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CH-2 STRUCTURE OF ATOM

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XI - Some Basic Concept
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XI - Some Basic Concept
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Any substance that has mass and occupies
space is called “Matter”. It is made up of
small particles called “atoms” that cannot
be further divided.
This idea was first proposed by the Greek philosopher
‘Democritus’ back in 460 B.C.
But due to lack of scientific evidence, Democritus’s ideas were considered mere
speculations and ignored for 2000 years until John Dalton proposed the Atomic
Theory of Matter in 1808.
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Dalton proposed that atom is an ultimate particle of an
atom.
Dalton’s atomic theory was able to explain laws of
chemical combinations.
However, the Dalton’s theory could not explain the results
of experiments conducted by various scientists in 19th and
20th century.
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Dalton could not explain the result of:
1932
1909
1897
1886
1879
1830
1808
• JAMES CHADWICK: DISCOVERY OF NEUTRON
• ROBERT MILLIKAN: CHARGE ON AN ELECTRON
• J.J. THOMSON: PROPERTIES OF CATHODE RAYS
• EUGEN GOLDSTEIN: DISCOVERY OF ANODE RAYS
• WILLIAM CROOKES: CONDUCTION OF ELECTRICITY
THROUGH GASES
• MICHAEL FARADAY: ELECTRIC NATURE OF MATTER
• JOHN DALTON: ATOMIC THEORY OF MATTER
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Michael Faraday found that when electricity is passed through an electrolyte
(solution that conducts electricity) chemical reactions resulting in deposition
of matter at the electrodes take place.
This experiment proves that electricity
consists of charged particles which in
turn indicated the electric nature of
matter.
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The structure of an atom was demystified only when William Crookes studied
electrical discharge through gases.
1932
1909
1897
1886
1879
1830
1808
• JAMES CHADWICK: DISCOVERY OF NEUTRON
• ROBERT MILLIKAN: CHARGE ON AN ELECTRON
• J.J. THOMSON: PROPERTIES OF CATHODE RAYS
• EUGEN GOLDSTEIN: DISCOVERY OF ANODE RAYS
• WILLIAM CROOKES: CONDUCTION OF ELECTRICITY
THROUGH GASES
• MICHAEL FARADAY: ELECTRIC NATURE OF MATTER
• JOHN DALTON: ATOMIC THEORY OF MATTER
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APPARATUS
The apparatus used for his experiment consisted of a glass discharge tube now called
Cathode Ray Tube or Crooke’s Tube and a high voltage source of 10,000 Volts. The
discharge tube was sealed at both the ends and was fitted with a thin piece of metal called
‘Electrode’ at each end.
It had a side tube fitted with a stop cork that
connected with a vacuum pump to control the
pressure of the gas or the air contained inside
the tube.
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OBSERVATION
Crooke observed that at a low pressure of 0.01 atmosphere when high
voltage of 10,000 Volts is applied across the electrodes of the discharge tube,
current starts flowing inside the tube.
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To further study the behaviour of this
current, he punctured the anode inside
the Cathode Ray Tube and coated the
glass tube behind the anode with the
fluorescent material such as zinc
sulphide.
And then he repeated the experiment and this time he noticed a bright spot
behind the anode.
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This confirmed two things:
• The current consisted of invisible particles some of which could pass
through perforated anode
• These rays were emitted out of the cathode and moved towards the anode.
THESE RAYS WERE NAMED CATHODE RAYS
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Some other facts regarding cathode rays were discovered by J.J. Thomson and other
scientists through series of experiments
1932
1909
1897
1886
1879
1830
1808
• JAMES CHADWICK: DISCOVERY OF NEUTRON
• ROBERT MILLIKAN: CHARGE ON AN ELECTRON
• J.J. THOMSON: PROPERTIES OF CATHODE RAYS
• EUGEN GOLDSTEIN: DISCOVERY OF ANODE RAYS
• WILLIAM CROOKES: CONDUCTION OF ELECTRICITY
THROUGH GASES
• MICHAEL FARADAY: ELECTRIC NATURE OF MATTER
• JOHN DALTON: ATOMIC THEORY OF MATTER
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EXPERIMENT NO. 1
If the pin wheel is placed in the path of cathode rays, the ray caused the pin
wheel to rotate.
This implied that cathode rays are made up of material particles that produce
a mechanical effect.
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EXPERIMENT NO. 2
If a metal foil is placed in the path of cathode rays, it becomes hot.
