Nuclear Physics Readings

CCOOMMPPUULLSSOORRYY

RREEAADDIINNGGSS11

1 According to the author of the module, the compulsory readings do not infringe known copyright.

Reading 1: STRUCTURE AND STATIC PROPERTIES OF NUCLEI. Complete reference :Nuclear Physics Lecture notes

From Department of Physic Addis Ababa University, Ethiopia.

Abstract : This is lecture notes given by the author of the Module at Ababa University. Rationale: This section has a well illustrated content on Structure and Static properties of the atomic nucleus. There are illustrative examples and exercises in the topic.

Reading 2: Radio Activity. Complete reference :Nuclear Physics Lecture notes From Department of Physic Addis Ababa University, Ethiopia.

Abstract :This is lecture notes given by the author of the Module at Addis Ababa University. Rationale: This section has a well illustrated content on Radioactivity. There are several illustrative numerical examples and exercises relevant to the second activity of this module.

Reading 3: INTERACTION OF IONIZING RADIATIONS WITH MATTER. Complete reference :Nuclear Physics Lecture notes From Department of Physic Addis Ababa University, Ethiopia. Abstract :This is lecture notes given by the author of the Module at Addis Ababa University. Rationale: This section has a well illustrated content on Interaction of Ionizing Radiation With Matter. There are several illustrative numerical examples and exercises relevant to the second activity of this module.

Reading 4: Nuclear Reacations Complete reference :Nuclear Physics Lecture notes From Department of Physic Addis Ababa University, Ethiopia. Abstract :This is lecture notes given by the author of the Module at Addis Ababa University. Rationale: This section has a well illustrated content on Nuclear Reaction There are several illustrative numerical examples and exercises relevant to the second activity of this module.

RREEAADDIINNGG ##11

NUCLEAR PHYSICS(Physics 481)

STUDY GUIDE & EXERCISES

TILAHUN TESFAYE (PHD)

xÄ!S xbÆ †n!vRSt

E

A

DD

IS A

B A B A U N IVE

RS

IT

Y

ADDIS ABABA UNIVERSITY

FACULTY OF SCIENCE

DEPARTMENT OF PHYSICS

2007/08 A.Y.

CONTENTS

1 STRUCTURE AND STATIC PROPERTIES OF NUCLEI 1-11.1 The Nuclear Hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-1

1.1.1 Early Atomic Theories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-1

1.1.2 Rutherford’s Scattering Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-2

1.2 Properties of the Nucleus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-7

1.2.1 Nomenclature: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-71.2.2 Size of the nucleus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-71.2.3 Nuclear Magnetic Moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-8

1.2.4 Parity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-8

1.2.5 Electric Quadruple Moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-8

1.2.6 Nuclear Spin and Magnetic Moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-9

1.3 Theories of Nuclear Composition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-9

1.3.1 Isotopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-9

1.3.2 electron-proton theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-12

1.3.3 proton neutron theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-13

1.4 Binding Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-14

1.5 Nuclear Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-161.5.1 Elementary Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-16

1.5.2 Yukawa’s Theory of Nuclear Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-17

1.6 Nuclear Structure Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-181.6.1 The Shell Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-191.6.2 The Liquid Drop Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-19

1.6.3 The Collective Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-20

2 Radioactivity 2-12.1 Basic Relations of Radioactivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-1

2.1.1 Law of Radioactive Disintegration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-1

2.1.2 Successive Radioactive Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-42.2 Alpha Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-12

2.2.1 Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-12

2.2.2 Absorption of α-particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-15

2.3 Beta Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-18

2.3.1 β− emission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-18

2.3.2 Absorption of β-particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-20

2.4 γ-Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-22

2.4.1 γ-ray emission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-22

2

2.4.2 Internal Conversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-232.4.3 Internal Pair production . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2-24

3 INTERACTION OF IONIZING RADIATIONS WITH MATTER 3-13.1 Interaction of X and γ-rays with matter. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-1

3.1.1 Interaction Mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-13.1.2 Gamma Ray Attenuation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-5

3.2 Interaction of Charged Particle with Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-7

3.2.1 Interaction Mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-73.2.2 Interaction of Heavy Charged Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-8

3.2.3 Interaction of Light Charged Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-9

3.3 Interaction of Neutrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-103.3.1 Energy Classification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-11

3.3.2 Neutron Sources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-113.3.3 Interaction Mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-113.3.4 Attenuation of neutrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-12

4 Nuclear Reactions 4-14.1 Nuclear Reactions In General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-1

4.1.1 Compound Nucleus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-1

4.1.2 Direct reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-24.1.3 Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-3

4.2 Nuclear Cross-section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-34.3 Classification of Nuclear Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-74.4 Fusion and Fission Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-10

4.4.1 Fusion in Stars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-114.4.2 Fission reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-11

4.5 Reactor Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4-12

UNIT 1

STRUCTURE AND STATIC PROPERTIES OF NUCLEI

Contents1.1 The Nuclear Hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-11.2 Properties of the Nucleus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-71.3 Theories of Nuclear Composition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-91.4 Binding Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-141.5 Nuclear Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-161.6 Nuclear Structure Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-18

Introduction

Searching for the ultimate building blocks of the physical world has always been a central theme in thehistory of scientific research. Many acclaimed ancient philosophers from very different cultures havepondered the consequences of subdividing objects into their smaller and smaller, invisible constituents.Many of them believed that eventually there would exist a final, inseparable fundamental entity of matter,as emphasized by the use of the ancient Greek word, ατoµoσ(atom), which means “not divisible."

Were these atoms really the long sought-after, indivisible, structureless building blocks of the physicalworld?

1.1 The Nuclear Hypothesis

1.1.1 Early Atomic Theories

Greek philosophers, who lived as early as the late 5th century BC, developed a theory of matter that wasnot based on experimental evidence, but on their attempts to understand the universe in philosophicalterms. According to this theory, all matter was composed of tiny, indivisible particles called atoms (fromthe Greek word atomos, meaning “indivisible”). If a sample of a pure element was divided into smallerand smaller parts, eventually a point would be reached at which no further cutting would be possible-thiswas the atom of that element, the smallest possible bit of that element.

Although the notion of atoms as tiny bits of elemental matter is consistent with modern atomic theory,the researchers of prior eras did not understand the nature of atoms or their interactions in materials. Asthere were no methods or technology to test theories about the basic structure of matter hence the viewof the ancient Greeks accepted for centuries.

