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At o m i c St r u c t u r e
CHEMI STRY
At omic St ruct ure
ATOMIC STRUCTURE
. LAWS OF CHEMICAL COMBINATIONS
Law of conservation of mass Laws of conservation of mass states that in a chemical reaction
the weight of
products is always equal to the weight of reactants.
Law of definite proportions Law of definite proportions states that the elemental composition
of a compound
always remains same if it is analysed form various sources e.g. water from a river or ditch or
pond either in India
or in USA would always give H O ratio as .
Law of Multiple Proportions Law of multiple proportion states that elements combine in
simple whole number
ratios to form various types of compounds e.g. The ratio of N O is , and in NO, NO
and N
O.
respectively.
. DALTONS THEORY OF ATOM
John Dalton developed his famous theory of atom is . The main postulates of his theory were
Atom was considered as a hard, dense and smallest indivisible particle of matter.
Each element consists of a particular kind of atoms.
The properties of elements differ because of differences in the kinds of atoms contained in
them.
This theory provides a satisfactory basis for the law of chemical combination.
Atom is indestructible i.e. it cannot be destroyed or created.
Drawbacks
It fails to explain why atoms of different kinds should differ in mass and valency etc.
The discovery of isotopes and isobars showed that atoms of same elements may have
different atomic
masses isotopes and atoms of different kinds may have same atomic masses isobars.
The discovery of various subatomic particles like Xrays, electrons, protons etc. during the th
century lead to the idea that the atom was no longer an indivisible and smallest particle of the
matter..
. DISCOVERY OF ELECTRON
William Crookes found that certain rays come out of cathode and gain lots of energy because
of high acceleration
potential before colliding with a gas molecule. This happens especially when the gas
pressure is low. These rays
are known as cathode rays.
Cathode
High Voltage
Gas at very low pressure
Cathode Anode
Suction pump
Fig. Cathode ray Tube experiment
Detailed study of cathode rays by J.J. Thomson led to the discovery of electrons. He
observed that
.Cathode rays always travel in straight lines
.They are negatively charged. Cathode rays turn towards positively charged plates.
.Charge on particles constituting rays was determined by Oil
Drop experiment by Millikan as
. . coulomb
.Specific charge e/m does not change when the gas inside discharge tube was changed
indicating that electron
is fundamental particle.
e/m for electron
. / Coulombs kg .
.Value of charge on electron
. Coulombs
.Mass of the electron
. kg
.
. ANODE RAYS DISCOVERY OF PROTON
It is wellknown fact that the atom is electrically neutral. The presence of negativelycharged
electrons in the
atom amphasized the presence of positivelycharged particles.
To detect the presence of positivelycharged particles, the discharged tube experiment was
carried out, in which
a perforated cathode was used. Gas at low pressure was kept inside the tube. On passing
high voltage between
the electrodes it was observed that some rays were emitted from the side of the anode.
These rays passed through
At omic St ruct ure
the holes in the cathode and produced green fluorescence on the opposite glass was coated
with ZnS. These rays
consist of positivelycharged particles known as protons.
ZnS
Coating
H gas inside
at low pressure
Perforated
cathode
For anode rays, e/m is not fundamental property as different gases used have different mass
on C scale.
Highest e/m is for hydrogen gas.
. DISCOVERY OF NEUTRON
After the discovery of electrons and protons, Rutherford had predicted the existence of a
neutral fundamental
particle. In , Bethe and Becker reported from Germany that if certain light elements, like
beryllium, were
expoesd to alpha rays from the naturally radioactive polonium, a very highly penetrating
radiation was obtained.
Similar results were obtained by Irene Cureia dn F. Jolit . Chadwick demonstrated that this
mysterious
radiation was a stream of fast moving particles of about the same mass as a proton but
having no electric charge.
Because of their electrical neutrality, these particles were called neutron.
The lack of charge on the neutron is responsible for its great penetraing power.
Thus, a neutron is a subatomic fundamental particle which has a mass
.g
approximately amu,
almost equal to that of a proton or a hydrogen atom but carrying no electric charge. The
/em
value of a neutron
is thus zero.
. CONSTITUENTS OF ATOM
Some of the well known fundamental particles present in an atom areprotons, electrons and
neutrons. Many
others were discovered later viz positron, neutrinos etc.
Subatomic Symbol Unit Charge Unit Mass Charge in Mass in AMU
particles Coulomb
Proton
p
.
.
Neutron
n
.
Electron
e
Negligible
.
.
Do you Know
Chandwick in discovered the neutron by bombarding elements like beryllium with fast moving
particles.
He observed that some new particles were emitted which carried no charged and had mass
equal to that of
proton.
. THEORIES OF ATOM
After the discovery of electrons, protons and neutrons several theories were proposed by
various scientists such
as
.Plum Pudding Model Thomson Model of Atom
.Rutherfords Model of Atom
.Plancks quantum theory
.Bohrs Atomic Model
Thomson Model of Atom
According to this model, electrons are embedded in uniform sphere of positive charge to
confer electrical neutrality.
This model was satisfactory to the extent that the electrostatic forces of repulsion among the
electron cloud is
balanced by the attractive forces between the positively charged mass and the electron. How
ever, this model
fails to explain the results of ionisation and scattering experiment and is, therefore discarded.
Rutherford Model of Atom
When particles are bombarded on thin foils
cm
thick of metals like gold, silver, platinum or copper, ,
At omic St ruct ure
most of them are passed through the metal foil with little or no deviation. However a small
proportion of
particles are scattered through large angles and even bounced back i.e. deflected through
. From these
observations, Rutherford drew the following conclusions
Gold foil Zinc sulphide
screen
Reflected oparticles
Lead plate Lead block
Fig. Rutherfords Experiment
a As most of the particles passed through the foil without undergoing any deflection, there
must be sufficient
empty space within the atom.
b As particles are positively charged, deflected by large angle there must be heavy small
positively charged
body present in the atom, which is called Nucleus.