This implied that cathode ray produced a heating effect.
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EXPERIMENT NO. 3
If a solid object is placed in the path of cathode ray, it produced a sharp
shadow of the object.
Therefore, he concluded that cathode ray travel in a straight line
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EXPERIMENT NO. 4
When he applied the electric field to the set up, the cathode ray deflected
towards the positive plate of the electric field.
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EXPERIMENT NO. 5
He observed the same result on applying the magnetic field.
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Therefore, he concluded that the cathode ray consisted of negatively charged
particles, which he named electrons.
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EXPERIMENT NO. 6
Through his experiments, Thomson also concluded that the properties of
cathode ray do not depend upon
• the material of electrodes and
• the nature of the gas present in the
cathode ray tube.
This led him to the conclusion that the electrons are the basic constituents of
all atoms.
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EXPERIMENT NO. 7
Thomson then tried to calculate the charge to mass ratio of electrons by
applying electric and magnetic field perpendicular to each other as well as to
the path of electrons flowing in a cathode ray tube.
Electric field
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Magnetic Field
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He observed that the amount of deviation of the electrons from their path
depends on three main factors:
• The magnitude of the negative charge on
the particle: the more the negative charge,
the greater the deflection
• The mass of the particle: the lighter the
particle, the greater the deflection
• The strength of the electrical and
magnetic field: the stronger the field, the
higher the deflection
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When only electric field is applied, the electrons deviate from their path and
hit the cathode ray tube at point A.
Similarly when only magnetic field is applied, electrons strike the cathode
ray tube at point C.
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By controlling the deviation of electrons, by varying the strength of electric
and magnetic field and accurately measuring the amount of resultant
deflection, he calculated the charge to mass ratio of an electron which equals
1.758820 × 1011 C Kg-1
c/m = 1.758820 × 1011 C Kg-1
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Modern Cathode Ray Tubes
Television
Computer Monitor
Cathode ray tubes pass electricity through a gas that is contained at a very
low pressure.
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However, till then the charge and mass of an electron remain unknown. Then Robert
Milliken devised a method to calculate these.
1932
1909
1897
1886
1879
1830
1808
• JAMES CHADWICK: DISCOVERY OF NEUTRON
• ROBERT MILLIKAN: CHARGE ON AN ELECTRON
• J.J. THOMSON: PROPERTIES OF CATHODE RAYS
• EUGEN GOLDSTEIN: DISCOVERY OF ANODE RAYS
• WILLIAM CROOKES: CONDUCTION OF ELECTRICITY
THROUGH GASES
• MICHAEL FARADAY: ELECTRIC NATURE OF MATTER
• JOHN DALTON: ATOMIC THEORY OF MATTER
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APPARATUS
• Transparent electrical condenser with one
metal plate at the top and the bottom of the
chamber.
• Plates connected to a battery such that the
upper plate is positively charged and the
lower plate is negatively charged.
• An atomiser to spray oil into the condenser
through a hole in the upper metal plate.
• A telescope to view the movement of oil droplets
• Source of X-Ray to ionise the air inside the electrical condenser.
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STEP 1
Using the atomiser, Milliken sprayed oil droplets into the electrical condenser.
As the droplets fell through A the upper plate hole he measured the rate of fall
and used it to calculate their mass.
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STEP 2
• He then ionised the air inside the condenser by passing the beam of X-Rays
through it.
• The X-Rays displaced electrons from the air molecules which negatively
charged the oil droplets.
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On applying voltage to the upper part of the plate, charged oil droplets got
attracted towards it against all gravitational and electrostatic forces.
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Milliken then varied the voltage to strike a balance between the acting forces
and to make oil drops stationery. He then calculated the charge on the droplet
from the mass of an oil droplet and the charge on the plate.
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He found that the magnitude of electrical charge ‘Q’ on the droplets is always
an integral multiple of the electrical charge ‘e’ i.e. Q = ne
Knowing the values of Q and n, Milliken calculated charge on an electron
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The mass of an electron me was then determined by combining the results of
Milliken’s Oil Drop Experiment and Thomson’s value of e/m ratio
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Just like the discovery of cathode rays led to the discovery of negatively
charged particles, electrons, the discovery of anode rays led to the discovery
of positively charged particles called protons.