In his book A New System of Chemical Philosophy (1808), Dalton made two assertions about atoms: (1)atoms of each element are all identical to one another but different from the atoms of all other elements,and (2) atoms of different elements can combine to form more complex substances.

Late in the nineteenth century, 1898, J.J. Thomson advanced the “plum-pudding” theory of atomic struc-ture, holding that negative electrons were like plums embedded in a pudding of positive matter.

1-1

1-2 1. STRUCTURE AND STATIC PROPERTIES OF NUCLEI

Rutherford scattering experiment was a forerunner to the present day nuclear model of an atom.

1.1.2 Rutherford’s Scattering Experiment

The atomic nucleus was first discovered by Rutherford from the scattering of α-particles. All the positivecharge and more than 99% of the mass of an atom is concentrated in the nucleus.

Theoretical and experimental studies in nuclear physics have prominent role in the 20th century physics.There is a reasonably good understanding of the properties of the nuclei and of the structure that isresponsible for those properties. Techniques learned and developed in nuclear physics have been appliedto:

• the study of interactions of matter at a fundamental level;• understand the processes that occurred just after the Big Bang;• perform diagnosis and therapy without recourse to surgery;• build fearsome weapons of mass destruction.

No other field of science comes readily to mind in which theory encompasses so broad a spectrum, fromthe most microscopic to the cosmic, nor is there another field in which direct applications of basic researchcontain the potential for the ultimate limits of good and evil.

Nuclear physics lacks a coherent theoretical formulation that would permit us to analyze and interpret allphenomena in a fundamental way; atomic physics has such a formulation in quantum electrodynamics,which permits calculations of some observable quantities to more than six significant figures. As a result,we must discuss nuclear physics in a phenomenological way, using a different formulation to describe eachdifferent type of phenomenon, such as α decay, β decay, direct reactions, or fission. Within each type, ourability to interpret experimental results and predict new results is relatively complete, yet the methodsand formulation that apply to one phenomenon often are not applicable to another. In place of a singleunifying theory there are islands of coherent knowledge in a sea of seemingly uncorrelated observations.

Some of the most fundamental problems of nuclear physics, such as the exact nature of the forces that holdthe nucleus together, are yet unsolved. In recent years, much progress has been made toward understand-ing the basic force between the quarks that are the ultimate constituents of matter, and indeed attemptshave been made at applying this knowledge to nuclei, but these efforts have thus far not contributed tothe clarification of nuclear properties.

In this course we adopt the phenomenological approach, discussing each type of measurement, the theo-retical formulation used in its analysis, and the insights into nuclear structure gained from its interpre-tation.

On the suggestion of Rutherford, his two associates Geiger and Marsden studied the scattering of theα-particles from a thin gold foil and obtained an important insight in to the structure of the atom. Anα-particle may be considered as doubly ionized helium atom.

Experimental Arrangement:

Rutherford’s alpha scattering experiment was carried out in a fairly good vacuum, the metal box beingevacuated through a tube T (see Figure 1.1). The αs came from a few milligrams of radium (to be precise,its decay product radon 222) at R in figure 1.1. The alpha beam was collimated by means of a diaphragmplaced at D, a narrow beam of alpha particles was directed normally on to the scattering foil F. By rotatingthe microscope M the alpha particles scattered in different directions could be observed on the screen S.

This experiment revealed that a very thin foil scatter α-particle with a large angle. It was totally unexpectresult and Rutherford own words:

“It was quite the most incredible event that ever happened to me in my life. It was almost asincredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hityou."

Department of Physics(Addis Ababa University)

Nuclear Physics (Phys 481)(Semester I 2007/08)

Instructor: Tilahun Tesfaye (PhD)[email protected]

1.1. THE NUCLEAR HYPOTHESIS 1-3

M

P

S

A

C

T

R B

F

D

L

Fig. 1.1: The Rutherford Scattering Experiment

Rutherford interpreted the observation in terms of a model in which the electrons of an atom filled asphere of atomic dimensions, approximately 10−10 m but their charge was neutralized by a central positivecharge on the nucleus of dimensions 10−14 m. In order to get a qualitative relationship between theangle of scattering and the parameters describing the motion of the alpha-particles, Rutherford made thefollowing assumptions.

• Both the α-particles and the nuclei are very small and can be considered as point masses and pointcharges.

• Electrostatic force of repulsion is the only force of interaction between the two

• The nucleus is relatively so heavy that it can be considered to be at rest during interaction.

b

Target Nucleus

(+Ze)

α-particle

(+2e)

π−φ

2

π−φ

2

φ

θ

∆P b

Fig. 1.2: Scattering of -particle in the vicinity of a nucleus





The derivation is as follows: Let

m = mass of the α-particle

b = the impact parameter

φ = angle of scattering# –

P1 = initial momentum of the α-particle# –

P2 = final momentum of the α-particle

θ = angle between ∆P and the force acting at a certain instant

∆#–

P =# –

P2 −# –

P1

= Impulse given to the particle by the nucleus

=

∫#–

F ∆t (1.1)

By hypothesis the nucleus is at rest. Hence the momentum and the kinetic energies of the α particle

remain unchanged in magnitude.

|# –

P1| = |# –

P2| = mv (1.2)

φ

π−φ

2

π−φ

2

∆P =P2+(-P

1)

-P1

P2

Fig. 1.3:

Applying the law of sines to the triangle at the left of fig 1.3, we have

|∆#–

P |

sin φ=

mv

sin(

π−φ2

) =mv

sin(

π2

)cos

(φ2

)− cos

(π2

)sin

(φ2

) = mv

cos(

φ2

)

we know that sin φ = 2 sin(φ2 ) cos(φ

2 )

∴ ∆P =mv

cos φ2

· 2 sin φ2 cos φ

2 = 2mv sin φ2

The change in momentum ∆P is in the same direction as the impulse given by the nucleus to the alphaparticle. Hence the impulse should be

∫Fdt =

∫F cos θdt

∴ 2mv sin(

φ2

)=

∫∞

0F cos θdt (1.3)




1.1. THE NUCLEAR HYPOTHESIS 1-5

From the trajectory shown in figure 1.2 it is seen

θ =−π − φ

2t = 0 i.e. the α is far away from the nucleus

=π − φ

2t = ∞ i.e. the α has gone far away from the nucleus

Hence: 2mv sin(

φ2

)=

∫(π−φ2 )

−(π−φ2 )

