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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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