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1

Structure of Atom

—Classical Mechanics

C H A P T E R
C O N T E N T S
DISCOVERY OF ELECTRON
MEASUREMENT OF E/M
FOR ELECTRONS
DETERMINATION OF THE
CHARGE ON AN ELECTRON
DEFINITION OF AN ELECTRON
POSITIVE RAYS
PROTONS
NEUTRONS
SUBATOMIC PARTICLES
ALPHA PARTICLES
RUTHERFORD’S
ATOMIC MODEL
MOSLEY’S DETERMINES
ATOMIC NUMBER
MASS NUMBER
COMPOSITION OF THE
NUCLEUS
QUANTUM THEORY
AND BOHR ATOM

ohn Dalton (1805) considered that all matter was composed
of small particles called atoms
...
At
the end of the nineteenth century there accumulated enough
experimental evidence to show that the atom is made of still
smaller particles
...
The number of subatomic particles now
known is very large
...
How these fundamental particles
go to make the internal structure of the atom, is a fascinating
story
...
J
...


1

2

1

PHYSICAL CHEMISTRY
Dalton (1805)
Thomson (1896) - Positive and negative charges

Rutherford (1909) - The Nucleus

Bohr (1913) - Energy levels

Schrödinger (1926) - Electron cloud model

Atomic Model : Timeline

CATHODE RAYS – THE DISCOVERY OF ELECTRON
The knowledge about the electron was derived as a result of the study of the electric discharge
in the discharge tube (J
...
Thomson, 1896)
...
1
...
Through a glass side-arm air can be drawn with a pump
...

The electric discharge passes between the electrodes and the residual gas in the tube begins to glow
...
The rays which
proceed from the cathode and move away from it at right angles in straight lines are called
Cathode Rays
...
1
Production of cathode rays
...
They travel in straight lines away from the cathode and cast shadows of metallic objects
placed in their path
...
Cathode rays cause mechanical motion of a small pin-wheel placed in their path
...

3
...

4
...

5
...

6
...

By counterbalancing the effect of magnetic and electric field on cathode rays
...
In SI units the value
of e/m of cathode particles is – 1
...
As a result of several experiments,
Thomson showed that the value of e/m of the cathode particle was the same regardless of both the
gas and the metal of which the cathode was made
...
Dutch Physicist H
...

Lorentz named them Electrons
...
These are also emitted as β-particles by radioactive substances
...


MEASUREMENT OF e/m FOR ELECTRONS
The ratio of charge to mass (e/m) for an electron was measured by J
...
Thomson (1897) using
the apparatus shown in Fig
...
2
...
Magnetic field is
applied first and causes the electrons to be deflected in a circular path while the spot is shifted to Y
...
An electrostatic
field of known strength is then applied so as to bring back the spot to its original position
...

Electrostatic
field plate
Beam of
electrons

X

Cathode

Y

Slit
Perforated
anode

Fluorescent
screen

Evacuated
glass tube

Figure 1
...


Electromagnet

4

1

PHYSICAL CHEMISTRY

Equating magnetic force on the electron beam to centrifugal force
...

e
v
=
m
Br


...

When the electrostatic field strength and magnetic field strength are counterbalanced,
Bev = Ee
where E is the strength of the electrostatic field
...
(2)

If E and B are known, v can be calculated and on substitution in equation (1), we get the value
of e/m
...
Using this
procedure, the ratio e/m works out to be – 1
...

or
e/m for the electron = – 1
...
A
...
The apparatus used by him is shown in Fig
...
3
...
An oil droplet falls through a hole in the
upper plate
...
Some of these electrons are captured by the oil droplet and it acquires a negative charge
...

He adjusted the strength of the electric field between the two charged plates so that a particular
oil drop remained suspended, neither rising nor falling
...
As the X-rays struck the air
molecules, electrons are produced
...
Thus,
Q = ne
where n = number of electrons and e = charge of the electron
...
60 × 10– 19 coulombs
...
3
Milikan's apparatus for the Oil-drop experiment
...

e/m = – 1
...
60 × 10– 19 coulomb (Milikan)
1
...
76 × 108
e/m
hence
m = 9
...
1 × 10– 31 kg
Mass of an Electron relative to H
Avogadro number, the number of atoms in one gram atom of any element is 6
...
From
this we can find the absolute mass of hydrogen atom
...
023 × 1023 atoms of hydrogen = 1
...
008
g
∴ Mass of a hydrogen atom =
6
...
67 × 10– 24 g
But mass of electron =
9
...
67 × 10−24
mass of H atom
∴
=
9
...
835 × 103 = 1835
Thus an atom of hydrogen is 1835 times as heavy as an electron
...

1835

6

1

PHYSICAL CHEMISTRY

DEFINITION OF AN ELECTRON
Having known the charge and mass of an electron, it can be defined as :
An electron is a subatomic particle which bears charge – 1
...
1 × 10– 28 g
...

Since an electron has the smallest charge known, it was designated as unit charge by Thomson
...
1
...
He observed
that while cathode rays were streaming away from the cathode, there were coloured rays produced
simultaneously which passed through the perforated cathode and caused a glow on the wall opposite
to the anode
...
He called them Positive rays
...
4
Production of Positive rays
...

(2) They are deflected by electric as well as magnetic field in a way indicating that they are
positively charged
...

(4) They possess mass many times the mass of an electron
...

How are Positive rays produced ?
When high-speed electrons (cathode rays) strike molecule of a gas placed in the discharge tube,
they knock out one or more electrons from it
...
When
electric discharge is passed through the gas under high electric pressure, its molecules are dissociated
into atoms and the positive atoms (ions) constitute the positive rays
...


PROTONS
E
...


→
H ⎯⎯ H+ + e–
proton
It was J
...
Thomson who studied their nature
...
672 × 10– 24 gram
...

(2) The electrical charge of proton is equal in magnitude but opposite to that of the electron
...
60 × 10–19 coulombs or + 1 elementary charge unit
...
Protons were also obtained in a variety of nuclear
reactions indicating further that all atoms contain protons
...

A proton is a subatomic particle which has one unit mass and one unit positive charge
...
He directed a stream of
4
alpha particles 2 He at a beryllium target
...
It has almost
the same mass (1
...


(

)

0

Beryllium

Neutrons

Charge detector
indicates no charge

Figure 1
...