F cos θ dtdθdt (1.4)

The angular momentum of the α-particle about the nucleus must remain constant = mvb

∴ mdθ

dtr2 = mvb Remember, L = Iω = (mr2)

(v

r

)= mvr

⇒ dt

dθ=

r2

vb(1.5)

Substituting equation (1.5) in (1.4) we obtain

2mv sin(

φ2

)=

∫F cos θ

(r2

vb

)dθ

=

∫ (2Ze2

4πε◦r2

)r2

vbcos θ

(r2

vb

)dθ

=Ze2

2πε◦vb

∫cos θdθ =

Ze2

2πε◦vb

[sin θ

](π−φ

2

)

−

(π−φ

2

)

=Ze2

πε◦vbcos

φ

2

⇒ b =ze2

πε◦ × 2mvcot

φ

2=

Ze2

4πε◦( 1

2mv2) cot

(φ

2

)

b = ze2

4πε◦T cot(

φ2

)(1.6)

Equation (1.6) shows

• all α-particles that approach the target within b are scattered at an angle φ or more.• α-particles any where within a circle of radius b are scattered at an angle φ ore more. The area πb2

is called the interaction cross-section and is denoted by σ.

Now let the foil of thickness t contain n atoms per unit volume. If the area of the foil over which theα-particles are incident is A, then the number of target nuclei encountered by the α particles is nta. Ifwe assume no overlap of nuclei and scattering of particles is entirely due to a single encounter, then thefraction f of α-particles that are scattered at an angle φ or more to the total number incident is:

fsingle =πb2

A=

σ

A→ for a single scattering center

∴ f =ntAσ

A= ntσ → for all scattering centers contained in (A× t)

= ntπb2

=aggregate cross section

area of target

=

(ntπZ2e4

(4πε◦)2T2

)cot2

(φ2

)by equation (1.6) (1.7)





The fraction of α-particles scattered between φ and φ + dφ is given by:

df = −

(ntπZ2e4

(4πε◦)2T2

)cot2

(φ2

)csc2

(φ2

)dφ (1.8)

The minus sign shows that φ increases as f decreases.

foilR sin φ

φ

R dφ

R

α

Fig. 1.4: Scintillating screen located at a distance R from the foil

α-particles scattered between φ and φ + dφ strike the screen inside an annular ring of radius R sin φ andwidth Rdφ. The area of the ring is :

(2πR sin φ)× Rdφ = 2πR2 sin φdφ

Hence the number of α particles scattered between angels φ and φ + dφ and striking unit area of theobservation screen is:

Nφ = Ni|df|

2πR2 sin φdφ= Ni ×

(ntπZ2e4

(4πε◦)2T2

)cot φ

2 csc2 φ2 dφ

2πR2 sin φdφ

where Ni is the number of incident alph particles.

Using sin φ = 2 sin φ2 cos φ

2 and cot φ2 =

cosφ2

sinφ2

we obtain

Nφ =NintZ2e4

64π2ε2◦R2T2 csc4 φ2 (1.9)

I EXAMPLE 1.1: Energy of α from distance of closest approach

An α-particle is traveling head-on towards a gold nucleus (Z = 79). Its distance of closest approach isr◦ = 3.5× 10−14 m. Calculate the energy of the α-particle in MeV.

Solution:At a distance of closest approach

Potential Energy = Total Kinetic Energy

But EP =1

4πε◦

q1q2

r◦=

14πε◦

2× 79× e2

r◦= 1.04× 10−12 J = 6.5 MeV

Notice that the kinetic energy is converted to electric potential energy r◦. J




1.2. PROPERTIES OF THE NUCLEUS 1-7

1.2 Properties of the Nucleus

1.2.1 Nomenclature:

A specific nucleus of an element X with Z number of protons and A number of nucleons (neutrons +protons) is denoted by

AZX

There is a one-to-one correspondence between Z and X. The number of neutrons N of this nucleus is AUZ.Often it is sufficient to specify only X and A, as in 235U, if the nucleus is a familiar one (uranium is wellknown to have Z=92). Symbol A is called the mass number

1.2.2 Size of the nucleus

The ideas about the shape and size of a nuclei and of the spatial distribution of nucleons in the nucleusar quite arbitrary since nucleons obey quantum laws. However, the mean distribution of nuclear matteris a quassi-classically defined concept.

Rutherford scattering can be used to estimate the nuclear radius using classical ideas. Suppose the goldnucleus in the scattering experiment be considered as a positive sphere of radius R. For the special case ofan alpha particle aimed directly at nucleus, the impact parameter b = 0, and the particle is momentarilyat rest at the distance of closest approach. i.e. all of its kinetic energy is converted to electrical potentialenergy

12

mαv2α =

14πε◦

2eZe

d

⇒ d =Ze2

πε◦mαv2α

= Z×(

e2

πε◦mαv2α

)= Z×

(2e2

4πε◦( 12mαv2

α)

)

= Z× 5.433× 10−16 m using 5.3 MeV for α particles from Po

= 4.3× 10−14 m for gold foil Z=79

= 2.6× 10−14 m for silver foil Z=47

= 0.7× 10−14 m for aluminum foil Z=13

This result is far less than the radius (≈ 10−10 m) of the space the atom occupies as computed from thedensity of the metals and the Avogadro constant or from the kinetic theory of matter. Furthermore, thedistance of nearest approach is only the upper limit.

Nucleon Mass(amu) Charge(Coulomb)

Proton 1.00758 1.602× 10−19

Neutron 1.00898 0

mass number (A) The total number of nucleons in a nucleus.

atomic number (Z) The number of protons in a nucleus.

Thus a nucleus of mass number A and atomic number Z contains (A-Z) neutrons and Z protons. Such anucleus is commonly represented as

AZX





where, X is the symbol of the element to which the nuclide belongs. The nuclide is identified with anelement by its value for Z. Some nuclide have the same value for Z and different values for A. Suchnuclides are referred to as the isotopes of the element whose atomic number is Z.

The volume of a nucleus is proportional to its mass number. The general relation for the radius of anucleus inte3rms of its mass number is

R = R0A1/3

Where R, is the radius of the nucleus and R0 ≈ 1.2× 10−15m is a constant.

