Download Module 2 ATOMIC STRUCTURE

Module 2
ATOMIC STRUCTURE
A brief historical outline
Atomic theory
Wave nature of light
Balmer series
Photoelectric effect
Discovery of electron
Blackbody radiation
Explanation of photoelectric effect
Paschen series
Discovery of nucleus
Bohr atom model
Wave-particle dualism
Wave nature of electrons
Uncertainty principle
Schrodinger equation
Discovery of neutrons
John Dalton
J.C Maxwell
Johann Balmer
Hertz
Sir J J Thomson
Max Planck
Albert Einstein
Paschen
H. Rutherford
Neil Bohr
de- Broglie
Davisson and Germer
Werner Heisenberg
Erwin Schrodinger
J. Chadwick
1803
1856
1885
1887
1897
1900
1905
1908
1911
1913
1924
1924
1927
1927
1932
2.1 DALTON`S ATOMIC THEORY
Based upon the known properties of matter, John Dalton (1809) regarded the atom as hard,
dense and smallest indivisible particle of matter. Although two centuries old, Dalton’s atomic
theory remains valid in modern chemical thought. According to him,
1.
2.
3.
4.
All matter is made of atoms. Atoms are indivisible and indestructible.
All atoms of a given element are identical in mass and properties.
Compounds are formed by the combination of two or more different kinds of atoms.
A chemical reaction is a rearrangement of atoms.
Limitation
1. Emission of charged particles from radioactive elements and gases (on the passage of
electricity at very low pressures) suggests that atom is divisible and consist of much smaller
fundamental particles.
2. The theory could not explain the existence of isotopes.
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When electric discharge from a high potential source is passed through a discharge tube
evacuated to pressures around 0.01mm or less, rays are emitted from the cathode. These rays
are called cathode rays.
 These rays travel in straight lines at right angles to the cathode surface.
 They produce mechanical motion in a small paddle wheel placed in their path indicating that
they are material particles.
 These particles are deflected from their path by electric and magnetic fields showing that
they are negatively charged.
 The negatively charged particles which constitute the cathode rays are called electrons.
2.2 THOMSON`S ATOM MODEL
With the discovery of electrons from the
discharge tube experiments (1897), J.J Thomson
proposed the plum pudding model of the atom in
1904 before the discovery of the atomic nucleus. In
Thomson's model, the atom is composed of electrons
(which Thomson called "corpuscles") surrounded by
a soup of positive charge to balance the electron`s
negative charges, like negatively charged "plums"
surrounded by positively charged “pudding”.
Fig.2.1Plum pudding model of the atom
In this model the electrons were assumed to be embedded in a ball of positive charge. He
determined the charge/mass ratio (e/m) of electron by subjecting the beam of electrons produced
in a discharge tube to magnetic as well as electric fields. He was awarded the Nobel Prize for
Physics in 1906 (e/m of electron = 1.75875 x 1011C kg-1)
Limitation
Although this model explained the main characteristics of the atom known at that time it
did not have any experimental support. Therefore, it was rejected by his co-scientists. However,
his prediction that an atom is electrically neutral and has no net charge is still accepted.
Robert Milliken determined the charge of an electron (1.6022 x 10-19C) by his famous oil drop
technique. Milliken was awarded the Nobel Prize for Physics in 1923.
Since e = 1.6022 x 10-19 C and e/m = 1.75875 x 1011C kg-1
𝑒
𝑒/𝑚
= mass of electron =
1.6022 𝑥 10−19 𝐶
1.75875 𝑥 1011 𝐶𝑘𝑔−1
= 9.1091 x 10-31kg
This is called the rest mass of the electron, i.e the mass which it possesses when it moved with a
velocity much smaller than that of light (2.9979 x 108 ms-1). On the atomic mass scale, the rest
mass of electron = 0.0005486 amu.
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Experiments with discharge tubes containing perforated cathode revealed the emission of another
type of radiations which originated from the anode and passed through the perforations of the
cathode. These radiations that carried positive charge were called positive rays or anode rays.
The positive particles were in fact, the positive residues of the gas left when one or more
electrons were knocked out of the atoms of the gas. The smallest positive particle given out by
the lightest element hydrogen is called proton. It is another subatomic particle whose charge is
equal in magnitude but opposite in sign to that of an electron. The mass of a proton is 1.6726 x
10-27 kg or 1.00722 amu.
2.3 RUTHERFORD’S ATOM MODEL
Ernest Rutherford (1911) directed a narrow beam of alpha particles at an extremely thin
sheet of gold (gold has maximum ductility among all metals) which was then made to strike a
fluorescent screen. His observations were,
 Most of the alpha particles passed through the metallic sheet without suffering any change in
their path indicating that the atom consists predominantly of empty space.
 An extremely small number of alpha particles got deflected through wider angles or were
even retarded, indicating the presence of a tiny heavy positively charged body at the centre of
each atom.
Based on the scattering experiments, Rutherford suggested that an atom consists of a tiny,
dense positively charged body, located at its centre, called the nucleus. Sufficient number of
light, negatively charged electrons is distributed around the nucleus to balance the positive charge
on the nucleus. Therefore, Rutherford proposed that the electrons are revolving round the nucleus
at extremely high speeds at great distances from the nucleus. The centrifugal force arising from
this motion balances the force of electrostatic attraction and therefore, the electrons do not fall
into the nucleus.