He named it neutron
...
Thus :
A neutron is a subatomic particle which has a mass almost equal to that of a proton and
has no charge
...

4
He
2

+ 9 Be ⎯⎯
→
4

12
C
6

+ 1n
0

SUBATOMIC PARTICLES
We have hitherto studied the properties of the three principal fundamental particles of the atom,
namely the electron, proton, and neutron
...
1
...
1
...
1 × 10– 28

–1

– 1
...
672 × 10– 24
1
...
60 × 10– 19
0

Nearly all of the ordinary chemical properties of matter can be examined in terms of atoms
consisting of electrons, protons and neutrons
...

Other Subatomic Particles
Besides electrons, protons and neutrons, many other subatomic particles such as mesons,
positrons, neutrinos and antiprotons have been discovered
...
With each discovery,
the picture of atomic structure becomes increasingly complex
...


ALPHA PARTICLES
Alpha particles are shot out from radioactive elements with very high speed
...
5 × 107 m/sec
...
Thus an alpha particle has 2+ charge and 4 amu mass
...

Conclusion
Though α-particle is not a fundamental particle of the atom (or subatomic particle) but because
2
of its high energy 1 mv , Rutherford thought of firing them like bullets at atoms and thus obtain
2
information about the structure of the atom
...
This showed the presence of protons in atoms other than
hydrogen atom
...


(

)

RUTHERFORD’S ATOMIC MODEL – THE NUCLEAR ATOM
Having known that atom contains electrons and a positive ion, Rutherford proceeded to perform
experiments to know as to how and where these were located in the atom
...
1
...
They directed a stream of very highly energetic α-particles from a radioactive
source against a thin gold foil provided with a circular fluorescent zinc sulphide screen around it
...


STRUCTURE OF ATOM – CLASSICAL MECHANICS

9

Flash of
light

Gold
foil

Slit

Figure 1
...


Rutherford and Marsden noticed that most of the α-particles passed straight through the gold
foil and thus produced a flash on the screen behind it
...
To their great astonishment, tiny flashes were also seen on other portions
of the screen, some time in front of the gold foil
...
Based on
these observations, Rutherford proposed a model of the atom which is named after him
...
According to it :
-Particles
Large
deflection

Figure 1
...


(1) Atom has a tiny dense central core or the nucleus which contains practically the
entire mass of the atom, leaving the rest of the atom almost empty
...
If the nucleus
were the size of a football, the entire atom would have a diameter of about 5 miles
...

(2) The entire positive charge of the atom is located on the nucleus, while electrons
were distributed in vacant space around it
...

(3) The electrons were moving in orbits or closed circular paths around the nucleus
like planets around the sun
...
8
Rutherford's model of atom ; electrons
orbiting around nucleus
...
9
Orbiting electron would radiate energy
and spiral into the nucleus
...
According to
the classical electromagnetic theory if a charged particle accelerates around an oppositely charged
particle, the former will radiate energy
...
This does not happen actually as then the
atom would be unstable which it is not
...


MOSLEY’S DETERMINATION OF ATOMIC NUMBER
The discovery that atom has a nucleus that carries a positive charge raised the question : What
is the magnitude of the positive charge? This question was answered by Henry Mosley in 1913
...
Mosley found that when cathode rays struck different elements used as anode targets
in the discharge tube, characteristic X-rays were emitted
...

Mosley plotted the atomic number against the square root of the frequency of the X-rays emitted
and obtained a straight line which indicated that atomic number was not a mere ‘position number’
but a fundamental property of the atom
...
The wavelength changed regularly as the element that came next in the Periodic Table had one
proton (one unit atomic mass) more than the previous one
...
10
Production of X-rays
...

Since the atom as a whole is electrically neutral, the atomic number (Z) is also equal to the
number of extranuclear electrons
...
This implies that it has a nucleus containing one proton (+ 1) and one
extranuclear electron (– 1)
...

Co, 27

Fe, 26

Atomic numbers

Mn, 25

Cr, 24

V, 23

Ti, 22

Sc, 21

Ca, 20
0

10

11

12

Frequency x

13

14

10 –8

Figure 1
...


WHAT IS MASS NUMBER ?
The total number of protons and neutrons in the nucleus of an atom is called the Mass Number,
A, of the atom
...
Thus mass number of an atom is
equal to the total number of nucleons in the nucleus of an atom
...
Since electrons have practically no
mass, the entire atomic mass is due to protons and neutrons, each of which has a mass almost exactly
one unit
...
For example,
the atomic mass of sodium and fluorine obtained by experiment is 22
...
9815 amu
respectively
...

Each different variety of atom, as determined by the composition of its nucleus, is called a
nuclide
...
By definition :

1

12

PHYSICAL CHEMISTRY

Atomic Number, Z = Number of protons
Mass Number, A = Number of protons + Number of neutrons
∴ The number of neutrons is given by the expression :
N = A–Z

SOLVED PROBLEM
...
029
...

SOLUTION
Atomic Number of uranium = 92
∴
Number of electrons = 92
and Number of protons = 92
Number of neutrons (N) is given by the expression
N = A–Z
Mass Number (A) is obtained by rounding off the atomic weight
= 238
...

The composition of nuclei of some atoms is given in Table 1
...


TABLE 1
...
COMPOSITION OF THE NUCLEUS OF SOME ATOMS
Atom

Mass Number (A)

Atomic Number (Z)

COMPOSITION
Protons = Z

Be
F
Na
Al
P
Sc
Au

9
19
23
27
31
45
197

4
9
11
13
15
21
79

Neutrons = A – Z

4
9
11
13
15
21
79

5
10
12
14
16
24
118

QUANTUM THEORY AND BOHR ATOM
Rutherford model laid the foundation of the model picture of the atom
...

Rutherford recognised that electrons were orbiting around the nucleus
...
Thus Rutherford model failed to explain why
electrons did not do so
...
He closely studied the behaviour of electrons, radiations
and atomic spectra
...
With his theoretical model he was able to explain as to why an orbiting electron did
not collapse into the nucleus and how the atomic spectra were caused by the radiations emitted when
electrons moved from one orbit to the other
...

Electromagnetic Radiations
Energy can be transmitted through space by electromagnetic radiations
...