1.2.3 Nuclear Magnetic Moment

Since the scale of observation of nuclides and nucleons is small principles, concepts and methods of quan-tum mechanics apply to nuclides and nucleons. As a result for nuclides and nucleons energy, angularmomentum, magnetic moment etc. are quantized. For example, for nucleons orbital angular momentumis an integral multiple of h

2π . Spin angular momentum is ± 12 ( h

2π ). Thus total angular momentum is givenby

ı = `± s

Where ` is an integral multiple of h2π and s = 1

2 ( h2π ). The total angular momentum is

I = L± S

Where L is orbital angular momentum of the nuclide and is an integral multiple of Ãh and S is spin angularmomentum of the nuclide and is a half integral multiple of Ãh. Thus I is an integral multiple of Ãh when Ais even and an odd half integral multiple when A is odd.

The nuclear magnetic momentum of a nucleus is given in terms of the angular momentum by:

µI = γIh

2πI (1.10)

Where γI is a factor that is referred to as the nuclear gyromagnetic ratio.

The statistics that is used for nucleons and nuclides is quantum statistics. For nuclides for which A iseven the statistics tat is used is the Bose-Einstein Statistics. For nucleons, and nuclides for which A isOdd it is the Fermi-Dirac statistics that is used.

1.2.4 Parity

A property that is of interest in connection with nuclides is the parity of a nuclide. The parity of anuclide is said to be odd when the spatial part of the wave function gives different values for (x, y, z) and(−x, −y, −z). The parity of a nuclide is said to be even when the spatial part of the wave function givesthe same values for (x, y, z) and (−x, −y, −z).

1.2.5 Electric Quadruple Moment

Another property that is of interest is the electric quadruple moment of a nuclide. In general the shapeof nuclide is spherical and this quantity is a measure of the deviation of the shape of a nuclide from thespherical shape. Quantitatively the electric quadruple moment of a nuclide is given by

Q =25

Z(b2 − a2) (1.11)

Where b and a are the axes of the ellipsoid of revolution which the nuclide is supposed to occupy.




1.3. THEORIES OF NUCLEAR COMPOSITION 1-9

2a

2b

Fig. 1.5: Electric quadruple moment

1.2.6 Nuclear Spin and Magnetic Moment

Nuclear angular momentum is often known as nuclear spin ÃhI; it is made up of two parts, the intrinsicspin of each nucleon and their orbital angular momenta. We call I the spin of the nucleus, which can takeon integral or half-integral values. The following is usually accepted as facts. Neutron and proton bothhave spin 1

2 (in unit of Ãh). Nuclei with even mass number A have integer or zero spin, while nuclei of oddA have half-integer spin. Angular momenta are quantized.

Associated with the spin is a magnetic moment µI, which can take on any value because it is not quan-tized. The unit of magnetic moment is the magneton

µn =|e|Ãh

2mpc=

µB

1836.09= 0.505× 10−23ergs/gauss where µB is the Bohr magneton. (1.12)

The relation between nuclear magnetic moment and nuclear spin is

µI = γÃhI (1.13)

where γ here is the gyromagnetic ratio (no relation to the Einstein factor in special relativity). Experi-mentally, spin and magnetic moment are measured by hyperfine structure (splitting of atomic lines due tointeraction between atomic and nuclear magnetic moments), deflations in molecular beam under a mag-netic field (Stern- Gerlach), and nuclear magnetic resonance (precession of nuclear spin in combined DCand microwave field).

1.3 Theories of Nuclear Composition

1.3.1 Isotopes

As mentioned in the first section Isotopes of nuclides are nuclides with the same atomic number, Z, butdifferent mass number A, i.e. isotopes are nuclides of the same value of Z. Nuclides of the same value ofA are called Isobars. Isotopes were first identified during positive ray analysis. In positive ray analysispositive ions were made to pass through electric and magnetic fields as shown in the diagram below.

The positive ions are deflected by both the electric field and the magnetic field of the setup. When theions that have passed through the electric and magnetic fields are made to fall on a screen, they describedparabolic curves. For substances with more than one isotopes more than one parabolas were obtaineddemonstrating the existence of isotopes. The fact that different parabolas would be obtained for differentmass numbers can be predicted theoretically as follows. When an ion is moving in the electric field itwould experience a constant force qE actin on it. At the same time, because the ion is moving in amagnetic field it would experience a constant force qvH acting on it.

The two forces are perpendicular to each other. So the motion of the ion is two motions (perpendicularto each other) with constant acceleration. If y ′ represents the deflection by the electric field and Z ′





Scr

een

(ph

otog

rap

hic

pla

te)

+

-

Slit

ChargedPlates

N

S

Insulation

Positiveion Source

Fig. 1.6: positive ray analysis

represents the deflection by the magnetic field. From the relation

S =12

a0t2 (1.14a)

y ′ =12

qE

M

(L

v

)2

(1.14b)

Z ′ =12

qHv

M

(L

v

)2

(1.14c)

where L is the distance moved. From equation 1.14b

Z ′ =12

qH

M

L2

v(1.14d)

⇒ v =12

qH

M

L2

Z ′(1.14e)

substituting this in to equation 1.14d gives:

y ′ =12

qE

M

L2

v2 =12

qE

ML2 4M2Z ′2

q2H2L4 (1.14f)

⇒ y ′ = 2M

q

E

H21L2 Z ′. (1.14g)

⇒ Z ′2 =L2

2q

M

H2

Ey ′ (1.14h)

⇒ Z ′2 = Cq

M

H2

Ey ′ (1.14i)

Which for a given values of H & E describes a parabola. Thus for a given values of H & E

Z ′2 = K ·( q

M

)y ′

This gives different parabolas for different values of qM . For a given substance (ie given q) this gives

Z ′2 = K1 ·1M

y ′

where K1 is a constant for a given substance and setup. So it is seen that for different values of M differentparabolas would be obtained.





Z'

y’ A

B

Parabolas for different mass numbes

Fig. 1.7: parabolas for different mass numbers

This was the way that the existence os isotopes was first detected. Thus many substances are mixturesof isotopes. The isotopes have the same chemical behavior but their behaviour in nuclear reaction differ.The relative abundance can be determined by what is referred to as a mass spectrograph. The massspectrograph (due to Aston) is a modification or improvement of the setup by Thomson to obtain greaterdispersion.

A schematic diagram of a simple spectrograph is as shown in figure 1.8:

MassSpectrograph

Ion source screen

magneticfield

Diaphram

Fig. 1.8: Mass Spectograph

Ion Current

Mass number or accelerating voltage

Fig. 1.9: Ion current vs mass

number or accelerating voltage

With this setup instead of parabolas, lines are obtained on the screen. Different lines are obtained fordifferent mass numbers. By comparing the position of the line for a given mass number with the positionof line of a known mass number the mass number of isotopes and their relative abundance are determined.

