Atoms & Nuclei

7/29/2019 Atoms & Nuclei

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Structure of Atom:

In order to find the actual structure of atom, Rutherford performed his Gold Foil Experiment.

Rutherford experiment:

In his experiment, he bombarded alpha particles on a very thin gold foil and observed the emerged alpha particles from

golden foil with the help of scintillations produced on a fluorescent screen. The experimental arrangement of his experiment isshown in figure.

On the basis of his experiment, Rutherford observed,

Most of the alpha particles pass through foil without being deviated. On the basis of this observation, he concluded

that most of the part of atom is empty.

Some particles experienced a deviation in their path and therefore Rutherford concluded that positive charge also

exist somewhere in the atom.

Some alpha particles reflected back along the path of their incidence. On the basis of this observation, Rutherfordconcluded that positive charge of the atom remains in a very small region inside the atom. This small region was named

Nucleus.

Rutherford Model:

On the basis of his alpha particle experiment, Rutherford proposed that atom consist of a rigid nucleus (of order of 1510

m) and revolving electrons such that nucleus is positively charged and revolving electrons have same negative charge.Electrons revolve in circular orbits around the nucleus and required centripetal force is observed from electrostatic attraction

between nucleus and electrons. The size of whole atom is of the order of 1010 m.

Failure of Rutherford Model: (Drawbacks of Rutherford Model):

Rutherford atomic model was failed because it was unable to explain

Stability of atom: In any atom, the revolving electrons will radiate energy in the form of electromagnetic waves and thereforeits energy should decrease continuously and hence the radius of its path should also going to be decreased. As a result of this,

finally the electrons should collapse into the nucleus. But in actual practice, it is not so.Hydrogen Spectrum: Since the electrons should revolve in all possible orbits, therefore it should have all frequencies. Due to

this, its spectrum should be continuous but it is not so.

Bohrs Model:

Neils Bohr gave his atomic model by making some changes in Rutherfords model as-(1) Electrons can revolve in some specified orbits in which the angular momentum of electron is integral multiple of

2

h

i.e. L = n.

2

h

where n = 1, 2, 3

(2) The revolving electron does not radiate any energy. The transfer of electron from one state to other state takesplace either by emission.

Impact Parameter (b): Impact parameter is defined as the perpendicular distance of the initial velocity vector of the alpha

particle from the central line of the nucleus, when the alpha particle is far away from the nucleus of the atom.Properties of Nucleus:

1) Size: On the basis of various experiments, it has been found that the volume of a nucleus is directly proportional to thenumber of nucleons in the nucleus.

i.e. Volume of Nucleus atomic mass


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1 1

3 3 3 30

4.

3R A R A R A R R A =

Where 0R is empirical constant which is nearly equal to151.2 10 m or 1.2 Fermi.

1 1

15 3 31.2 10 Meter 1.2 FermiR A R A= =

2) Estimation of Size of the Nucleus(Distance of nearest Approach)When alpha particles are bombarded on the nucleus present in the golden foil then we can estimate the size of the

nucleus.

As the alpha particles move towards nucleus, its kinetic energy begins to change in potential energy and finally we obtain

that point where the whole kinetic energy of the alpha particle is changed into potential energy. This point, where kinetic

energy becomes zero is the point of nearest approach to the nucleus and the distance of this point from the centre of this

nucleus is called the distance of nearest approach.

Let Q is the point of nearest approach then at point Q; the whole kinetic energy of alpha particle appears in the form of itspotential energy.

i.e. ( . .) ( . .)P QK E P E= 2 2

0

0 0 0 0 0

1 (2 )( ) 1 2 1 2

4 4 4

e Ze Ze ZeK K r

r r K = = =

On the basis of his experiment Rutherford observed for gold atom that distance of the closest approach of alpha particles is142.5 10 m, and therefore he made a conclusion that the size of the nucleus should be of the order of 1510 m.

3) Shape: The nucleus is generally assumed to be spherical however it is violated from its spherical shape.

4) Nuclear Density: We know that, Density =mass

volume. (1)

Mass of nucleus = Mass of nucleons = Mass of neutrons and protons

If m is average mass of each nucleon and A is the total mass of the nucleons,Then, M = m.A (2)

Volume of the nucleus,1

3 3 330 0

4 4 4( . )3 3 3

V R R A R A = = = . (3)

Nowneq (1) becomes 33 0

0

. 3

4 4

3

m A m

RR A

= =. (4)

This is average nuclear density.