Electromagnetic radiations are so named because they consist of waves which have electrical
and magnetic properties
...
Such a vibrating particle causes an intermittent disturbance which constitutes
a wave
...
The wave travels at right
angle to the vibratory motion of the object
...
12
Illustration of wave motion caused by a vibrating source
...

The stone makes the water molecules vibrate up and down and transmit its energy as waves on water
surface
...

A wave may be produced by the actual displacement of particles of the medium as in case of
water or sound waves
...
Thus vibratory motion of electrons would cause a wave train of oscillating electric field
and another of oscillating magnetic field
...

Characteristics of Waves
A series of waves produced by a vibrating object can be represented by a wavy curve of the type
shown in Fig
...
12
...
Waves are
characterised by the following properties :
Wavelength
The wavelength is defined as the distance between two successive crests or troughs of a
wave
...
It is expressed in centimetres or metres
or in angstrom units
...
It is also expressed in nanometers
(1nm = 10– 9 m)
...

Frequency is denoted by the letter ν (nu) and is expressed in hertz (hz)
...


14

1

PHYSICAL CHEMISTRY

Speed
The speed (or velocity) of a wave is the distance through which a particular wave travels
in one second
...
If the speed of a wave is c cm/sec,
it means that the distance travelled by the wave in one second is c cm
...
As
evident from Fig
...
13, all types of radiations travel with the same speed or velocity
...

(a)

1 Second

24 cm

(b)

12 cm

(c)

6 cm

Figure 1
...

In all three cases; velocity = x = 120 cm/sec
...
This is reciprocal of the
wavelength and is given the symbol ν (nu bar)
...
Its units are cm–1
or m–1
...
The wavelength of a violet light is 400 nm
...


SOLUTION
...
0 × 108 m sec– 1; λ = 400 nm = 400 × 10– 9 m

frequency, v =

ν =

c 3
...
5 × 1014 sec– 1

=

Also, wave number

ν =
ν =

1
λ

1
400 × 10−9 m

= 25 × 105 m–1

SOLVED PROBLEM
...
09 × 1014 sec– 1
...

c
SOLUTION
...
0 × 108 m sec– 1
ν = 5
...
0 × 108 m sec−1
5
...
09
= 589 × 10–9 m

=

= 589 nm

[∵ 1 nm = 10–9 m]

SPECTRA
A spectrum is an array of waves or particles spread out according to the increasing or
decreasing of some property
...


THE ELECTROMAGNETIC SPECTRUM
Electromagnetic radiations include a range of wavelengths and this array of wavelengths is
referred to as the Electromagnetic radiation spectrum or simply Electromagnetic spectrum
...
1
...


16

1

PHYSICAL CHEMISTRY

Figure 1
...
Wavelength boundaries
of each region are approximate
...
It is composed of
light waves in the range 4000-8000 Å
...
When a beam of white
light is passed through a prism, different wavelengths are refracted (or bent) through different
angles
...
This series of bands that form a continuous
rainbow of colours, is called a Continuous Spectrum
...
15
The continuous spectrum of white light
...
The red component has longer wavelengths (6500 – 7500 Å) and lower frequencies
...


ATOMIC SPECTRA
When an element in the vapour or the gaseous state is heated in a flame or a discharge tube, the
atoms are excited (energised) and emit light radiations of a characteristic colour
...

4900

Blue

5750 5850

Yellow

4250

Indigo

Violet

4000

Green

o

7500 A

6500

Orange

Red

Figure 1
...


For example, a Bunsen burner flame is coloured yellow by sodium salts, red by strontium and
violet by potassium
...
If we
examine the emitted light with a Spectroscope (a device in which a beam of light is passed through
a prism and received on a photograph), the spectrum obtained on the photographic plate is found to
consist of bright lines (Fig
...
18)
...
The emission spectra of some elements are shown in Fig
...
17
...

K

Na

Li

H
3500

4000

4500

5000

5500

6000

6500

7000

o

Wavelength, A

Figure 1
...


When white light composed of all visible wavelengths, is passed through the cool vapour of an
element, certain wavelengths may be absorbed
...
The spectrum obtained in this way consists of a series of dark lines which is
referred to as the Atomic Absorption spectrum or simply Absorption spectrum
...
The absorption
spectrum of an element is the reverse of emission spectrum of the element
...


18

1

PHYSICAL CHEMISTRY

Since the atomic spectra are produced by emission or absorption of energy depending on the
internal structure of the atom, each element has its own characteristic spectrum
...
The most important consequence of the discovery of spectral lines of hydrogen and
other elements was that it led to our present knowledge of atomic structure
...
The light radiation emitted is then examined
with the help of a spectroscope
...
1
...

In 1884 J
...
Balmer observed that there were four prominent coloured lines in the visible hydrogen
spectrum :
(1) a red line with a wavelength of 6563 Å
...

(3) a blue line with a wavelength 4340 Å
...

Photographic
film
o

6563 A
o

4861 A

Hydrogen
discharge tube

o

4340 A
o
4102 A

Glass prism

Slit

Lens

Figure 1
...


The above series of four lines in the visible spectrum of hydrogen was named as the Balmer
Series
...
The Balmer Equation is

1 ⎞
⎛ 1
1
= R⎜ 2 − 2 ⎟
λ
n ⎠
⎝2
where R is a constant called the Rydberg Constant which has the value 109, 677 cm– 1 and n = 3, 4,
5, 6 etc
...

Red

Blue-green

Blue

Violet

6563

4861

4340

4102

Figure 1
...


In addition to Balmer Series, four other spectral series were discovered in the infrared and
ultraviolet regions of the hydrogen spectrum
...
Thus in all
we have Five Spectral Series in the atomic spectrum of hydrogen :

STRUCTURE OF ATOM – CLASSICAL MECHANICS

Name
(1)
(2)
(3)
(4)
(5)

Lyman Series
Balmer Series
Paschen Series
Brackett Series
Pfund Series

19

Region where located
Ultraviolet
Visible
Infrared
Infrared
Infrared

Balmer equation had no theoretical basis at all
...
However, in 1913 Bohr
put forward his theory which immediately explained the observed hydrogen atom spectrum
...


