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CHAPTER 1BASIC CONCEPTS OF NUCLEAR PHYSICS

1.1 Historical survey:The origin of nuclear physics and the progress after can be understand from the historical review as follow:1895 – The discovery of X-Ray by Rontgen1896 – Becquerel discovers radioactivity of Uranium

1897 – J.J. Thompson discovers the electron

1900 – The discovery of black body radiation formula by Planck1905 – The development of the theory of special relativity by Enstein

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1911 – Rutherford discovers the atomic nucleus

1913 – Boher’s theory of Hydrogen atom

1932 – Chadwick discovers the neutron, Heisenberg formulates the 'isospin hypothesis

1935 Yukawa's meson hypothesis as intermediary of

the strong force

Where 1896 and 1911 is the beginning of nuclear physics

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1939 – Discovery of nuclear fission by Hahn, Strassmann, Meitner

1946 Discovery of the pion by Occhialini and Powell

1948 The shell model, discovered by Jensen and Goeppert Mayer

1951-1954 Development of the collective model by A. Bohr and B. Mottelson

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THOMSON

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Matter consists of atoms, which is the smallest part as known by Greeks , then electrons discovered , Nucleus, protons and neutrons discovered latter , so atoms consists of Nucleus and electrons.

But Quarks have been discovered which is the smallest part of matter , protons and neutrons consists of Quarks up, down, strange , charm , bottom, and top (u,d,s,c,b,t) protons consist of three QuarksP (uud), neutrons n (udd) , where charge of (u,c,t) is

+2/3e while (d,s,b) is -1/3e

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Objectives: After completing this module, you should be able to:

• Define and apply the concepts of mass number, atomic number, and isotopes.

• Define and apply concepts of radioactive decayand nuclear reactions.

• Calculate the mass defect and the binding energy per nucleon for a particular isotope.

• State the various conservation laws, and discuss their application for nuclear reactions.

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Composition of MatterAll of matter is composed of at least three fundamental particles (approximations):

Particle Fig. Sym Mass Charge Size

The mass of the proton and neutron are close, but they are about 1840 times the mass of an electron.

Electron e- 9.11 x 10-31 kg -1.6 x 10-19 C ∼

Proton p 1.673 x 10-27 kg +1.6 x 10-19 C 3 fm

Neutron n 1.675 x 10-31 kg 0 3 fm

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The Atomic Nucleus

Beryllium Atom

Compacted nucleus:

4 protons5 neutrons

Since atom is electri-cally neutral, there must be 4 electrons.

4 electrons

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Modern Atomic TheoryThe Bohr atom, which is sometimes shown with electrons as planetary particles, is no longer a valid representation of an atom, but it is used here to simplify our discussion of energy levels.

The uncertain position of an electron is now described as a probability distribution—loosely referred to as an electron cloud.

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DefinitionsA nucleon is a general term to denote a nuclear particle - that is, either a proton or a neutron.

The atomic number Z of an element is equal to the number of protons in the nucleus of that element.

The mass number A of an element is equal to the total number of nucleons (protons + neutrons).

The mass number A of any element is equal to the sum of the atomic number Z and the number of neutrons N : A = N + Z

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Symbol NotationA convenient way of describing an element is by giving its mass number and its atomic number, along with the chemical symbol for that element.

[ ]

A Mass numberZ Atomic numberX Symbol=

For example, consider beryllium (Be): 94 Be

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Example 1: Describe the nucleus of a lithium atom which has a mass number of 7 and an atomic number of 3.

Lithium Atom

N = A – Z = 7 - 3

A = 7; Z = 3; N = ?

Protons: Z = 3

neutrons: N = 4

Electrons: Same as Z

73 Li

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Isotopes of ElementsIsotopes are atoms that have the same number of protons (Z1= Z2), but a different number of neutrons (N). (A1 ≠ A2)

Helium - 4

42 He

Helium - 3

32 He Isotopes

of helium

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NuclidesBecause of the existence of so many isotopes, the term element is sometimes confusing. The term nuclide is better.

A nuclide is an atom that has a definite mass number A and Z-number. A list of nuclides will include isotopes.

The following are best described as nuclides:32 He 4

2 He 126 C 13

6 C

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Atomic Mass Unit, uOne atomic mass unit (1 u) is equal to one-twelfth of the mass of the most abundant form of the carbon atom--carbon-12.

Atomic mass unit: 1 u = 1.6606 x 10-27 kg

Common atomic masses:

Proton: 1.007276 u Neutron: 1.008665 u

Electron: 0.00055 u Hydrogen: 1.007825 u

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Exampe 2: The average atomic mass of Boron-11 is 11.009305 u. What is the mass of the nucleus of one boron atom in kg?