From these observations, Rutherford proposed the following model of atom
a Atom is composed of a positively charged nucleus where the whole mass of the atom is
concentrated and the
electrons are present in relatively large volume around the nucleus. The total positive charge
carried by nucleus
must be equal to the total negative charge carried by electrons and electroneutrality of atom
is maintained.
Fig Thomson Model
electron
positive sphere
b Electron are constantly moving around the nucleus in different orbits.
The centrifugal force arising from this motion balances the electrostatic attraction between
the nucleus and the
electron. Therefore the electron dont fall into the nucleus.
Drawbacks in Rutherford Model
a The most fundamental objection arises from the electromagnetic theory of radiation which
predicts that when
a charged body moves in a circular path, it should radiate energy continuously. As the
electron is a negatively
charged particle revolving around the nucleus, it should radiate energy continuously. As a
result, the electron
should fall into the nucleus.
b As the electron is continuously radiating energy, the spectra should be continuous.
Actually, the spectra of
atom is a line spectra.
. ATOMIC NUMBER OF AN ELEMENT
Total number of protons present in the nucleus
Total number of electrons present in the atom
Atomic number is also known as proton number because the charge on the nucleus
depends upon the number
of protons.
Since the electrons have negligible mass, the entire mass o the atom is mainly due to
protons and neutrons only.
Since these particles are present in the nucleus, therefore they are collectively called
nucleons.
As each of these particles has one unit mass on the atomic mass scale, therefore the sum of
the number of
protons and neutrons will be nearly equal to the mass of the atom.
Mass number of an element No. of protons No. of neutrons.
The mass number of an element is nearly equal to the atomic mass of that element.
However, the main difference
between the two is that mass number is always a whole number whereas atomic mass is
usually not a whole
number.
The atomic number Z and mass number A of an element X are usually represented
alongwith the symbol
of the element as
At omic St ruct ure
A
Z
Mass Number
Atomic Number
Symbol of the
element
x
e.g.
, Na Cl and so on.
.Isotopes
Such atoms of the same element having same atomic number but different mass numbers
are called isotopes.
, H H and H and named as protium, deuterium D and tritium T respectively. Ordinary
hydrogen is
protium.
. Isobars
Such atoms of different elements which have same mass numbers and of course different
atomic numbers are
called isobars.
e.g.
, , Ar K Ca .
. Isotones
Such atoms of different elements which contain the same number of neutrons are called
isotones.
e.g.
,,CKO.
. Isoelectronics
The species atoms or ions containing the same number of electrons are called isoelectronic.
For Example,
, , , , , O F Na Mg Al Ne
all contain electrons each and hence they are isoelectronic.
Illustration.
Complete the following table
Particle Mass No. Atomic No. Protons Neutrons Electrons
Nitrogen atom
Calcium ion
Oxygen atom
Bromide ion
Solution.
For nitrogen atom.
No. of electron given
No. of neutrons given
No. of protons Z
atom is electrically neutral
Atomic number Z
Mass No. A No. of protons No. of neutrons
For calcium ion.
No. of neutrons given
Atomic No. Z given
No. of protons Z
No. of electrons in calcium atom Z
But in the formation of calcium ion, two electrons are lost from the extranuclear part
according to the equation
Ca Ca e
but the composition of the nucleus remains unchanged.
No. of electrons in calcium ion
Mass number A No. of protons No. of neutrons .
For oxygen atom.
Mass number A No. of protons No. of neutrons Given
Atomic No. Z Given
No. of protons Z ,
At omic St ruct ure
No. of electrons Z
No. of neutrons A Z
Some important characteristics of a wave
Crest Crest
Trough Trough
a
a
Wavelength of a wave is defined as the distance between any two consecutive crests or
troughs. It is represented
by
and is expressed in or m or cm or nm nanometer or pm picometer.
cm m
, nm m pm m
Frequency of a wave is defined as the number of waves passing through a point in one
second. It is represented
by v nu and is expressed in Hertz Hz or cycles/sec or simply sec
or s
.
Hz cycle/sec
Velocity of a wave is defined as the linear distance travelled by the wave in one second. It is
represented by v and
is expressed in cm/sec or m/sec ms
.
Amplitude of a wave is the height of the crest or the depth of the trough. It is represented by
a and is expressed
in the units of length.
Wave number is defined as the number of waves present in cm length. Evidently, it will be
equal to the
reciprocal of the wavelength. It is represented by
v
read as nu bar.
v
If
is expressed in cm,
v
will have the units cm
.
Relationship between velocity, wavelength and frequency of a wave. As frequency is the
number of waves
passing through a point per second and
is the length of each wave, hence their product will give the velocity
of the wave. Thus,
vv
Cosmic rays lt rays lt Xrays lt Ultraviolet rays lt Visible lt Infrared lt Micro waves lt Radio
waves.
. PLANCKS QUANTUM THEORY
It states
.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 which is proportional to
frequency of radiation.
E hv
Where, h Plancks constant
. sec Joule
.
v Frequency of the light radiation
.A body can emit or absorb radiations only in whole multiples of quantum i.e.
E nhv
where , , ...... n
c
v
where c velocity of light
wavelength
At omic St ruct ure
hc
E
a. Some Important Formulae
.A Z N Number of neutrons
.dynamic mass of particle
/
//mmvc
.Radius of nucleus
/
,.RRARm
.
cv
.wave number / v
.
/ E hv hc hcv
.
./
qq
F K K Nm C
r
.
E hv E E A
Illustration .
Calculate number of photon coming out per sec. from the bulb of watt. If it is efficient and
wavelength
coming out is nm.
Solution.
energy J
energy of one photon
..
hc
no. of photon
.
.
Illustration .
Certain sun glasses having small of AgCl incorporated in the lenses, on exposure to light of
appropriate wavelength
turns to gray colour to reduce the glare following the reactions
hv
AgCl Ag Gray Cl
If the heat of reaction for the decomposition of AgCl is kJ mol
, what maximum wavelength is needed to
induce the desired process
Solution.
Energy needed to change
/ J mol
If photon is used for this purpose, then according to Einstein law one molecule absorbs one
photon. Therefore,
A
hc
N
...
.
m
.
