E1  0   13.6  13.6 eV
I  E2  E  E2   0   3.4  3.4 eV

I  E3  E  E3  0    1.51 eV    1.51 eV

14. A certain metal when irradiated with light ( = 3.2 × 1016 Hz) emits photo electrons with twice kinetic energy
as did photo electrons when the same metal is irradiated by light ( = 2.0 × 1016 Hz). Calculate 0 of
electron?
(1) 1.2 × 1014 Hz (2) 8 × 1015 Hz (3) 1.2 × 1016 Hz (4) 4 × 1012 Hz
Sol. Answer (2)
K. E. = h( – 0)
K.E. of photoelectrons when  = 3.2 × 1016 Hz
K. E1 = h (3.2 × 1016 – 0)
K. E. of photoelectron when  = 2.0 × 1016 Hz
K. E2 = h(2.0 × 1016 – 0)
According to question K  E1  2K  E2
 h(3.2 × 1016 – 0) = 2h(2.0 × 1016 – 0)
3.2 × 1016 – 0 = 4.0 × 1016 – 20
0 = 4.0 × 1016 – 3.2 × 1016 = 0.8 × 1016 Hz = 8 × 1015 Hz = 8 × 1015 Hz

15. En = –313.6/n2 kcal/mole. If the value of E = –34.84 kcal/mole, to which value does ‘n’ correspond?
(1) 4 (2) 3 (3) 2 (4) 1
Sol. Answer (2)

313.6
En  E = – 34.84
n2
313.6 313.6
 – 34.84 = n2  9
n2 34.84
n 9 3
n3
Solution of Assignment (Set-2) Structure of Atom 29
16. Which transition of Li2+ is associated with same energy change as n = 6 to n = 4 transition in He+?
(1) n = 3 to n = 1 (2) n = 8 to n = 6 (3) n = 9 to n = 6 (4) n = 2 to n = 1
Sol. Answer (3 )
In He ; n = 6 the corresponding energy level in Li2+ ion will be
⎡For He ⎤ 1312 1312 1312
  2  
2
⎢ ⎥  E6  4 
⎣z  2 ⎦ 6 2 36 9

Li2 1312 1312 2

  2  3  
2
Corresponding energy level for n  81 n  9
( z  3) n 9

In He n = 4; the corresponding energy level in Li2+ ions

For He 1312 1312

E4   4  
 z  2  4 2 4

Corresponding energy level for Li 1312 1312
 3  
2

 z  3 n 2 4
n2  36 n6
Shortcut
1 2⎡ 1 1⎤
 R   2 ⎢ 2  2 ⎥ ......(1)
He2  ⎣4 6 ⎦

1 2⎡ 1 1⎤
 R  3 ⎢ 2  2 ⎥ ......(2)
Li2  ⎢⎣ n1 n2 ⎥⎦
For same energy equation (1) must be equal to equation (2) which only possible when
n1 = 6 and n2 = 9

17. Zeeman effect refers to the

(1) Splitting of the spectral lines in a magnetic field
(2) Splitting up of the spectral lines in an electrostatic field
(3) Emission of electrons from metals when light falls on it
(4) Random scattering of -particles by gold foil
Sol. Answer (1)
Splitting of line in magnetic field is known as Zeeman effect

18. Number of spectral lines in Balmer series when an electron return from 7th orbit to 1st orbit of hydrogen atom
are
(1) 5 (2) 6 (3) 21 (4) 15
Sol. Answer (1)
As only visible lines have to be calculated i.e. Balmer lines n=7
n=6
n=5
Visible lines when ground state = 2
n=4
n1 = 2 n=3
Possible arrangements 7  2, 6  2, 5  2, 4  2, 3  2 n=2
Total 5 lines n=1

All have visible region because n1  2

30 Structure of Atom Solution of Assignment (Set-2)

19. If kinetic energy of a proton is increased nine times, the wavelength of the de-Broglie wave associated with
it would become

1 1
(1) 3 times (2) 9 times (3) times (4) times
3 9
Sol. Answer (3)

1 1
mv 2  K.E. ⇒ m2 v 2  m  K  E
2 2

h
m2 v 2  2m  K  E  substitute the value of mv
mv
h 1
mv  2m  K  E   ...... (1)
2m  K  E K E
1
From equation (1) when K.E. of the electron increased 9 times. The de-Broglie wavelength decreased by
3
times.
1 1
 
9 3

20. The number of waves in the third orbit of H atom is

(1) 1 (2) 2 (3) 4 (4) 3
Sol. Answer (4)
Number of waves in third orbit = 3
Number of waves = Number of shell

Circumference
Number of waves = 
de Broglie X
h

mv
2r  mv
Number of waves =
h
nh
mvr =
2
Number of waves = n

21. The de-Broglie wavelength of an electron travelling with 10% of velocity of light is equal to
(1) 242.4 pm (2) 24.2 pm (3) 2.42 pm (4) 2.424 pm
Sol. Answer (2)
v = 10% of velocity of light
10
  3.0  10 m/s  3.0×107 m/s
100

m  9.1 1031 kg

h 6.6  1034 J-s

   24.2  1012 m  24.2 pm
mv 9.1 1031  3  107
Solution of Assignment (Set-2) Structure of Atom 31
22. The wavelength associated with a ball of 200 g and moving with a speed of 5 m/hour is of the order of
(1) 10–10 m (2) 10–20 m (3) 10–30 m (4) 10–40 m
Sol. Answer (3)
5
v = 5 m/hour = m/s
3600
m = 200 g = 0.2 kg

6.6  1034 J-s  3600


0.2  5
= 23760 × 10–34
= 2.3 × 10–30 m

 1030 m

23. The momentum of a particle which has a de-Broglie wavelength of 0.1 nm is

(1) 3.2 × 10–24 kg ms–1 (2) 4.3 × 10–22 kg ms–1
(3) 5.3 × 10–22 kg ms–1 (4) 6.62 × 10–24 kg ms–1
Sol. Answer (4)
h
 momentum P = mv
mv
h 6.6  10 34 kgm2 s1
P  
 0.1 10 9 m