The relative abundance of the isotopes of a substance can also be measured with what is referred to asa mass spectrometer. Just as with the mass spectrograph, in a mass spectrometer (due to Dempster) thepositive ions are made to pass through an electric field and a magnetic field consecutively. But in massspectrometer, the ions that have passed through the electric and magnetic fields are connected with anelectrometer, hence the name mass–spectrometer.

For the motion of an ion through the electric field, qV = 12Mv2. For the motion of the ion in the mag-

netic field Hqv = mv2

R ie magnetic force is equal to centrifugal force. From the first relation, v =√

2qvm ,

substituting this in to the second relation and rearranging gives





R =

√2V

H2q·√

M

So for a given substance only the nuclides of a certain mass number will pass through the opening to theelectrometer and be counted. So by varying the voltage between the two plates the number of nuclei ofthe different isotopes of a substance can be counted. In this way when a plot of the ion current versusmass number or accelerating voltage is made plots of the types shown below are obtained

The points where there are peaks of the ion current represent the positions of the peaks are proportionalto the relative abundances of the isotopes.

In these method and also other methods the isotopes of the elements have been identified.

When the nuclear model of the atom was advanced the composition of the nucleus became a crucialproblem of nuclear physics. An answer to this question could only be given after the discovery of variousproperties of the nucleus, notably nuclear charge Z, nuclear mass, and nuclear spin. The nuclear chargewas found to be defined by the sum of the positive charges it contains. Since an elementary positive chargeis associated with the proton, the presence of protions in the nucleus appeared to be beyond any doubtfrom the outset.

Two more facts were also established, namely:

1. The masses of the isotopes (except ordinary hydrogen), expressed in proton mass units, were foundto be numerically greater than their nuclear charges expressed in elementary charge units, thisdifference growing with increases in Z. For the elements in the middle of the periodic Table theisotopic masses (in amu) are about twice as great as the nuclear charge. The ratio is still greater forthe heavier nuclei. Hence one was forced to think that the protons were not the only particles thatmake up the nucleus.

2. The masses of the isotopic nuclei of all chemical elements suggested two possibilities, either the par-ticles making up the nucleus had about the same mass, or the nucleus contained particles differingin mass to a point where the mass of some was negligible in comparison with that of the others, thatis, their mass did not contribute to the isotopic mass to any considerable degree

1.3.2 electron-proton theory

The electron proton model nicely fits with the second possibility mentioned above. Further the the nucleusmight contain electrons seemed to follow from the fact that natural beta-decay is accompanied by theemission of electrons. The proton-electron model also explained the fact why the isotopic atomic weightswere nearly integers. According to this model, the mass of the nucleolus should be partially equal to themasses of the protons that make it up, because the electronic mass is about 1/2000th that of the proton.The number of electrons in the nucleus must be such that the total charge due to the positive protons andthe negative electrons is the true positive charge of the nucleus.

For all its simplicity and logic, the proton-electron model was refuted by advances in nuclear physics. Infact, it ran counter to the most important properties of the nucleus.

If the nucleus contained electrons, the nuclear magnetic moment would be of the same order of magnitudeas the electronic Bohr magneton. Notice that the nuclear magnetic moment is defined by the nuclearmagneton which is about 1/2000th the electronic magneton.

Data on nuclear spin also witnessed against the proton-electron model. For example, according to thismodel the beryllium nucleus,94Be, would contain nine protons and five electrons so that the total chargewould be equal to four elementary positive charges. The proton and the electron have each a half-integralspin, h/2. The total spin of the nucleus made up of 14 particles (nine protons and five electrons) wouldhave to be integral. Actually, the beryllium nucleus, 9

4Be, has half-integral spin of magnitude 3h/2. Manymore examples might be cited.





Last but not least, the proton-electron model conflicted with the Heisenberg uncertainty principle. If thenucleus contained electrons, then the uncertainty in the electron position, ∆x would be comparable withthe linear dimensions of the nucleus, that is, 10−14 or 10−15m. Let us choose the greater value, From theHeisenberg uncertainty relation for the electron momentum we have

∆P ≈ h/∆x ≈ 10−14 = 10−19kgm/s

The momentum P is directly related to its uncertainty, that is P ≈ ∆P. Once the momentum of the electronis known, one can readily find its energy. Since in the above example, P >> mec = 10−30kg× 3× 108m/s

one should use the relativistic relation for energy and momentum:

E2 = c2p2 + m2ec4

Then we can get

E = c√

p2 + mec2 = 3× 108√

10−38 + (10−30 × 3× 108)2

≈ 2× 108 eV = 200 MeV

This figure is greatly in excess of that (7-8 MeV)found for the total binding energy by experiment andis many times the energy of electrons emitted in beta-decay. If, on the other hand, the electrons in thenucleus were assumed to have the energy comparable with that associated with the particles emittedin beta-decay (usually a few MeV), then the region where the electrons must be localized, that is, thesize of the nucleus as found from the uncertainty relations would be much greater than that found byobservation.

1.3.3 proton neutron theory

A way out, of the inconsistencies posed by electron proton theory, was found when in 1932 Chadwickdiscovered a new fundamental particle. From an analysis of the paths followed by the particles producedin some nuclear reactions and applying the law of conservation of energy and momentum, Chadwickconcluded that these paths could only be followed by a particle with a mass slightly greater than that ofthe proton and with a charge of zero. Accordingly, the new particle was called the neutron.

Soon after Chadwick’ a discovery, hypothesis was forwarded that the atomic nucleus should consist solelyof protons and neutrons. A similar hypothesis was advanced by Heisenberg.

Before long these views met with general acceptance and served as a basis for the present-day theory ofthe atomic nucleus. According to the latest views, the mass number A represents the total number ofprotons and neutrons in the nucleus. The charge number Z, a multiple of the proton charge, defines thenumber of neutrons so that the difference A − Z = N defines the number of neutrons in the nucleus of agiven isotope.

The proton-neutron model of the nucleus accounts for both the observed values of isotopic masses and,the magnetic moments of the nuclei. For, since the magnetic moments of the proton and the neutron areof the same order of magnitude as the nuclear magneton, it follows that a nucleus built up of nucleonsshould have a magnetic moment of the same order as the nuclear magneton. Therefore, with protons andneutrons as the building blocks of nuclei, the magnetic moment should be of the same order of magnitude.Observations have confirmed this.