Limitation
Niels Bohr pointed out that Rutherford`s atom should be highly unstable as an electric
charge subjected to acceleration, should continuously emit radiation losing energy. Its orbit
should become smaller and smaller and finally it should drop into the nucleus.
Bohr solved this problem on the basis of the quantum theory of radiation put forth by
Max Planck. (See section 2.6)
Chadwick discovered, in 1932, that when beryllium or boron was bombarded by alpha particle,
a new particle which carried no charge but mass almost equal to that of a proton, were emitted.
These particles were called neutrons. Chadwick won the Nobel prize for Physics in 1935 for the
discovery of neutron. Mass of neutron = 1.6749 x 10-27kg or 1.00866 amu
Mass of a proton is almost the same as the mass of a neutron and it is taken as unit mass. Mass
of an electron is only 1/1836 of the mass of a proton.
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2.4 CLASSICAL MECHANICS: ITS SUCCESS AND FAILURE
Classical mechanics formulated by Newton in 17th century and further developed in 18th
and 19 century could adequately describe the motion of macroscopic bodies. According to
classical mechanics a particle can possess any amount of energy between zero and infinity. Also,
the position and velocity of the particle can be determined simultaneously. Newton’s law of
motion and Maxwell’s electromagnetic wave theory successfully explained most of the
phenomena known then. But when experiments were conducted at lower temperatures and at
different wavelengths of radiation, phenomena such as low-temperature heat capacities of solids,
photoelectric effect, black body radiation, atomic spectra etc were observed. The observations
indicated that the system can take up energy not continuously but only in discrete amounts
(quanta). This could not be explained by the classical laws. By 1925, a new mechanics - the
quantum mechanics was formulated, which successfully explained all these phenomena.
th
2.5 BLACK BODY RADIATION
A black body is defined as an object that absorbs and emits all the radiations falling on it.
The energy density (the amount of energy radiated per unit volume) by a black body depends
upon the temperature at which the body is kept. The correlation between energy density and
frequency at different temperatures usually referred to as the black body radiation curves is
depicted in figure 2.2.
It can be noticed that,
1. For each temperature, there is a particular
frequency at which the energy radiated is the
maximum.
2. The position of the maximum shifts towards
higher frequency with increase in
temperature.
3. The higher the temperature, the more
pronounced is the maximum.
Fig.2.2 Black body radiation curves
2.6 PLANCK`S QUANTUM THEORY
The explanation for black body radiation was given by Max Planck as follows,
 Radiant energy is emitted or absorbed discontinuously in the form of tiny bundles of energy
called quanta.
 Each quantum is associated with a definite amount of energy (E), proportional to the
frequency (ν) of radiation. i.e E α ν or E = hν where h is a fundamental constant known as
Planck`s constant. The numerical value of h is 6.626 x 10-34 Js.
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 A body can emit or absorb energy only in whole number multiples of quantum i.e E = nhν
where n = 1,2,3…. This means that a body can emit or absorb energy equal to 1hν, 2hν,
3hν,……. This is known as quantization of energy.
Max Planck, known as father of quantum theory was awarded Nobel Prize for physics in
1918. Based on this theory, Planck obtained the expression for energy density, ρ(𝜆, 𝑇) of black
body radiation as
𝜌(λ,T) =
8πhc
hc
(
)
λ5 e λkT −1
where h is Planck`s constant, c is velocity of light, λ is wave length of radiation coming from
black body, k is Boltzmann constant, T is temperature in absolute scale.
2.7 PHOTOELECTRIC EFFECT
Sir J.J Thomson (1905) observed that when light of a certain frequency strikes the surface
of a metal, electrons are ejected from the metal. The phenomenon of ejection of electrons from
the surface of a metal when light of a suitable frequency strikes on it is known as photoelectric
effect. The ejected electrons are called photoelectrons. It may be noted that only a few metals
show this effect under the action of visible light but many more show it under the action of more
energetic ultraviolet light. Cesium with the lowest ionization energy is the metal from which
electrons are ejected easily. This metal is therefore, used largely in photoelectric cells.
According to classical theory, the intensity of radiation is proportional to the square of
the amplitude of the electric field. The electrons at the surface of the metal should oscillate along
with the field and so, as the intensity increases, the electrons oscillate more violently and
eventually breakaway from the surface with a kinetic energy that depends on the intensity of the
field. Furthermore the classical theory predicts that the photoelectric effect should occur for any
frequency of light as long as the intensity is sufficiently high.
However, it was experimentally
noticed that the kinetic energy of the ejected
electrons is independent of the intensity of the
incident radiation, but dependant on
frequency of the incident radiation. There is a
minimum frequency, termed threshold
frequency (ν0), characteristic of the metallic
surface, below which no electrons are ejected
whatever be the intensity of radiation. Above
ν0 the kinetic energy of the ejected electrons
varies linearly with the frequency, ν.
Fig.2.3 Plot of kinetic energy with frequency of an electron
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To explain these results, Einstein used Planck’s quantum theory. When light is absorbed
by a metal, the total energy of the photon, hν is given to a single electron within the metal. If this
energy is sufficiently large, the electron may penetrate the potential barrier at the surface of the
metal and still retain some energy as kinetic energy. The kinetic energy retained by the electron
depends on the frequency of the photon that ejected it.