. BOHRS ATOMIC MODEL
The postulates of Bohrs atomic theory regarding stability of electrons of an atom are as
follows
i The electrons is an atom revolve around the nucleus only in certain selected circular orbits.
These orbits are
known as energy levels or stationary states. An electron can be excited from a lower state to
higher state with the
absorption of a quantum of energy, or can come down from a higher to lower state with
emission of a radiation
of energy as shown in figure equal to energy to quantum
E E E hv A .
E and
E are energies of the
electron associated with stationary orbits.
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Absorption of AE
Emission of
radiation with energy
AEE E
E
Fig. Bohrs Atomic Model
ii The stability of the circular motion of an electron requires that the electrostatic force due to
the attraction
between the nucleus and the electron provides the necessary centripetal force for the motion
of electron.
./
Ze e
mv r
r
... i
where Z atomic number
e charge on electron
permittivity of the charge in vacuum
r distance between positive charge amp electron
iii Angular momentum of electron is quantised i.e. electron can revolve only in those orbits
where its angular
momentum is an integral multiple of
/h
/ mvr nh ... ii
where, v velocity of electron
m mass of electron
h
Plancks constant
, , ........ n are known as Principal quantum number..
Bohrs Atomic Radius
from i ampii we have
from i
/ v ze mr ... iii
amp from ii
/vnhmr
... iv
Equating iii and iv, we have
/ / Ze mr n h m r
or,
nh
r
mZe
e
r
is called Bohrs radius.
Where
, , , h e m and are constants, Thus,
/rKnZ
Putting the values of
, , , h e m and ,
./rmnZ
Illustration .
For hydrogen atom
Z
, therefore radius of the first orbit
./m
For
He
ion
Z
, therefore radius of the first orbit
./m
.m
Solution.
Velocity of electrons in various orbits
At omic St ruct ure
Substituting value of r in equation no. ii
/
Ze
vKZn
nh
e
The energy of an electron in an orbit
E Kinetic energy Potential energy
r
Ze
mv
.
Putting
mv from equation i, we have
Ze Ze Ze
E
rrr
Now, putting the value of
r
, we have
/./
mZ e
EKZnJZn
nh
The energy difference between two energy levels n
and n
is given by
//
nn
mZ e
EEEnn
h
A
.JZ
nn
.
It terms of wavenumber, we have
//
H
vRZnn
a. Bohrs Model for Hydrogen like Atoms
.
h
mvr n
.
./.
n
Ezz
E z J atom eV
nnn
me
E
n
.
.
n
nhn
r
ZZem
.
.
/
ze z
vms
nh n
.Revolutions per sec
.
/
Z
vr
n
.Time for one revolution
.
/
n
rv
Z
.........
n
EKEPEPEKE
.
../,..
ze
K E mv P E
r
At omic St ruct ure
Illustration .
What is the principal quantum number of H atom orbital is the electron energy is . eV Also
report the
angular momentum of electron.
Solution.
E
for H . eV
Now,
n
E
E
n
.
.
n
n
Now, Angular momentum mur
.
. . sec
.
h
nJ
.
Illustration .
Calculate the energy, velocity and radius of electron in Li
ion.
Solution.
For Li
ion, z , n , then
radius
...
n
z
velocity
. / sec . / sec.
z
cm cm
n
. / sec. cm
Energy
..
z
ev ev
n
. ev .
. DEFINITION VALID FOR SINGLE ELECTRON SYSTEM
i Ground state
lowest energy state of any atom or ion is called ground state of the atom.
Ground state energy of Hatom . eV
Ground state energy of He
ion . eV
ii Excited state
State of atom other than the ground state are called excited states
n first exited state
n second exited state
n third exited state
nnn
th
exited state
iii Ionisation energy IE
Minimum energy required to move an electron from ground state to
n is called ionisation energy of the atom or ion.
Ionisation energy of Hatom . eV
Ionisation energy of He
ion . eV
Ionisation energy of Li
ion . eV
iv Ionisation Potential I.P
Potential difference through which a free electron must be accelerated from rest, such that its
kinetic energy
becomes equal to ionisation energy of the atom is called ionisation potential of the atom.
I.P. of H atom . V
I.P of He
Ion . V
At omic St ruct ure
v Excitation Energy
Energy required to move an electron from ground state of the atom to any other state of the
atom is called
excitation energy of that state.
excitation energy of
nd
state excitation energy of
st
state
st
excitation energy . eV.
vi Excitation Potential
Potential difference through which an electron must be accelerated from rest to so that its
kinetic energy become
equal to excitation energy of any state is called excitation potential of that state.
excitation potential of third state excitation potential of second excitate state seconds
excitations potential
. v.
vii Binding Energy or Separation Energy
Energy required to move an electron from any state to n is called binding energy of that
state.
Binding energy of ground state I.E. of atom or Ion.
Illustration.
A single electron system has ionisation energy kJ mol
. Find the number of protons in the nucleus of the
system.
Solution.
I.E.
.
Z
J
n
.
.
Z
I
Z
. HYDROGEN SPECTRUM
. Study of Emission and Absorption Spectra
An instrument used to separate the radiation of different wavelengths or frequencies is called
spectroscope or
a spectrograph. Photograph or the pattern of the emergent radiation recorded on the film is
called a spectrogram
of simply a spectrum of the given radiation. The branch or science dealing with the study of
spectra is called
spectroscopy.
Emission Spectra
When the radiation emitted from some source e.g. from the sun or by passing electric
discharge through a gas at
low pressure or by heating some substance to high temperature etc, is passed directly
through the prism and then
received on the photographic plate, the spectrum obtained is called Emission spectrum.
Depending upon the sources of radiation, the emission spectra are mainly of two types
i Continuous spectra
When white light from any source such as sun, a bulb or any hot glowing body is analysed by
passing through a
prism it is observed that it splits up into seven different wide bound of colours from violet to
red. These colours
are so continuous that each of them merges into the next. Hence the spectrum is called
continuous spectrum.