Also with protons and neutrons as the constituents of nuclei, the uncertainty principle leads to reasonablevalue of energy for these particles in a nucleus, in full agreement with the observed energies per particle.

Finally, with the assumption that nuclei are composed of neutrons and protons, the difficulty arising fromnuclear spin has likewise been resolved. For if a nucleus contains an even number of nucleons (an evenmass number A), it has integral spin (in units of h). With an odd number of nucleons (an odd mass numberA), its spin will be half-integral (in units of h).





???????????????? Exercises: 1.1

1. Is the average angle of deflection of α-particles by a thin gold foil predicted by Thompson modelmuch less, about the same, or much greater than that predicted by Rutherford’s model?

2. Outline in brief Rutherford’s model of the nuclear atom. What were the facts which could not beexplained by Rutherford’s model of the atom?

3. During the measurement of the mass of a nuclide with a mass spectrometer, if the charge of the ionis q = 1.602× 10−19 C, the accelerating voltage v = 1000 V, the magnetic intensity, H = 1000 G andthe radius of the path of the ion is 18.2 cm determine the mass of the nuclide

4. atoms of 238U and 235U are

(a) isotopes(b) isobars(c) isomers(d) isotones

5. As the photon is a quantum in electromagnetic field theory, which one of the following is consideredto be the quantum in nuclear field

(a) neutrino(b) electron(c) meson(d) neutron

6. Which element has a nuclear radius about twice that of the carbon nucleus?

1.4 Binding Energy

It has been observed experimentally that nuclei with certain numbers for their protons or neutron i.e.,with certain values of Z or (A-Z) are more stable than other nuclei. This numbers are referred to as magicnumbers. The magic numbers are 2, 8, 20, 50, 82, 126. Nuclides with number of protons or neutronsequal to any one of the magic numbers are stable. Nuclides are made up of neutrons and protons, but themass of a nuclide is not the same as the sum of the mass of the neutrons and protons of which it consists.The mass of a nuclide is less than the sum of the masses of the protons and neutrons of which it is madeup. The difference in the masses is referred to as the mass defect.

The mass defect(∆M) is given by

∆M = (A − Z)mn + Zmp = MN (1.15)

where MN is the mass of a nuclide, mn is the mass of a neutron, mp is the mass of a proton. By the energymass relation this is related to the binding energy by:

B.E. = ∆MC2 (1.16)

This energy is the energy that is necessary to break the nuclide apart. From experimental observationbinding energy per nucleon varies with mass number as shown in figure 1.10




1.5. NUCLEAR FORCE 1-15

020

4060

80100

120140

160180

200220

240

Mass Number (A)

10

9

8

7

6

5

4

3

2

1

0

H-1

He-3

Li-6

B-10

He-4

Al-27

Cu-63Sn-120

Pt-195U-238

Th-232

Bin

din

g E

ne

rgy p

er

Nu

cle

on

(M

eV

)

Fig. 1.10: Dependence of Binding Energy Per nucleon on mass number

???????????????? Exercises: 1.2

1. Calculate the binding energy per nucleon of 3517Cl nucleus (its mass being 34.9800 a.m.u.) Given mass

of 10n=1.008665 a.m.u., mass of 1

1H=1.007825 a.m.u.

2. Start from

∆M = ZmH + (A − Z)mn − MZ,A,

show that the average binding per nucleon for all but the lightest elements is close to 8 MeV.

3. Show that the distance of closest approach d, in Rutherford scattering leading to an angle of deflec-tion φ is given by

d =p

2(1 + csc

φ

2)

where p is the distance of closest approach for zero impact parameter.[Use conservation of energy and angular momentum]

4. Assuming the kinetic energy T of a nucleon in a nucleus to be equal to the binding energy per nucleonεb = 8 MeV, estimate the de Broglie wavelength λ of a nucleon and compare it with the size of thenucleus.

5. Determine the minimum energy Emin (in MeV) required for splitting a 12C nucleus into three α-particles.





1.5 Nuclear Force

1.5.1 Elementary Particles

The discussion and explanation of nuclear force is connected with the physics of elementary particles.Among the particles that are of importance in nuclear physics are the ones given in table 1.1.

Table 1.1: Characteristic Properties of Elementary Particles (Fenyves and Haiman, 1969)

Group Particle Anti- Decay Mean

life time

Rest

mass

particle (sec) (in terms

of me )

Photons Photon γ 0

2 Neutrino ν 0

3 Electron β− 1

4 Muon µ− 207

5 Pion (pi Meson) π+, π0 273

6 Charged K Meson k+ 966

7 Neutral k Meson k0 966

8 proton p 1836

9 Neutron n 1836

10 Lambda hyperon λ0 2181

11 positively charged sigma hyperon Σ+ 2328

12 Neutral Sigma Hyperon Σ0 2335

13 Negatively charged sigma hyperon Σ− 2343

14 Neutral Xi Hyperon Ξ 2584

15 charged Xi hyperon Ξ− 2584

Many of these particles have their anti-matter counterpart. For example there is p− for p, for β− thereis β+, for k+ there is k− for Ξ− there is Ξ+ etc. When a particle and its antiparticle meet they annihilateeach other.

Particle are in general classified into two types according to the statistics they obey.

I. Fermions:

• Obey the FD statistics

• have half integral spin i.e. Ãh2 , 3Ãh

2 , 5Ãh2 , . . .

• β−, n, p, λ, Σ are examples of Fermions

Fermions are further classified as Baryons (Fermions of massm > mp) and Leptons (Fermions ofmass m < mp).

II. Bosons:

• Obey the BE statistics

• have integral spin i.e. Ãh, 2Ãh, 3Ãh, . . .

• γ, π, κ are examples of Fermions




1.5. NUCLEAR FORCE 1-17

Bosons are further classified as Photons (Bosons of zero rest mass ) and mesons (Bosons of non-zerorest mass)

Mesons and baryons, which interact strongly with nuclei (Nucleons) are also referred to in general ashadrons. On the other hand leptons and photons do not interact strongly with nuclei.

1.5.2 Yukawa’s Theory of Nuclear Forces

In covalent bonding, molecules are held together by sharing (exchanging) electrons. In 1936, Yukawaproposed a similar mechanism to explain nuclear forces.

