There is no denying the fact that landmark in the evolution
of atomic structure may be stated as follows with the theory as explored in the
years as mentioned below:
1805 Dalton’s
atomic theory
1896 Thomson discovery of electron and
proton
1909 Rutherford’s nuclear atom
1913 Bohr’s atomic model
1921 Bohr-Bury scheme of electronic
arrangement
1932 Chadwick’s discovery of the neutron
Atomic Theory of Dalton:
(1805)
1.
All matter is composed of tiny particles called atom
which can not be created, destroyed or splitted.
2.
All atoms of any one element are identical, have
same mass and chemical properties.
3.
A compound is a type of matter composed of atoms
of two or more elements.
4.
A chemical reaction consists of rearranging
atoms from one combination to another.
British Chemist John Dalton provided the basic theory:
all matter- whether element, compound, or mixture- is composed of small
particles called atoms.
Limitations: (1) Atom can be divided into subatomic
particle namely electron, proton and neutron. (2) All atoms of any elements are
not identical, have different mass and chemical properties. This property is
known as isotopes.
Rutherford’s
Model of Atom: (1909)
1. Atom has a tiny dense center core or the NUCLEUS,
which contains practically the entire mass of the atom, leaving rest of the
atom almost empty.
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.
(1)
Newton’s Laws of motion and gravitation can only be applied to
neutral bodies such as planets and not to charged bodies such as tiny electrons
moving round a positive nucleus. The analogy does not hold good since the
electrons in an atom repel one another, whereas planets attract each other
because of gravitational forces. Besides, there is electrostatic attraction in
a nuclear atom model.
(2)
According to Maxwell’s
theory, and charged body such as electrons rotating in an orbit must radiate
energy continuously thereby losing kinetic energy. Hence the electron must
gradually spiral in towards the nucleus. The radius of the electron will
gradually decrease and it will ultimately fall into the nucleus, thus
annihilating the atom model.
(3)
Since the process of
radiating energy would go on continuously, the atomic spectra should also be
continuous and should not give sharp and well-defined lines.
Contribution: Rutherford
laid the foundation of the model picture of atom.
Three
subatomic particles or principal fundamental particles: An atom
is the smallest particle of an element having its own chemical identity and
properties. But the experiments described before indicate that atoms can be
subdivided into smaller subatomic particles known as fundamental particles. The
most important three subatomic particles and there short description are given
below.
Particle |
Symbol |
Mass
|
Charge
|
||
amu
|
grams
|
Units
|
Coulombs
|
||
Electron
|
e-
|
1/1835
|
9.1´10-28
|
-1
|
-1.6´10-19
|
Proton
|
p+
|
1
|
1.672´10-24
|
+1
|
+1.6´10-19
|
Neutron
|
n or no
|
1
|
1.672´10-24
|
0
|
0
|
Quantum Theory and Bohr
Atom: (1913)
To understand the Bohr theory, we need to learn
-
the nature of electromagnetic
radiations
-
the atomic
spectra
Electromagnetic Radiations:
Energy can be transmitted through space by electromagnetic
radiations. Some forms of radiant energy are
v
radio waves
v
visible light
v
infrared light
v
ultraviolet light
v
x-rays etc.
 |
Characteristics of Waves:
Wavelength (l, lambda): The wavelength
is defined as the distance between two successive crests or troughs of a wave.
Units: cm, m or Å (angstrom).
1
Å = 10-8 cm = 10-10 m, 1 nm = 10-9 m
Frequency (n, nu): The frequency is
the number of waves which pass a given point in one second.
Units:
hertz (hz), one cycle per second.
A
wave of high frequency has a shorter wavelength, while a wave of low frequency
has a longer wavelength.
Speed: The speed (or velocity) of a wave is the
distance through which a particular wave travels in one second.
Speed = Frequency ´
Wavelength
Wave Number: This is reciprocal of the wavelength and
is given the symbol 
(nu bar).
(nu bar). = 
Problem-1: The wavelength
of a violet light is 400 nm. Calculate its frequency and wave number.
Problem-2: The frequency of strong yellow line in the spectrum
of sodium is 5.09
1014 sec-1. Calculate the wavelength of
the light in nanometers.