1
hν = ϕ + 2 mv2
where ϕ is called the work function of the metal. The minimum frequency that will eject an
electron or threshold frequency, ν0 is just the frequency required to overcome the work function
of the metal. Then, hν0 = ϕ. Substituting the value of ϕ, we can write
1
2
mv2 = h(ν-ν0)
The number of electrons ejected depends on the number of incident photons and therefore the
intensity of light. Thus photoelectric effect supported the particle nature of light.
Problem : Work function for Na is 2.92 x 10-19 J. Calculate the threshold frequency of Na.
Solution :
ϕ = hν0
ν0 =
2.92 x 10 −19 J
6.626 x 10 −34 Js
= 4.40 x 1014 Hz
Arthur Compton found that if monochromaic X-rays are allowed to fall on carbon or some other
light elememt, the scattered X-rays have wave lengths larger than the incident rays. Since
scattering is caused by electrons, it is evident that some interaction between X-rays and electrons
has taken place and this has resulted in decrease in energy of the former. This decrease in
energy or increase in wave length of X-rays after scattering from the surface of an object is
known as the Compton effect(shows the particle nature of electron).
2.8 ATOMIC SPECTRA OF HYDROGEN AND HYDROGEN LIKE ATOMS
Every atom, when subjected to high temperatures or an electrical discharge, emits
electromagnetic radiation of characteristic frequencies. Each element has a characteristic
emission spectrum. The emission spectra of atoms consist of only certain discrete frequencies,
and are called line spectra. Hydrogen, the lightest and simplest atom has the simplest spectrum.
Since atomic spectra are characteristic of the atoms involved, it is reasonable to suspect that the
spectrum depends on the electron distribution in the atom. A detailed analysis of the hydrogen
spectrum was a major step in the elucidation of the electronic structure of atoms.
Johann Balmer (1885) discovered that the wavelength, λ of the lines in the visible region
of the emission spectrum of hydrogen atoms could be expressed by a simple relation as,
1
λ
1
1
= ̅𝜈 = R( 22 – 𝑛2 ) where n = 3,4,5….
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23
̅𝜈 is the wavenumber, the reciprocal of wavelength, λ and R is the Rydberg constant with value
1.097 x 107m-1 for hydrogen. This equation is called Balmer formula.
The success of the Balmer formula led to further exploration and other series of lines
were discovered in the atomic hydrogen spectrum that could be represented by the equation
1
1
̅𝜈 = R ( 𝑛2 – 𝑛2 )
1
2
where n1 = 1,2,3…. and n2 = 2,3,4… and the condition is that n2 > n1. Table 2.1 shows the
different spectral series of hydrogen with their n1 and n2 values. The shortest wavelength line any
series occurs when n2 is infinitely large so that 1/n22 tends to zero. That is, if n2 = ∞, then
Table 2.1 Spectral series of hydrogen
Name of series
Value of n1
Value of n2
Region
Lyman
Balmer
Paschen
Brackett
Pfund
Humphrey
1
2
3
4
5
6
2, 3, 4……
3, 4, 5……
4, 5, 6……
5, 6, 7……
6, 7, 8……
7, 8, 9……
ultraviolet
visible
infrared
infrared
infrared
infrared
Fig.2.4 Atomic spectra of hydrogen
1
𝑛22
=0
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24
It is important to note that every line in the spectrum can be defined as the difference of
two terms, R/n12 and R/n22 for one electron system like He+, Li2+ etc. The spectra of atoms with
more than one electron (multielectron systems) are much complicated.
2.9 BOHR MODEL OF ATOM
In 1913, Danish physicist Niels Bohr presented a theory of the hydrogen atom that gave a
simple explanation of the hydrogen spectrum. Bohr made his contribution by arbitrarily
incorporating Planck`s quantum hypothesis to classical mechanics.
The postulates of Bohr theory of atom are:
1. The electron in an atom revolves around the nucleus in certain states called stationary states
(orbits) with a definite amount of energy. In these states they never radiate energy even if it
has acceleration.
2. Energy is emitted only when an electron makes a transition between two stationary states.
3. The frequency of the radiation emitted during the transition between two orbits is such that
the difference in energy ΔE = hν
4. The angular momentum of an electron in an orbit can have
only certain values that are integral multiples of
=
nh
2π
h
2π
. i.e mvr
where n = 1,2,3……
5. The centrifugal force,
mv2
when the electron moves in an
r
Ze2
orbit is exactly balanced by the electrostatic force,
4πε0 r2
.
Fig.2.5 Visualization of Bohr model
Consider a nucleus charged Ze and an electron with mass, m and charge, e revolving at a
distance, r from the nucleus, with a velocity v. The electrostatic force of attraction between the
nucleus and the electron =
Ze2
4πε0 r2
where ε0 is permittivity of vacuum (8.85418 x 10-12 C2 N-1 m-2).
The centrifugal force acquired by the electron =
mv2
r
.
In the circular orbit, the electrostatic force of attraction balances the centrifugal forces,
mv2
r
=
Ze2
4πε0 r2
The velocity of electron,
According to Bohr, mvr =
nh
2π
v=
Ze2
4πε0
where n = 1,2,3….
v=
Ze2
2ε0 nh
This equation gives the velocity of an electron in a circular orbit.
x
1
mvr
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25
The radius of the orbit, r also can be calculated as mvr =
nh
2π
𝑛ℎ
r = 2πmv
substituting the value of v,
r=
n2 h2 ε0
Zmπe2
Thus the radii of the allowed orbits or Bohr orbits are quantized and are proportional to
n . The electron can move around the nucleus in circular orbits with radii given by the above
equation only. The orbit with the smallest radius is one with n =1 for which
2
r1
=
(8.8541 x 10-12C2N-1m-2) (6.626 x 10-34Js)2
3.14 (9.109 x 10-31kg) (1.6022 x 10-19C)2
= 5.292 x 10-11m = 0.529Ǻ.