Photographic
Plate
Prism
Beam
White light
Slit
V
I
B
G
Y
O
R
ii Line Spectra
When some volatile salt e.g., sodium chloride is placed in the Bunsen flame or an electric
discharge is passed
through a gas at low pressure light emitted depends upon the nature of substance.
Photographic
Plate
Prism
Beam
Slit
Platinum wire
Two yellow lines
At omic St ruct ure
It is found that no continuous spectrum is obtained but some isolated coloured lines are
obtained on the
photographic plate separated from each other by dark spaces. This spectrum is called Line
emission spectrum
or simply Line spectrum.
. Absorption spectra
When white light from any source is first passed through the solution or vapours of a
chemical substance and
then analysed by the spectroscope, it is observed that some dark lines are obtained in the
otherwise continuous
spectrum. These dark lines are supposed to result from the fact that when white light
containing radiations of
many wavelengths is passed through the chemical substance, radiation. of certain
wavelengths are absorbed,
depending upon the nature of the element.
Photographic
Plate
Prism
NaCl
Solution
Slit
V
I
B
G
Y
O
R White light
Dark lines in yellow region of
continuous spectrum
Photographic
Plate
Prism
Beam
Slit
EMISSION SPECTRUM OF HYDROGEN
a. HAtom Spectrum
When an electric discharge is passed through hydrogen gas at low pressure, a bluish light is
emitted. When a ray
of this light is passed through a prism, discontinuous line spectrum of several isolated sharp
lines is obtained as
shown in figure.
n
n
Energy
Lymann series
Balmer series
Paschen series
Bracket series
Pfund series
Energy levels of Hatom
All these lines observed in the hydrogen spectrum can be classified into the series as is
tabulated in the table.
The hydrogen Spectrum
Region Spectral lines n
n
UV Lyman series ,,,.....
Visible Balmer series ,,,.....
I R Paschen series ,,,.....
farI.R Brackett series ,,,.....
At omic St ruct ure
farI.R. P fund series ,,,.....
Illustration
Calculate the highest wavelength of line spectra of Hatom when the electron is situated in
rd
excited state.
Solution.
Highest wavelength means lowest energy difference of electronic transition from one energy
level to other
energy level.
Hence, lowest energy transition will be n to n .
..
E ev ev
..
E ev ev
. . E E E ev A
...
hc
ev J
.
..
mm
. DEBROGLIE RELATIONSHIP
According to deBroglie matter has dual character i.e. wave as well as particle, if the
wavelength of matter be
having mass m moving with velocity v , then,
h
mv
where
h
is Plancks constant
. sec Joul
.
wave in phase wave out of phase
Case I Case II
For electron moving around a nucleus in a circular path, two possible waves of different
wavelengths are possible.
In case I, the circumference of the electron orbit is an integral multiple of wavelength.
In case II, wave is destroyed by interference and hence, does not exist.
Therefore, the necessary condition for a stable orbit of electron of radius r is,
rn
when , , , n etc .
As
h
mv
,
nh nh
r or mvr
mv
This is simply the original Bohr condition for a stable orbit. Hence, the Bohrs model of Hatom
is justified by
deBroglie relationship.
.a
DeBroglie Relations
hh
mc p
At omic St ruct ure
deBroglie pointed out that the same equation might be applid to material particle by suing m
for the mass of the
particle instead of the mass of photon and replacing c, the velocity of the photon, by v, the
velocity of the
particle.
..
hh
mv m K E
From the deBroglie equation it follows that wavelength of a particle decreases with increase
in velocity of the
particle. Moreover, lighter particles would have longer wavelengths than heavier paticles,
provided the velocity
is equal.
If a charged particle Q is accelerated through potential difference V from rest then debroglie
wavelength is
h
mQV
deBroglie concept is more isgnificant for microscopic or submicroscopic particles whose
wavelength can be
measured.
The circumference of the nth orbit is equal to n times the wavelength of the electron.
n
rn
Wavelength of electron is always calculated using Debroglie calculation.
Illustration.
Calculate the wavelength of a body of mass kg moving with a velocity of m sec
.
Solution.
We know,
h
mv
Substituting the values, m , mg
kg, v m sec
.
and
. h kg m s
.
.
m
Illustration.
. eV is needed for ionisation of a hydrogen atom. An electron in a hydrogen atom in its
ground states absorbs
. times as much energy as the minimum energy required for it to escape from the atom. What
is the wavelength
of the emitted electron
.,.,..
e
m kg e coulomb h J s
Solution.
. times of . eV, i.e., . eV, is absorbed by the hydrogen atom out of which . eV . . is
converted to kinetic energy.
KE . eV .
. coulomb
volt
.J
.
Now,
KE mv
or,
.
./.
.
KE J
vms
m kg
..
..
../
hJs
metres
mv kg m s
. HEISENBERGS UNCERTAINTY PRINCIPLE
If subatomic particles have wave nature then we cant pinpoint where exactly a particle is.
The idea was defined
by Heinsenberg as There is a limit to the precision to which the position and momentum of a
particle may be
determined simultaneously
/ x p h A A gt
At omic St ruct ure
Where
xA
uncertainty in position of an electron.
p A uncertainty in momentum of an electron, i.e., when we try to determine position for a
subatomic
particle correctly, uncertainly in momentum will be very large or when we try to determine the
momentum
correctly then uncertainly in position will be large.
ORIGIN OF QUANTUM THEORY
When solid body heated it emit radiations in the forms of waves. The wave nature of light can
be explained by
diffraction interference etc. But some other observable properties such as photoelectric
effect, compton effect
could not be explained from wave nature. Hence a different theory is needed to explain these
facts.
Illustration.
Calculate the uncertainty in the velocity of a wagon of mass kg whose position is known to an
accuracy of
m.
Solution.
Uncertainty in position, x m A
Mass of the wagon, m kg
According to Heinsenbergs principle,
.
h
xmv
A
.
.
..
h kgm s
v ms
m x kg m
A
A
.
a. Black Body Radiation
When radiation falls on an object, a part of it is reflected, a part is absorbed and the
remaining part is transmitted
because no object is a perfect absorber. But the black body e.g., a metallic hollow sphere
with a small hole,
blackened on the inside surface absorbs completely all the radiations falls on it by
successive reflections inside
the enclosure.
