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Chitkara University Himachal Pradesh Presentation

6511.Nuclear Physics PPTs_A

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Chitkara University

Himachal Pradesh

Presentation

Nuclear Physics

Elements and atoms

Nuclear Composition

Atomic Number and Atomic Mass

Isotopes, Isotones, Isobars & Mirror Nuclei

Mass Defect

Binding Energy

Binding Energy per Nucleon

The atom consists of two parts:

1. The nucleus which contains:

2. Orbiting electrons.

protons

neutronsNucleons

The term “nucleon” refers to either a proton or a neutron in the

nucleus

The term “nuclide” refers to a nucleus with a specific number of

protons and neutrons.

All matter is made up of elements (e.g. carbon,

hydrogen, etc.).

Atom of different elements contain different numbers of

protons.

The mass of an atom is almost entirely due to the

number of protons and neutrons.

……

XA

Z

Mass number

Atomic number

Element symbol

= number of protons + number of neutrons

= number of protons

XA

Z

A = number of protons + number of neutrons

Z = number of protons = number of electron

A – Z = number of neutrons

Number of neutrons = Mass Number – Atomic Number

Inside Atoms: neutrons, protons, electrons

Carbon (C )

Gold (Au)

Atomic number Z=6 (number of protons)

Mass number A=12 (number of protons + neutrons)

# electrons = # protons(count them!)

(atom is electrically neutral)

Atomic number Z = 79

Mass number A = 197

#electrons = # protons(trust me!)

Further layers of substructure:

If each proton were 10 cm across, each quark would be

.1 mm in size and the whole atom would be 10 km wide.

www.cpepweb.org

u quark: electric

charge = 2/3

d quark:

electric charge = -1/3

Proton = uud

electric charge = 1

Neutron = udd

electric charge = 0

Atomic Mass• Atomic mass = 1/12 of the mass of a neutral

Carbon atom (126C) or

• The mass of 1 carbon atom =

Because 1 mole of carbon means that 12 g

of atomic carbon contain Avogadro’s

number (N) of atoms.

Therefore

1 a.m.u.=1.66 x 10-27 kg

a.m.u.1N

12g

12

1

Unit of EnergyThe unit in which nuclear energy is expressed is

called electron volt.

The amount of energy acquired by an electron when

accelerated through a potential difference of 1

volt is called an electron volt (eV)

1 electron volt (or eV) = charge on electron x 1 volt

=1.6 x 10-19C x 1V

1eV =1.6 x 10-19 J

The bigger unit of nuclear energy is mega electron volt (MeV)

1MeV= 106 eV

Relation between a.m.u. and Energy

• According to Einstein’s equation

E= mc2

Here, m = 1a.m.u.=1.66 x 10-27 kg

and c = 3 x 108 ms-1

E = 1.484 x 10-10 J

E= 931.49 MeV

Energy Equivalent:

E = mc2 E = 931.49 MeV

The energy corresponding to 1 a.m.u. is 931.49 MeV

Particle Mass (kg) Mass (a.m.u.) Mass (MeV/c2) Electric Charge

(e)

Proton 1.6726 x 10-27 1.007276 938.28 +1

Neutron 1.6750 x 10-27 1.008665 939.57 0

Electron 9.1095 x 10-31 5.486 x

10-4

0.511 -1

11H atom 1.6736 x 10-27 1.007825 938.79 0

Properties of nucleons with different mass scale used

• Units:– The electric charge of an electron is -1 in these units.

– Mass units are “million electron volts” where 1 eV is a typical

energy spacing of atomic electron energy levels.

Nuclear Size

Nuclear diameter ~10-15 m

Atomic diameter ~10-10 m

AR3

3

4

3

1

AR 3

1

0 ARR

Where R0 is constant for all nuclei

and has a value = 1.2 fm (where 1fm = 10-15m)

If nucleus is spherical then

Isotopes,

The number of protons (atomic number) determines which element it belongs to. Atoms with the same atomic number but different number of neutrons are called isotopes.

Examples:(1) 1H, 2H and 3H (2) 35Cl and 37Cl

(3) 235U and 238U.