P  6.6  10 24 kg ms1

24. The uncertainty in velocity of an electron present in the nucleus of diameter 10–15m hypothetically should be
approximately
(1) 10–11 m/s (2) 108 m/s (3) 1011 m/s (4) 10 Å/s
Sol. Answer (3)
x = 10–15 m

h
m = 9.1 × 10–31 kg x  mv 
4

6.6  10 34 kg m2 s 1 6.6

⇒ v  15 31
  1012 m/s
10  9.1 10  4  3.14 114.296

= 0.05  1012 m/s

App. = 0.5  1011 m/s

25. The set of quantum numbers not applicable to an electron

1 1 1 1
(1) 1, 1, 1,  (2) 1, 0, 0,  (3) 1, 0, 0,  (4) 2, 0, 0, 
2 2 2 2
Sol. Answer (1)
The value of l can never be equal to n
n=1 l=1 not possible
32 Structure of Atom Solution of Assignment (Set-2)

26. The principal and azimuthal quantum number of electrons in 4f orbitals are
(1) 4, 2 (2) 4, 4 (3) 4, 3 (4) 3, 4
Sol. Answer (3)
⎡Principal quantum number (n) = 4 ⎤
For 4f electron ⎢ ⎥
⎣ Azimuthal quantum number (l) = 3 ⎦
⎡l  0 s ⎤
⎢ l  1 p⎥
⎢ ⎥
⎢l  2 d⎥
⎢ ⎥
⎣l  3 f ⎦

1
27. How many 3d electrons can have spin quantum number  ?
2
(1) 5 (2) 7 (3) 8 (4) 10
Sol. Answer (1)
For 3d number of electrons will be 10

1
5 e– have clockwise  
2
1
5 e– have anti clockwise  
2
For l = 2 m = –2, –1, 0, +1, +2
⎡5 orbital = 10e ⎤
⎢ ⎥
⎢half are clockwise ⎥
1 1 1 1 1 1 1 1 1 1 ⎢and half are anticlockwise ⎥
          ⎣ ⎦
2 2 2 2 2 2 2 2 2 2

28. The correct order of increasing energy of atomic orbital is

(1) 5p < 4f < 6s < 5d (2) 5p < 6s < 4f < 5d
(3) 4f < 5p < 5d < 6s (4) 5p < 5d < 4f < 6s
Sol. Answer (2)
More the (n + l) value ; more will be the energy
5p 4f 6s 5d
(n + l) (5 + 1) (4 + 3) (6 + 0) (5 + 2)
6 7 6 7
According to Aufbau principle
Smaller the (n + l); smaller will be energy
For same (n + l); smaller the value of n ; lesser will be energy

 increasing order 5 p  6s  4f  5d

29. Which shell would be the first to have ‘g’ sub-shell?

(1) L (2) M (3) N (4) O
Sol. Answer (4)
For g-subshell l = 4; the value of n will be = (l + 1) = 5
K, L, M, N, O
n 1 n2 n3 n4 n5
 For n = 5 corresponding is 'o' shell and it contain 'g' subshell.
Solution of Assignment (Set-2) Structure of Atom 33
30. For which one of the following set of quantum numbers an electron will have the highest energy?

1 1 1 1
(1) 3, 2, 1, (2) 4, 2,  1, (3) 4, 1, 0,  (4) 5, 0, 0,
2 2 2 2
Sol. Answer (2)
Larger the value of (n + l) ; larger will be the energy

1
(1) 3, 2, 1, (n + l) = 3 + 2 = 5
2

1
(2) 4, 2, –1, (n + l) = 4 + 2 = 6 (Max energy)
2

1
(3) 4, 1, 0,  (n + l) = 4 + 1 = 5
2

1
(4) 5, 0, 0, (n + l) = 5 + 0 = 5
2

31. The energies of orbitals of H-atom are in the order

(1) 3s < 3p < 4s < 3d < 4p (2) 3s < 3p < 3d < 4s < 4p
(3) 3s = 3p = 3d < 4s = 4p (4) 3s = 3p = 3d < 4s < 4p
Sol. Answer (3)
As 'H' have 1 electron
 Orbitals are not degenerated
3s = 3p = 3d < 4s = 4p
same energy same energy

32. Which of the following set of quantum number is possible?

1
(1) n = 4, l = 2, m = –2, s = –2 (2) n = 4, l = 4, m = 0, s =
2
1
(3) n = 4, l = 3, m = –3, s = (4) n = 4, l = 0, m = 0, s = 0
2
Sol. Answer (3)
Option 1 not possible because s can never have –2 value
Option 2 not possible because n and l cannot have same value
Option 4 not possible because s cannot have zero value
 Correct answer = 3
1
n=4 l=3 m = –3 s=
2

33. The maximum number of electrons in an atom which can have n = 4 is

(1) 4 (2) 8 (3) 16 (4) 32
Sol. Answer (4)
Number of electrons = 2n2 (n = shell number)
For 4th shell = 2 × (4)2 = 32 electrons
34 Structure of Atom Solution of Assignment (Set-2)

34. In the presence of magnetic field, the possible number of orientations for an orbital of azimuthal quantum
number 3, is
(1) Three (2) One (3) Five (4) Seven
Sol. Answer (4)
When magnetic field is applied subshell will give orbital i.e.,

l=3 m = –3, –2, –1, 0, +1, +2, +3

(Total 7 orbitals are possible)

35. Assuming the velocity to be same, the wavelength of the waves associated with which of the following particles
would be maximum?
(1) An electron (2) A proton (3) An -particle (4) A deutron
Sol. Answer (1)

h 1
  for same velocity
mv m
Mass of electron is minimum than proton, deutron and -particle

i.e., me = 9.1 × 10–31 kg mp = 1.67 × 10–27 kg mD = 1 unit mparticle = 4 unit

 e– will have minimum mass and maximum wavelength

36. For a ‘p’ electron, the orbital angular momentum is

(1) 6 (2) 2 (3)  (4) 2

Sol. Answer (2)

h
Orbital angular momentum = l  l  1  l  l  1 
2

For p-electron value of l = 1

 Orbital angular momentum  l 1  1   2 

37. Which of the following electronic level would allow the hydrogen to absorb a photon but not emit a photon?
(1) 3s (2) 2p (3) 2s (4) 1s
Sol. Answer (4)
1s-orbital is the ground state
Further emission is not possible i.e. de excitation not possible

38. Which of the following transition will emit maximum energy in hydrogen atom?
(1) 4f  2s (2) 4d  2p
(3) 4p  2s (4) All have same energy
Sol. Answer (4)
Transition energy depends upon the shell number i.e. value of principle quantum number 'n' in all the case
transition is between 4th energy level to 2nd level
 All have same energy
Solution of Assignment (Set-2) Structure of Atom 35
39. In an atom, which has 2K, 8L, 18M and 2N electrons in the ground state. The total number of electrons having
magnetic quantum number, m = 0 is
(1) 6 (2) 10 (3) 7 (4) 14
Sol. Answer (4)
Total number of e– = 30. Therefore, e– configuration will be
1s2 2s2 2p6 3s2 3p6 4s2 3d10
n=1 n=2 n=2 n=3 n=3 n=4 n=3
l=0 l=0 l=1 l=0 l=1 l=0 l=2
m=0 m=0 m = –1, 0, +1 m=0 m = 0, –1, +1 m=0 m = –2, –1, 0, +1, +2
For s-subshell 1 orbital have m=0
For p-subshell 1 orbital have m=0
For d-subshell 1 orbital have m=0