T
T
T
Wave length
energy
TgtTgtT
The black body is not only a perfect absorber of radiation energy, but also an ideal radiation,
i.e., when the black
body is heated, it radiates the maximum amount of energy. The energy which is radiated is
dependent on the
temperature of the black body and is independent of the nature of the interior material.
The curves represent the distribution of radiation from a black body at different temperatures.
The shape of the
curves couldnt be explained on the basis of classical electromagnetic theory in which it was
assumed that the
body radiates energy continuously. So, the intensity of radiation should increase continuously
without limits as
the frequency increases. But the experimental observations are contrary to the classical
view. For each temperature,
there is a maximum in the curve corresponding to a particular wavelength, indicating the
maximum radiation of
energy. At higher temperature, the position of the maximum in the curve shifts towards
shorter wavelength and
becomes more pronounced. To explain this black body radiation, Max Planck put forward
new quantum theory.
Planck Quantum Theory
a Radiation energy is not emitted or absorbed continuously but discontinuously in the form of
tiny bundles of
energy, called quanta.
b Each quanta is associated with a defined amount of energy E which proportional to the
frequency of
radiation i.e,
Ev
or,
E hv
where h is Plancks constant
. J sec
.
c A body can emit or absorb energy only in whole number multiples of quanta i.e.
E nhv
where , , . n etc
. PHOTOELECTRIC EFFECT
Sir J.J. Thomson has discovered this phenomenon of ejection of electron from the surface of
a metal when light
of suitable frequency of strikes on it.
At omic St ruct ure
Only few metals show this effect under the action of visible light, but many more show it
under the action of
more energetic u.v. light. For every metal, there is a minimum frequency of incident radiation
necessary to eject
electron from that metal surface, is known as Threshold frequency
v . This
v varies metal to metal.
The number of ejected electrons from the metal surface depends upon the intensity of the
incident radiation.
Greater the intensity, the larger is the number of ejected electrons.
Hence, according to quantum theory, when a photon of light of frequency
v v gt strikes on an electron in a
metal, it imparts it entire energy to the electron. Then some of its energy equal to binding
energy of electron with
the nucleus is consumed to separate the electron from the metal and the remaining energy
will be imparted to the
ejected electron.
hv hv mv where
hv is the binding energy of work function of the electron and
mv is the kinetic
energy of electron. Alkali metals are mainly used for photoelectric effect. Cesium, amongst
alkali metals, has
lowest threshold energy and used largely in photoelectric cell.
. SHAPES OF ORBITALS
sorbital they do not have directional character. They are spherically symmetrical. The sorbital
of higher
energy levels are also spherically symmetrical. They are more diffused and have spherical
shells within them
where probability of finding the electron is zero.
z
x
y
Node
s orbital
In the sorbital, number of nodes is n
porbital p orbital has a dumbbell shape and it has a directional character.
The two lobes of a porbital are separated by a plane that contains the nucleus and is
perpendicular to the
corresponding axis. Such plane is called a nodal plane because there is no probability of
finding the electron.
px
z
y
py
z
y
xx
pz
z
y
x
In the absence of an external electric or magnetic field, the three porbitals of a particular
energy level have same
energy and are degenerate. In the presence of an external magnetic field or electric field this
degeneracy is
removed.
dorbitals For dorbitals five orientations are possible viz., d
xy
,d
yz
,d
xz
,
xy
d
,
z
d . All these five orbitals in the
absence of magnetic field are equivalent in energy and are degenerate.
The shapes of the orbitals are as follows
At omic St ruct ure
dxy
x
y
dxz
x
z
dyz
x
z
These three d orbitals are similar. The maximum probability of finding the electron is in lobes
which are
directed in between the axes. Nodal region is along the axes.
dz
z
dx
y
y
x
These two dorbitals are similar. Probability of finding the electron is maximum along the axes
and the nodal
region is in between the axes.
. QUANTUM NUMBERS
These are used to determine the region of probability of finding a particular electron in an
atom.
a Principal quantum number n
This denotes the energy level or the principal or main shell to which an electron belongs. It
can have only
integral values , , etc. The letter K, L, M ..... are also used to designate the value of n. Thus,
an electron in the
K shell has n , that is L shell has n and so on.
Illustration.
The principal quantum number of selectron is
Solution.
n
b Azimuthal quantum numbers l
This denotes the orbital Sublevel to which an electron belongs. It gives an idea about the
shape of the orbital.
l can have any value from to n , for a given value of n,
i.e. l , , , ..... n
Value of l
Sub Shell s p d f
The value of orbital angular momentum of the electron for a given value of l is
h
ll
c Magnetic quantum number m
It gives us the idea about the orientations an orbital can have in space in the presence of
magnetic field. The
values of m depends on l orbital quantum number.
Total value of m l and it varies l to l.
For example, for l the value of magnetic quantum number m is also equal to zero, i.e. sorbital
can have
only one orientation in space in presence of magnetic field.
d Spin quantum number s
The electron while moving round the nucleus in an orbit also rotates or spins about its own
axis either in a
At omic St ruct ure
clockwise direction or in an anticlockwise direction. Its value is
or
corresponding to clockwise or
anticlockwise spin.
The value of spin angular momentum for a given value of s is
h
ss
The spin magnetic moment of electron excluding orbital magnetic momentum is given by
effective
n n BM Where n Number of unpaired electrons.
. DISTRIBUTION OF ELECTRONS IN AN ATOM
The filling up of orbitals with electrons takes place according to certain rules which are given
below
i The maximum number of electrons in a main shell is equal to n
, where n is the principal quantum number.
ii The maximum number of electrons in a subshell like s, p, d, f is equal to l , where l is the
azimuthal
quantum number for the respective orbitals. Thus s, p, d, f can have a maximum of , , and
electrons
respectively.
a Afbau Principle
According to the principle, Electrons are added progressively to the various orbitals in the
order of increasing
energy.