According to Yukawa’s theory (also known as meson theory) all nucleons consist of identical cores sur-rounded by a cloud of one or more mesons and each nucleon continuously emitting and absorbing pions.i.e. the force between nucleons is explained as being the exchange of elementary particles by nucleons byone of the following processes.

p À p + π0

n À n + π0

p À n + π+

n À p + π−

These equations violet the law of conservation of energy. A proton of mass equivalence of 938 MeV becomesaneutron with 939.55 MeV and ejects a pion with 139.58 MeV! This energy conservation violation canhappen only if the violation exists for such short time that it can not be measured or observed by theHeisenberg’s uncertainty principle:

∆E∆t > Ãh

so the violation can exist only if

∆E∆t 6 Ãh ⇒ ∆t 6 h

∆E=

h

mπc2

during this time, even if the pion moves with the speed of light, the distance that it can move is

∆r = c∆t

the range of nuclear force. i.e. the distance within which the exchange of pions by nucleons takes place.

⇒ ∆t =1.5× 10−15m

3× 108 m≈ 0.3× 10−23 sec

⇒ ∆E =Ãh∆t

= 3.5152× 10−11 J = 145.57 MeV

This is close to the measured value of pion mass. Therefore Yukawa’s theory (the meson theory) satisfiesthe two important characteristics of nuclear forces.

(i) Nuclear force is the same between any two nucleons. i.e. p-p; p-n and n-n forces are the same. Thisis satisfied by the meson theory sice there are three types of mesons with the same mass.

(ii) Exchange of π meson (a particle of non-zero rest mass) by nucleons satisfies the short range natureof nuclear forces. As reasoned above, the energy violation can happen only if the the exchange tookplace with in the limits of nuclear dimension.





Potential

distance

2

re

Vr

µ

γ−

= −

Fig. 1.11: Yukawa Potential

The potential for the π meson field is approximately given by

V(r) = −γ2 e−µr

r(1.17)

where γ is a constant and µ = mπCÃh . This is commonly referred to as Yukawa Potential.

The attractive force between nucleons does not exist for distance between nucleons below a certain lim-iting distance. For distances less than a limiting distance, the force between nucleons is a very strongrepulsive force. The limiting distance is about 0.5 F. This repulsive force is believed to be due to exchangeof π mesons. The repulsion is often taken to be a hard core, i.e., a region where the potential goes toinfinity. Thus the general nature of the force between nucleons is as shown in the diagram shown below.

???????????????? Exercises: 1.3

1. From the Fact that the diameter of nuclei is ≈ 10−14 m using Heisenberg’s uncertainty principle inthe form (∆p)(∆x) > h estimate the mass of a π meson.

1.6 Nuclear Structure Models

In the last section the forces which keep nucleons together in nuclei have been discussed. Nuclei are madeup of many nucleons. The behavior of each nucleon contributes to the resultant behavior of nuclei. Theapproximations of the resultant behavior of nuclei due to the behaviors of nucleons comprising them arereferred to as nuclear models.

The development of nuclear models is based on experimental observations in connection with the behav-iors of nuclei. As it is done with the development any theoretical model, first a theoretical model is putforward; then predictions are made with the model about some observable phenomenon. If the predictionagrees with the observed phenomenon the model is good, if it disagrees it is incorrect; and if it agrees withobservation to a limited degree it is correct or good to the degree to which it agrees with observation.

The phenomena that have been used as the measure for the testing of nuclear models among many are:

i) the binding energy per nucleon of nuclei

ii) nuclear reactions including fission reaction

iii) the stability of nuclei.

The development of nuclear models is connected with the above two observations i.e. the stability ofnuclides with the number of protons or neutrons equal to any one of the magic number and the relation




1.6. NUCLEAR STRUCTURE MODELS 1-19

between binding energy and mass number have been used as the tests for the validity of the models. Withthis in mind there are three nuclear models that are of importance. These are

i) The shell model or independent particle model

ii) The liquid drop model

iii) The collective model

1.6.1 The Shell Model

In shell model nucleons are treated as individual particles existing within the potential created by thenucleons of a nuclide. Hence the shell model is also referred to as the independent particle model. This issimilar to the treatment of the electrons of atoms in atomic physics. In the shell model of nuclear physicsthough, the potential is due to both electromagnetic potential and nuclear potential with this in mind, thepotential that a nucleon finds itself in a nuclide is

V(r) = Vn(r) =−V0

1 + e−(r−R)/afor a neutron

V(r) = Vn(r) + Ve(r) =−V0

1 + e−(r−R)/a+ Ve(r) for a proton

V(r) =

Ze2

4πε0Re

[1 + 1

2

(1 −

(r

Re

)2)]

, for r < Re;

Ze2

4πε0r , for r > Re.

where

V0 = 57± 27(A − 2Z)

AMeV

+ for protons and - for neutrons

R = 1.25 A4/3F, a constant for a nuclide a = 0.65 F, a constant

With this relation for the potential and the assumption that for nucleons their is a strong spin-orbitcoupling. Solving Schrödinger’s equation for nucleons in nuclides predicts the fact that for Values of Zor (A-Z)=2,8,20,28,50,82,126,and 184 there would be closed shells, i.e., stable nuclides in agreement withwhat is observed experimentally.

1.6.2 The Liquid Drop Model

In the liquid drop model of nuclei, nuclei are considered to behave like drop of incompressible liquid,i.e. like drops of very high density (density of the order of 1014 kg/m3) with this point of view and usingconcepts from classical physics (i.e. physics of continua) concepts like surface tension and surface energy,volume energy etc predictions are made about the overall behavior of nuclides. One of the predictions, asmentioned above, is about the relation between the binding energy and mass number for nuclides.

Using the liquid drop point of view, the binding energy of a nuclide would be the resultant of five energies,the volume energy (Ev), the surface energy (Es), the energy due to asymmetry (ie, deviation from a stableconfiguration Ea), the energy due to even-odd combination of nucleons in a nuclide, Eδ and coulomb energy(for protons) Ec. With this the total binding energy for a nuclide is

B.E = Ev + Es + Ea + Eδ + Ec





From the above point of view and other simple reasoning the relation for Ev, Es, Ea, Eδ and Ec are

Ev = CvA

Es = −CsA2/3

Ea =−Ca [(A − Z) − Z]2

A=

−Ca (A − 2Z)2

A

Eδ =

δ/2a, for even-even nuclides;

0, for even-odd or 0dd-even nuclides;

−δ/2a, for odd-odd nuclides.