1014 sec-1. Calculate the wavelength of
the light in nanometers.
Spectra: A spectrum is an array of waves
or particles spread out according to the increasing or decreasing of some
property.
An increase in frequency or a decrease in wavelength
represents an increase in energy.
Continuous Spectrum: White
light is radiant energy coming from the sun or from incandescent lamps. It is
composed of light waves in the range 4000-8000 Å.
When a beam of white light is
passed through a prism, a continuous series of colour bands (rainbow: violet,
indigo, blue, green, yellow, orange and red; VIBGYOR) is received on a screen
with different wavelengths called Continuous Spectrum.
The violet component of
the spectrum has shorter wavelengths (4000-4250Å) and higher frequencies.
The red component has
longer wavelengths (6500-7500Å) and lower frequencies.
The invisible region beyond the
violet is called ultraviolet region and the one below the red is called infrared
region.
Atomic Spectra:
When an element in the vapor or the gaseous state is
heated in a flame or a discharge tube, the atoms are excited and emit light
radiations of a characteristic colour. The colour of light produced indicates
the wavelength of the radiation emitted. The spectrum obtained on the
photographic plate is found to consists of bright lines.
 |
Atomic Spectrum of Hydrogen: Balmer Equation
and Rydberg constant
The emission line spectrum of hydrogen can be obtained
by passing electric through the gas contained in a discharge tube at low
pressure. The light radiation emitted is then examined with the help of a
spectroscope.
In 1884 J. J. Balmer observed the following four
prominent coloured lines in the visible hydrogen spectrum:
(1) a red line with a wavelength of 6563 Å
(2) a blue-green line with a wavelength of 4861 Å
(3) a blue line with a wavelength of 4340 Å
(4) a violet line with a wavelength of 4102 Å
The above series of four lines in the visible spectrum
of hydrogen is known as Balmer series.
 |
Balmer was able to give an equation which relate the
wavelengh (l) of the observed lines.
The Balmer equation is,
where R is a constant called Rydberg constant
which has the value 109,677 cm-1 and n = 3, 4, 5, 6, etc.
Five spectral series: In addition to Balmer series, four other spectral
series were discovered in the infrared (ir) and ultraviolet (uv) regions
of the hydrogen spectrum. These bear the
names of discoverers.
(1) Lyman series (uv)
(2) Balmer series (visible)
(3) Paschen series (ir)
(4) Brackett series (ir)
(5) Pfund series (ir)
Quantum Theory of Radiation:
(1) When atoms or molecules absorb or emit radiant energy,
they do so in separate ‘units of waves’ called quanta or photons.
 |
Continuous
Wave
 |
|||||
 |
|||||
 |
|||||
Photons or
quanta
individual
photon
(2) The energy, E, of a quantum or photon is given by the
relation.
E = hn (h, Planck’s constant = 6.62 ´ 10–27 erg sec. Or 6.62 ´ 10-34 J sec.)
c = ln (c = velocity of radiation)
Therefore, E =
(3) An atom or molecule can emit (or absorb) either one
quantum of energy (hn) or any whole number multiple of this unit.
Bohr Model of the Atom:
(1)
Electrons travel around the nucleus in specific
permitted circular orbits and in no others.
(2)
While in these specific orbits, an electron does not radiate
(or lose) energy.
(3)
An electron can move from one energy level to another
by quantum or photon jumps only.
 |
(4)
The angular momentum (mvr) of an electron
orbiting around the nucleus is an integral multiple of Planck’s constant
divided by 2p.
Calculation of Radius of Orbits:
Consider an electron of charge e revolving around a nucleus of charge Ze, where Z is the atomic number and e the charge on a proton. Let m be the mass of the electron, r the radius of the orbit and v the tangential velocity of the revolving electron.
The electrostatic force of attraction between the
nucleus and the electron (Coulomb’s Law),
 |
=
The centrifugal force acting on the electron
=
 |
Bohr assumed that these two opposing forces must be
balancing each other exactly to keep the electron in orbit. Thus,
=
For hydrogen Z=1, therefore,
= --- --- ---
(1)
Multiplying both sides by r
= mv2 ---
--- --- (2)
According to Bohr’s theory, angular momentum of the
revolving electron is given by the expression:
mvr = 
or v = 
--- ---
--- (3)
--- ---
--- (3)
Substituting the value of v in equation (2)
= 
Solving for r,
---
--- --- (4)
Since the value of h, m and e had been
determined experimentally, substituting these values in (4), we have
--- --- --- (5)
where n is the principal quantum number and hence the
number of the orbit.