The radius of the first Bohr orbit (r1) is often denoted by a0.
In general we can write, rn = 0.529
𝐧𝟐
𝐙
Ǻ.
Problem: Calculate the radius ratio for 2nd orbit of He+ ion and 3rd orbit of Be+++ ion.
Solution: r1 (radius of 2nd orbit of He+ ion) = 0.529 x
r2 (radius of 3rd orbit of Be+++ ion) = 0.529
𝑟1
𝑟2
=
0.529 x 22
2
0.529 x 32
22
Ǻ.
2
32
x
Ǻ.
4
8
= 9 = 0.89
4
2.9.1 Energy of electron in hydrogen and hydrogen-like atoms
Energy of an electron is the sum of its kinetic energy and potential energy.
In the circular orbit,
mv2
r
=
mv2 =
Kinetic energy = ½ mv2 =
Ze2
4πε0 r2
Ze2
4πε0 r
Ze2
8πε0 r
r
Potential energy of an electron at a distance, r from the nucleus = ∫∞
=Therefore, total energy of the electron =
1
2
x(
Ze2
4πε0 r2
𝑍𝑒 2
4𝜋𝜀0 𝑟
Ze2
8πε0 r
)-
Ze2
4πε0 r
=-
Ze2
8πε0 r
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26
Substituting the value of r,
Energy of an electron, E =
-
Ze2 Zmπe2
8πε0 n2 h2 ε0
E=-
mZ2 e4
8ε20 n2 h2
where n is the principal quantum number, which can have values 1,2,3…..
The negative sign in this equation
indicates that the energy states are in bound
states. n = 1 corresponds to the state of
lowest energy. This energy is called the
ground state energy.
At ordinary
temperature, hydrogen atom as well as
most other atoms and molecules are found
almost exclusively in their ground
electronic states. The states of higher
energy are called excited states and are
unstable with respect to the ground state.
An atom or a molecule in an excited state l
relaxes back to the ground state and give
off the energy as electromagnetic radiation.
Figure 2.6 Variation in energy with distance
2.9.2 Spectrum of hydrogen atom
Bohr assumed that the observed spectrum of the hydrogen atom is due to transition of
electron from one higher allowed energy state to another of lower energy. Thus, the simple
explanation for the line spectrum is that the atoms are constrained to emit quanta of only a few
specific energies depending on the energy difference between the orbits involved in the transition
of electrons. Energetically excited atoms are not able to emit light of continuously varying
wavelengths and therefore a continuous spectrum is not produced from atoms.
According to Bohr, the spectrum arises
when the electron in the initial higher energy state of
principal quantum number, n2 shifts to the final
lower energy state of principal quantum number, n1
so that the difference in energies associated with
these levels is emitted as photon of frequency, ν (Fig
2.7).
ΔE = hν = E2 – E1 = -
mZ2 e4
mZ2 e4
8ε20 n22 h
8ε20 n21 h2
- (2
)
Atomic Structure
ν=
27
mZ2 e4
8ε20 h3
1
𝜈̅ = 𝜆 =
(
1
𝑚𝑍 2 𝑒 4
8𝜀02 𝑐ℎ 3
𝜈̅ = R (
1
𝑛12
-
1
-
n21
(
1
n22
1
-
𝑛12
)
1
) since c = νλ.
𝑛22
Fig 2.7 Origin of spectral series from Bohr model
)
𝑛22
where R is Rydberg constant with value,
mZ2 e4
8ε20 ch3
= 1.097 x 107m-1 for H atom.
Using this equation, the wave numbers of photons of the various spectral series can be calculated.
Problem 1:
Calculate the ionization energy of the hydrogen atom in the ground state.
𝜈̅ = R (
Solution :
1
𝑛12
-
1
𝑛22
hc
E = hν =
λ
n1 =1, n2 = ∞
)
= 6.602 x 10-34 Js x 3 x108 ms-1 x 1.097 x 107m-1
= hc𝜈̅
= 2.179 x 10-18 J
Problem 2 :
Solution :
Calculate the value of the Rydberg constant of hydrogen atom in eV.
R = 10973731.6 m-1 for H atom and 1eV = 8.0655 x 105m-1
R=
10973731.6
8.0655 x105
= 13.6 eV
Problem 3: Calculate wavelength for second line of Balmer series of He+ ion.
Solution:
For He+,
1
𝜆
= R x 22 (
1
𝑛12
-
1
𝑛22
) (since Z = 2)
For second line of Balmer series, n1 = 2, n2 = 4
1
𝜆
= 1.097 x 107 x 22 (
λ=
4
3 x 10973731.6
1
𝑛12
-
1
𝑛22
)
= 1.215 x 10-7 m
Merits of Bohr model of atom
 Bohr theory explains well the features of hydrogen spectrum and the spectra of hydrogen like
atoms He+, Li2+ etc.
 Bohr’s formula for the Rydberg constant is in perfect agreement with experimental data.
 It helps to calculate the energy and radii of various energy levels.
 The velocity of electrons in various orbits can be calculated.
Defects of Bohr model of atom
Though, Bohr model explains the stability of atoms and the origin of line spectrum, the
model is not satisfactory for many observations.