 Total 7 orbital have m = 0 in above configuration. Therefore, total number of electron = 7 × 2 = 14

40. The probability density curve for 2s electron appears like

2 2
R
2 2
R R R
(1) (2) (3) (4)

r r r r
Sol. Answer (1)
Graph is not correct [because 2s have only one node]
Correct graph will be

R2

41. A p-orbital can accommodate upto

(1) Four electrons (2) Six electrons (3) Two electrons (4) Eight electrons
Sol. Answer (3)
In any orbital maximum two electrons can accommodate
[A p-orbital can accommodate upto two electrons]

42. If the uncertainty in the position of electron is zero, the uncertainty in its momentum would be

h h
(1) Zero (2) Greater than (3) Less than (4) Infinite
4 4
Sol. Answer (4)

h
x  P  if x = 0
4
h h
P   
4  x 4  0
36 Structure of Atom Solution of Assignment (Set-2)

43. The number of radial nodes in 4s and 3p orbitals are respectively

(1) 2, 0 (2) 3, 1 (3) 2, 2 (4) 3, 2
Sol. Answer (2)

Number of radial nodes =  n  l  1

⎪⎧For 4s n  4 l  0  4  0  1  3
⎨
⎪⎩For 3 p n = 3 l  1  3  1  1  1
44. Which of the following orbital is with the four lobes present on the axis?

(1) d z2 (2) dxy (3) dyz (4) d x 2  y 2

Sol. Answer (4)

dx2  y2 (all the lobes are present on axis)

X
d
x2  y 2

45. Which of the following statement concerning the four quantum number is incorrect?
(1) n gives the size of an orbital
(2) l gives the shape of an orbital
(3) m gives the energy of the electron in orbital
(4) s gives the direction of spin of electron in the orbital
Sol. Answer (3)
m = represents the orientation of orbital in magnetic field.
m = orbitals

46. Which of the following has maximum number of unpaired electrons?

(1) Mg2+ (2) Ti3+ (3) Fe2+ (4) Mn2+
Sol. Answer (4)
Number of unpaired
Mg2+ = 10 = 1s2, 2s2, 2p6 0
Ti3+ = 19 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s0, 3d1 1
Fe2+ = 24 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s0, 3d6 4
Mn2+ = 23 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s0, 3d5 5 maximum number

47. Two electrons in K shell will not have

(1) Same principal quantum number (2) Same azimuthal quantum number
(3) Same magnetic quantum number (4) Same spin quantum number
Sol. Answer (4)
As K shell is the 1st shell and have maximum two electron. Therefore, to Pauli's exclusion principal two
electrons can't have the same value of all the four quantum number.
Therefore, can't have same spin quantum number
Solution of Assignment (Set-2) Structure of Atom 37
48. Which of the following electronic configuration is not possible?
(1) 2p3 (2) 2d5 (3) 4s1 (4) 5f 8
Sol. Answer (2)
Value of l cannot be greater or equal to n.
For 2d n = 2, l=2 not possible

49. The orbital diagram in which both Pauli’s exclusion principle and Hund’s rule are violated is
2s 2p
(1) (2)

(3) (4)
Sol. Answer (1)

2s 2p
According to Pauli number two electrons have same value of all the four quantum numbers i.e. pair in 2p orbital
have same spin not possible.
According to Hund's rule electrons are firstly clockwise arranged then pairing is done which is not in 2p
subshell.

50. Which of the following electronic configuration is incorrect?

(1) 1s2, 2s2, 2px2, 2py2, 2pz2, 3s2, 3px1 (2) 1s2, 2s1, 2px1, 2py1, 2pz1
(3) 1s2, 2s2, 2p6, 3s2, 3p6, 3d5, 4s2 (4) 1s2, 2s2, 2p6, 3s2, 3px1, 3py1, 3pz1
Sol. Answer (2)
Before 2s completely filled electrons are not further added into higher energy level.
Therefore Option 2 incorrect.

SECTION - B
Objective Type Questions
1. What will be the longest wavelength line in Balmer series of spectrum of H-atom?
(1) 546 nm (2) 656 nm (3) 566 nm (4) 556 nm
Sol. Answer (2)
All the wavelength are in visible region i.e. between 400 nm to 760 nm. Therefore maximum wavelength line
will be 656 nm.

2. The uncertainty in momentum of an electron is 1 × 10–5 kg-m/s. The uncertainty in its position will be
(h = 6.62 × 10–34 kg-m2/s)
(1) 5.27 × 10–30 m (2) 1.05 × 10–26 m (3) 1.05 × 10–28 m (4) 5.25 × 10–28 m
Sol. Answer (1)
P = 10–5 kgms–1
h
x  P 
4

6.6  1034 J
x  5
 5.2  1030 m
10  4  3.14
38 Structure of Atom Solution of Assignment (Set-2)

3. In hydrogen atom, energy of first excited state is –3.4 eV. Then find out KE of same orbit of hydrogen atom
(1) +3.4 eV (2) +6.8 eV (3) –13.6 eV (4) +13.6 eV
Sol. Answer (1)
KE
 1
ETotal
1
mv 2
KE 2
   1
Total E 1
 mv 2
2
1
KE = mv 2 ....... (1)
2
Ze2
PE =  ...... (2)
r
Electrostatic force = centrifugal force

Ze2 mv 2
 
r2 r

PE  mv 2  mv 2 By comparing (2)

1 1
Total energy = mv 2  mv 2   mv 2
2 2
Total energy = –3.4 eV (Given)
 KE = –(–3.4 eV) = +3.4 eV

4. Maximum number of electrons in a subshell with l = 3 and n = 4 is

(1) 10 (2) 12 (3) 14 (4) 16
Sol. Answer (3)

n  4, l  3 means 4f

for l = 3, m = –3, –2, –1, 0, 1, 2, 3 = 7 orbital

Therefore, maximum 14 electrons are present.