Problem: Calculate the first five Bohr radii.
Energy of Electron in each Orbit:
For hydrogen
atom, the energy of the revolving electron, E is the sum of its kinetic energy 
and potential energy 
.
E = 
--- ---
--- (6)
--- ---
--- (6)
From equation (1)
Substituting the value of mv2 in (6),
E = 
E = 
--- ---
--- (7)
--- ---
--- (7)
Substituting the value of r from equation (4) in (7),
E = 
= 
--- ---
--- (8)
= --- ---
--- (8)
Substituting the values of m, e and h
in (8),
E = 
,
,
or E = 
--- ---
--- (9)
--- ---
--- (9)
By using proper integer for n, we can get the energy for
each orbit.
Problem: Calculate the five lowest energy levels of
the hydrogen atom.
Bohr Explanation of Hydrogen
Spectrum:
Ø 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).
Ø When energy is supplied to hydrogen gas in the discharge
tube, the electron moves to higher energy levels viz., 2, 3, 4, 5, etc.,
depending on the quantity of energy absorbed.
Ø From these high energy levels, the electron returns by
jumps to one or other lower energy level.
Ø In doing so the electron emits the excess energy as a
photon.
Ø This gives an excellent explanation of the various
spectral series of hydrogen.
Lyman series is obtained
when the electron returns to the ground state i.e., n = 1 from higher levels (n2
= 2, 3, 4, 5, etc.). 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 Figure.
Table: Spectral
series of hydrogen
Series
|
n1
|
n2
|
Region |
Wavelength l (Å)
|
Lyman
|
1
|
2, 3, 4, 5, etc.
|
Ultraviolet
|
920-1200
|
Balmer
|
2
|
3, 4, 5, 6, etc.
|
Visible
|
4000-6500
|
Paschen
|
3
|
4, 5, 6, 7, etc.
|
Infrared
|
9500-18750
|
Brackett
|
4
|
5, 6, 7
|
Infrared
|
19450-40500
|
Pfund
|
5
|
6, 7
|
Infrared
|
37800-75000
|
 |
Values 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:
E2
= 
The difference of energy between
the levels n1 and n2 is:
--- --- --- (10)
According to Planck’s equation
---
--- --- (11)
where 
is wavelength of photon
and c is velocity of light. From equation (10) and (11), we can write
is wavelength of photon
and c is velocity of light. From equation (10) and (11), we can write
= 
or, 
= 
=
or, = 
--- --- --- (12)
= --- --- --- (12)
where R is Rydberg constant. The
value of R can be calculated as the value of e, m, h and c are known. 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).
Calculation of wavelengths of the spectral lines of hydrogen in the visible region
These lines constitute the Balmer
series when n1 = 2. Now the equation (12) above can be written as
Thus the wavelengths of the
photons emitted as the electron returns from energy levels 6, 5, 4 and 3 were
calculated by Bohr. The calculated values corresponded exactly to the values of
wavelengths of the spectral lines already known. This was, in fact, a great
success of the Bohr atom.
Problem-3: Find the
wavelength in Å of the line in Balmer series that is associated with drop of
the electron from the fourth orbit. The value of Rydberg constant is 109,676 cm-1.
Problem-4: Find the
wavelength in Å of the third line in Balmer series that is associated with drop
of the electron. (Rydberg constant =109,676 cm-1).
Shortcoming of the Bohr Atom:
(1) It is
unsuccessful for every other atom containing more than one electron,
(2) In
view of modern advances, like dual nature of matter, uncertainty principle etc.
any mechanical model of the atom stands rejected.
(3) Bohr’s
model of electronic structure could not account for the ability of atoms to
form molecules through chemical bonds.
(4) Bohr’s
theory could not explain the effect of magnetic field (Zeeman effect) and
electric field (Stark effeck) on the spectra of atoms.
No comments:
Post a Comment