 It cannot explain the fine spectrum with additional lines observed with high resolution
instrument.
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28
 It cannot explain the splitting of the spectral lines when the source emitting the spectral lines
was placed in a magnetic field (Zeeman effect) or electric field (Stark effect).
 The model could not explain why atoms react to form compounds.
2.10 SOMMERFELD MODIFICATION OF BOHR MODEL
Sommerfeld modified the Bohr model by adding elliptical orbits in addition to the
circular orbits to explain the occurrence of fine spectral lines. In an elliptical orbit, there will be a
major axis and a minor axis. As the orbit broadens, the lengths of the two axes become closer
and they become equal when the orbit becomes circular. Thus, the circular orbit is only a special
case of elliptical orbits.
The electron travelling in an elliptical orbit will have its angular momentum. This
momentum must be quantized and thus can have only a limited number of values given by kh/2π
where k is an integer known as the azimuthal quantum number. It can be shown that n, the
principal quantum number, used by Bohr and k, the azimuthal quantum number used by
Sommerfeld, are related to one another as,
n
k
=
length of major axis
length of minor axis
Thus, for any given value of n
(except 1) k can have more than a single
value. When k = n, the orbit must be circular.
But, as k becomes smaller, the orbits become
elliptical with greater eccentricity. It may be
noted that k cannot be zero because in that
case minor axis would be zero which would
imply linear motion of the electron passing
through the nucleus. The number of possible
value of k is equal to the principal quantum
number n. Thus if n is 4, k can have four
values, 4, 3, 2, and 1. When n is 1, k can have
only one value.
Fig. 2. 8 Sommerfeld elliptical orbitals
According to Sommerfeld`s extension of Bohr theory, referred to as Bohr – Sommerfeld
model, the energy depends not only on the principal quantum number, n but also, to some extent,
on the azimuthal quantum number, k. Hence, the shift of an electron from an energy level n2 to
another energy level n1 will be slightly different as it would also depend upon the various possible
values of k in the two states. This explains the occurrence of fine spectrum under high resolution.
2.11 DUAL NATURE OF MATTER AND RADIATION (de-BROGLIE MATTER WAVES)
Scientist have always had trouble describing the nature of light. In many studies such as
diffraction and interference light shows a definite wave like character, but in many others as
Atomic Structure
29
explained in the case of photoelectric effect and Compton effect, light seems to behave as a
stream of photons. This disparity is referred to as the wave-particle duality of light. In 1924, a
young French scientist, Louis de Broglie reasoned that if light could display this wave-particle
duality, then matter, which certainly appears particle like, might also display wave like properties
under certain conditions. de Broglie was able to put his idea into a quantitative scheme, by
combining the known equations,
E = mc2 (Einstein’s equation ) and E = hν (Planck’s equation)
Equating the above equations
hν = mc2
or
c
h = mc2
λ
or
h
λ
= mc
or
λ=
h
mc
.
Replacing c by the velocity of electron,
λ=
h
mv
known as de Broglie equation. The importance of de Broglie equation is that, it connects the
particle nature and wave nature. de Broglie argued that both light and matter obey this equation.
He was awarded Nobel Prize for Physics (1929). de Broglie equation is applicable to moving
neutrons, protons, α particles, atoms, molecules etc.
The wave nature of objects that we come across in everyday life (a moving foot ball or
cricket ball etc) is not significant since wavelength is inversely proportional to mass. Thus de
Broglie wavelength of the foot ball or cricket ball (or any macroscopic body) is so small as to be
completely undetectable and is of no practical significance.
2.12 ELECTRON DIFFRACTION (WAVE NATURE OF ELECTRONS)
When a beam of X-ray is directed at a crystalline solid, the beam is scattered in a definite
manner characteristic of the atomic structure of the solid. This phenomenon called X-ray
diffraction occurs because the interatomic spacing in the crystal is in the same range as the wave
length of the X-ray. If we use an electron beam instead of X-ray, it behaves in the same manner.
This similarity shows that both X-rays and electrons do indeed behave in analogous manner. The
wave like property of electron is being used in the design of electron microscopes.
Davisson and Germer proved the wave nature of electrons from an experiment conducted
on the scattering of slow moving electrons reflected from the surface of a nickel crystal. The
intensities of electron waves scattered by the nickel plate at different angles were measured and
the wave length of electron waves were calculated. This provided an independent experimental
model for determining wavelength of electron waves. Davisson and Germer found the value of λ
to be very close to that obtained by de-Broglie`s equation and thus offered support to de Broglie
equation.
Atomic Structure
30
An interesting historical aside in the concept of the wave-particle duality of matter is that the
English physicist J. J Thomson was awarded the Noble Prize for showing that the electron is a
particle (1906) and G.P Thomson, his son, was awarded the Noble Prize for showing that the
electron is a wave (1937).
Problem: What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10 m/s?
Solution: According to de Broglie equation,
λ=
h
mv
=
6.626 x 10−34 Js
0.1kg x 10m/s
= 6.626 x 10-34m
2.13 HEISENBERG UNCERTAINTY PRINCIPLE
We have seen the duality of light and radiations. Let’s consider a measurement of the
position of an electron. If we wish to locate the electron within a distance Δx, then we must use a
light having wavelength of the order of λ ≈ Δx. The photon has a momentum, p = h/λ and during
the collision, some of this momentum will be transferred to the electron. The very act of locating
the electron leads to a change in its momentum. If we wish to locate the electron more
accurately, we must use light with a smaller wavelength. Consequently, the photons in the light
beam will have greater momentum because of the relation p = h/λ. Since some of the photon`s
momentum must be transferred to the electron in the process of locating it, the momentum change
of the electron becomes greater.