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Anode rays, also called Canal Rays were discovered by
1932
1909
1897
1886
1879
1830
1808
• JAMES CHADWICK: DISCOVERY OF NEUTRON
• ROBERT MILLIKAN: CHARGE ON AN ELECTRON
• J.J. THOMSON: PROPERTIES OF CATHODE RAYS
• EUGEN GOLDSTEIN: DISCOVERY OF ANODE RAYS
• WILLIAM CROOKES: CONDUCTION OF ELECTRICITY
THROUGH GASES
• MICHAEL FARADAY: ELECTRIC NATURE OF MATTER
• JOHN DALTON: ATOMIC THEORY OF MATTER
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When E. Goldstein repeated the cathode ray experiment conducted by
Crooke’s with the perforated cathode instead of a perforated anode. He
observed that the discharge tube containing perforated cathode also emit a
glow at the cathode end.
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Goldstein concluded that in addition to the already known cathode rays, there
is another ray that travels in a opposite direction. Since these rays passed
through the holes or canals in the cathode, he called them canal rays.
They consisted of positive ions whose
charge and mass values depended on
the gas inside the discharge tube.
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Hydrogen gas produced the smallest and the lightest positive ions with a
magnitude of charge same as that of the electron but with a positive charge
and mass that was found to the similar to that of a hydrogen atom.
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The mass of the positive particle originating from other gases in the discharge
tube was found to be whole number multiples of the mass of those originating
from hydrogen gas.
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Therefore, the positive particles from hydrogen gas carrying one positive
charge and mass equivalent to a hydrogen atom were taken as the fundamental
particle of any atom and were named as protons
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By 1913 after performing a considerable amount of research on the charge
and mass of negative and positive sub-atomic particles, scientists tried
calculating the atomic mass. It was determined that the atomic number of an
element is equal to the number of protons present in its nucleus.
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It was thought that since each proton has one unit mass, on the atomic
scale the mass of an atom must be the number of protons present in the
nucleus as electrons have negligible mass. However, it was discovered
that the mass of all protons in an atom put together is much less than the
actual mass of an atom.
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It was thus concluded that the excess mass was due to the presence of another
particle present in an atom that has considerable mass but no charge
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Presence of heavy neutral particles through a series of scattering experiments
1932
1909
1897
1886
1879
1830
1808
• JAMES CHADWICK: DISCOVERY OF NEUTRON
• ROBERT MILLIKAN: CHARGE ON AN ELECTRON
• J.J. THOMSON: PROPERTIES OF CATHODE RAYS
• EUGEN GOLDSTEIN: DISCOVERY OF ANODE RAYS
• WILLIAM CROOKES: CONDUCTION OF ELECTRICITY
THROUGH GASES
• MICHAEL FARADAY: ELECTRIC NATURE OF MATTER
• JOHN DALTON: ATOMIC THEORY OF MATTER
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In one of his experiments, on bombarding the Beryllium plate with alpha
particles, he observed the emission of neutral particle whose mass was
equivalent to that of a proton.
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DISCOVERY OF NEUTRON
He named this particle as “Neutron”
After all the sub-atomic particles were discovered, it was concluded that
an atom is made up of protons or positively charged particles, electrons
or negatively charged particles and neutrons that are neutral particles.
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SUMARRY OF PROPERTIES OF SUB-ATOMIC PARTICLES
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STRUCTURE OF ATOM
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Through various experiments it was established that an atom consists of subatomic particles
such as protons, electrons and neutrons.
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Now let us look at the arrangement of these subatomic particles in an atom different
models were proposed to explain the distribution of subatomic particles in an atom.
Of all these models, the models proposed by J.J. Thompson and Ernest Rutherford out of
great significance as they paved the way for modern structure of an atom.
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Thompson's Atomic Model
J.J. Thompson in 1898 proposed the atomic model called
Thompson's atomic model soon after the discovery of Electrons.
According to this model an atom is a sphere of radius 10-10 m with
uniform distribution of mass and positive charge with negatively
charged particles called electrons embedded in it.
In this model the negatively charged particles that are embedded in
evenly spread positive charge can be visualized as plums or raisins
or seams that are embedded in the pudding or watermelon therefore
this model is also called plum pudding/raisin pudding or
watermelon model.
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This model successfully explained the overall neutrality of the atom.
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DRAWBACKS
✓ It could not explain the results of the scattering
experiment conducted by Ernest Rutherford.
✓ Also it couldn't explain the stability of an atom that is
how the positively charged particles are shielded from
the negatively charged particles without getting
neutralized.