QUANTUM THEORY OF RADIATION
The wave theory of transmission of radiant energy appeared to imply that energy was emitted
(or absorbed) in continuous waves
...
According to him, light radiation was produced
discontinuously by the molecules of the hot body, each of which was vibrating with a specific frequency
which increased with temperature
...
The ‘unit wave’ or ‘pulse of energy’ is
called Quantum (plural, quanta)
...
These light
quanta are called photons
...
Thus light radiations obtained from energised or
‘excited atoms’ consist of a stream of photons and not continuous waves
...
20
A continuous wave and photons
...
(1)
where ν is the frequency of the emitted radiation, and h the Planck’s Constant
...
62 × 10– 27 erg sec
...
62 × 10– 34 J sec
...
(2)
Substituting the value of ν from (2) in (1), we can write

20

1

PHYSICAL CHEMISTRY

hc
λ
Thus the magnitude of a quantum or photon of energy is directly proportional to the
frequency of the radiant energy, or is inversely proportional to its wavelength, λ
...

Thus radiant energy can be emitted as hν, 2hν, 3hν, and so on, but never as 1
...
27 hν,
5
...
e
...


E =

SOLVED PROBLEM
...
8 Å
...
0 × 1010 cm sec −1
ν = λ =
6057
...
952 × 1014 sec–1
(b) Calculation of Energy :
E = hν = (6
...
952 × 1014 sec–1 )
= 3
...
This phenomenon is known as Photoelectric
effect and the ejected electrons Photoelectrons
...
21, the photoelectric effect occurs
...
21
Apparatus for measuring the photoelectric effect
...
This current can be measured with the help of an ammeter
...
It merely increases their rate of emission
...
1
...
If the frequency is decreased below a certain critical value
(Threshold frequency, ν0), no electrons are ejected at all
...
Thus it fails to explain the above
observations
...


Kinetic energy of photoelectrons

(1) A photon of incident light transmits its energy (hν) to an electron in the metal surface which
escapes with kinetic energy 1 mv 2
...
This increases the rate
of emission of electrons, while the kinetic energy of individual photons remains unaffected
...
22
Kinetic energy of photoelectrons plotted
against frequency of incident light
...
The energy of a
photon (hν) is proportional to the frequency of incident light
...
For frequency less than ν0, no electrons will be emitted
...
Thus,
hν = hν 0 +

1 2
mv
2


...
23
It needs a photon(h ) to eject an electronwith energy

m 2
...
hν0 is constant for a
2
particular solid and is designated as W, the work function
...
(2)
2
This is the equation for a straight line that was experimentally obtained in Fig
...
22
...
The value of h thus found came out to be the same as was given by
Planck himself
...
What is the minimum energy that photons must possess in order to produce
photoelectric effect with platinum metal? The threshold frequency for platinum is 1
...

SOLUTION
The threshold frequency (ν0) is the lowest frequency that photons may possess to produce the
photoelectric effect
...

E = hν0
= (6
...
3 × 1015 sec–1)
= 8
...
Calculate the kinetic energy of an electron emitted from a surface of
potassium metal (work function = 3
...
5 × 10–8 cm
...
0 × 1010 cm sec –1
λ
λ = 5
...
0 × 1010 cm sec −1
= 5
...
5 × 10−8 cm
1 2
mv = hν – W
2
= (6
...
5 × 1017 sec–1) – 3
...
63 × 10– 9 erg – 3
...
63 × 10– 9 erg
Thus the electron will be emitted with kinetic energy of 3
...


STRUCTURE OF ATOM – CLASSICAL MECHANICS

23

COMPTON EFFECT
In 1923 A
...
Compton provided one more proof to the quantum theory or the photon theory
...
He
demonstrated that : When X-rays of wavelength λ ' struck a sample of graphite, an electron
was ejected and the X-rays scattered at an angle θ had longer wavelength λ
...
Thus he argued that light radiation (X-rays) consisted of particles
(photons), as a continuous wave could not have knocked out the electron
...
This process could not have occurred unless light
radiation consisted of particles or photons
...
24
Compton scattering of X-rays
...

mc

where h is Planck’s constant, m the mass of an electron, c the velocity of light and θ the angle of
scattering
...
Given the wavelength of a photon, one can calculate the
momentum of the electron ejected
...
It did not say anything as to how and where those electrons were
arranged
...
In 1913 Niels Bohr proposed a new model of atom which explained some of these things
and also the emission spectrum of hydrogen
...

Postulates of Bohr’s Theory
(1) Electrons travel around the nucleus in specific permitted circular orbits and in no
others
...
The
orbits are given the letter designation n and each is numbered 1, 2, 3, etc
...
) as the
distance from the nucleus increases
...

Therefore in each of these orbits the energy of an electron remains the same i
...
it neither loses
nor gains energy
...

(3) An electron can move from one energy level to another by quantum or photon jumps
only
...
When an electron is supplied energy, it
absorbs one quantum or photon of energy and jumps to a higher energy level
...

Orbit numbers
Nucleus

4
3
2

Electrons
permitted in
circular orbits

1

Electrons not
allowed between
orbits

Figure 1
...


The quantum or photon of energy absorbed or emitted is the difference between the lower and
higher energy levels of the atom
ΔE = Ehigh – Elow = hν

...

(4) The angular momentum (mvr) of an electron orbiting around the nucleus is an integral
multiple of Planck’s constant divided by 2π
...
(2)
2π
where m = mass of electron, v = velocity of the electron, r = radius of the orbit ; n = 1, 2, 3, etc
...

By putting the values 1, 2, 3, etc
...

2π 2π 2π
There can be no fractional value of h/2π
...

