12th physics - atoms molecules and nuclei for jee main 2014
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12th Physics - Atoms Molecules and Nuclei for JEE Main 2014TRANSCRIPT
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Atoms Molecules and Nuclei
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J. J. Thomson
1. Thomson’s atomic model :-
His model was based on two facts that
i. electrons are the constituents of all the atoms
and
ii. since atom as a whole electrically neutral, the
positive and negative charges in the atom must
be equal.
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Geiger Marsden experiment
When a thin film of gold was
bombarded with α particles,
it was found that 99.9% of
the particles went through
the foil and remaining (about
1 in 20,000) were deviated through different angles.
Some of them were reflected through angles even
grater than 90o, up to 180o. This could not be
explained on the basis of interaction between heavy,
fast moving α particles and light weight, tiny electrons
distributed randomly in the atomic space. So, this
model was discarded.
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2. Rutherford’s atomic model (1909):-
planetary model of atom.
The radius of the nucleus is of
order of 10-15m
Radius of the atom: 10-10m.
Rutherford’s model gave
satisfactory explanation of scattering of α particles from
the thin metal foil. But it was failed to explain the
following facts :
A. According to classical electromagnetic theory of
radiation, an accelerated charge always radiates
energy. So, Electron has to radiate energy
continuously. Due to the loss of energy by radiation,
its radius of orbit would go on decreasing.
Hence, the electron would move along a spiral path
till a stable state is reached when it falls to the
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nucleus. Hence, Rutherford’s atom is unstable,
whereas, in practice, atom is the most stable
structure.
B. the frequency of the waves radiated by the revolving
electron should be same as the frequency of
revolution of the electron. As the electron spiral
inwards, its frequency of revolution would increase
continuously. In a gas different atoms are in
different states, emitting different frequencies.
Hence, continuous spectra should have been
observed. But actually a line spectra of fixed
frequency is observed.
Rutherford thus failed to propose a stable
atomic model which could explain the origin of line
spectra. Hence, his model was also discarded
when Neils Bohr proposed a foolproof model.5
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3. Bohr’s atomic model :-
Base: Plank’s quantum theory
Postulate 1:-
The electron in a hydrogen atom revolves in a
circular orbit round the nucleus, situated at the
centre of the orbit. The centripetal force required for
this circular motion is provided by the electrostatic
force of attraction between the positively charged
nucleus and negatively charged electron
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Postulate 2 :-
The electron can revolve, without radiating energy,
only in those orbits for which the angular
momentum of the electron is equal to an integral
multiple of h/2π, where h is Plank’s constant.
The M.I. of an electron about its axis of rotation is
given by, I = mr2
If ω is the angular velocity of the electron, v = r ω∴ Angular momentum of electron
= Iω = mr2 × v/r = mvrThus, according to second postulate,
Where n can have integral values,
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i.e. n = 1,2,3,.... etc. i.e. n = K,L,M,.... etc.(Priciple
quantum number)
Postulate 3 :-
An electron radiates energy only when it jumps
from higher energy level to lower energy level or
from an orbit of higher energy to an orbit of lower
energy.
If En is energy of electron in higher orbit and Ep is
that in the lower orbit, the energy radiated by an
electron while falling through these orbits is given
as En - Ep = hν
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Radius of Bohr orbit
The quantum selected stationary orbit in which electron
do not radiate energy is called as Bohr orbit.
According to Bohr’s first postulate,
According to Bohr’s second postulate,
Squaring both sides, we get,
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Comparing the R.H.S. of equations (1) and (2)
Every term in the bracket is constant.
∴ r ∝ n2
Thus, radius of Bohr orbit is directly proportional to the
principal quantum number.
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Energy of an electron in Bohr orbit
Kinetic energy due to its motion round the nucleus,
which can be obtained from Bohr’s first postulate as,
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It also possesses potential energy due to its position in
the electric field produced by the positively charged
nucleus and due to electrostatic force of attraction
between them.
Electric potential V at a point at distance r from nucleus
of charge +e is given as
This is the potential energy at that point per unit
charge. So, the P.E. of an electron of charge -e placed
at this point will be,
P.E. = V (-e)
Total energy of the electron is sum of these energies.
i.e. T.E. = P.E. + K.E.
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Substituting value of r as r = we get,
This gives the total energy of an electron revolving in
an orbit of principal quantum number n.
The negative sign indicates that with this much amount
of energy that electron is bound to the nucleus.
as every term in the bracket is constant T.E. ∝ 1 / n2
Substituting the values of m, e, ε0 and h, we get,
T.E. = (-13.6 / n2) eV
Thus, substituting values of n, we get energy of
electrons in successive orbits as,13
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E1 = (-13.6 ) eV, E2 = (-13.6 / 4 ) eV,
E3 = (-13.6 / 9 ) eV, E4 = (-13.6 / 16) eV,
and so on...
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