U235

92U

238

92

There are many types of uranium:

A

Z

Number of protons

Number of neutrons

A

Z

Number of protons

Number of neutrons

U235

92U

238

92

There are many types of uranium:

Isotopes of any particular element contain the same

number of protons, but different numbers of neutrons.

A 235

Z 92

Number of protons 92

Number of neutrons 143

A 238

Z 92

Number of protons 92

Number of neutrons 146

Isobars,

The nuclei of an atom having the same atomic mass but different number of protons are called isobars.

Example:40Ar and 40Ca.

Ar40

18Ca

40

20A 40

Z 18

Number of protons 18

Number of neutrons 22

A 40

Z 20

Number of protons 20

Number of neutrons 20

Isobars of any particular element contain the same number of nucleon,

but different numbers of protons.

Isobars,

Isotones,

The nuclei having the same number of neutrons are called isotones. e.g., 15N and 16O.

Example:15N and 16O.

N15

7O

16

8A 15

Z 7

Number of protons 7

Number of neutrons 8

A 16

Z 8

Number of protons 8

Number of neutrons 8

Isotones of any particular contain the same number of neutons.

Isotones,

• The nuclei having the number of proton

in one nucleus is equal to number of

neutrons in other nucleus or vice versa.

Mirror Nuclei

Example:

73Li and 74Be

Mass Defect

The nucleus is formed by bringing protons and neutrons together,

the mass of the nucleus so formed is less than the sum of the

masses of the constituent protons and neutrons.

Mass of 11H atom = 1.007825 a.m.u.

+ mass of neutron = 1.008665 a.m.u.

Expected mass of 21H atom= 2.016490 a.m.u.

However,

The measured mass of 21H atom=2.014102 a.m.u

Difference ΔM = 0.002388 a.m.u.

This mass difference is called mass defect and is

denoted as ΔM

Mass Defect ΔM = [Zm(11H)+ Nm(n)]-m( A

ZX) in a.m.u.

Hydrogen

atommH = 1.007825 a.m.u.

neutron mn = 1.008665 a.m.u.

Deuterium

atommD = 2.014102 a.m.u.

= 2.016490

The mass of a deuterium atom is less than the sum of the masses of a

hydrogen atom and a neutron

Mass Defect Cont…….

ΔM = 0.002388 a.m.u.

0 50 100 150 200 250 300-20

0

20

40

60

80

100

16

8O

Pack

ing

Fra

cti

on

Mass Number A

Packing Fraction

It is observed that the atomic masses are very close to

whole numbers, but they invariably differ from integers

By a small amount

This deviation of the atomic

mass from whole number is

expressed in the form of a

quantity called Packing

Fraction

Definition: It is defined as

the mass defect per nucleon

Packing

fraction A

AX

A

mf

A

Z

Packing Fraction Cont…….Packing fraction measures the comparative stability of the nucleus.

The smaller the value of f, the larger is the stability of the nucleus.

From Figure:

1. For very light nuclei f is maximum and hence these are

unstable nuclei.

2. As the value of A increses, f goes on decreasing till it becomes zero

for A=16. Therefore the packing fraction of 168O is zero.

3. Beyond A=16 with the increase of A the value of f decreases

(-ve value) and becomes maximum with further increase of mass

number , the value of f starts increasing (becoming less –ve) till

it again becomes zero at A>240. Thus, one can conclude that

the nuclei of the middle range are stable whereas the heavy nuclei

(A>240) are unstable.

4. All the nuclieds have the tendency to move from region of higher

f to the lower f. This is possible for lighter nuclei by means of fusion

and for heavy nuclei by means of fission

Binding Energy

E = mc2

The energy equivalent to the mass defect is called the binding energy

of the nucleus.

Definition: It can be defined as the energy required to separate a

Nucleus into its constituents particles.

Mass Defect ΔM = [Zm(11H ) + Nm(n)]- m( A

ZX) in

a.m.u.

EBE = Δmc2

= [{Zm(11H )+ Nm(n)}-m( A

ZX)] c2

= [{Zmp+ (A-Z)mn}- AZX] a.m.u.