5. The total number of subshells in fourth energy level of an atom is

(1) 4 (2) 8 (3) 16 (4) 32
Sol. Answer (1)

6. For which of the following sets of four quantum numbers, an electron will have the highest energy?
n l m s
(1) 3 2 1 +1/2
(2) 4 2 –1 +1/2
(3) 4 1 0 –1/2
(4) 5 0 0 –1/2
Solution of Assignment (Set-2) Structure of Atom 39
Sol. Answer (2)
Energy of an electron depends upon (n + l) value
More the (n + l) value more will be the energy
n l m s (n + l)
(1) 3 2 1 +1/2 5
(2) 4 2 –1 +1/2 6 Max. (n + l). max. energy
(3) 4 1 0 –1/2 5
(4) 5 0 0 –1/2 5

7. A transition element X has a configuration (Ar)3d 4 in its +3 oxidation state. Its atomic number is
(1) 22 (2) 19 (3) 25 (4) 26
Sol. Answer (3)
Total number of e– in X+3 = [Ar] 3d4
= 18 + 4 = 22

 Number of electrons in X = 22 + 3 = 25

Atomic number = 25

8. Among the following which one is not paramagnetic? [Atomic numbers; Be = 4, Ne = 10, As = 33, Cl = 17]
(1) Ne2+ (2) Be+ (3) Cl– (4) As+
Sol. Answer (3)
Ions having all the electron paired will be non-paramagnetic or diamagnetic

Ne+2 = 8 = 1s2, 2s2, 2p4 2 unpaired e–

Be = 3 = 1s2, 2s1 1 unpaired e–

2s

Cl– = 18 = 1s2, 2s2, 2p6, 3s2, 3p6 0 unpaired e–

3p

As = 33 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p3 3 unpaired e–

9. Isoelectronic species are

(1) CO, CN–, NO+, C22– (2) CO–, CN, NO, C2–
(3) CO+, CN+, NO–, C2 (4) CO, CN, NO, C2
Sol. Answer (1)
Isoelectronic species have same number of electrons

⎡CO = 14 e  NO  14 e  ⎤
⎢  
⎥ [All have same number of electrons]
⎣⎢CN = 14 e C22   14 e  ⎦⎥
40 Structure of Atom Solution of Assignment (Set-2)

10. Consider the following sets of quantum number

n I m s
(i) 3 0 0 +1/2
(ii) 2 2 1 +1/2
(iii) 4 3 –2 –1/2
(iv) 1 0 –1 –1/2
(v) 3 2 3 +1/2
Which of the following sets of quantum number is not possible?
(1) (i), (ii), (iii) and (iv) (2) (ii), (iv) and(v) (3) (i) and (iii) (4) (ii), (iii) and (iv)
Sol. Answer (2)
(ii), (iv), and (v) are not possible
(ii) n = 2 l=2 m=1 s = +1/2 l not equal to n not possible
(iii) n = 1 l=0 m = –1 s = –1/2 Not possible because m = –1 where l = 0
(iv) n = 3 l=2 m=3 s = +1/2 Not possible because m = 3 is not for l = 2

11. Two particles A and B are in motion. If the wavelength associated with particle A is 5 × 10–8 m; calculate the
wavelength associated with particle B if its momentum is half of A.
(1) 5 × 10–8 m (2) 10–5 cm (3) 10–7 cm (4) 5 × 10–8 cm
Sol. Answer (2)

h h 1
A  B  PB  PA (Given)
PA PB 2

h
 A PA PB 1
 
B h PA Putting PB = 2 PA
PB

 A 1 PA

B 2 PA

B = 2A [A = 5 × 10–8m]

B = 2 × 5 × 10–8

= 10 × 10–8m ∵ 1 m = 100 cm
= 10–7 m = 10–5 cm

12. Any f-orbital can accommodate upto

(1) 2 electrons with parallel spin (2) 6 electrons
(3) 2 electrons with opposite spin (4) 14 electrons
Sol. Answer (3)
Any orbital have maximum of two electrons with opposite spin.
Solution of Assignment (Set-2) Structure of Atom 41
13. Total number of spectral lines in UV region, during transition from 5th excited state to 1st excited state
(1) 10 (2) 3 (3) 4 (4) Zero
Sol. Answer (4)
As 1st excited state means n1 = 2
For 5th excited state means n2 = 6
 e– will transit between 6th level to 2nd level
No transition will be upto 1st level. Because no line will appear in Lyman series i.e. UV region.

14. For principal quantum number n = 5, the total number of orbitals having l = 3 is
(1) 7 (2) 14 (3) 9 (4) 18
Sol. Answer (1)
For l = 3 m = –3, –2, –1, 0, +1, +2, +3
i.e., 7 orbitals are present

15. The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at

5R 3R 7R 9R
(1) cm 1 (2) cm1 (3) cm1 (4) cm1
36 4 144 400
Sol. Answer (1)
st
1 line in the Balmer series means n1 = 2, n2 = 3

1 ⎡ 1 1⎤
v   R ⎢ 2  2 ⎥ z2 for H z=1
 ⎣⎢ n1 n2 ⎥⎦
1 ⎡1 1⎤ ⎡ 1 1⎤ 5R
v   R ⎢ 2  2 ⎥  12  R ⎢  ⎥  cm1
 ⎣2 3 ⎦ ⎣4 9⎦ 36

16. The four quantum numbers of valence electron of potassium are

1 1 1 1
(1) 4, 0, 1, (2) 4, 1, 0, (3) 4, 0, 0, (4) 4, 1, 1,
2 2 2 2
Sol. Answer (3)
K = 19 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s1 last e–
Last electron 4s1
1
n=4 l=0 m=0 s= 
2

17. In a hydrogen atom, if the energy of electron in the ground state is –x eV., then that in the 2nd excited state
of He+ is

4 9
(1) –x eV (2)  x eV (3) +2x eV (4)  x eV
9 4
Sol. Answer (2)
⎧Eground  x eV given
⎪⎪
Eground state
En   z2 ⎨n  3 because 2nd excited state
n 2 ⎪z  2 because
⎪⎩

x 4
  2    x eV
2

3 2 9
42 Structure of Atom Solution of Assignment (Set-2)

18. In the ground state, an element has 13 electrons in its M-shell. The element is
(1) Manganese (2) Cobalt (3) Nickel (4) Iron
Sol. Answer (1)
M shell means 3rd orbit
2 2 6 2 6 2 5
Mn = 25 = 1s , 2s , 2p , 3s , 3p , 4s , 3d total 13 e– in 3 orbit
8 5
Co = 27 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d7 total 15 e– in 3 orbit
Ni = 28 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d8 total 16 e– in 3 orbit
Fe = 26 = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d6 total 14 e– in 3 orbit

19. The wavelength of radiation emitted, when in He+ electron falls from infinity to stationary state would be
(R = 1.097 × 107 m–1)
(1) 2.2 × 10–8 m (2) 22 × 10–9 m (3) 120 m (4) 22 × 107 m
Sol. Answer (1)

n1 = 1 For He z = 2
n2 =  given
1 ⎡1 1⎤
 R ⎢ 2  2 ⎥   2
2
He ⎣1  ⎦

1
 109678  4 cm1
He

1 1
He    2.2  10 6 cm
109678  4 438712
 2.2  108 m

20. In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of
the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen?
(1) 3  1 (2) 5  2 (3) 2  5 (4) 3  2
Sol. Answer (2)
Third line means third excited state
i.e. n1 = 2 Balmer series (visible region)
n2 = 5 Third line
 Third line will appear when electron comes from 5th energy level to 2nd level.