Fromneq (4), we observe that the nuclear density is independent to the mass number and therefore nuclear density of all

nuclei is nearly same.

We have,15

0 1.2 10R=

M =27

1.67 10

kg , Then,

2717 3

15 3

3 1.67 10

2 104 3.14 (1.2 10 ) kg m

= = This is the average nuclear density.

5) Isotopes: The atoms of an element whose nuclei have same protons but different neutrons are called Isotopes. e.g.1 1 1

1 2 3, ,H H H

6) Isobars: The nuclei which have the same number of nucleons but different number of protons and neutrons are called

Isobars. e.g.3 3 14 14 17 17

1 1 6 7 8 9, ,H and He C and N O and F


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7) Composition of nucleus:

Proton Electron Hypothesis: According to this hypothesis, a nucleus consists of protons and electrons. The positivecharge of the nucleus is the combined result of electrons and protons. The mass of the electrons is negligible in

comparison to the mass of proton. Therefore proton contributes to the mass of the nucleus.

e.g.4

2He have 3 electrons and 4 protons

Mass of4

2He = 4 times to the mass of proton

Charge of4

2He = +2e (Charge of 2 protons)

Drawbacks:

If electrons are confined in such a small space then energy of the electron should be of the order of 100

million eV but the energy of particles emitted from nucleus is of the order of 2.3 million eV.

If electron exist inside nucleus then magnetic moment of nucleus cannot be less than the magnetic momentof electrons but it is not so.

Proton Neutron Hypothesis: According to this hypothesis, nucleus contains protons and neutrons (called nucleons). Thepositive charge of a nucleus is a result of positive charge of protons. The total number of protons and neutrons is called

Atomic mass A. The number of protons is called Atomic number z. Therefore any atom is symbolized byA

Z X .

e.g. mass of Helium nucleus is nearly 4 times to the mass of a proton and its charge is 2 times to the charge of proton.

i.e. a Helium nucleus consists of two protons and two neutrons.

8) Nuclear Forces: All the nucleons remain together in the nucleus due to strong nuclear force acting on them. The basicproperties of the nuclear forces are-

Nuclear forces are primarily attractive.

Nuclear forces are not gravitational forces.

Nuclear forces are extremely strong.

Nuclear forces are spin dependent.

Nuclear forces are saturated forces.

9) Stability of nucleus: In small nuclei, the strong nuclear forces completely overcome to the electrical repulsive forces

acting between proton-proton and therefore lighter nucleis elements are stable.As we move towards heavier elements, the number of neutrons and the protons inside the nucleus increases. The nuclear

attractive forces are active only between nucleons, which are extremely close to each other. Therefore the total electricalrepulsive force rises rapidly. Therefore the stability of the nucleus going to decreased.

10) Yuckawa Meson Theory: In order to explain the nuclear forces, in 1935 Japanese Scientist Yuckawa predicted Mesontheory. According to this theory, Mesons are elementary particles having rest mass between the rest mass

of electron and proton. On the basis of charge mesons are of three types-

i) + ii) iii) 0Yuckawa assume that all the nucleons are surrounded by these mesons and a continuous exchange occurs between

nucleons due to which they continue to be converting into one another. Due to exchange of Mesons, strong nuclearforces are to be produced; these forces are known as exchange forces.

According to Yuckawa-

When a + meson jumps from a proton to a neutron then the proton is converted into neutron and neutron is convertedinto proton.

P - + = n & n + + = P

When a meson jumps from neutron to proton then neutron converts into proton and proton is converted into neutron.n - = P & P + = n


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A continuous exchange of 0 mesons becomes between proton-proton and neutron-neutron such that proton remainsproton and neutron remain neutron.

P - 0 = P & P + 0 = P

n - 0 = n & n + 0 = n

11) Mass Defect: The difference between the sum of mass of the nucleons and the mass of nucleus is defined as massdefect.

Mass defect = total mass of Nucleons Rest mass of nucleus

m = {mass of protons + mass of neutrons} rest mass of nucleusm = [{ ( )p nZm A Z m+ } M]

Where Z is atomic number, pm is rest mass of proton, M is rest mass of nucleus, (A Z) is number of neutrons in the

nucleus.