= [{Zmp+ (A-Z)mn}- AZX](931.49 MeV/a.m.u)

Einstein’ mass energy equation:

Note: All masses are in atomic mass units

Hydrogen

atommH = 1.007825 a.m.u.

neutron mn = 1.008665 a.m.u.

Deuterium

atommD = 2.014102 a.m.u.

= 2.016490

The energy equivalent of the missing mass is called the binding

Energy of the nucleus The greater is the binding energy of a nucleus

more is the energy required to break it

Binding Energy Cont…….

EBE = Δmc2

= 0.002388 x 931.49 MeV= 2.224 MeV

Deuterium nucleus

=

Proton

Neutron2.224-MeV

gamma ray

The binding energy of the deuterium nucleus is 2.224 MeV.

A gamma ray whose energy is 2.224 MeV or more can split

A deuterium nucleus into protons and neutrons

Note: A gamma ray whose energy is less than 2.224 MeV

cannot do this.

Range of BE from 2.224 MeV for 21H (deuterium)

To 1640 MeV for 20983Bi (an isotope of metal bismuth)

Numerical

.

Q1. The binding energy of the neon isotope 2010Ne is 160.647 MeV

find its atomic mass. [Ans=19.992 a.m.u.]

Q2. The mass number of 3517Cl is 34.9800 a.m.u. calculate its binding

energy. Given

Mass of =1.007825 a.m.u.

Mass of proton = 1.007276 a.m.u.

Mass of neutron = 1.008665 a.m.u.

Binding Energy per Nucleon

Definition: The amount of energy required to release a

nucleon from the nucleus is called binding energy per nucleon.

Binding energy per nucleon

(BE/A) = Binding Energy/ Total number of nucleons in the nucleus

BE/A = [{Zm(11H )+ Nm(n)}-m( A

ZX)] c2 /A

= [{Zmp+ (A-Z)mn}- AZX]/A a.m.u.

= [{Zmp+ (A-Z)mn}- AZX](931.49 MeV/a.m.u)/A

Binding energy per nucleon of a nucleus determines

The stability of the nucleus.

If BE/A of a nucleus is less, the nucleus is less stable

whereas the nucleus is more stable if its BE/A is higher.

The left figure shows the

binding energy per nucleon

as a function of mass

number.

Iron is the most stable

element.

Large amount of energy

(yield) is released by fusion

of light elements into

heavier elements or by

fission of heavy elements

into lighter elements.

The BE/A of 21H is 2.2

MeV/2 =1.1 MeV/nucleon

The BE/A of 20983Bi is

1640 MeV/209 =7.8

MeV/nucleon

Binding Energy per Nucleon Cont…….

Fusion

+ + Energy+ + Energy

Fission

1. BE of all stable nuclei

is positive

2. Peaks in BE at A=4n

for light nuclei

corresponds to 42He, 8

4Be, 126C, 16

8O, 2010Ne

and 2412Mg.

3. In the medium mass

region, i.e. between

A=25 to A=125 the

average BE/A is

constant.

4. There is slight decrease

in the BE/A for A>125

Binding Energy per Nucleon Cont…….

Fusion

+ + Energy+ + Energy

Fission

Iron (Fe) has most binding energy/nucleon. Lighter have too

few nucleons, heavier have too many.

BIN

DIN

G E

NE

RG

Y i

n M

eV/n

ucl

eon

92238U

10

Binding Energy Plot

Fission

Fusion = Combining small atoms into large

Fission = Breaking large atoms into small

9/26/2011 33

Binding Energy per Nucleon Cont…….

Two remarkable conclusions

1. If we can somehow split a heavy nucleus into two medium size one,

each of the new nuclei will have more binding energy per nucleon

than the original nucleus did. The extra energy will be given off

and it can be a lot.

For Example: If the uranium nucleus 23592U is broken into two smaller

Nuclei, the binding energy difference per nuclei is about 0.8MeV

Therefore, the total energy given off

(0.8 MeV/nucleon)(235 nucleons) = 188MeV.