The integer n in equation (2) can be used to designate an orbit and a corresponding energy level n is
called the atom’s Principal quantum number
...
26
An electron absorbs a photon of light while it jumps from a lower to a higher energy
orbit and a photon is emitted while it returns to the original lower energy level
...
The wavelengths calculated by this method
were found to be in excellent agreement with those in the actual spectrum of hydrogen, which was a
big success for the Bohr model
...
Let m be the mass of the electron, r
the radius of the orbit and ν the tangential velocity of the
revolving electron
...
27
r
Forces keeping electron in orbit
...
Thus,
=

Ze 2

=

mv 2
r

=

mv 2
r


...
(2)

2

r
For hydrogen Z = 1, therefore,
e2
r
Multiplying both sides by r

2

26

1

PHYSICAL CHEMISTRY

According to one of the postulates of Bohr’s theory, angular momentum of the revolving electron
is given by the expression
nh
mvr =
2π
nh
or
ν = 2π mr

...
(4)
4π2 me 2
Since the value of h, m and e had been determined experimentally, substituting these values in
(4), we have
r = n2 × 0
...
(5)
where n is the principal quantum number and hence the number of the orbit
...
529 × 10–8 cm = α0

...
This value is reasonably consistent with other information on the
size of atoms
...
, the value of the second and third orbits of hydrogen comprising
the electron in the excited state can be calculated
...
Calculate the first five Bohr radii
...
529 × 10–8 cm
n = 1 ; r = 12 × 0
...
529 × 10– 8 cm
n = 2 ; r = 22 × 0
...
12 × 10– 8 cm
n = 3 ; r = 32 × 0
...
76 × 10– 8 cm
n = 4 ; r = 42 × 0
...
46 × 10– 8 cm
n = 5 ; r = 52 × 0
...
2 × 10– 8 cm
Energy of electron in each orbit
For hydrogen atom, the energy of the revolving electron, E is the sum of its kinetic energy

⎛1 2⎞
⎜ mv ⎟ and potential energy
⎝2
⎠

⎛ e2 ⎞

...
(7)

27

STRUCTURE OF ATOM – CLASSICAL MECHANICS

E =

1 e2 e2
−
2 r
r

e2
2r
Substituting the value of r from equation (4) in (8)
E = −

or

E = −
= −


...
(9)

n2 h2
Substituting the values of m, e, and h in (9),

E =
or

E =

or

E =

or

E =

− 2
...
179 × 10−18
n2

erg/atom
J per atom

− 2
...
02 × 1023
n2
− 1311
...
3


...


or

E =

SOLVED PROBLEM
...

SOLUTION
From equation (10)
E =

− 2
...
179 × 10−11
2

1

− 2
...
179 × 10−11
2

3

− 2
...
179 × 10−11
52

= − 2
...
8 kJ mol – 1
= − 0
...
9 kJ mol – 1
= − 0
...
5 kJ mol – 1
= − 0
...
0 kJ mol – 1
= − 0
...
44 kJ mol – 1

28

1

PHYSICAL CHEMISTRY

Significance of Negative Value of Energy
The energy of an electron at infinity is arbitrarily assumed to be zero
...
When an electron moves and comes under the influence of nucleus, it does some
work and spends its energy in this process
...
e
...

Bohr’s Explanation of Hydrogen Spectrum
The solitary electron in hydrogen atom at ordinary temperature resides in the first orbit (n = 1)
and is in the lowest energy state (ground state)
...
, 2, 3, 4, 5, 6, 7, etc
...
From these high energy levels, the electron returns by jumps to one or
other lower energy level
...
This gives an
excellent explanation of the various spectral series of hydrogen
...
e
...
Similarly, Balmer, Paschen, Brackett and Pfund series are
produced when the electron returns to the second, third, fourth and fifth energy levels respectively
as shown in Fig
...
28
...
28
Hydrogen spectral series on a Bohr atom energy diagram
...
3
...

3, 4, 5, 6, etc
...

5, 6, 7
6, 7

ultraviolet
visible
infrared
infrared
infrared

920-1200
4000-6500
9500-18750
19450-40500
37800-75000

29

STRUCTURE OF ATOM – CLASSICAL MECHANICS

6
5
4

4102 A

3
2
1

4861 A

o
o

4340 A
o

o

6563 A

Figure 1
...


Value of Rydberg’s constant is the same as in the original empirical Balmer’s equation
According to equation (1), the energy of the electron in orbit n1 (lower) and n2 (higher) is
En = −

2 π2 me 4
2
n1 h 2

1

En = −

2 π2 me 4
2
n2 h 2

2

The difference of energy between the levels n1 and n2 is :
ΔE = En − En =
2

1

2π 2 me 4 ⎡ 1
1⎤
⎢ 2 − 2⎥
2
h
n2 ⎥
⎢ n1
⎣
⎦


...
(2)
λ
where λ is wavelength of photon and c is velocity of light
...
(3)

where R is Rydberg constant
...
It comes out to be 109,679 cm– 1 and agrees closely with the value of Rydberg constant in the
original empirical Balmer’s equation (109,677 cm– 1)
...
Now the equation (3) above can be
written as
⎡1
1
1⎤
= 109679 ⎢ 2 − 2 ⎥
λ
n2 ⎥
⎢2
⎣
⎦
Thus the wavelengths of the photons emitted as the electron returns from energy levels 6, 5, 4
and 3 were calculated by Bohr
...
This was, in fact, a great success of the Bohr atom
...
Find the wavelength in Å of the line in Balmer series that is associated
with drop of the electron from the fourth orbit
...

SOLUTION
The wavelengths of lines in Balmer series are given by

1 ⎞
1
⎛ 1
= R⎜ 2 − 2⎟
λ
n ⎠
⎝2
where λ = wavelength, R (Rydberg constant) = 109,676 cm– 1 ; n = 4
...
561 × 10−5 cm
λ =
109676 × 5
λ in Å = 6
...
But it was spectacularly unsuccessful for every other atom containing
more than one electron
...
In
fact, in view of modern advances, like dual nature of matter, uncertainty principle, any
mechanical model of the atom stands rejected
...
Today we only accept Bohr’s views regarding
quantization as nobody has explained atomic spectra without numerical quantization
and no longer attempted description of atoms on classical mechanics
...


SOMMERFELD’S MODIFICATION OF BOHR ATOM
When spectra were examined with spectrometers, each line was found to consist of several
closely packed lines
...
Sommerfeld modified Bohr’s theory as follows
...
An ellipse has a
major and minor axis
...
The
angular momentum of an electron moving in an elliptic orbit is also supposed to be quantized
...
It is further assumed that the angular momentum can be
an integral part of h/2π units, where h is Planck’s constant
...
The two quantum numbers n and k are related by
the expression :

angular momentum =

length of major axis
n
= length of minor axis
k
n = 3, k = 3
n = 3, k = 2

n = 3, k = 1

Nucleus

Figure 1
...


The values of k for a given value of n are k = n – 1, n – 2, n – 3 and so on
...
When n = k, the orbit will be
circular
...
However,
calculations based on wave mechanics have shown that this is incorrect and the Sommerfeld’s
modification of Bohr atom fell through
...