A careful analysis of this process was carried out in the mid-1920s by the German
physicist Werner Heisenberg, who showed that it is not possible to determine exactly how much
momentum is transferred to the electron. This difficulty means that if we wish to locate an
electron to within a region Δx, there will be an uncertainty in the momentum of the electron.
Heisenberg was able to show that if Δp is the uncertainty in the momentum of the electron, then
(Δx). (Δp) ≥
h
4π
where h is Planck`s constant.
Heisenberg uncertainty principle (1927) states that, ‘It is impossible to determine the
exact velocity and position of a subatomic particle simultaneously’. It is a fundamental
principle of nature. If we wish to locate any particle to within a distance Δx, then we
automatically introduce an uncertainty in the momentum of the particle. Note that this
uncertainty does not stem from poor measurement or experimental technique but are inherent in
quantum mechanics. Because of this uncertainty the results of quantum mechanical calculations
are expressed in terms of probabilities.
Heisenberg uncertainty principle can also be stated as, Δt. ΔE ≥
h
4π
Problem:
A golf ball with mass of 40g moves with speed of 45 ms-1. If the speed can be
measured within accuracy of 2% , calculate the uncertainty in the position.
Solution : The uncertainty in the speed (Δv) is 2% , i.e 45 x
2
100
= 0.9 ms-1
Atomic Structure
31
h
4π mΔv
6.626 x 10−34 Js
Δx =
=
4 x 3.14 x 40x 10−3 0.9 ms−1
= 1.46 x 10-33m
For large particles, the uncertainty principle sets no meaningful limit to the precision of
measurements.
2.14 SCHRODINGER THEORY AND THE CONCEPT OF ORBITALS
In Schrö dinger’s wave model of an atom (1926), the discrete energy levels or orbits
proposed by Bohr are replaced by the mathematical function ψ, known as wave function. The
state of any system at any instant can be specified by a state function or wave function, ψ. It is a
function of position and time.
The energy of a particle is the sum of kinetic energy and potential energy possessed by it.
The energy, E and the associated wave function, ψ of a particle for a single particle of mass, m
moving in space can be calculated by the Schrö dinger equation expressed as,
[−
h2
(
4πm
d2
d2
d2
dx
dy
dz2
[−
defined as ∇2 = (
−
h2
4πm
(
d2
dx2
+
2
h2
4πm
+
d2
dy2
d2
d2
d2
dx
dy
dz2
+
2
+
2
+
2
) +V(x,y,z)] ψ = E ψ
∇2 + V(x,y,z) ]ψ = E ψ
+
d2
dz2
where ∇2 is Laplacian operator
)
) is the operator for kinetic energy and V(x,y,z) is the operator for
potential energy of a particle moving in space.
The Schrö dinger equation can generally be expressed as,
Ĥψ = Eψ where Ĥ is known as Hamiltonian operator, an operator for energy which
consists of two parts kinetic energy part −
h2
4πm
∇2 and potential energy part v(x,y,z)
2.14.1 Significance of ψ and ψ2
The Schrö dinger equation is not derived but is a postulate of quantum mechanics and
must be judged on its success in describing experiments. The interpretation of ψ was provided by
Born who suggested that the probability of finding the particle between x and (x + dx) is given by
ψ*(x) ψ(x) dx (or ψ2dx) where ψ* is the complex conjugate of ψ. This means that ψ*(x) ψ(x) dx
(or ψ2dx) represents the probability density of finding the particle between x and (x + dx).
With this interpretation of ψ, the probability of finding the particle anywhere on the x axis must
be,
+∞
∫−∞ ψ∗ (x) ψ(x) dx = 1.
Atomic Structure
32
This equation called normalization condition of ψ, put some restriction to the behavior of ψ.
Though Schrö dinger equation can have several solutions, only those solutions are significant
which give acceptable values of total energy, E. The solution of Schrö dinger equation implies
finding out appropriate ψ which satisfies the equation.
An acceptable (well behaving) wave function for Schr𝑜̈ dinger equation should be single
valued, finite, continuous and normalized.
The concept of operators is very important in Quantum mechanics just like √ ,
𝑑
,∫
𝑑𝑥
that in
mathematics . An operator is a symbol for a well defined procedure to get a new function from a
given function. Corresponding to every measurable quantity (like momentum, energy etc) there is
a mathematical operator. The operator for energy is known as Hamiltonian operator Ĥ which
consists of the kinetic energy part −
h2
4πm
∇2 and potential energy part v(x,y,z). The operator
h
for momentum is -i 2π ∇.
2.15 QUANTUM NUMBERS
The solution of the Schrö dinger wave equation for hydrogen atom yields three quantum
numbers; the principal quantum number (n), subsidiary or azimuthal quantum number (l) and
magnetic quantum number (m). A fourth quantum number, the spin quantum number (s) was
derived from the magnetic character of electron as demonstrated by Stern- Gerlach experiment.
1. Principal quantum number (n)
The principal quantum number determines to a large extent the energy of the electron. It
also determines the average distance of an electron from the nucleus. As the value of n increases,
the electron gets farther away from the nucleus and its energy increases. n can have values from
one to infinity.