This is the energy produced in a single atomic event.

Note: In ordinary chemical reaction involve the rearrangement of

the electrons in atoms and liberate only a few eV per reacting atom.

Here the splitting of heavy nucleus which is called Fission, involves

100 million times more energy per atom then say the burning of coal

and oil

Fission ProcessHeavy elements are split into lighter ones.

Binding Energy per NucleonCont…….

2. If we join two light nuclei together to give a single nucleus of

medium size. It also means more binding energy per nucleon

in the new nucleus.

For Example: If two 21H deuterium nuclei combine to form a 4

2He helium nucleus over 23 MeV energy is released Such a process is

called nuclear fusion.

Note: Nuclear fusion is the main source of energy in sun and

other stars.

Fusion ProcessLight elements are combined into more heavy ones

Numerical

Q1. The mass number of 3517Cl is 34.9800 a.m.u.at is the BE/A Given

Mass of =1.007825 a.m.u.

Mass of proton = 1.007276 a.m.u.

Mass of neutron = 1.008665 a.m.u.

Q2. (a) Find the energy needed to remove a neutron from the nucleus

of the calcium isotope 4220Ca.

(b) Find the energy needed to remove a proton from 4220Ca

nucleus. Why are these energies different?

Subatomic particles interact by exchanging integer-spin

“boson” particles. The varied interactions correspond

to exchange of bosons with different characteristics.

Force Strength Carrier Physical effect

Strong nuclear 1 Gluons Binds nuclei

Electromagnetic .001 Photon Light, electricity

Weak nuclear .00001 Z0,W+,W- Radioactivity

Gravity 10-38 Graviton? Gravitation

Nuclear ForcesNuclear Forces are attractive in nature. This is true as result

of positive BE for all nuclei. If the BE is –ve, then the nuclei

would not be stable and the repulsive forces would be existing in

the nucleus. The nuclear forces are very strong in nature

Meson Theory of Nuclear forces

In 1936, A Japanese Scientist Yukawa predicted that

The nuclear forces must be arising because of some

constant exchange of particles between the nucleons.

According to him, all nucleon i.e. protons and neutrons

could be held together if they share these particles.

The particles were named as mesons.

Later these particles were detected in1947, and were

found to have a mass 270 times mass of electron

Meson Theory of Nuclear forces

According to this theory:

1. All nucleons consists of identical cores surrounded

by a cloud of one or more mesons.

2. Mesons may be neutral π0 or carry either charge (π+or π-).

3. The difference between proton and neutron is supposed to

be in the composition of their respective meson clouds.

4. The forces that act between one neutron and another neutron

and proton-proton are the result of the exchange of neutral

meson (π0) between them. The forces between neutron and

proton is due to the exchange of charged mesons (π+or π-)

between them.

Meson Theory of Nuclear Forces

Neutron Proton

π+ π- and π0

Meson

Cloud

-+

+

+

+

--

-+

--

-+

++

http://en.wikipedia.org/wiki/Nuclear_force

1. A neutron emits π- meson and convert into proton.

n p + π-

then, the proton absorb this π- meson and convert into neutron.

i.e. p + π- n

Similarly

p n + π+

and n + π+ p

Thus the protons and neutrons continuously exchange their

Nature by absorbing and emitting π mesons.

π0 is exchanged not between neutrons and protons but between

Proton-proton or neutron-neutron as explained below.

p p + π0

p + π0 and

n n + π0

n + π0

Analogy

Imagine two interacting nucleons as two dogs and mesons as

a piece of a bone which each of them is trying to snatch from

each other. The bone can be considered to be rapidly exchanged

between the two dogs as neither of them can part with it. Thus the

bone keeps the two dogs bound together.

Similarly, the exchange of π meson keep the two nucleon bound together in the nucleus.

Properties of nuclear forces

1. Nuclear Forces are short range.

2. Nuclear forces are charge independent.

3.Nuclear forces are strongest forces in nature.

4. Nuclear forces are spin dependent.

5. Nuclear forces are saturated forces.

6. Nuclear forces are non-central.

7.Exchange character.