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In order to validate Thompson's atomic model in 1911 Rutherford and his students Hans
Geiger and Ernest Marsden conducted an experiment called the Gold foil experiment or
Alpha particle scattering experiment.
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Before we proceed to alpha particle scattering experiment, let us acquaint ourselves with the
term alpha particle, its source, other particles emitted from the same source and the few
characteristics.
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We know that certain elements emit radiation on their own and this phenomenon is called
radioactivity and the elements are radioactive elements.
The Alpha rays are emitted by these radioactive elements along with beta and gamma rays.
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ALPHA RAYS
Alpha rays consist of high-energy
particles carrying two units of positive
charge and four units of atomic mass
hence these particles are di positive
helium nuclei.
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BETA RAYS
The beta rays are negatively charged
particles similar to electrons.
GAMMA RAYS
The gamma rays are high energy
radiations like x-rays. Unlike alpha and
beta rays gamma rays do not contain
particles and are neutral in nature.
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The alpha particles have the least penetrating power among the three.
The penetrating power of beta particles is nearly 100 times and the penetrating power of
gamma rays is 1,000 times more than that of the alpha particles.
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Alpha particle scattering experiment
In this experiment a stream of high-energy
alpha particles from a radioactive source
was bombarded on a very thin gold foil.
The thin gold foil had a circular
fluorescent zinc sulphide screen around
it.
Whenever alpha particles struck the zinc sulphide screen a tiny flash of light was produced
at that point.
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If Thompson's model is correct then all the alpha particles should pass through the gold foil
almost undeflected. This is because the entire mass and positive charge is uniformly spread
throughout an atom.
However the observations are quite contrary to these expectations.
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OBSERVATIONS
The following were the unexpected observations
made in this experiment:
✓ Most of the alpha particles passed through the
gold foil undeflected.
✓ A small fraction of the alpha particles were
deflected by small angles.
✓ A few alpha particles were deflected by large
angles
✓ A very little amount of alpha particles, one in
20,000 were bounced back to the source i.e.
nearly by 180 degrees.
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INTERPRETATION
Based on these observations, Rutherford concluded:
✓ As most of the alpha particles passed undeflected, most of the
space inside the atom is empty.
✓ Only a small number of deflected particles suggested that the
positive charge of the atom is not spread throughout the atom
as Thompson had presumed rather it is concentrated at the
center in a very small volume.
✓ The alpha particles that are deflected by larger angles and by
180 degrees indicated that the atom has a dense positive
charge with entire mass concentrated at the center. As the
alpha particles approach much nearer to the massive
positively charged center they get deflected by larger angles
and as they directly hit the massive positively charged center
they get deflected nearly by 180 degrees. It is due to the
increase in enormous repulsive forces from the massive
positively charged center.
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Rutherford named the positively-charged center present in an atom as nucleus.
Calculation by Rutherford showed that
the volume of the nucleus is negligible
as compared to the total volume of the
atom.
For an atom of radius 10-10 meter, the radius of
the nucleus is about 10-15 meter.
That means if a cricket ball represents
a nucleus, the radius of the atom would
be about five kilometres.
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Based on the observations and conclusions of the gold foil experiment,
Rutherford proposed a new atomic model called Rutherford's nuclear
model of atom immediately after the discovery of protons.
According to the Rutherford’s nuclear model of atom, atom consists of a
tiny dense positively charged center called Nucleus.
Entire positive charge and most of the atomic mass of an atom is
concentrated in the nucleus.
Electrons are found outside the nucleus. They revolve around the nucleus
with high velocities and circular parts so as to counterbalance the
electrostatic forces of attraction between the protons and electrons.
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Electrostatic forces of attraction hold the electrons and the nucleus together in an atom. As
electrons move in circular paths around the nucleus just like planets revolve around the Sun
in the solar system, Rutherford's atomic model is also called Planetary model.
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Rutherford's model could account for the presence of nucleus and electrons outside the
nucleus. However it failed to explain the stability of the atom.
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According to the Maxwell's theory of
electromagnetic radiation, a charged particle
in circular motion, emits energy continuously.
Hence an electron revolving round nucleus
also loses energy.
As a consequence of this, the electron
follows the spiral path towards the
nucleus and this ultimately results in
collapse of an atom
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STRUCTURE OF ATOM
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ISOTOPES AND ISOBARS
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Atoms of the same element may have the same number of protons that is the atomic number
but may have different numbers of neutrons. In other words the atoms have the same atomic
number but different mass numbers,
such atoms are called isotopes.