12) Binding Energy: Binding energy of nucleus is defined as the maximum energy required separating its nucleonsand placing them at rest, at infinite distance apart.

i.e. B.E. = mass defect X 2c = [{ ( )p nZm A Z m+ } M] X 2c13) Binding energy per nucleon: Binding energy increases with increase in atomic mass A. to compare the stability we

have to define average binding energy per nucleon.The total binding energy of nucleus upon atomic mass of that nucleusis defined as binding energy per nucleon.

B.E. per nucleon =. .B E

atomic mass=

E

A

=

2{ ( ) } ]p nZm A Z m M c

A

+

14)Binding energy curve: The variation of binding energy per nucleon with increase in mass number is shown here. Theobserved curve is called Binding Energy Curve. On the basis of this curve, the following conclusions can be made-

The curve is roughly flat from A= 50 to A= 80, having average binding energy per nucleon of about 8.5million eV which suggest that nuclei having mass number 50 to 80 are most stable.

From curve we observe that the binding energy per nucleon from 56Fe is maximum (nearly equal to 8.5MeV) hence we conclude that Fe is most stable.

For the element having mass number greater than 50, the binding energy curve goes to decrease and is drops

at 7.6 MeV for235

U . For nuclei having mass number below 50, the binding energy per nucleon decreases and for the nuclei

having mass number less than 20, the binding energy per nucleon decreases sharply which suggest that elements havingmass number below 20 are less stable.

We observe sharp peaks for some elements in the binding energy curve. These peaks suggest that thesenuclei are most stable than their neighbors.

From this curve, we observe that very light nuclei have a lower binding energy per nucleon than the nucleiof intermediate masses. Therefore if two very light nuclei be combine and form comparatively heavier nucleus than

energy is released in this process and this process is called nuclear fusion.

If a heavier nucleus ( 238U ) is spitted into two lighter nuclei than energy is released. The above writtenstatement can be explained by the curve such that binding energy per nucleon of heavier nucleus is less than the

intermediate nuclei.

15)Packing Fraction: Packing fraction of a nucleus is defined as the mass excess per nucleon.

i.e. Packing fraction =mass excess

mass number=

M A

A

Where M is actual weight of a nuclei on the physical atomic weight scale.

16) Calculation of Energy

Atomic mass unit(a.m.u.):1 Amu is defined as the 1/12th of the mass of the 126C atom.


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i.e. 1 Amu(1u) =12

6

1( )

12mass of C atom =

1 12

12 N

1u = 23

1

6.023 10=

241.6605 10 gm = 271.6605 10 Kg

Energy Equivalence to 1u: According to E = 2m c E (1u) = 27 8(1.6605 10 ) (3 10 ) = 101.4924 10 Joule

Energy in eV 1u =

10

19

1.4924 10

1.6 10

=6931.25 10 eV

Energy in MeV = 931 Mev

RADIOACTIVITY

Some heavy elements disintegrate spontaneously and the emission of , and rays takes place from their nucleus. Thisphenomenon is called Radioactivity and these types of substances are called Radio Active Substances.

The phenomenon of radioactivity was discovered in 1896 by Henry Bequerel.

Radium, thorium, actinium, polonium etc are some radioactive substances.

(1) Emission of Radiation:During radioactivity, the emission of radiation of three types takes place-

a) rays, b) rays, c) rays Rays: The ionized helium is consideredas particle and therefore the rays consisting of these particles are called rays.Properties:

An alpha particle has charge equal to helium nucleus. Mass of particle is nearly equal to the four times of hydrogen atom or is equal to the mass of helium

nucleus.

The velocity of particles depends on their source of emission. Its range is 7 7 11.4 10 2.1 10to ms . Particles have least penetrating power due to having large mass. They can easily stop by an aluminum

sheet of 0.02 mm thickness.

Particles have large ionizing power. At normal pressure in air, the particle ranges from 3 to 8 cm. Particles can produce fluorescence in certain substances like Barium, platino cyanide and zinc sulphide. Rays affect photographic plates. Particles are deflected by electric and magnetic fields. Particles cause burning effect on human body and they produce heating effect by stopping them.

Rays: The rays consist of fast moving electrons called particles.Properties:

A particle has a unit negative charge. Mass of particle is same as that of electron. The velocity range of particles varies from 33% to 99% of the velocity of light. Particles have very large penetrating power. They can easily pass through a few mm of aluminum. The ionizing power of particles is less than the paricles. The range of particles in air is several meters. Particles can produce fluorescence and affect photographic plates. They are deflected by electric and magnetic fields.