Langmuir Scheme
We are indebted to Langmuir for putting forward the first elaborate scheme of the arrangement
of extranuclear electrons in 1919
...
Since helium
has two planetary electrons, the first orbit is considered fully saturated with 2 electrons
...
Argon with atomic number 18 will similarly

K 19

2, 8, 8, 1

Ar 18

2, 8, 8

2, 8, 8, 2

Ca 20

2, 8, 2

Mg 12

Sr 38

Fr 87

2, 8, 18,
32, 18, 8, 1

Rn 86

2, 8, 18,
32, 18, 8

2, 8, 18, 32,
18, 8, 2

Ra 88

Hg 80
2, 8, 18,
32, 18, 2

2,8,18,18,8,2

Ba 56

Au 79

2,8,18,18,8,1

2,8,18,18,8

2, 8, 18,
32, 17, 2

Cs 55

Xe 54

Cd 48
2, 8, 18, 18, 2

2, 8, 18, 8, 2

Ag 47

2, 8, 18, 8

2, 8, 18, 17, 2

Rb 37

2, 8, 18, 8, 1

Kr 36

Zn 30

2, 8, 1

2, 8

2, 8, 18, 2

Na 11

Ne 10

2, 2

Cu 29

2, 1

2, 8, 17, 2

Li 3

2

Be 4

Ac 89
2, 8, 18, 32,
18, 9, 2

2, 8, 18,
32, 18, 3

TI 81

2, 8, 18, 18, 9, 2
to 2,8,18,32,9,2

La and Rare
Earths (57-71)

2, 8, 18, 18, 3

In 49

2, 8, 18, 9, 2

Y 39

Ga 31
2, 8, 18, 3

2, 8, 9, 2

2, 8, 3

Al 13

2, 3

2, 8, 18, 32,
19, 9, 2

Th 90

2, 8, 18,
32, 18, 4

Pb 82

2, 8, 18,
32, 10, 2

Hf 72

2,8,18,18,4

Sn 50

2,8,18,10,2

Zr 40

2, 8, 18, 4

Ge 32

2, 8, 10, 2

Ti 22

2, 8, 4

Si 14

2, 4

C6

2, 8, 18, 32,
20, 9, 2

Pa 91

2, 8, 18,
32, 18, 5

Bi 83

2, 8, 18,
32, 11, 2

Ta 73

2,8,18,18,5

Sb 51

2,8,18,11,2

Nb 41

2, 8, 18, 5

As 33

2, 8, 11, 2

V 23

2, 8, 5

P 15

2, 5

N7

Group 4
Group 5
B A
B A
B

B5

Group 3

Sc 21

A

2, 8, 18, 32
21, 9, 2

U 92

2, 8, 18,
32, 18, 6

Po 84

2, 8, 18,
32, 12, 2

W 74

2,8,18,18,6

Te 52

2, 8,18,12,2

Mo 42

2, 8, 18, 6

Se 34

2, 8, 12, 2

Cr 24

2, 8, 6

S 16

2, 6

O8

Group 6
A
B

2,8,18,18,7

2, 8, 18,
32, 18, 7

At 85

2, 8, 18,
32, 13, 2

Re 75

2,8,18

Ru 44

2,8,14,2

Fe 26

2,8,18,
32,14,2

Os 76

I 53 14,2

2,8,18,13,2

Tc 43

2, 8, 18, 7

Br 35

2, 8, 13, 2

Mn 25

2, 8, 7

Cl 17

2, 7

F9

Group 7
A
B
H1

B
1

Group 2

1

A

H1

Group 1
A
B

Ni 28

Pd 46

Pt 78

16,2

2,8,18

2,8,18, 2,8,18,
32,15,2 32,16,2

Ir 77

15,2

2,8,18

Rh 45

2,8,15,2 2,8,16,2

Co 27

Group 8

1

He 2

Group 0

ELECTRONIC CONFIGURATION OF ELEMENTS
(ATOMIC NUMBERS ARE GIVEN AFTER THE SYMBOLS OF THE ELEMENTS)

32
PHYSICAL CHEMISTRY

STRUCTURE OF ATOM – CLASSICAL MECHANICS

33

have the similar arrangement 2, 8, 8
...
Langmuir’s scheme although quite correct for the first few elements, failed
to explain the behaviour of higher elements
...
At about the same time as Bury developed
his scheme on chemical grounds, Bohr (1921) published independently an almost identical scheme of
the arrangement of extra-nuclear electrons
...
Bohr-Bury scheme as it may be called, can be summarised as follows :

+
4p
5n

+
6p
6n

+
8p
8n

Beryllium
Z=4
A=9

Carbon
Z=6
A = 12

Oxygen
Z=8
A = 16

+
10p
10n

+
11p
12n

+
12p
12n

Neon
Z = 10
A = 20

Sodium
Z = 11
A = 23

Magnesium
Z = 12
A = 24

+
13p
14n

+
14p
14n

+
17p
18n

Aluminium
Z = 13
A = 27

Silicon
Z = 14
A = 28

Chlorine
Z = 17
A = 35

Figure 1
...


34

1

PHYSICAL CHEMISTRY

Rule 1
...

The first orbit can contain 2 × 12 = 2 ; second 2 × 22 = 8 ; third 2 × 32 = 18 ; fourth 2 × 42 = 32,
and so on
...
The maximum number of electrons in the outermost orbit is 8 and in the next-to-the
outermost 18
...
It is not necessary for an orbit to be completed before another commences to be formed
...

Rule 4
...

According to Bohr-Bury scheme the configuration of the inert gases is given in the table below :

TABLE 1
...
ELECTRON CONFIGURATION OF INERT GASES
Inert Gas

Atomic
Number

1st (K)

2nd (L)

Helium (He)
Neon (Ne)
Argon (Ar)
Krypton (Kr)
Xenon (Xe)
Radon (Rn)

2
10
18
36
54
86

2
2
2
2
2
2

Electron Orbits
3rd (M)
4th (N)

–
8
8
8
8
8

–
–
8
18
18
18

–
–
–
8
18
32

5th (O)
–
–
–
–
8
18

6th (P)
–
–
–
–
–
8

A complete statement of the electron configuration of elements elucidating the various
postulates of Bohr-Bury scheme is given in the table on page 31 for ready reference
...
It is called
the Zeeman effect after the name of the discoverer
...
1
...