2.Subsidiary quantum number or azimuthal quantum number (l)
This quantum number defines the three dimensional shape of the orbital in which the
electron is located. It indicates the sub energy level and describes the energy associated with the
angular momentum of the electron by orbital motion around the nucleus. l may have all possible
whole number values from 0 to n-1. The various sub energy levels are designated as s, p, d, f …
according to the value of l = 0, 1, 2, 3 respectively.
3. Magnetic quantum number (m)
Since an electron has an angular momentum, its motion creates a magnetic field. This
field can interact with an external magnetic or electric field. As a result, the electrons in a given
energy sublevel orient themselves in certain specific regions of space around the nucleus. These
regions of space are called orbitals. i.e The wave function for an electron in an atom is called
atomic orbital. The number of orbitals in a given sub energy level within a principal energy level
Atomic Structure
33
is given by the magnetic quantum number, m. The number of values allowed to m depends on the
values of l, i.e the possible values of m range from –l through 0 to +l, thus, making a total of
(2l+1) values.
Fig.2.8. Different atomic orbitals
When l =0 , m = 0
l = 1, m = -1, 0, 1
l = 2, m = -2, -1, 0, 1, 2
l = 3, m = -3, -2, -1, 0, 1, 2, 3
Thus there are three p orbitals (px, py and pz), five d orbitals (dxy, dxz, dyz, dx2 – y2, dz2 ) and
seven f orbitals, they are fx3 , fxz2 , fyz2, fxyz, fx(x2-3y2), fy(3y2 – x2) and fz(x2- y2)
4. Spin quantum number (s)
The spinning electron behaves like a tiny bar magnet as demonstrated by the Stern –
Gerlach experiment and it possesses an intrinsic angular momentum or spin, specified by the
1
2
quantum number s = . In an applied magnetic field, the spin of an electron can orient in two
Atomic Structure
34
1
1
directions and the spin quantum number, specified as ms, can have two values, + 2 and - 2. The
1
+ 2 component has an upward orientation (↑) and -
1
2
component has a downward orientation (↓).
2.16 PAULI’S EXCLUSION PRINCIPLE
The four quantum numbers define completely the position of an electron in a given atom.
They give its position in the major energy level (n), the sub energy level (l), the orientation in the
sub energy level (m) and the direction of spin (ms). It is thus possible to identify an electron in an
atom completely by stating the values of its four quantum numbers. They serve as an ‘address’
for the electron.
The Pauli`s exclusion principle states that, it is impossible for any two electrons in the
same atom to have all the four quantum numbers same.
This principle is very useful in determining the maximum number of electrons that can
occur in any shell. For K-shell, since n=1, l can have only one value (= 0) and m can also have
1
2
1
+2
1
2
only one value (= 0). Hence, ms can be either +
n=1
l=0
m=0
ms =
n=1
l=0
m=0
ms =
or -
1
2
. Thus, we have
This shows that in the K-shell, there is only one sub shell l = 0 and in this only two electrons of
opposite spins can be accommodated.
For the L-shell, since n = 2, l can have two values (0 and 1), m can have three values (0,-1
and +1) and ms can have two values (+
1
2
and -
1
2
). These possibilities give rise to eight
combinations of the four quantum numbers, keeping in view the exclusion principle.
1
n=2
l=0
m= 0
ms = + 2
n=2
l=0
m= 0
ms = - 2
n= 2
l=1
m = -1
ms = + 2
n= 2
l=1
m = -1
ms = - 2
n=2
l=1
m= 0
ms = + 2
n= 2
l=1
m= 0
ms = -
n= 2
l=1
m=1
ms =
n= 2
l=1
m=1
ms =
1
1
1
1
1
2
1
+2
1
-2
Thus, L-shell can accommodate 8 electrons, 2 in the l = 0 sub shell (s-orbital) and 6 in the
l =1 sub shell (p- orbital). Similarly, it can be shown that the M-shell can have 18 electrons, 2 in
the l = 0 sub shell (s-orbital) and 6 in the l =1 sub shell (p- orbital) and 10 in the l = 2 sub shell (dorbital) and so on. The maximum number of electrons in a particular energy level is given by
Atomic Structure
35
2n2. W. Pauli, the Austrian physicist was awarded the 1945 Physics Nobel Prize for his
discovery of the exclusion principle.
2.17 HUND`S RULE OF MAXIMUM MULTIPLICITY
Hund`s rule states that electrons are filled in a degenerate level in such a way that
maximum number of unpaired electrons with same spin are placed before pairing. For
example, of the three orbitals (px, py, pz) of a principal energy level, each must get one electron of
parallel spin before any one of them gets the second electron of opposite spin. This can be
illustrated with specific examples.
(i) Consider nitrogen atom (atomic number 7). According to Hund`s rule, the electronic
configuration will be;
1s2, 2s2, 2px1, 2py1, 2pz1
(ii) For oxygen atom (atomic number 8), the electronic configuration will be
1s2, 2s2, 2px2, 2py1, 2pz1
Logical explanation of Hund`s rule may be stated with two factors.
(i) Electrostatic repulsion
If two electrons with opposite spin are placed in the same orbital, the electrostatic
repulsion would be greater than when they are placed in separate orbitals. Electrons can
minimize the repulsive forces between themselves by occupying different orbitals and by having
parallel spins. Therefore pairing takes place only when it is essential.