21. The correct order of energy difference between adjacent energy levels in H atom
(1) E2 – E1 > E3 – E2 > E4 – E3 (2) E2 – E1 > E4 – E3 > E3 – E2
(3) E4 – E3 > E3 – E2 > E2 – E1 (4) E3 – E2 > E4 – E3 > E2 – E1
Sol. Answer (1)
In H atom

1312 1312 1312 1312 1312

E1 = E2 = E3 = E4 = E5 =
12 4 9 25 36
 (E2 – E1) > (E3 – E2) > (E4 – E3) .......
[Alternatively as the distance from the nucleus increases the value of E (energy difference between two shell)
decreases]
Solution of Assignment (Set-2) Structure of Atom 43
22. Which of the following electronic in a transition hydrogen atom will require the largest amount of energy?
(1) n = 1 to n = 2 (2) n = 2 to n = 3 (3) n = 1 to n =  (4) n = 3 to n = 5
Sol. Answer (3)
Largest amount of energy is required for the transition between 1 

1 ⎡1 1⎤
E  hc  ⇒ hcR ⎢ 2  2 ⎥ [Large the difference between n and n large will be the value of E]
 ⎣⎢ n1 n2 ⎥⎦
1 2

23. Which combinations of quantum numbers n, l, m and s for the electron in an atom does not provide a
permissible solutions of the wave equation?
1 1 1 1
(1) 3, 2,  2, (2) 3, 3, 1,  (3) 3, 2, 1, (4) 3, 1, 1, 
2 2 2 2
Sol. Answer (2)
n = 3 l = 3 [Not possible because value of l can never be equals to n]

24. The orbital angular momentum of electron in 4s orbital is

1 h h h
(1) . (2) Zero (3) (4) (2.5)
2 2 2 2
Sol. Answer (2)
h
Orbital angular momentum = l  l  1
4
For 4s electron the value of l = 0  [orbital angular momentum = zero]

25. Radial nodes present in 3s and 3p-orbitals are respectively

(1) 0, 2 (2) 2, 1 (3) 1, 1 (4) 2, 2
Sol. Answer (2)
Radial nodes = (n – l – 1)
for 3s (3 – 0 – 1) = 2 ; For 3p (3 – 1 – 1) = 1

26. Quantum numbers for some electrons are given below

A : n = 4, l = 1 B : n = 4, l = 0
C : n = 3, l = 2 D : n = 3, l = 1
The correct increasing order of energy of electrons
(1) A < B < C < D (2) D < C < B < A
(3) D < B < C < A (4) C < B < A < D
Sol. Answer (3)
Energy = (n + l)
A=n=4 l=1 =4+1=5
B=n=4 l=0 =4+0=4
C=n=3 l=2 =3+2=5
D=n=3 l=1 =3+1=4
According to Pauli exclusion principle
(1) Larger the (n + l); larer will be energy
(2) Same value of (n + l) ; smaller n ; more will be energy

 DBC A
44 Structure of Atom Solution of Assignment (Set-2)

27. The number of lobes in most of the d-orbitals are

(1) 6 (2) 8 (3) 10 (4) 4
Sol. Answer (4)

28. For which of the following options m = 0 for all orbitals?

(1) 2s, 2px, 3dxy (2) 3s, 2 pz , 3d z2 (3) 2s, 2 pz , 3d x 2  y 2 (4) 3s, 3px, 3dyz

Sol. Answer (2)

Value of m = 0 for 3s, 2Pz and 3d z2

29. Which is the correct graphical representation based on photoelectric effect?

No. of photons
K.E. K.E. K.E.
I. II. III. IV.

 0  Intensity of light Intensity of light

(1) I & II (2) II & III (3) III & IV (4) II & IV
Sol. Answer (4)
For photoelectric effect
KE = h( – 0)
KE = h – h0
No. of photons 
KE 

0  Intensity of light 
(II)
(IV)
0 = Threshold frequency

 [KE of e– increases after crossing. Threshold frequency]

30. The time taken by the electron in one complete revolution in the nth Bohr’s orbit of the hydrogen atom is

(1) Inversely proportional to n2 (2) Directly proportional to n3

h n
(3) Directly proportional to (4) Inversely proportional to
2 h
Sol. Answer (2)

circumference 2r n3
Time period =   2  1.5  1016 seconds
velocity vn z
Time period  n3
Solution of Assignment (Set-2) Structure of Atom 45
31. What will be the ratio of the wavelength of the first line to that of the second line of Paschen series of H
atom?
(1) 256 : 175 (2) 175 : 256 (3) 15 : 16 (4) 24 : 27
Sol. Answer (1)
First time of paschen series n1 = 3, n2 = 4
1 ⎡1 1 ⎤
 R⎢  ⎥
1 ⎣ 9 16 ⎦
1 7R 144
 1 
1 144 7R
Second line of paschen series n1 = 3, n2 = 5
1 ⎡1 1 ⎤
 R⎢  ⎥
2 ⎣ 9 25 ⎦
1 16R 225
 2 
 2 225 16R
1 144 16R 2304 256
   
2 7R 225 1575 175

32. In any sub-shell, the maximum number of electrons having same value of spin quantum number is
(1) l (l  1) (2) l + 2 (3) 2l + 1 (4) 4l + 2
Sol. Answer (3)
Total number of electron in subshell = 2(2l + 1) l = angular quantum number
2  2l  1
Number of electrons having same spin =   2l  1
2
[Because half e– have clockwise and half e– have anti clockwise spin]