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Rays: rays are electromagnetic radiation consists of highly energetic photon.Properties:

rays are neutral and are therefore not deflected by electric and magnetic fields. The rest mass of a ray photon is zero. rays travel with the speed of light.

rays have very high energy and very large penetrating power. rays have least ionizing power. rays can produce fluorescence in a substance like willimite. rays affect a photographic plate mare strongly than particles. rays can produce nuclear reactions.

(2) Laws of Radioactive Decay

Radioactivity is a spontaneous process which does not depend on external factors. During disintegration, the emission of and particles does not take place simultaneously. i.e. at a particulartime either the emission of particles or emission of particles takes place.

Emission of particles from any atom will change it into a new atom as the charge number is reduced by 2 andmass number is reduced by 4.

4 4

2 2

A A

Z ZX He X

The emission of particles from any atom will change it into a new atom whose charge number is raised by 1without any change in its mass number.

0

1 1

A A

Z ZX e X + Radioactive Decay Law: According to this law, The rate of disintegration of radioactive atoms at any instant isdirectly proportional to the number of radioactive atoms actually present in the sample at that instant.

i.e. R N R = NWhere N is number of atoms left at that instant, is Decay constant and R is Rate of disintegration

(3) Expression for Radioactive Decay: Let we have a radioactive substance at t = 0, having 0N atoms. After a time

t, let the number of atoms left in the sample are N remains undecayed.

During this process, for any small disintegration dN in a time dt, the rate of disintegration

R =dN

dt . (1)

Negative sign indicates that the undecayed atoms are decreasing.According to radioactive decay law,

R = N (2)From eq. (1) and (2),

R =dN

dt = N . (3)

From (3),

dN dN

N dtdt N = =

On integrating both side within limits,

0

tdN

dtN

= loge N t C= + (4)

At t = 0, N = 0N Therefore, 0loge N C=


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neq (4) Becomes, 0log loge eN t N= +

00

log log loge e eN

N N t tN

= =

By taking antilog 0

0

t tN e N N e

N

= =

The above relation shows that the radio active decay is exponential term.

Disintegration Constant:We have, 0

tN N e=

By substituting t =1

We have,

1

1 0

0 0

NN N e N N e N

e

= = =

i.e the disintegration constant of a radioactive element is the reciprocal of time at the end of which, the number of atoms

left undecayed in a radioactive sample reduces to1

etimes, the original number of atoms in the sample.

(4) Half Life of a Radioactive Element: The time in which a radioactive element becomes half to the initially present atoms in

a sample, is called half life of the radioactive element.

We have, 0tN N e

=

At half life,0

2

NN = and t = T

0

0

12

2 2

T T TN N e e e = = =

Taking log, 2 2Te e e elog e log T log e log = =

0.6932

2.303 0.3010 0.6932TT

= = =

i) let initially any sample contains 0N atoms, then

a) After one half life ,

1

0

0

1

2 2

NN N

= =

b) After two half life,

1 2

0 0

0 0

/ 2 1 1

2 4 4 2

N NN N N

= = = =

c) After three half life,

32

0

0

(1/ 2) 1

2 2

NN N

= =

d) Similarly for n half life, 01

2

n

N N

= ii) Activity: The rate of disintegration of a sample of a radioactive material is called activity of the sample. The

activity is directly proportional to the number of atoms left undecayed in the sample.

We have,0

1

2

n

N N =

But t = n x T n = t / T


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Therefore0

1

2

t

T

N N =

Again we have, A N

Therefore,0 0

1

2

t

TA N

A N = = Where A = Activity and 0A = Instantaneous activity

(5) Average / Mean life of Radioactive Element: The average life times the total life time of all the atoms of theelement upon the total number of atoms present initially in the sample of element.

i.e. Average Life =Total life time

Initial number of atoms present

The average life of a radioactive element is inversely proportional to the decay constant of that material.

1

= (2)

But we have,0.6932 1

1.440.6932

TT

T

= = =

1

1.44T

= =

i.e. Average life of a radioactive material is 1.44 times to its half life.