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Isotopes of the same element occupy the same place in the
periodic table because they have the same atomic number.
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EXAMPLES OF ISOTOPES
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The chemical properties of an element depend on the number of electrons present and their
configuration within an atom and not on the number of neutrons. Hence as the isotopes have
the same number of electrons they exhibit similar chemical properties. However as the mass
of isotopes differ due to the different numbers of neutrons present in an atom their physical
properties differ from each other
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Atoms of different elements having different atomic numbers but the same mass numbers are
called Isobars. Isobars have different chemical properties because they have different atomic
Numbers.
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EXAMPLES OF ISOBARS
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ISOTONES
Isotone, any of two or more species of atoms or nuclei that have the same number of
neutrons.
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Questions
Que1
Which of the following conclusions could not be derived from
Rutherford’s α-particle scattering experiment?
(a) Most of the space in the
atom is empty
(c) Electrons move in a
circular path of fixed
energy called orbits
Que2
(b) The radius of the atom is about
10–10 m while that of nucleus is
10–15
(d) Electrons and the nucleus are
held together by electrostatic
forces of attraction.
Which one of the following pairs constitutes isotones?
(a)
6C
13
and 6C14
(b)
6C
13
and 7N14
(c)
14
7N
and 9C19
(d)
14
7N
and 7N15
Ques 1 (c)
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Ques 2 (b)
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NCERT Exercise:
2.1 (i) Calculate the number of electrons which will together weigh
one gram.
(ii) Calculate the mass and charge of one mole of electrons.
Ans. (i)
(ii)
2.2
1.098 × 1027
mass = 5.486 x 10-7 kg
charge = 9.65 × 104 C
(i) Calculate the total number of electrons present in one mole
of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons
in 7 mg of 14C. (Assume that mass of a neutron = 1.675 × 10–27 kg).
(i) 6.022 × 1024
(ii) (a) 2.4092 × 1021
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(b) 4.0352 × 10–6 kg
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NCERT Exercise:
2.4
Write the complete symbol for the atom with the given
atomic number (Z) and atomic mass (A)
(i)
Z = 17 , A = 35
(ii)
Z = 92 , A = 233
(iii)
Z= 4, A= 9
Ans.
(i)
2.22
Which of the following are isoelectronic species i.e., those
having the same number of electrons?
Na+, K+, Mg2+, Ca2+, S2–, Ar
Ans.
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(ii)
(i) Na+ , Mg2+
(iii)
(ii) K+ , Ca2+ ,S2–, Ar
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NCERT Exercise:
2.40
In Rutherford’s experiment, generally the thin foil of heavy
atoms, like gold, platinum etc. have been used to be
bombarded by the α-particles. If the thin foil of light atoms
like aluminium etc. is used, what difference would be
observed from the above results ?
A thin foil of lighter atoms will not give the same results as given with
the foil of heavier atoms. Lighter atoms would be able to carry very little
positive charge. Hence, they will not cause enough deflection of αparticles (positively charged).
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Limitation of Rutherford’ s model:
Charged
Particle
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Accelerated
motion
Radiate/
loose
energy
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DRAWBACKS OF RUTHERFORD MODEL
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PREDICTED
OBSERVED
As the electron is emitting the energy continuously the spectra of such an atom also should be continuous but the
spectra of elements indicate that atoms emit energy discontinuously and hence the Rutherford atomic model
couldn't explain the observed spectra of atoms.
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These drawbacks inspired several scientists to revisit the historical results about the
behaviour of electromagnetic radiation and its interaction with matter in order to
improve the Rutherford’s atomic model
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The improvised outcome of the Rutherford’s atomic model is the Bohr's atomic model.
Bohr's theory could successfully explain the stability of an atom as well as the discontinuous
spectra for hydrogen atom to understand the Bohr's atomic model.
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Thank You
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Atomic Structure
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Electromagnetic Waves
Energy can be transmitted through space by electromagnetic
radiations. These electromagnetic radiations are the best describe in
the terminology of waves.
Let's now understand the general characteristics of waves an object
produces a series of waves, when it vibrates continuously.
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However the vibratory motion of charged particles like electrons
produce a wave train of oscillating electrical field and magnetic field,
such waves which are produced due to the periodic motion of charged
particles such as electrons are called electromagnetic waves.