33. For the transition from n = 2  n = 1, which of the following will produce shortest wavelength?
(1) H atom (2) D atom (3) He+ ion (4) Li2+ ion
Sol. Answer (4)
1 ⎡1 1⎤
 R ⎢ 2  2 ⎥ z2 as n  n are constant
 ⎣⎢ n1 n2 ⎦⎥
1 2

1
  more the nuclear charge smaller will be the 
z2
H=z=1
D=z=1  Li2+ have shorter wavelength

He = z = 2
Li2+ = z = 3

34. If each orbital can hold a maximum of 3 electrons. The number of elements in 2nd period of periodic table
(long form) is
(1) 27 (2) 9 (3) 18 (4) 12
Sol. Answer (4)
For 2nd period electronic configuration = 2s2, 2p6
If each orbital have 3e– then electronic configuration = 2s3, 2px3, 2py3, 2pz3
Total 12 e– will present
46 Structure of Atom Solution of Assignment (Set-2)

SECTION - C
Previous Years Questions
1. What is the maximum numbers of electrons that can be associated with the following set of quantum numbers?
n = 3, l = 1 and m = –1
(1) 6 (2) 4 (3) 2 (4) 10
Sol. Answer (3)
Value of m = –1 one value represents one orbital. Therefore maximum number of electrons will be two

2. Maximum number of electrons in a subshell with l = 3 and n = 4 is

(1) 10 (2) 12 (3) 14 (4) 16
Sol. Answer (3)
n=4 l = 3 represents 4f subshell having 7 orbitals

 Total number of electrons = 14

3. The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) is
1 1 1 1
(1) 5, 0, 0, + (2) 5, 1, 0, + (3) 5, 1, 1, + (4) 6, 0, 0, +
2 2 2 2
Sol. Answer (1)
Rb = 37 = [Ar] 4s2, 3d10, 4p6, 5s1 last e–
1
 n = 5, l = 0, m = 0, s = 
2

4. The total number of atomic orbitals in fourth energy level of an atom is

(1) 4 (2) 8 (3) 16 (4) 32
Sol. Answer (3)
2
Number of orbitals = x n = number of orbit

= 42 = 16

5. The energies E1 and E2 of two radiations are 25 eV and 50 eV respectively. The relation between their
wavelengths i.e. 1 and 2 will be
1
(1) 1   (2) 1   2 (3) 1  2 2 (4) 1  4 2
2 2
Sol. Answer (3)

hc hc
1  2  E1 = 25 eV E2 = 50 eV
E1 E2

hc
1  ...... (1)
25

hc
2  ...... (2)
50
hc
1 25 1
 2  2 1  2 2
 2 hc 2
50
Solution of Assignment (Set-2) Structure of Atom 47
6. If n = 6, the correct sequence of filling of electrons will be

(1) ns  (n  1)d  (n  2)f (2) ns  (n  2)f  (n  1)d  np

(3) ns  (n  1)d  (n  2)f  np (4) ns  (n  2)f  np  (n  1)d

Sol. Answer (2)

Putting the value of n and calculating the (n + l) value
6s < 4f < 5d < 6p
6  0  4  3 5  2  6  1
6 7 7 7
(lower energy)  high energy 

7. The value of Planck's constant is 6.63 × 10–34 Js. The speed of light is 3 × 1017 nm s–1. Which value is closest
to the wavelength in nanometer of a quantum of light with frequency of 6 × 1015 s–1?
(1) 25 (2) 50 (3) 75 (4) 10
Sol. Answer (2)

c 3.0  1017 nms1 1

    102 nm
v 6  1015 s1 2

 0.5  102 nm = 50 nm

8. According to the Bohr Theory, which of the following transitions in the hydrogen atom will give rise to the least
energetic photon?
(1) n = 6 to n = 5 (2) n = 5 to n = 3 (3) n = 6 to n = 1 (4) n = 5 to n = 4
Sol. Answer (1)
Because (E2 – E1) > (E3 – E2) > (E4 – E3) > (E5 – E4) > (E6 – E5)
As the difference is of one energy levels
 (E6 – E5) have less energy
{Alternatively value of E [difference between two succisive energy level decreases] as the distance from the
nucleus increases.}

⎛ Z2 ⎞
9. Based on equation E = –2.178 × 10–18 J ⎜⎜ 2 ⎟⎟ certain conclusions are written. Which of them is not correct?
⎝n ⎠
(1) Larger the value of n, the larger is the orbit radius
(2) Equation can be used to calculate the change in energy when the electron changes orbit
(3) For n = 1, the electron has a more negative energy than it does for n = 6 which means that the electron
is more loosely bound in the smallest allowed orbit
(4) The negative sign in equation simply means that the energy of electron bound to the nucleus is lower than
it would be if the electrons were at the infinite distance from the nucleus
Sol. Answer (3)
In (n = 2) 1st shell e– is tightly held compared to n = 6 (6th shell)
48 Structure of Atom Solution of Assignment (Set-2)

10. Uncertainty in position of an electron (mass = 9.1 × 10–28g) moving with a velocity of 3 × 104 cm/s accurate
upto 0.001% will be (Use h/(4) in uncertainty expression where h = 6.626 × 10–27 erg-s)
(1) 5.76 cm (2) 7.68 cm (3) 1.93 cm (4) 3.84 cm
Sol. Answer (3)

3  10 4  0.001
v  cm/second
103  100
h 6.6  1031 J-s
x    1.93 cm
4mv 0.001
4  3.14  9.1 1028   3  104
100

11. The radius of hydrogen atom in the ground state is 0.53 Å. The radius of Li2+ ion (atomic number = 3) in a similar
state is
(1) 0.53 Å (2) 1.06 Å (3) 0.17 Å (4) 0.265 Å
Sol. Answer (3)

0.53   3 
2
r0  n2
rn     0.53  3  Å
z 3

1.59 Å  1.7 Å
n = 3 orbit
z = 3 Li2+

12. In a Bohr’s model of an atom, when an electron jumps from n = 1 to n = 3, how much energy will be emitted
or absorbed?
(1) 2.389 × 10–12 ergs (2) 0.239 × 10–10 ergs (3) 2.15 × 10–11 ergs (4) 0.1936 × 10–10 ergs
Sol. Answer (4)