(6) Units of Radioactivity:

i) Curie:The activity of a radioactive sample is said to be curie if 103.7 10 disintegrations take place persecond. i.e. 1Ci = 103.7 10 decays / sec

ii) Becquerel: If one decay per second takes place from any radioactive material, then the Activity of thematerial is one Becquerel.

i.e. 1Bq = 1 decay / sec

iii) Rutherford:If 610 decays take place in one second from a radioactive material, then the activity of thatmaterial is said to be 1 Rutherford.

i.e. 1 Rd = 610 decay / sec

(7) Artificial Radioactivity and Induced Radioactivity: The phenomenon of disintegration of a nucleus by bombarding it with suitable projectile is called Artificial

Radioactivity.

If in artificial radioactivity, the disintegration of nucleus keeps on even after the bombardment is stopped. This

phenomenon is called Induced Radioactivity.

e.g.

10 4 14 13 1

5 2 7 7 0

13 13 0

7 6 1

B He N N n

N C e

+ +

+

Radioactive Decay


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(1) Alpha Decay: The phenomenon of emission of particle from a radioactive nucleus is called Alpha decay. When analpha decay takes place from a nucleus then its mass number is decreased by 4 and charge number decreased by 2.

4 4

2 2

A A

Z ZX X He Q

+ +Where Q is the amount of energy released during the alpha decay.

The total mass of daughter nucleus and alpha particle is less than the mass of the parent nucleus. Thisreduced mass appears in the form of energy in accordance to 2E m c = .

Q =2{ }x y HeE m m m c =

Speed of Alpha particle: Let the parent nuclei are at rest and yv a n d v are speeds of and daughternuclei respectively when parent nuclei undergo decay.Then according to conservation of angular momentum, we have

0y y

y y y y

m vm v m v m v m v v

m

= = =

Energy of

particle: Energy during

decay appears in the form of kinetic energy of

particle anddaughter nuclei

2 21 1Q = ( . ) ( . )2 2

Y y yK E K E m v m v + = +

2 2

2

2

1 1

2 2y y

y y

m v m vBut v therefore Q m v m

m m

= = +

( )2 2 2 21 1 1

2 2 2y y y y

m Q m m v m v m Q m v m m = + = +

( ) ( )2 21 1

( . .)2 2

y y

y y

m Q m Q

m v Or K E m vm m m m

= = =+ +

(2) Beta Decay: The phenomenon of emission of an electron from a radioactive element/nucleus is called Betadecay. When beta decay takes place, then the atomic number is decreased by one and mass number remains same.

0

1 1

A A

Z ZX X + +

Violation of Energy conservation in emission: The graphbetween the energy of emitted particles and their number isobserved as

From here, we observe that

a) Energies of particles vary from zero to end point energy (max. energy Q ).

b) Most particles are emitted with smaller energies than the end point energy.

c) Since the energy of emitted particles range from zero to Q but for each emission, the energy of nucleus is

reduced by Q , that is why the Energy Conservation should be violated in decay.

d) In order to explain this violation in 1931, Pauli suggested a hypothesis called Neutrino Hypothesis.

Neutrino Hypothesis: According to this hypothesis,


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a) decay is always accompanied by Neutrino.

b) Neutrino is an elementary particle of zero rest mass and zero charge and spin1

2 2

h

.

c) The total energy of particle and neutrino remains constant and become equal to end point energy.

d)According to this hypothesis, the emission of particle takes place as-

0

1 1

A A

Z ZX X Q + + + +(3) Gamma Decay: The phenomenon of emission of Gamma ( ) ray photon from a radioactive nucleus is called decay. This decay occurs due to the transition of nucleus from upper energy state to lower energy state Since, photon donot have any charge and have zero rest mass, therefore during decay, atomic mass and atomic number remains same.

A A

Z ZX X +

Nuclear Reactions and Stellar Energy

(1) The reactions in which any stable nucleus is going to be changed in other stable nuclei after bombardment ofsuitable high energy particles.

4 4 3 1

2 2 1 1

A A A

Z Z ZX He X Y H Q+ +

+ ++ + +

(2) Examples of Nuclear Reactions:

7 1 8 4 4

3 1 4 2 2

11 1 12 11 1

5 1 6 6 0

12 1 13 12

6 1 7 7

14 1 15 14 1

7 0 7 6 1

2 1 1

1 1 0

)

)

)

)

) ( sin )

i Li H Be He He

ii B H C C n

iii C H N N

iv N n N C H

v H H n Photo Di tegration

+ +

+ +

+ +

+ +

+ +(3) Chain Reaction: A continuous reaction which would start and a huge amount of energy is released in a shorttime, is called Chain Reaction.

e.g.238 1 141 92 1

92 0 56 36 03U n Ba Kr n Q+ + + +It is clear from above relation that in fission of 238U nucleus, three neutrons are produced. These three neutrons will

disintegrate three more uranium nuclei and will create nine neutrons. This process will be continuous and therefore a huge

amount of energy will be released in a short time. This process is called Chain Reaction.