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Electromagnetic Wave Theory
Electromagnetic Wave Theory/ classical theory/ Maxwell’s
theory/ Wave nature of light
• Given by James Clark Maxwell
(1870)
• Light is a wave;
electromagnetic wave (EMW)
• When a charged particle,
undergoes periodic motion, it
produces electromagnetic
waves.
• EMW has alternating electric
and magnetic fields
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Electromagnetic Wave Theory
Features:
1) EMW consist of electric and
magnetic
fields
oscillating
perpendicular to each other and
both are perpendicular to the
direction of propagation of
radiation.
2) EMW do not require any
medium for propagation.
3) All EMW travel with speed of light (i.e. 3.0 × 108 m sec–1).
4) There are many types of electromagnetic waves (spectrum).
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Electromagnetic Wave Theory
1) EMW consist of electric and magnetic fields oscillating perpendicular to
each other and both are perpendicular to the direction of propagation of
radiation.
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Characteristics of EMW(5):
Wavelength
wave
Frequency
Velocity/speed
Amplitude
Wave number
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Characteristics of EMW:
Wave length (λ):
• It is the distance between two
neighbouring crests or troughs of the
wave.
• It is denoted by Greek letter Lambda (λ)
and is measured in Angstrom (Å) or
nanometre (nm).
• 10-10 m = 1 Angstrom or 1 m = 1010 Å
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1 centi meter
10-2 m
1 Milli-meters
10-3 m
1 micro-meter
10-6 m
1 nano-meter
10-9 m
1 Angstrom
10-10 m
Picometer
10-12 m
Fermi
10-15 m
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Characteristics of EMW:
Frequency (ν):
Frequency =
1
Time period
• It is defined as the number of waves which pass through a given
point in one second.
• It is denoted by Greek letter nu (ν) and is expressed in units of
cycles per second (cps) or Hertz (Hz).
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Characteristics of EMW:
Velocity/speed:
• The distance travelled by a wave in one second is called velocity
of the wave. It is denoted by letter c.
• All EMW travel with speed of light (i.e. 3.0 × 108 m sec–1).
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Characteristics of EMW:
Amplitude (a):
It represent the height of the crest or depth of the trough of a wave.
It is denoted by the letter ‘a’ and determines the intensity or
brightness of radiation.
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Characteristics of EMW:
Wavenumber:
It is the number of waves per unit distance.
It is denoted by the letter ν and determines the intensity or
brightness of radiation.
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Formulae from Wave theory:
Relationship between velocity, wave length and frequency of a wave
Frequency(ν) ∝
1
Wavelength (λ)
Frequency(ν) = Velocity of light (c)
Wavelength (λ)
Wavenumber =
1
Wavelength (λ)
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TIME PERIOD
A time period (denoted by 'T' ) is the time taken for one complete cycle of
vibration to pass a given point. As the frequency of a wave increases, the time
period of the wave decreases. The unit for time period is 'seconds'.
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NCERT Exercise:
2.5
Yellow light emitted from a sodium lamp has a wavelength (λ)
of 580 nm. Calculate the frequency (ν) and wavenumber
of the yellow light.
Frequency = 5.17 x 1014 Hz
Wave number = 1.72 x 106 m-1
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NCERT Exercise:
2.7
Calculate the wavelength, frequency and wavenumber of a
light wave whose period is 2.0 × 10–10 s.
Frequency =
1
Time period
Wavelength = 6 x 10-2 m, Frequency = 5 x 109 Hz
Wave number = 16.66 m-1
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Electromagnetic spectrum:
The complete range of electromagnetic waves is called electromagnetic
spectrum.
Frequency(ν) ∝
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1/Wavelength (λ)
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Electromagnetic spectrum:
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Limitations of Electromagnetic wave theory :
Electromagnetic wave theory successfully explained
the
properties of light such as interference and diffraction but it
could not explain
1. The phenomenon of black body radiations.
2. Photoelectric Effect.
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Atomic Structure
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The wave nature of electromagnetic radiation successfully explains
the phenomena of Interference and Diffraction.
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Limitations of Electromagnetic wave theory :
But the phenomenon of blackbody radiation, photoelectric effect and
the spectra of atoms were not explained.
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These things could be explained only if the radiation is assumed to be
a stream of photons i.e. by considering particle nature.
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First let us understand the phenomena of the blackbody radiation
which was given by Max Planck in 1900.
In general when radiation strikes any surface of
the body part of it is reflected part of it is
absorbed and part of it is transmitted.