It consists of electromagnets capable of producing
strong magnetic field with pole pieces through which
holes have been made lengthwise
...
When the spectral lines are
viewed axially through the hole in the pole pieces i
...
,
parallel to the magnetic field, the line is found to split up
into two components, one having shorter wavelength
(higher frequency) and the other having higher
wavelength (shorter frequency) than that of the original
spectral line, which is no longer observable
...
32
Zeeman effect
...
When viewed in a direction perpendicular to the applied
field the lines split up into three, the central one having the same wavelength and frequency as that
of the original line and the other two occupying the same position as observed earlier
...
The motion of the electron in an orbit is equivalent
to a current in a loop of wire
...
The correct values of the energies are obtained if the components of the
angular momentum of the electron along the direction of the magnetic field are restricted to the
value
h
= m×
2π
where m = 0, ± 1, ± 2,
...
Corresponding to these values of m, a given line splits into as
many lines
...
This is, in fact, the cause of Zeeman
Effect
...
33
Splitting of the D2 and D1 lines in the sodium spectrum
by a weak magnetic field (Illustration of Zeeman effect)
...
The
equation can also be written as
e
4πcd λ
= ±
m
H λ2

The validity of the above equation can be tested experimentally by observing the Zeeman shift dλ for
a given light source of known λ (say D-line of sodium) for a magnetic field of known strength H and
calculating the value of e/m for the above equation
...


36

1

PHYSICAL CHEMISTRY

EXAMINATION QUESTIONS
1
...


3
...


5
...


7
...


9
...


11
...

13
...


Define or explain the following terms :
(a) Neutrons
(b) Nucleons
(c) Atomic number
(d) Mass Number
(e) Photoelectric effect
(f) Threshold energy
Give an account of the experiment which led Rutherford to conclude that every atom has a positively
charged nucleus which occupies a very small volume
...
Derive an expression for the nth orbit of
a hydrogen atom
...

(b) Calculate the energy of transition involving n1 = 6 to n2 = 3 in a hydrogen atom, given that Rydberg
constant R = 109737
...
63 × 10–34 J sec
...
(b) 1
...
Show how it successfully explains the spectra of hydrogen atom
...
(h = 6
...
529Å; m = 9
...
(b) 2
...

(b) Give the defects of Rutherford’s model of atom
...
(h = 6
...
529Å;
m = 9
...
8 × 10–10 esu)
Answer
...
763 × 10–8
Calculate the wavelength of the first line in Balmer series of hydrogen spectrum
...
1215 Å
(a) How does Bohr’s theory explain the spectrum of hydrogen atom?
(b) Calculate the wavelength associated with an electron moving with a velocity of 1 × 108 cm sec–1
...
1 × 10–28 g
Answer
...
28 × 10–8 cm
A line at 434 nm in Balmer series of spectrum corresponds to a transition of an electron from the nth to
2nd Bohr orbit
...
n = 5
(a) Explain Rutherford’s atomic model
...

(c) Calculate the radius of third orbit of hydrogen atom
...
625 × 10–27 erg sec; m = 9
...
8 × 10–10 esu)
(d) Calculate the wavelength of first line in Balmer series of hydrogen spectrum
...
(c) 4
...

Write Rutherford’s experiment of scattering of α-particles and give the drawbacks of atomic model
...


STRUCTURE OF ATOM – CLASSICAL MECHANICS
15
...


17
...


19
...


21
...

23
...


25
...


27
...


37

Give an account of Bohr’s theory of atomic structure and show how it explains the occurrence of spectral
lines in the atomic spectra of hydrogen
...
7 × 10–12/n2 ergs
...
What is the longest wavelength (in cm)
of light that can be used to cause this transition?
Answer
...
42 × 10–12 erg; 3
...
Find out the frequency and wavelength
of the spectral line
...
1025
...
42×10–12 erg and
–2
...
Calculate the wavelength of the emitted radiation when the electron drops
from third to second orbit
...
6600 Å
Calculate the wavelength in Å of the photon that is emitted when an electron in Bohr orbit n = 2 returns
to the orbit n = 1 in the hydrogen atom
...
17 × 10–11 erg per atom
...
1220 Å
What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition
n = 4 to n = 2 of He+ transition?
(Baroda BSc, 2001)
Answer
...

(b) Give any four limitations of Bohr’s theory of an atom
...
How was it improved by Bohr?
(Arunachal BSc, 2002)
Atomic hydrogen is excited to the 4th energy level from the ground state
...
(RH = 109677 cm–1)
Answer
...
55 Å
(Vidyasagar BSc, 2002)
Radius of the first Bohr orbit of H-atom is 0
...
Find the radii of the first and second Bohr orbit of
Li2+ ion
...
(a) 0
...
7053 Å
(Vidyasagar BSc, 2002)
If the energy difference between the ground state of an atom and its excited state is 4
...
4
...
(R = 1
...
625 × 10–34 Joule sec and
c = 2
...
9 × 10–23 kJ
Answer
...
11 × 10
The energy transition in hydrogen atom occurs from n = 3 to n = 2 energy level
...
097 × 107 m–1)
...

(b) Will this electron be visible?
(c) Which spectrum series does this photon belong to?
(Jadavpur BSc, 2003)
Calculate the energy emitted when electrons of 1
...
1 × 107 m–1; c = 3 × 108 m sec–1;
h = 6
...
182
...
In hydrogen atom the energy of the electron in first Bohr's orbit is – 1312 × 105 J mol–1
...


(Burdwan BSc, 2005)

Answer
...
84 × 105 J mol–1
Calculate the wavelength in Å of the photon that is emitted when an electron in Bohr orbit n = 2
returns to the orbit n = 1 in the hydrogen atom
...
17 × 10–11 erg per atom
...
1220 Å
A line at 434 nm in Balmer series of spectrum corresponds to a transition of an electron from the nth to

31
...
What is the value of n ?

(Gulbarga BSc, 2006)

32
...
n = 5
The energy transition in hydrogen atom occurs from n = 3 to n = 2 energy level
...
097 × 107 m–1)
...