Chain reactions are of two types-

Controlled Chain Reaction: When the energy produced in a chain reaction is controlled by controlling the access

number of neutrons. This process is called Controlled chain reaction. The energy released in this reaction is used for

constructive purposes. E.g. Nuclear Reactor

Uncontrolled chain reaction:When the energy produced in a chain reaction can not be controlled by any means, then

the reaction is called uncontrolled chain reaction. E.g. Atom Bomb

(4) Conservation Laws In Nuclear Reactions: In a nuclear reaction, the following conservation laws are obeyed Total energy conservation

Conservation of linear momentum

Conservation of number of nucleons

Conservation of charge number

(5) Types of Nuclear Reactions:


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Nuclear Fission:When an energetic particle is bombarded on a heavy nucleus, then heavy nucleus splits up into small

nuclei. This phenomenon is called Nuclear Fission.

238 1 141 92 1

92 0 56 36 03U n Ba Kr n Q+ + + +The mass of resultants in fission reaction is always less than the mass of reactants. This loss in mass appears in the form

of energy.

The mass defect achieved in above reaction is nearly equal to 0.2153 amu and therefore the energy released in reaction is

nearly equal to 0.2153 X 931 M ev. E = 200.4 Mev

Nuclear Fusion: When two or more light nuclei fuse together and forms comparatively heavy nucleus, then in thisprocess huge amount of energy is released and the process is called Nuclear Fusion.

1 1 2

1 1 1

2 2 3 1

1 1 2 0

2 2 3 1

1 1 1 1

)

) 2.37

) 4.03

i H H H e Q

ii H H He n Mev

iii H H H H Mev

++ + + +

+ + +

+ + + Nuclear fusion can not be controlled. The process of nuclear fusion takes place at very high temperature and pressure. In a fusion reaction, the mass of resultants is less than the mass of reactants which is the cause of energy

produced during fusion. The source of stellar energy is fusion reactions occurring inside the sun and stars.

(6) Stellar Energy: The energy obtained continuously from the sun and stars is called Stellar Energy.The source ofstellar energy is fusion reactions occurring inside the sun and stars. The reactions responsible for stellar energy are of

following types-

Proton-Proton cycle:In this process, the reactions involved are

1 1 2 0

1 1 1 1 1

1 2 3

1 1 2 2

3 3 4 1

2 2 2 1 3

2( ) 2( ) 2( ) 2( )

2( ) 2( ) 2( )

2( )

H H H e Q

H H He Q

He He He H Q

++ + +

+ +

+ + +

Net =

1 4 0

1 2 14( ) 2( )H He e Q+ + +Where 1 2 3Q Q Q Q= + + = 26.7 mev

Carbon-Nitrogen Cycle

12 1 13

6 1 7 1

13 13 0

7 6 1 2

13 1 14

6 1 7 3

14 1 15

7 1 8 4

15 15 0

8 7 1 5

15 1 12 47 1 6 2 6

C H N Q

N C e Q

C H N Q

N H O Q

O N e Q

N H C He Q

+

+

+ +

+ +

+ +

+ +

+ +

+ + +Net =

1 4 0

1 2 14 2( )H He e Q+ + +

Where 1 2 3 4 5 6Q Q Q Q Q Q Q= + + + + + = 26.7 Mev

ATOMIC PHYSICS


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Rutherford Theory: According to Rutherford, the required centripetal force for the circulation of electron in its orbit is

provided by electrostatic force between nucleus and revolving electron.

i.e.