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For example: when an iron rod is heated at first it turns red, as the
temperature is further increased it becomes yellow then white
and finally turns blue.
This actually shows that
the radiation is emitted in
the order of increasing
frequency.
Red light being of lower frequency and blue light being of higher
frequency of visible region.
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On the contrary an ideal blackbody is
a perfect absorber and emitter of
radiation.
A blackened metallic surface or a hollow sphere
blackened insight with a small opening behaves
almost as a blackbody any type of radiation that
enters this small opening gets reflected multiple
times within the sphere until all the
energy is absorbed.
Thus, a blackbody is not only a perfect
absorber of radiant energy but also a perfect
radiator.
Black body radiates a maximum amount of
energy at a given temperature.
Note that it radiates the same amount of
energy it has absorbed.
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From the graph shown here
it is visible that at a given temperature,
the intensity of radiation emitted
increases with decrease in wavelength,
reaches a maximum value at a particular
wavelength and then starts decreasing
with further decrease in wavelength.
We have seen that as a given temperature, wavelength at which the maximum
intensity of radiation is emitted but the position of this maximum intensity shifts
towards higher wavelengths with decrease in temperature.
These observations could not be explained by the
Electromagnetic wave theory of light.
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To account for these experimental observations, Max Planck
in 1900 gave Planck’s Quantum Theory.
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He proposed that radiant energy is emitted or absorbed discontinuously in the
form of discrete packets or bundles of energy and not continuously.
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Planck called these discrete quantities or packets as quanta. Quantum is the
smallest quantity of energy absorbed or emitted in the form of electromagnetic
radiation.
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This means that the light radiations absorbed or emitted by atoms all molecules
comprise a stream of quanta but not continuous waves as shown.
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CHAPTER 2 [ PART-I ]
Structure Of Atom
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Photoelectric Effect :
Photoelectric effect, may be defined as the phenomenon of instant
ejection of electrons from the surface of metal when light of particular
frequency/wavelength strikes it.
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Heinrich Rudolf Hertz studied two parameters closely:
1. Frequency of Light : related to colour of light
2. Intensity of Light : related to brightness
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Threshold frequency is defined as the minimum frequency of light which
causes electrons to be emitted from a metal surface. If no electrons are ejected,
this means that the frequency of the light is less than the threshold frequency.
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EFFECT OF INTENSITY
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Properties of K.E. of photoelectrons
K.E. Of
Photoelectrons
❖ K.E. of photoelectrons is independent of the intensity of
radiations.
K.E. Constant
Intensity of Incident
Radiation (with
constant Frequency)
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Properties of K.E. of photoelectrons
K.E. Of
Photoelectrons
❖ Number of ejected electrons ∝ intensity of incident light.
❖ K.E. of photoelectrons ∝ frequency of incident light.
Slope = h
ν0
Frequency of
Absorbed Photons
❖ This effect supports the particle nature of light as only the light of
suitable frequency can bring about the ejection of photoelectrons.
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SUMMARY
Minimum frequency for photoemission = ʋ0 = Threshold Frequency
Frequency ↑ → KE ↑ for photoelectron
Intensity ↑ → Photocurrent ↑ (no. of electrons ejected ↑ ) → KE remains same
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These observations could not be explained using the laws of classical physics.
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EINSTEIN’S EXPLANATION
Einstein explained these observations of the photoelectric effect for which he received the
Nobel Prize in 1921 he has taken into account the Planck's quantum theory of
electromagnetic radiation for explaining the photoelectric effect.
Note that the quanta were called photons by Einstein
Light consists of
photons
Each photon can eject only
one electron from the metal
Each photon has an energy
equal to hʋ
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GOVERNING EQUATION
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Questions
Que1 In photoelectric effect, the kinetic energy of the photoelectrons
increases linearly with the
Frequency of incident
(b)
(a) Wavelength of incident
light
light
(c)
Velocity of incident
(d) Atomic mass of the
element
light
(b)
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Electrons with a kinetic energy of 6.023 x 104 J/mol are evolved from the surface
of a metal, when exposed to a radiation of wavelength of 600 nm (photoelectric
effect). The minimum amount of energy required to remove an electron from the
metal atom is :
a) 2.3125 x 10-19 J
b) 3 x 10-19 J
c) 6.02 x 10-19 J
d) 6.62 x 10-34 J
Logic:
When a radiation is passed on to the surface of a metal, some amount of energy is used in
overcoming the attraction force on the electron and knocks it out from the atom. The
remaining part is converted to kinetic energy.