(ii) Exchange energy
Electrons in orbitals are not placed permanently and migrate from one degenerate orbital
into another. This sort of exchange releases energy and the atom become more stable. The more
the energy given out, the more stable the atom is.
On the basis of exchange energy calculations it can be shown that chromium and copper have
different actual electronic configurations than expected.
V (atomic number = 23) : 1s2, 2s2, 2p6, 3s2, 3p6, 3d3, 4s2
Cr (atomic number = 24) : 1s2, 2s2, 2p6, 3s2, 3p6, 3d5, 4s1
Mn (atomic number = 25) : 1s2, 2s2, 2p6, 3s2, 3p6, 3d5, 4s2
The configuration 3d4,4s2 has 10 sets and the configuration 3d5,4s1 has 15 sets for exchange.
𝑛!
(Number of possible sets = 2(𝑛−2)! where n is number of unpaired electrons). Therefore half
filled states are more stable than partially filled state.
Atomic Structure
36
Similar argument show that Cu (atomic number = 29) has a configuration 1s2, 2s2, 2p6,
3s2, 3p6, 3d10, 4s1 (high exchange energy) and not 1s2, 2s2, 2p6, 3s2, 3p6, 3d9, 4s2 . A
completely filled state is also much more stable than a partially filled state.
2.18 AUFBAU PRINCIPLE
The term ‘aufbau’ (German word meaning building up) stands for a method for filling the
atomic orbitals. Aufbau principle states that ‘the orbitals are occupied with electrons in the
increasing order of their energies’. As a working rule, a new electron enters the orbital where
(n+l) is the minimum. When (n+l) has the same value for two or more orbitals, the new electron
enters the orbital where n is minimum. Using this rule the most usual sequence in which the
electrons occupy various orbitals is shown in figure 2.9.
Fig.2.9 The diagonal rule for electron filling order
(n + l ) rule
The relative order of energies of various sub shell in a multi electron atom can be
predicted with the help of (n + l) rule. The sub shell with lower value of (n + l) has lower energy
and it should be filled first.
Example:
3d
4s
(n + l) = 3+2 = 5
(n + l) = 4+0 = 4
Therefore 4s sub shell will be filled before 3d sub shell.
If two sub shells have same value of (n + l) then the sub shell with lower value of n has lower
energy and it should be filled first.
Atomic Structure
37
3d
4p
(n + l) = 3+2 = 5
(n + l ) = 4+1 = 5
Therefore 3d sub shell is filled before 4p sub shell.
Very few exceptions for (n + l) rule occur.
 A 5d electron is added before any 4f orbital is occupied. The remaining nine 5d electrons
enter this sub shell after the 4f sub shell has been completely filled.
 One or more electrons enter the 6d sub shell before any electron occupies the 5f sub shell.
QUESTIONS
Section A
1.
2.
3.
4.
5.
6.
Which metal is used in the photoelectric cells? Why?
Atomic spectra are line spectra. Why?
Why spectral lines split in a magnetic field?
Cr has 5 electrons in the 3d orbitals. Why?
What are the quantum numbers of the valance electrons in the magnesium atom?
Do all p-electrons in nitrogen atom have same energy? What is the effect of magnetic field on the
energies of these electrons?
7. What are the possible values of magnetic quantum number for an electron in 4f sub-shell?
8. What are degenerate orbitals? How many such orbitals are present in 5d sub-shell?
9. What is the value of Planck`s constant?
Section B
1. What is photoelectric effect?
2. State Heisenberg`s uncertainty principle.
3. Give the de-Broglie relation.
4. Calculate the momentum of a particle which has a de-Broglie wavelength of 0.1nm.
5. Calculate the wave length associated with a radiation frequency 2000 Hz.
6. What are the n, l and m values for 2px and 3pz electrons?
7. What type of metals is used in photoelectric effect? Justify.
8. Write the electronic configuration of ground state carbon atom.
9. Define threshold frequency. Name a metal which shows photoelectric effect in visible light.
10. What do you understand by wave function ψ? What is its significance?
11. Write the electronic configuration of elements with atomic number 10. 17 and 25.
Section C
1.
2.
3.
4.
5.
6.
7.
Discuss briefly the photoelectric effect. How did Einstein explain it?
Briefly discuss the salient features of the Bohr theory of hydrogen atom. What are its defects?
Derive de Broglie equation
State and explain Heisenberg`s uncertainty principle.
Define Pauli`s exclusion principle. Explain.
Explain Hund`s rule of maximum multiplicity.
Calculate the de Broglie wave lengths of a 1g bullet with velocity 300 m/s (Ans: 2.21 x 10-33 m)
Atomic Structure
38
8. Sketch the shapes of different d orbitals.
9. Write the electronic configuration of Cr (Z = 24) and Cu (Z = 29)
10. Calculate the wavelength of a photon emitted when an electron in H-atom makes a transition from n =
2 to n =1.
(Ans λ = 4/3R)
11. What is the wavelength of a ball of mass 0.1kg moving with a velocity of 10 ms-1? (Ans: 6.626x10-34
m)
Section D
1.
2.
Calculate the wave length in nm of the first line of Brackett series in hydrogen spectrum where n1 = 4
Calculate the de Broglie wave length of an electron moving with a kinetic energy 2.8 x 10 -25 J. (Ans:
λ = 9.245 x 10-7m)
3. What are quantum numbers? Explain.
4. Account the origin of line spectrum of hydrogen with Bohr model of atom.