2 2

2 2

0 0

1 ( )( ) 1 ( )

4 4

mv Ze e Ze

r r r = =

Radius of the path of revolving electron:

2 22

2 2

0 0 0

1 ( )( ) 1 ( )( ) 1

4 4 4

mv Ze e Ze e Zemv r

r rr mv = = =

Velocity of revolving electron: According to Rutherford,

2 2

2

0 0

1 ( )( )

4 4

mv Ze e Zev

r mrr = =

Energy of revolving electron:

Kinetic energy:

2 2 22

2

0 0 0

1 ( )( ) 1 1 1 1

4 2 2 4 8

mv Ze e Ze Zemv K

r r rr

= = =

Potential energy:

2

0 0

1 ( )( ) 1

4 4

Ze e ZeU

r r

= =

Total energy:

2 2 2

0 0 0

1 1 1E = K + U

8 4 8

Ze Ze Ze

r r r = =

Bohr Theory: According to Bohr Theory, electron can revolve in some specified orbits in which angular momentum of

electron is an integral multiple of

2

h

Angular momentum of electron =

2 2

h nhn

=

. (A)

And required centripetal force for revolution is provided by electrostatic force between nucleus and proton.

2 22

2

0 0

1 ( )( ) 1. .

4 4

mv Ze e Zei e mv

r rr = = (B)

Radius of the path of revolving electron: By dividing the square of neq (A) from neq (B), we have

2

2 2 2 22

0 0

2 2 2 2

0

( ) 2

1

4

nh

n h n hmvrmr r

mv Ze Ze mZe

r

= = = .. (C)

Fromneq (C), we have

2r n The radius of allowed orbits will be as

1 2 3: : ............ 1: 4 :9...............r r r =

For hydrogen atom(Z=1), the radius is


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2 22 110

25.29 10

n hr n m

me

= =

Velocity of electron in orbit: According to Bohr,

v v

2 2

nh nhm r

mr

= =

2 2

0

2But

n hr

me

=

2

2 200

2

1Then, v v

22

nh Ze

h nn hm

mZe

= =

For H-Atom, Z = 1 & for n = 1, We have6 1v 2.19 10X ms=

6

8

v 2.19 10 1v

137 1373 10

X c

c X= = =

Energy in stationary orbits:

Kinetic energy:

2 2 22

2

0 0 0

1 ( )( ) 1 1 1 1

4 2 2 4 8

mv Ze e Ze Zemv K

r r rr

= = =

Potential energy:

2

0 0

1 ( )( ) 1

4 4

Ze e ZeU

r r

= =

Total energy:

2 2 2

0 0 0

1 1 1E = K + U

8 4 8

Ze Ze Ze

r r r = =

2 2

0

2

2 2 4

2 2 22 200

0 2

But,

1

88

n hr

m e

Ze Z e mTherefore E

h nn h

me

=

= =

As19 31 341.6 10 , 9.1 10 , 6.6 10e m h

= = = 2 4

2

2 2 2 2

0

1 13.6,

8

Z e mTherefore E Z eV

h n n

= =

For hydrogen atom, Z = 1

2

13.6E eV

n

=

Energy of orbital electron in terms of Rydberg Constant


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2 4 2 4 2

2 2 2 2 3 2 2

0 0

1 1

8 8

Z e m Z e mhc Z RhcE

h n h c n n

= = =

4

2 3

0

Where8

e mR

h c=

For hydrogen atom, 2

RhcE

n=

7 1 34 81.097 10 , 6.6 10 , 3 10But R m h c = = =

2

13.6Therefore E eV

n

=

Frequency of electron in Bohr stationary orbit: The number of revolutions completed per second by the revolvingelectron in a stationary orbit is called frequency.

2

2 22 200

2

vv v

For revolving electron, v (2 ) 2 22

m Z e

r r r n hn h

mZe

= = = = =

For

hydrogen atom, (for Ist Orbit)15 16.57 10 ( )round s Hz =

Energy Levels:

The exchange of energy in an atom is not continuous. It takes place by photon of energy h , i.e. the energy emitted orabsorbed by any atom is discreet which suggest that the atom has discreet energy levels.

Energy Level Diagram:

Experimental Verification: In order to explain the different energy levels, Frank and Hertz performed an experiment on

mercury vapours as-

i) E = 0 4.86E eV < , then 'E E=ii) E = 4.86 eV, then 'E E= or ' 0E =iii) E > 4.86 eV, then 'E E= or ' 4.86E E=


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iv) E = 6.67 eV, then 'E E= or ' 0E = or ' 4.86E E= v) E > 6.67, then 'E E= , ' 6.67E E= or ' 4.86E E= vi) E = 10.4 eV, number of electron get increased.

On the basis of above observations, they conclude that any atom absorb some definite amounts of energy, which suggest that

the atom has different energy levels.