Therefore:
Eradiation = Φ + KE
Where:
Φ = work function or ionization energy required to remove the electron from the atom.
KE = kinetic energy
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Calculation of KE for one electron:
In the question, KE for one mole of electrons is given. However, in the options, the energy
values are given for one electron. Hence we have to find the KE value for one electron.
From the data:
KE for 6.023 x 1023 electrons (one mole) = 6.023 x 104 J
KE for one electron = 6.023 x 104 / 6.023 x 1023 = 1 x 10-19 J
Calculation of Eradiation:
Eradiation = hc/λ = 6.626 x 10-34 J s x 3.0 x 108 m s-1/600 x 10-9 m = 3.313 x 10-19 J
Solution:
Φ = Eradiation - KE = (3.313 x 10-19 J) - (1 x 10-19 J) = 2.313 x 10-19 J
Conclusion:
The correct option is "a".
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The work function (Φ) of some metals is listed below. The number of metals which
will show photoelectric effect when light of 300nm wavelength falls on the metal is:
Metal Li
Φ (in
2.4
eV)
Na
K
Mg
Cu
Ag
Fe
Pt
W
2.3
2.2
3.7
4.8
4.3
4.7
6.3
4.75
Logic & solution:
The energy of radiation must be equal to or greater than the work functions of metals to show
photoelectric effect.
We need to convert wavelength of radiation into energy expressed in eV units.
Eradiation = hc/λ = 6.626 x 10-34 J s x 3.0 x 108 m s-1/300 x 10-9 m = 6.626 x 10-19 J
Now convert this value into eV.
We know that:
1 J = 6.24 × 1018 eV
Therefore:
6.626 x 10-19 J = 6.626 x 10-19 x 6.24 × 1018 eV = 4.134 eV
Conclusion:
Since the work functions of only Li, Na, K and Mg fall below 4.134 eV, only these metals can show
photoelectric effect upon exposure of radiation of 300nm wavelength.
The number of metals that can show photoelectric effect = 4.
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NCERT Exercise:
2.6
Find energy of each of the photons which
(i)
correspond to light of frequency 3 × 1015 Hz.
(ii)
have wavelength of 0.50 Å
(i) E = 1.988 × 10–18 J
(ii) E = 3.98 × 10–15 J
2.8
What is the number of photons of light with a wavelength
of 4000 pm that provide 1J of energy?
2.012 × 1016photons
2.10
Electromagnetic radiation of wavelength 242 nm is just
sufficient to ionize the sodium atom. Calculate the ionization
energy of sodium in kJ mol–1.
1
= 494.5 kJ mol–1
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NCERT Exercise:
2.11
A 25 watt bulb emits monochromatic yellow light of
wavelength of 0.57µm. Calculate the rate of emission of
quanta per second.
(Given)Power of bulb, P = 25 Watt = 25 Js–1
We know, Energy of one photon, E = hν λ
Substituting the values in the given expression of E:
E = 34.87 × 10–20 J
Rate of emission of quanta per second is given by R = P/ E, Where R is the rate of
emission, P is the power & E is the energy
Substituting the values in the equation we get
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NCERT Exercise:
2.46
Nitrogen laser produces a radiation at a wavelength of 337.1
nm. If the number of photons emitted is 5.6 × 1024, calculate the
power of this laser.
3.33 × 106 J
2.47
Neon gas is generally used in the sign boards. If it emits
strongly at 616 nm, calculate (a) the frequency of emission, (b)
distance travelled by this radiation in 30 s (c) energy of
quantum and (d) number of quanta present if it produces 2 J of
energy.
(a) 4.87 × 1014 s–
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(b) 9.0 × 109 m
(c) 32.27 ×10–20 J
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147
NCERT Exercise:
2.48
In astronomical observations, signals observed from the distant
stars are generally weak. If the photon detector receives a total
of 3.15 × 10–18 J from the radiations of 600 nm, calculate the
number of photons received by the detector.
9.51 ≈ 10
2.49
Lifetimes of the molecules in the excited states are often
measured by using pulsed radiation source of duration
nearly in the Nano second range. If the radiation source has
the duration of 2 ns and the number of photons emitted
during the pulse source is 2.5 × 1015, calculate the energy of
the source.
8.282 × 10–10 J
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