What does the word Afbau mean
Afbau is a German term which means building up or construction.
The energy of various orbitals increase in the order given below
s lt s lt p lt s lt p lt s lt d lt p lt s lt d lt p lt s lt f lt d lt p lt s lt f lt d lt p lt s lt ......
i A new electron enters the orbitals for which n l is minimum, e.g. if we consider d and s
orbitals, the
electron will first enter sorbitals in preference to d.
This is because the value of n l for sorbitals is less than that for dorbital
ii In case where n l values are the same, the new electron enters the orbitals for which n is
minimum, e.g.
in a choice between d and p for which n l values are same , , the electron will prefer to
go to the dorbital, since n is lower for this orbital.
b Paulis Exclusion Principle
It states that it is impossible for two electrons in a given atom to have same set of quantum
numbers.
Example
an,l,m,s/
n,l,m,s/
bn,l,m,s/
n,l,m,s.
n,l,m,s/
n,l,m,s/
n,l,m,s/
n,l,m,s/
c Hunds Rule of Maximum Multiplicity
According to this rule, electrons enter the orbitals e.g. s, p
x
,p
y
,p
z
... in the same sublevel in such a way as to
At omic St ruct ure
give maximum number of unpaired electrons. In other words it means that pairing begins with
the introduction of
the second electron in the sorbital, the fourth in p, etc.
What is the electronic configuration of Cu Z
s
s
p
s
p
d
s
Exceptional Electronic Configuration
Some atoms such as copper and chromium exhibit exceptional electronic configuration.
For example
CrZ has an electronic configuration
s
s
p
s
p
d
s
It is because of the extra stability associated with the halffilled and completely filled orbitals.
CONCEPT BUILDING EXAMPLES
Example .
Find out the energy of the electron in the first excited state in an Hatom.
Solution.
Energy of an electron in Hlike atom is given by
.
Z
EJ
n
.
where z is the number of protons and n is
number of shell in which electron is present.
For the first excited state,
n
.
EJ
.
.EJ
Example .
The K.E. of a moving electron is
J
Calculate its velocity and the wavelength.
Solution
..
K E mv
.
.
.
KEJ
velocity v ms
m kg
Now, wavelength
h
mv
.
.
..
Js
m
kg ms
Example .
If the electron of the hydrogen atom has been excited to a level corresponding to . electron
volts, what is the
wavelength of the line emitted when the atom returns to its ground state
Solution.
EEEA
E hv A
hc
E
A
. E eV A
..EJ
A
At omic St ruct ure
.
..
m
.
..
nm
.nm .
Example .
An electron jumps from fourth excited state to the ground stable in Li
ion and the energy released in the form
of photon is allowed to strike a metal x surface whose work function is
J
. What is the K.E. amp
velocity of the electron ejected.
Solution.
The amount of energy released is given by
.EJz
nn
A
.
where z is the atomic number of Hlike atom.
..
EJJ
A
.
Now, . . hv K E
...KEJ
.
mv J
.
.
.
v ms ms
Example .
Ionisation energy of hydrogen atom is . eV. What will be the ionisation energy of He
and Li
ions
Solution.
As we know ionisation energy for one electron system is given by
...
z
I P eV
n
For hydrogen atom, z , n , So, I.P . eV
For He
ion z and n .
So, I.P .
.
i.e., . . . I P eV For Li
, z amp n I.P.
.
i.e., . . I P eV .
Example .
A hydrogen like atom atomic number Z is in a higher excited state of quantum number n .
This excited atom
can make a transition to the first excited state by successively emitting two photons of
energies . eV and
eV respectively. Alternatively, the atom from the same excited state can make a transition to
the second excited
state by successively emitting two photons of energy . eV and . eV respectively. Determine
the values of
n and
z
ionisation energy of hydrogen atom . eV.
Solution.
Total energy liberated during transition of electron from n
th
shell to first excited state i.e.,
nd
shell
. . eV
. . erg
At omic St ruct ure
H
hc
R Z hc
n
.
..
H
R Z hc
n
.
... i
Similarly, total energy liberated during transition of electron from nth shell to second excited
state i.e.
rd
shell
. . .eV
. . erg
..
H
R Z hc
n
.
... ii
Dividing eq. i by eq. ii
n
On substituting the value of n in eqns. i or ii
Z
.
Example .
mol of He
ion excited. Spectral analysis showed the existence of ions in
rd
level, in
nd
level and
remaining in ground state. Ionisation energy of He
is . eV calculate total energy evolved when all the
ions return to ground state.
Solution.
of He
ions are already in ground state, hence energy emitted will be from the ions present in
rd
level and
nd
level.
E IP
nn
A
.
per ion or atom.
.
N
E
A
for
N
ions falling to ground state
.
N
eV
and
.
N
E
A
for
N
ions falling to ground state.
.
N
eV
Hence total energy
.
N
..
eV
...
J
.J
Example .
Estimate the difference in energy between the st and nd Bohr orbit for a hydrogen atom. At
what minimum
atomic number, would a transition from n to n energy level result in the emission of Xrays
with
.m
. Which hydrogen atom like species does this atomic number correspond to
Solution.
For hydrogen atom, the expression for energy difference between two energy level is,
H
E R hc
nn
A
At omic St ruct ure
So,
..
EEE
A
.J
For hydrogen like species, the same expression is
H
zR
nn
or,
H
zR
nn
So,
.
z
.
or, z
So, the species is He
because z .
Example .
What is the degeneracy of the level of the hydrogen atom that has the energy
aR
H
,bR
H
/
Solution.
/
nH
ERn
a,
H
ER
n
when
n
,
l
,
e
m
The level is degenerate.
b/
H
ER
n
when
n
, ,, l
when
l
,
e
m sorbital
when
l
,,,
e
m porbital
when
l
, , , ,, me dorbitals
This is states in all. The degeneracy is .
Example .
Calculate the uncertainty in the velocity of a ball of mass g if uncertainty in this position is
. h Js
Solution.
From Heisenberg uncertainty principle we know that
hhh
x p or xm v or v
mx
AAAAA
A
Substituting value
. , m g kg x m
A
.