Atomic Transitions:

When energy from external agency is supplied to an atom then its orbital electron jumps to higher energy exited state (In thiscondition atom is said to be excited).In excited state ,atom remain up to 108 Seconds and then return to its ground state. When

an excited atom jumps from higher energy state to lower energy state, then it emits a photon, having energy equal to the energy

difference of both orbits. The frequency and wavelength of this emitted photon are determined as,

Energy in higher state =

2 4

2 2 2 2

0 2

1

8

Z e mE

h n

=

Energy in lower state =

2 4

1 2 2 2

0 1

1

8

Z e mE

h n

=

Energy difference =

2 4 2 4 2 4

2 1 2 2 2 2 2 2 2 2 2 2

0 2 0 1 0 1 2

1 1 1 1

8 8 8

Z e m Z e m Z e mE E

h n h n h n n

= =

i.e. when electron jumps from higher energy state to lower energy state, then it emits of energy equal to the energy difference of

both orbits.

2 4 2 4

2 1 2 2 2 2 2 3 2 2

0 1 2 0 1 2

1 1 1 1

8 8

Z e m Z e mh E E

h n n h n n

= = =

2 4 2 4

2 3 2 2 2 3 2 2

0 1 2 0 1 2

1 1 1 1 1

8 8

c Z e m Z e mor

h n n h c n n

= =

2 42

2 3 2 2 2 2

0 1 2 1 2

1 1 1 1 1

8

Z e mZ R

h c n n n n

= =

47 1

2 3

0

Where 1.097 108

e mR m

h c

= =

Excitation Potential and Ionization Potential:

That minimum accelerating potential which provides energy to electron such that by getting this energy, the

electron can excite to an atom by striking, is calledExcitation Potential.

That minimum accelerating potential, which provides sufficient energy to electron such that by getting this

energy, electron can ionize the atom, is calledIonization Potential.

Energy Level diagram of hydrogen atom: A diagram which represents the total energies of electron in different stationary

orbits of an atom is called the Energy Level Diagram of that atom.

We have, that energy of electron in hydrogen atom in any orbit is

2

13.6nE eV

n

= , Then

if n = 1, 1 13.6E eV=


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If n = 2, 213.6

3.44

E eV= =

If n = 3, 313.6

1.519

E eV= =

If n = 4, 4

13.6

0.8516E eV= =

If n = 5, 513.6

0.5425

E eV= =

-------------------------------------------

-------------------------------------------

If n = ,13.6

0nE = =

Hydrogen Spectrum

On the basis of study of hydrogen spectrum of visible region, in 1885 Balmer found that the spectrum contains bright lines on

dark background. The distance and intensity of these lines decreases from one side to the other as shown in figure,

Balmer gave an experimental formula for the wavelength of this spectrum,

2 21 1 1

2R

n

=

Where, n = 3, 4, 5, 6 &7 11.097 10R m=

After this, various other series were also observed in the visible regions such that the wavelengths of these series can be

formulated as-


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i) Lyman Series: When an electron jumps in the Ist orbit from any other orbit, then the Lyman series is obtained. Itis observed in Ultra Violet region.

2 21 1 1

2,3, 4..........1

R Where nn

= = =

ii) Balmer Series: When an electron jumps in the IInd orbit from any other orbit, then the Balmer series isobtained. It is observed in visible region of spectrum.

2 21 1 1

3,4,5..........2

R Where nn

= = =

iii) Paschen Series: When an electron jumps in the IIIrd orbit from any other orbit, then the Paschen series isobtained. It is observed in Infra Red region of spectrum.

2 2

1 1 14,5,6,.........

3R Where n

n

= = =

iv) Brackett Series:When an electron jumps in the IVth orbit from any other orbit, then the Brackett series isobtained. It is observed in Infra Red region of spectrum.

2 2

1 1 1, 5,6,7..........

4R Where n

n

= = =

v) Pfund Series:When an electron jumps in the Vth orbit from any other orbit, then the Pfund series is obtained. Itis observed in Infra Red region of spectrum.

2 21 1 1

, 6,7,8..........5

R Where nn

= = =

Limitations of Bohr Theory


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Bohr Theory is applicable only for simple atoms like Hydrogen. Bohr Theory was unable to explain the existence of elliptical orbits. Bohr Theory was unable to explain the fine structure of spectral lines. Bohr Theory was unable to explain the relative intensities of spectral lines.

Atoms & Nuclei

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