1312
Energy of electron when n = 1 E1   kJ/mol
12
1312 1312
Energy of electron when n = 3 E3    kJ/mol
3 2 9
E  E3  E1

1312 ⎛ 1312 ⎞
 ⎜ ⎟  1166 kJ
9 ⎝ 1 ⎠
= 1166 × 103 J
= 1166 × 10+10 erg
Alternatively

hc ⎡1 1⎤
E   hc  R ⎢ 2  2 ⎥
 ⎣⎢ n1 n2 ⎦⎥
⎡1 1⎤
 3.0  108  6.6  1034  8.314 ⎢ 2  2 ⎥ J
⎣1 3 ⎦
⎡8 ⎤
 3.0  108  6.6  1034  8.314 ⎢ ⎥
⎣9 ⎦
Solution of Assignment (Set-2) Structure of Atom 49
13. The electronic configuration of gadolinium (Atomic No. 64) is
(1) [Xe]4f 3 5d 5 6s2 (2) [Xe]4f 6 5d 2 6d 2 (3) [Xe]4f 8 5d 9 6s2 (4) [Xe] 4f 7 5d 1 6s 2
Sol. Answer (4)
Gd have exceptional configuration e– will enter in 5d because 4f have 7 electrons and have half filled stability

Gd = [Xe]54 4f 7 5d 1 6s 2

14. The ion that is isoelectronic with CO is

(1) CN– (2) N2+ (3) O2– (4) N2–
Sol. Answer (1)
Isoelectronic means same number of electrons
CO = Number of electrons = 14
CN– = 6 + 7 + 1 = 14

15. The Bohr orbit radius for the hydrogen atom (n = 1) is approximately 0.530 Å. The radius for the first excited state
(n = 2) orbit is (in Å)
(1) 4.77 (2) 1.06 (3) 0.13 (4) 2.12
Sol. Answer (4)

r0  n2
rn  n = Number of orbit, z = charge on nucleus
z
n2
Ist excited state for H =
z 1
0.53   2 
2

1
 0.53  4  2.12 Å

16. The position of both, an electron and a helium atom is known within 1.0 nm. Further the momentum of the electron
is known within 5.0 × 10–26 kg ms–1. The minimum uncertainty in the measurement of the momentum of the
helium atom is
(1) 8.0 × 10–26 kg ms–1 (2) 80 kg ms–1 (3) 50 kg ms–1 (4) 5.0 × 10–26 kg ms–1
Sol. Answer (4)

h
Δx electron  Pelectron 
4
h
xelectron 
4 Pelectron
x electron  nHe  1.0 nm
h
xHe  PHe 
4
h
xHe 
4PHe
h h
 
4Pelectron 4PHe

PHe  5.0  10 26 kg ms 1

50 Structure of Atom Solution of Assignment (Set-2)

17. Which of the following configuration is correct for iron?

(1) 1s 22s 22p63s 23p64s23d 7 (2) 1s22s22p63s23p64s23d 5
(3) 1s22s22p63s 23p63d 5 (4) 1s 22s 22p63s 23p64s23d 6
Sol. Answer (4)
Fe = 26 = 1s 2, 2s 2, 2p6, 3s 2, 3p6, 4s 2, 3d 6

18. Which of the following has maximum number of unpaired d-electrons?

(1) N3+ (2) Fe2+ (3) Zn+ (4) Cu+
Sol. Answer (2)
N3+ = 4 = 1s 2, 2s 2 Zero unpaired

Fe2+ = 24 = 1s 2, 2s 2, 2p6, 3s 2, 3p6, 3d 6 Four unpaired

Zn = 29 = 1s 2, 2s 2, 2p6, 3s 2, 3p6, 4s 2, 3d 10 One unpaired


Cu = 28 = 1s 2, 2s 2, 2p6, 3s 2, 3p6, 4s 0, 3d 10 Zero unpaired

19. Who modified Bohr’s theory by introducing elliptical orbits for electron path?
(1) Rutherford (2) Thomson (3) Hund (4) Sommerfield
Sol. Answer (4)

1 2
n length of major axis 3

k length of minor axis

Number of elliptrical orbit in shell = (n – 1)

20. The de Broglie wavelength of a particle with mass 1 g and velocity 100 m/s is
(1) 6.63 × 10–35 m (2) 6.63 × 10–34 m (3) 6.63 × 10–33 m (4) 6.65 × 10–35 m
Sol. Answer (3)

h
 m = 1 g = 0.001 kg v = 100 m/s
mv

6.6  1034 kg m2 s2


0.001 kg  100 m/s

  6.63  1033 m

21. The following quantum numbers are possible for how many orbitals : n = 3, l = 2, m = +2?
(1) 1 (2) 2 (3) 3 (4) 4
Sol. Answer (1)
As the value of m = + 2
i.e. one value
Therefore one orbital is represented

22. The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen
atom will be (Given ionization energy of H = 2.18 × 10–18 J atom–1 and h = 6.625 × 10–34 Js)
(1) 1.54 × 1015 s–1 (2) 1.03 × 1015 s–1 (3) 3.08 × 1015 s–1 (4) 2.00 × 1015 s–1
Solution of Assignment (Set-2) Structure of Atom 51
Sol. Answer (3)

c ⎡1 1⎤
  c  R ⎢ 2  2 ⎥ z2
 ⎣⎢ n1 n2 ⎥⎦
⎡1 1⎤
  3.0  1010  109678 ⎢ 2  2 ⎥  1
2

⎣1 4 ⎦
10 ⎡15 ⎤
  3 10
 109678 cm1 ⎢ ⎥  3.09  1015 s1
10 ⎣16 ⎦
Alternatively

I  E = E – E1

2.18  1018  E  E1

E1  2.18  10 18 J

2.18  1018 2.18  1018

E4     0.136  1018 J
 4 2 16

E = E 4  E1 ⎡⎣ 0.136   2.18  ⎤⎦  10 18

hc
 2.04  10 18  .... (1)
E
c
v .... (2)

Put (1) in (2)
c E
v  E 
hc h
2.04  10 18
v  0.309  10 16  3.09  1015 s 1
6.6  10 34

23. Given : The mass of electron is 9.11 × 10–31 kg, Planck constant is 6.626 × 10–34 Js, the uncertainty involved
in the measurement of velocity within a distance of 0.1 Å is
(1) 5.79 × 105 ms–1 (2) 5.79 × 106 ms–1 (3) 5.79 × 107 ms–1 (4) 5.79 × 108 ms–1
Sol. Answer (2)
h h
x  m v = v 
4 x  m  4  

6.6  10 34 J-s

v  6
0.1  10 10 m  9.1  10 31 kg  4  3.14  5.799  10 m/s

24. Which of the following is not permissible arrangement of electrons in an atom?

(1) n = 3, l = 2, m = –2, s = –1/2 (2) n = 4, l = 0, m = 0, s = –1/2
(3) n = 5, l = 3, m = 0, s = +1/2 (4) n = 3, l = 2, m = –3, s = –1/2
Sol. Answer (4)
1
n=3 l=2 m = –3 s= 
2
Value of m (orbital) depends upon l i.e., it cannot be more than 'l '. Therefore is wrong.
52 Structure of Atom Solution of Assignment (Set-2)