..
v
A
. v ms
A.
Subjective Solved Examples
Exampe.
Calculate the waelength and wave number of the spectral line when an electron in Hatom
falls from higher
energy state
n
to a state
n
. Also determine the energy of a photon to ionise this atom by removing the
electron from nd Bohrs orbit. Compare it with the energy of photon required to ionise the
atom by removing
the electron from the ground state.
Solution.
At omic St ruct ure
e
n
n
Photon emitted
e
n
n
Photon absorbed
n
First calculate the energy E A between the To ionise the atom from
n
, the
Bohr orbits
n
and
n
using responsible transition will be
nn
.
.EZJ
nn
A
.
.
EJ
A
.
.
EJZ
A
.
.J
.J
To ionise the atom from ground state
Now this energy difference is the energy of the n , the transition is
.
photon emitted.
.
E
A
.
Photon
hc
E hv hcv
.J
..
hc
and
.vm
Example
A hydrogen atom in the ground state is hit by a photon by a photon exciting the electron to rd
excited state.
The electron then drops to nd Bohr orbit. What is the frequency of radiation emitted and
absorbed in the
process
Solution.
Energy is absorbed when electron moves from ground state n to rd excited state n .
e
n
Photon absorbed
n
n
n
First calcuate the energy difference btween
n
and
n
.
Use
.EZ
nn
A
.
.
At omic St ruct ure
Here,
, , Z n n Put
n and
n in the expression of
EA
, to get
.
EJ
A
.
.
EJ
A
.
.J
.J
This is the energy of the photon absorbed. This is the energy of the photon emitted.
Use
.
Photon
E hv J
to get Use
.
Photon
E hv J
. v Hz
. v Hz
Similarly, when electron jumps from
n
to
n
, energy is emitted and is given by the same relation.
Example
A hydrogen like ion, He Z
is exposed to electromagnetic waves of . . The excited electron gives
out induced radiations. Find the waelength of the indicated radiations, when electron
deexcites back to the
ground state. R cm
.
Solution
He
ion contains only one electron, so Bohrs From
n
, the electron can fall back to the
medhot is applicable here. It absorbs a photon of ground state in three possible ways
transitions
wavelength
.
. Assume the electron to , ,
be in ground state initially. Let it jumps to an Hence three possible radiations are emitted.
excited state n
. Find the wavelengths corresponding to these
v RZ
nn
.
transitions.
Substitute for
. . cm
, The wavelength
for transition,
will
be same i.e., . . Find
for
and
using the same relation.
, R Z for He
ion, . , .
n and find
n..
.n
.
n
n
n
n
Example
Hydrogen gas when subjected to photondissociation, yields one normal atom and one atom
possessing .
eV more energy than normal atom. The bond dissociation energy of hydrogen moleucle into
normal atoms is
kcals mol
. Compute the wave length of effective photon for photon dissociation of hydrogen molecule
in
the given case.
Solution
HHH
where
H
is normal
H
atom and
H
is excited Hatom. So the energy requird to dissociate
H in this
manner will be greater than the usual bond energy of
H molecule.
E absorbed dissociation energy of
H extra energy of excited atom.
Energy required to dissociated in normal manner
At omic St ruct ure
cal per mol given
.
./
J atom
The extra energy possessed by excited atom is . eV
...JJ
E absorbed
..J
.J
Now calculate the wavelength of photon corresponding to this energy.
.
Photon
hc
E
Example
An electron in the first excited state of Hatom absorbs a photon and is further excited. The de
Broglie wavelength
of the electron in this state is found to . . find the wavelength of photon absorbed by the
electron in .
Also find the longest and shortest wavelength emitted when this electron deexcites back to
ground state.
Solution
Note The energy state
n
is known as Ground State
The energy state
n
is known as First Excited State
The energy state
n
is known as Second excited
State and so on.
e
n
nn
Photon
The electron from
n
absorbs a photon and is further excited to a higher energy level let us say n .
The electron in this energy level n has a de Broglie wavelength .
e
ee
h
mv
and
.
n
Z
v ms
n
v
n
is the velocity of e
in n
th
Bohr orbit
.
h
v
mn
.
.
..
.
n
n
Now find the wavelength of the photon responsible for the excitation from n to n .
Using the relation
.EZ
nn
A
.
.J
n
,n
,Z
.
hc
E
A
.
The Longest wavelength emitted when this electron
.E
A
At omic St ruct ure
from
n
falls back to the ground state will corresponds
.
Photon
hc
EEJ
A
to the minimum energy transition.
.
The transition corresponding to minimum energy will Shortest wavelength
be
.
.
E
A
.
Note The transition corresponding to maximum energy
.J
will be
. Photon
hc
EE
A
. Energy diff Photon
hc
E E hv
A
.
Photon
Photon
E or E v
AA
Using the same relation
.EZ
nn
A
.
,,nnZ
Example
A single electorn orbits around a stationary nucleus of charge
Ze
, where
Z
is a constant and e is the
magnitude of electronic charge. It requires . eV to excite the electron from second Bohr orbit
to the third
Bohr. Find
a the value of
Z
b the energy required to excite the electron from n to n .
c the wavelength of radiation required to remove electron from nd Bohrs orbit to infinity
d the kinetic energy, potential energy and angular momentum of the electron in the first orbit.
e the ionisation energy of above one electron system in eV.
Solution.
Since the nucleus has a charge
Ze
, the
atomic number of the ion is
Z
.
a The transition is
n n by absorbing a photon of energy . eV .
. E eV A
Using the relation
. E Z eV
nn
A
.
..
ZZ
.
b The required transition is
n n by absorbing a photon of energy
EA
.
Find
EA
by using the relation
. E Z eV
nn
A
.
.
E eV
A
.
. E eV A
c The required transition is
n n by absorbing a photon of energy
EA
.
At omic St ruct ure
Find
EA
by using the relation
.
E E eV
AA
.
Find
of radiation corresponding to energy eV..
.