Electron: 0.00055 u115 B = 11.009305

The mass of the nucleus is the atomic mass less the mass of Z = 5 electrons:

Mass = 11.009305 u – 5(0.00055 u)

1 boron nucleus = 11.00656 u

-271.6606 x 10 kg11.00656 u1 u

m

=

m = 1.83 x 10-26 kg

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2 8; 3 x 10 m/sE mc c= =

Mass and EnergyRecall Einstein’s equivalency formula for m and E:

The energy of a mass of 1 u can be found:

E = (1 u)c2 = (1.66 x 10-27 kg)(3 x 108 m/s)2

E = 1.49 x 10-10 J Or E = 931.5 MeV

When converting amu to energy:

2 MeVu931.5 c =

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Example 3: What is the rest mass energy of a proton (1.007276 u)?

E = mc2 = (1.00726 u)(931.5 MeV/u)

Proton: E = 938.3 MeV

Similar conversions show other rest mass energies:

Electron: E = 0.511 MeV

Neutron: E = 939.6 MeV

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The Mass DefectThe mass defect is the difference between the rest mass of a nucleus and the sum of the rest masses of its constituent nucleons.

The whole is less than the sum of the parts!Consider the carbon-12 atom (12.00000 u):

Nuclear mass = Mass of atom – Electron masses = 12.00000 u – 6(0.00055 u) = 11.996706 u

The nucleus of the carbon-12 atom has this mass.(Continued . . .)

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Mass Defect (Continued)Mass of carbon-12 nucleus: 11.996706

Proton: 1.007276 u Neutron: 1.008665 u

The nucleus contains 6 protons and 6 neutrons:

6 p = 6(1.007276 u) = 6.043656 u

6 n = 6(1.008665 u) = 6.051990 u

Total mass of parts: = 12.095646 u

Mass defect mD = 12.095646 u – 11.996706 u

mD = 0.098940 u

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The Binding Energy

The binding energy EB of a nucleus is the energy required to separate a nucleus into its constituent parts.

EB = mDc2 where c2 = 931.5 MeV/u

The binding energy for the carbon-12 example is:

EB = (0.098940 u)(931.5 MeV/u)

EB = 92.2 MeVBinding EB for C-12:

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Binding Energy per NucleonAn important way of comparing the nuclei of

atoms is finding their binding energy per nucleon:

Binding energy per nucleon

MeV = nucleon

BEA

For our C-12 example A = 12 and:

MeVnucleon

92.2 MeV 7.6812

BEA

= =

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Formula for Mass DefectThe following formula is useful for mass defect:

( )D H nm Zm Nm M= + − Mass defect

mD

mH = 1.007825 u; mn = 1.008665 u

Z is atomic number; N is neutron number; M is mass of atom (including electrons).

By using the mass of the hydrogen atom, you avoid the necessity of subtracting electron masses.

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Example 4: Find the mass defect for the nucleus of helium-4. (M = 4.002603 u)

( )D H nm Zm Nm M= + − Mass defect

mD

ZmH = (2)(1.007825 u) = 2.015650 uNmn = (2)(1.008665 u) = 2.017330 u

M = 4.002603 u (From nuclide tables)

mD = (2.015650 u + 2.017330 u) - 4.002603 u

mD = 0.030377 u

42 He

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Example 4 (Cont.) Find the binding energy per nucleon for helium-4. (mD = 0.030377 u)

EB = mDc2 where c2 = 931.5 MeV/u

EB = (0.030377 u)(931.5 MeV/u) = 28.3 MeV

MeVnucleon

28.3 MeV 7.07 4

BEA

= =

A total of 28.3 MeV is required To tear apart the nucleons from the He-4 atom.

Since there are four nucleons, we find that

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Binding Energy Vs. Mass Number

Mass number ABind

ing

Ener

gy p

er n

ucle

on

50 100 150 250200

2

6

8

4

Curve shows that EB increases with A and peaks at A = 60. Heavier nuclei are less stable.

Green region is for most stable atoms.

For heavier nuclei, energy is released when they break up (fission). For lighter nuclei, energy is released when they fuse together (fusion).

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Stability Curve

Atomic number ZN

eutr

on n

umbe

r N Stable

nuclei

Z = N

20 40 60 80 100

40

100

140

20

60

80

120

Nuclear particles are held together by a nuclear strong force.

A stable nucleus remains forever, but as the ratio of N/Z gets larger, the atoms decay.

Elements with Z > 82 are all unstable.

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Nuclear Length:A length of one femtometer (fm) = 10-15 m

Nuclear size:Range from 1 fm for single nucleon to 7 fm heavy nuclei

Nuclear Energies:Measured in Million of electron volts

1 eV =1.602x10-19 JNuclear Mass:

Measured in terms of atomic mass units