25. The energy absorbed by each molecule (A2) of a substance is 4.4 × 10–19 J and bond energy per molecule
is 4.0 × 10–19 J. The kinetic energy of the molecule per atom will be
(1) 4.0 × 10–20 J (2) 2.0 × 10–20 J (3) 2.2 × 10–19 J (4) 2.0 × 10–19 J
Sol. Answer (2)
KE = Energy observed by molecule – Energy required to break one bond

4.4  1019 J  4.0  1019 J

KE =
2
⎡ 0.4  1019 ⎤ 19 20
KE per atom  ⎢ ⎥  ⎡⎣0.2  10 J⎤⎦  ⎡⎣2  10 J⎤⎦
⎣⎢ 2 ⎦⎥

26. Which one of the following ions has electronic configuration [Ar]3d6?
(1) Co3+ (2) Ni3+ (3) Mn3+ (4) Fe3+
(At. nos. Mn = 25, Fe = 26, Co = 27, Ni = 28)
Sol. Answer (1)
Electronic configuration [Ar] 3d6 represents 24 electrons

i.e. Co3  24 e

Ni3+ = 28 – 3 = 25 e–
Mn3+ = 25 – 3 = 22 e–
Fe3+ = 26 – 3 = 23 e–

27. A 0.66 kg ball is moving with a speed of 100 m/s. The associated wavelength will be
(1) 6.6 × 10–32 m (2) 6.6 × 10–34 m (3) 1.0 × 10–35 m (4) 1.0 × 10–32 m
Sol. Answer (3)

h 6.6  1034 kgm2 s2  s

   1.0  1035 m
mv 0.66 kg  100 m/s

28. Which of the following is not among shortcomings of Bohr’s model?

(1) Bohr theory could not account for the fine lines in the atomic spectrum
(2) Bohr theory was unable to account for the splitting of the spectral lines in the presence of magnetic field
(3) Bohr theory failed for He atom
(4) It did not give information about energy level
Sol. Answer (4)
Bohr's model explain the energy level i.e. Energy of electron in each orbital is quantized.

1312
En  z2 kJ/mol
x2

29. Number of spectral lines falling in Balmer series when electrons are de-excited from nth shell will be given as
(1) (n – 2) in UV (2) (n – 2) in visible region
(3) (n – 3) in near IR (4) (n – 3) in far IR
Sol. Answer (2)
Solution of Assignment (Set-2) Structure of Atom 53
30. The ratio of the energy required to remove an electron from the first three Bohr’s orbits of hydrogen is
(1) 3 : 2 : 1 (2) 9 : 4 : 1 (3) 36 : 9 : 4 (4) 1 : 4 : 9
Sol. Answer (3)
1312 x
En  2

x x2
x x x
E1   E2  E3 
2
1 2 2
32
1 1 1 36 : 9 : 4
: :   36 : 9 : 4
1 4 9 36

31. What is the maximum number of orbitals that can be identified with the following quantum numbers?
n = 3, l = l, ml = 0
(1) 1 (2) 2 (3) 3 (4) 4
Sol. Answer (1)

32. Calculate the energy in joule corresponding to light of wavelength 45 nm: (Planck's constant
h = 6.63 × 10–34 Js; speed of light c = 3 × 108 ms–1)
(1) 6.67 × 1015 (2) 6.67 × 1011 (3) 4.42 × 10–15 (4) 4.42 × 10–18
Sol. Answer (4)

hc 6.63  10 34  3  108 6.63

E    10 17 = 4.42 × 10–18
 45  10 9 15

SECTION - D
Assertion - Reason Type Questions
1. A : Orbital angular momentum of (1s, 2s, 3s etc.) all s electrons is same.
R : Orbital angular momentum depends on orientation of orbitals.
Sol. Answer (3)

2. A : Energy of electron is taken negative.

R : Energy of electron at infinity is zero.
Sol. Answer (1)

3. A : Bohr’s orbits are also called stationary states.

R : Electrons are stationary in an orbit.
Sol. Answer (3)

4. A : K.E. of two subatomic particles, having same de-Broglie’s wavelength is same.

R : de-Broglie’s wavelength is directly related to mass of subatomic particles.
Sol. Answer (4)

5. A : Electronic energy for hydrogen atom of different orbitals follow the sequence :
1s < 2s = 2p < 3s = 3p = 3d.
R : Electronic energy for hydrogen atom depends only on n and is independent of ‘l’ & ‘m’ values.
Sol. Answer (1)
54 Structure of Atom Solution of Assignment (Set-2)

6. A : Wavelength for first line of any series in hydrogen spectrum is biggest among all other lines of the same
series.
R : Wavelength of spectral line for an electronic transition is inversely related to difference in the energy levels
involved in the transition.
Sol. Answer (1)

7. A : Zn(II) salts are diamagnetic.

R : Zn2+ ion has one unpaired electron.
Sol. Answer (3)

8. A : In third energy level there is no f-subshell.

R : For n = 3, the possible values of l are 0, 1, 2 and for f-subshell l = 3.
Sol. Answer (1)

9. A : The charge to mass ratio of the particles in anode rays depends on nature of gas taken in the discharge
tube.
R : The particles of anode rays carry positive charge.
Sol. Answer (2)

10. A : Angular momentum of an electron in an atom is quantized.

R : In an atom only those orbits are permitted in which angular momentum of the electron is a natural number
h
multiple of .
2
Sol. Answer (1)

11. A : The radius of second orbit of He+ is equal to that of first orbit of hydrogen.
R : The radius of an orbit in hydrogen like species is directly proportional to n and inversely proportional to
Z.
Sol. Answer (4)

12. A : The orbitals having equal energy are known as degenerate orbitals.
R : The three 2p orbitals are degenerate in the presence of external magnetic field.
Sol. Answer (3)

13. A : In a multielectron atom, the electrons in different sub-shells have different energies.
R : Energy of an orbital depends upon n + l value.
Sol. Answer (1)

14. A : Isotopes of an element have almost similar chemical properties.

R : Isotopes have same electronic configuration.
Sol. Answer (1)

15. A : The number of angular nodes in 3d z2 is zero.

R : Number of angular nodes of atomic orbitals is equal to value of l.
Sol. Answer (2)