.
hc
E
.
d If energy of electron be E
n
, then KE E
n
and PE E
n
..
n
Z
E eV
n
KE eV eV
PE eV eV
Angular momentum
h
ln
.
.
l
.
.Js
e The ionisation energy IE is the energy required to remove the electron from ground state to
infinity. So
the required transition is
. The ionisation energy
. IE E Z eV
. IE eV
Example
With what velocity should an alpha particle travel towards the nucleus of a copper atom so as
to arrive at
a distance
m from the nucleus of the copper atom
Solution.
As particle appraoches towards the Cu nucleus, it decelerates due to repulsion from it and
finally its
velocity will become zero at point A which is the turning point. After that, particle will move in
the left
direction and accelerating
A
Vo
Cu Nucleus
e
r
oparticle
Vo
To arrive at a distance r
from the nucleus, the kineticc energy of alpha particle should be equal to the
electrostaic potential energy of it, i.e., KE EPE
N
Kq q
mv
r
m
mass of particle
. kg
v
velocity of particle
/KNmC
At omic St ruct ure
q
charge on particle
.C
r
Minimum distance of approach
Note ,
p
qemm
particle is He
ion or He Nucleus
q
N
charge in Cu nucleus Ze .
C
d distance from nucleus
m
N
Kq q
v
mr
Substituting the given values, we get,
./.vms
Note This is a simple cases where velocity of particle is directed towards the centre of the
Cunucleus.
A Cu
r
oparticle
Vo
Note When there is a difference between the velocity vector of particle and the Cutarget
nucleus, the
trajectory is more complicated.
Target oparticle
Example
Find the energy required to excite . litre of hydrogen atoms gas at . atm and K to the first
excited
state of atomic hydrogen. The energy requied for the dissociatio of HH bonds is kJ/mol. Also
calculate
the minimum frequency of a photon to break this bond.
Solution.
Let us, first find the number of moles of hydrogne atoms.
.
.
.
H
PV
n
RT
Thus the energy required to break . moles of H
HH bond . . . kJ
Now calculate the energy needed to excite the Hatoms to first excited state i.e., to
n
First excited state is
referred to
n
.
./
E J atom
A
.
. / J atom
No. o H atoms No. of H
molecules
...
At omic St ruct ure
The energy required to excited the given number of Hatom
. . . J kJ
So the total energy required
. . .kJ
Now the energy required to break to single
HH bond
.
.
Energy supplied by the photon
. . hv v
. v Hz
Example
Estimate the differnce in energy between st and nd Bohrs orbit for a Hatom. At what
minimum atomic
number Z, a transition from
n
to
n
energy level would result in the emission of radiatio with wavelength
.m
Which hydrogen atom like species this atomic number corresponds to How much ionisation
potential is needed to ionise this species
.Rm
Solution.
The difference in energy is given by
EA
./
E J atom
A
.
. . . J ergs eV
For a Hlike atom,
.m
.
.
EZJ
A
.
Photon
hc
E
Solve to get Z
Hence the Hlike atom is He
ion.
To ionise, He
ion, ionisation energy IE E
. . IE eV
The ionisation potential IP is the voltage difference required to generate this much energy.
. IE qV e IP eV
required . IP Volt
Example
A stationary He
ion emits a photon correspondings to the first line H
of Lyman series. The photon thus
emitted, strikes a Hatom in the ground state. Find the velocity of the photoelectrons ejected
out of the hydrogen
atom. The value of
..Rm
Solution.
The difference in energy E A will be equal to the energy of the photon emitted.
First line in Lyman series corresponds to the transition
.
./
E J atom
A
.
.J
At omic St ruct ure
The photon of this much energy strikes a Hatom in the ground state. Note that the ionisation
energy of Hatom
is
.J
. This will be the work function of Hatom. Using the Einsteins photoelectric equation
iee
KE E W m v E
i
Incident energy
i
e
e
EW
v
m
..
.
e
v
./
e
vms
We can also calculate the wavelength of electron ejected out
..m
.
.
..
e
ee
h
m
mv
Example
An electron in a hydrogen like species, makes a transition from nth Bohr orbit to next outer
Bohr
n
.
Find an approximate relation between the dependence of the frequency of the photon
absorbed as a function of
n . Assume n to be large value
n gtgt
.
Solution.
.
nn
energy difference
E hv Z J
nn
A
.
.
n
hv Z J
nn
.
.
Since
n gtgt
given
nnnn
.
n
hv Z J
n
vn
.
At omic St ruct ure
MIND MAP
. According to the quantum
theory, the radiant energy is
emitted by atoms amp molecules
in small discrete amounts
quantam rather than over a
continuous rante. The energy of
each quanta is given by E hv.
. According to Bohr model, the
angular mometum of an electron
is an integral multiple of
/h
.
Bohrs model is applicable single
electron species hydrogen like
species.
. The radius of an orbit is given
by
/ . r n h kZms
The
velocity of an electron in an orbit
is given by
/ v nh mr
and the
energy of an electron in an orbit
is given by
/ E pk Z ms n h .
. In phot oelectric effect,
electrons are elected from the
surface of certain met al
exposed to light of at least a
certain minimum frequency.
. . hv hv K E
. In bohr model, an electron
emits a photon when it drops
from a higher energy state to a
lower energy state.
. Four quant um numbers
characterise each electron in an
atom. The principal quantum
numbern indentifies the main
energy level, t he angular
quantum number l indicates
shapes of orbital, the magnetic
orientation of orbital in space
and the spin quant um
numbers indicat es t he
direction of the electrons spin
on its axis.
. The emission spectra of
hydrogen is obtained when
electron from an ecited state is
deexcited to the ground state.
The release of specific amounts
of energy in the form of photons
accounts for the lines in the
hydrogen spectrum. v of each
line in the spectrum can be
given by
/ / / Ryz n n
. An orbital may be defined as
a region in space around the
nucleus where the probability
of finding the electron is
maximum.
. De Broglie exended
Einsteins wave particle
descrition of light to all matters
in motion. The wavelength of
a moving particle of mass m
and velocity v is given by de
Broglie equation,
/ . h mv
ATOMIC
STRUCTURE