Answer
...
42 × 10–12 erg
and – 2
...
Calculate the wavelength of the emitted radiation when the electron

spectrum series does this photon belong to ?

drops from third to second orbit
...
6600 Å

MULTIPLE CHOICE QUESTIONS
1
...
(c)
2
...
(c)
3
...
(a)
4
...
(d)
5
...

(a) neutrons
(b) protons
(c) both the neutrons and protons
(d) electrons
Answer
...
The mass number of an atom is equal to the number of _______ in the nucleus of an atom
(b) neutrons
(a) protons

STRUCTURE OF ATOM – CLASSICAL MECHANICS

7
...


9
...


11
...


13
...


15
...
(d)
If Z is the number of proton and A the number of nucleons, then the number of neutrons is an atom is given
by
(b) A – Z
(a) A + Z
(c) Z – A
(d) none of these
Answer
...
The number of
neutrons in the nucleus is
(b) 16
(a) 15
(c) 31
(d) 46
Answer
...
(c)
Which of the following is not correct for electromagnetic waves?
(a) the wavelength is the distance between two successive crests
(b) the frequency is the number of waves which pass a given point in one second
(c) the velocity of a wave is the distance covered by the particular wave in one second
(d) all electromagnetic waves have equal wavelengths
Answer
...
(b)
The unit in which wave number is measured
(a) hertz
(b) sec–1
(c) nanometer
(d) cm–1
Answer
...
(a)
The Balmer series in the spectrum of hydrogen atom falls in
(b) visible region
(a) ultraviolet region
(c) infrared region
(d) none of these
Answer
...
(b)
When a beam of light of sufficiently high frequency is allowed to strike a metal surface in vacuum,
electrons are ejected from the metal surface
...
(b)

(c)

16
...
In photoelectric effect, the kinetic energy of the photoelectrons increases linearly with the
(a) wavelength of the incident light
(b) frequency of the incident light
(c) velocity of the incident light
(d) none of these
Answer
...
The kinetic energy of the photoelectrons emitted from the metal surface is given by the relation (vo is the
threshold frequency and v is the frequency of incident light)
(b) ½ mv2 = hv + hvo
(a) ½ m v2 = hv – hvo
2= hv
(c) ½ mv
(d) ½ mv2 = hvo
Answer
...
In Bohr’s model of atom, the angular momentum of an electron orbiting around the nucleus is given by
the relation
h
nh
(a) m ν r = 2 π
(b) m ν r = 2 π
nh
n2 h 2
(c) m ν r =
(d) m ν r = 4 π
4π
Answer
...
The radius of first orbit in hydrogen atom according to Bohr’s Model is given by the relation
h
h2
(a) r =
(b) r =
2
4 π2 m e 2
4 π m e2

h2
h2
(d) r =
2
4πme
4 π m e4
Answer
...
The radius of first orbit in hydrogen atom is 0
...
The radius of second orbit is given by
(b) 2 × 0
...
529 Å
(c) 4 × 0
...
529 Å
Answer
...
The energy of an electron in the first orbit in hydrogen atom is –313
...
The energy of the
electron in 3rd orbit is given by the relation
(c)

r=

(a) E3 =

24
...


26
...
6
kcal mol −1
2

−313
...
6 × 3 kcal mol−1
9
Answer
...
(a)
A line in Pfund series is obtained when an electron from higher energy levels returns to
(b) 3rd orbit
(a) 1st orbit
(c) 5th orbit
(d) 6th orbit
Answer
...
(c)
When an electron drops from a higher energy level to a lower energy level, then
(a) the energy is absorbed
(b) the energy is released
(c) the nuclear charge increases
(d) the nuclear charge decreases
Answer
...


−313
...
The spectrum of hydrogen atom is similar to that of
(b) He+ ion
(a) H+ ion
(c) Li+ ion
(d) Na+ ion
Answer
...
If r is the radius of first orbit, the radius of nth orbit of hydrogen atom will be
(b) n r
(a) n2 r
(c) n/r
(d) r/n
Answer
...
The ratio of radii of second and first orbit of hydrogen atom according to Bohr’s model is
(a) 2:1
(b) 1:2
(c) 4:1
(d) 1:4
Answer
...
The spectrum of helium is expected to be similar to that of
(b) Li atom
(a) H-atom
(d) Na+ ion
(c) Li+ ion
Answer
...
Electromagnetic radiations with minimum wavelength is
(b) X-rays
(a) ultraviolet
(c) infrared
(d) radiowaves
Answer
...
Which of the following statements is false?
(a) electrons travel around the nucleus in specific permitted circular orbits
(b) an electron does not lose energy as long as it moves in its specified orbits
(c) an electron can jump from one energy level to another by absorbing or losing energy
(d) the angular momentum of an electron is not quantised
Answer
...
The idea of stationary orbits was first given by
(b) JJ Thomson
(a) Rutherford
(c) Niels Bohr
(d) Max Planck
Answer
...
The maximum number of electrons that can be accommodated in an orbit is
(a) 2n
(b) n2
2
(c) 2n
(d) 2n + 1
Answer
...
The maximum number of electrons is the outermost orbit is
(b) 8
(a) 2
(c) 18
(d) 32
Answer
...
When the source emitting lines is placed in a strong magnetic field the spectral lines are split into its
components
...
(b)
37
...
no
...
(a)
38
...
(c)

42

1

PHYSICAL CHEMISTRY

39
...
(d)
40
...
(c)
41
...
(a)
42
...
6 eV
(b) –13
...
36 eV
(d) –1
...
(b)
43
...
6 eV
...
53 eV
(b) – 2
...
51 eV
(d) none of these
Answer
...
En = –1311
...
If the value of E is –52
...
(d)
45
...
It corresponds to the radiation emitted by an electron jumping
from higher energy sates to
(b) second energy state
(a) first energy state
(c) third energy state
(d) fifth energy state
Answer
...
The ground state of an atom corresponds to a state of
(b) minimum energy
(a) maximum energy
(c) zero energy
(d) negative energy
Answer
...
Balmer series in the spectrum of hydrogen atom lies in
(a) ultraviolet region
(b) visible region
(c) infrared region
(d) none of these
Answer
...
The spectrum of H-atom is expected to be similar to that of
(b) Na+
(a) Li+
(c) He+
(d) K+
Answer
...
An atom of Calcium (at
...
20) contains _______ electrons in the third energy level
...
(b)
50
...
(d)



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