Ec =−4CcZ(Z − 1)

A1/3

where

Cv ≈ 14 MeV

Cs ≈ 13.1 MeV

Ca ≈ 19.4 MeV

δ ≈ 270 MeV

Cc ≈ 14 MeV

From this the mass of a nuclide is given by

Mn = (A − Z)mn + Zmp −B.EC2

where the B.E. is as given in 1.16 This relation is referred to as the semi-empirical mass relation. Thesepredictions by the liquid drop model and other predictions such as predictions about the fission of nuclidesare in agreement with observation.

1.6.3 The Collective Model

The term collective model is not a specific and clear term. In much of the literature on nuclear model theterm is used to imply any model that deals only with the collective behavior of nucleons. With this pointof view even the liquid drop model can be looked at as a collective model. In some parts of the literaturethe term is used with any model that takes collective effects in to account. Here it is the latter usage thatis being taken. With this in mind in the collective models when the overall behavior of a nuclide is whatis of concentration of some concepts from classical physics are used except energy is quantized.

Nuclide can have rotational energy or vibrational energy. In both cases the energies will be integermultiples of a phonon hνλ. With this in the overall modeling of the structure of nuclei, first makingcertain assumptions about the nature of nuclei the Hamiltonian for a certain model is derived. Then theHamiltonian is solved and the wave function for the nuclide or the nucleon of interest is determined. Thenpredictions with the wave function are compared with experimental observation. From this the modelis evaluated depending on the degree of agreement. and disagreement. The experimental observationthat played an important role in the development of collective nuclear models is the observation aboutphoton-nuclear reaction, namely the observation about giant resonances in photonuclear reactions. Forthe collective behavior of nucleons, nuclides are considered to consist of two fluids. A proton fluid and aneutron fluid.

The proton and neutron liquids could undergo rotational and vibrational motion at their surfaces. Inaddition to this in the presence of electromagnetic fields there could be density fluctuations of the densityof proton ρp(r, t), and the density of neutron ρn(r, t) and resulting dipole, quadruple etc resonances.




1.6. NUCLEAR STRUCTURE MODELS 1-21

This is due to the fact that electromagnetic fields (& Photons) react only with the protons.

In addition to these two collective behaviors the structure of a nuclide can be affected by the individualmotion of the individual motion of the individual nucleons comprising it.

So putting these three factors together the Hamiltonian for a nucleus would be:

H = Hsurface effects + Hgiant resonance + Hinteraction (1.18)

where

Hs, the hamiltonian due to surface effects

Hgr, the hamiltonian due to giant resonance

Hint, the hamiltonian due to the interaction between individual motion

and collective motion

The Hamiltonian for the nucleus is

Hs = Hs + Hgr + Hint&Hs = Hvib + Hrot

ThereforeH = Hvib + Hrot + Hgr + Hint

From the point of view of the individual particle (nucleon) the Hamiltonian would be the sum of theHamiltonian of the collective motion, the Hamiltonian of the individual particle and the Hamiltonianthat takes into account the interaction between the individual motion and the collective motion. Thus fora particle,

H = Hpart + Hcoll + Hint

so in collective models individual particles move in deformed shell potential and nuclei as a whole behavelike incompressible fluid with their motions (vibratory and rotational) being influenced and affected bythe motion of the individual particles inside the nuclei.

???????????????? Exercises: 1.4

1. On the basis of classical theory Rutherford’s atomic model was

(a) Unstable(b) stable(c) semi stable(d) metastable

2. Rutherford and Geiger and Marsden’s experiments on scattering alpha particles from metal foilssuggested

(a) The electrons formed a hard, impenetrable shell about the nucleus(b) That an extremely small, positively charged nucleus existed(c) The positive charges in an atom are evenly distributed throughout the atom(d) Neutrons exist in the nucleus

3. What is the radius of iodine atom (At No. 53 mass number 106)?

(a) 2.5× 10−11 m





(b) 2.5× 10−9 m(c) 7× 10−9 m(d) 7× 10−6 m

4. In Rutherford scattering experiment, what will be correct angle of scattering for an α particle ofimpact parameter b=0?

(a) 90◦

(b) 270◦

(c) 0◦

(d) 180◦

5. An α-particle of energy 6 MeV is projected toward a nucleus of atomic number 50. The distance ofnearest approach is

(a) 2.4× 10−10 m(b) 2.5× 10−12 m(c) 2.5× 10−14 m(d) 2.5× 10−20 m




RREEAADDIINNGG ##22

UNIT 2

RADIOACTIVITY

Spontaneous transformation of the atomic nucleus, leading to a change in the composition and/or internalenergy of the nucleus, is known as radioactivity. Among well known radioactive processes are alphadecay, beta decay, gamma radiation, spontaneous fission of heavy nuclei, emission of delayed protons andneutrons and delayed fission.

2.1 Basic Relations of Radioactivity

When a radioactive disintegration occurs with the emission of α-, and β-particles, the original atom calledthe parent atom changes into something else called the daughter.

2.1.1 Law of Radioactive Disintegration

Decay Law

Radioactivity is a property of the nucleus and consequently, the probability (λ)of radioactive decay perunit time is constant for a given nucleus in a given energy state. i.e. the number dN of nuclei that decayradioactively per unit time is proportional to the number N of nuclei.

dN(t)

dt= −λN(t) (2.1)

where λ is the disintegration probability also known as disintegration (decay) constant. Solving thisequation shows the number of nuclei of radioactive nuclide decreases with time exponentially.

N(t) = N(0)e−λt = N◦e−λt (2.2)

If we treat N as a continuous variable, which is justified as long as N is very large compared to dN,and taking the statistical nature of radioactivity we can obtain the law of radioactive decay from purestatistical arguments as follows.

Let P be the probability for a nucleus to decay. Further let P be independent of the age or past history ofa nucleus and it is the same for all. Therefore P depends only on ∆t

⇒ P = λ∆t

Probability Q1, that nucleus will not decay in ∆t is

Q1 = 1 − P = 1 − λ∆t

The probability Q2, that a nucleus will not decay in 2∆t is

Q2 = (1 − P)(1 − P) = (1 − P)2

2-1

2-2 2. RADIOACTIVITY

Ingeneral, the probability Qn, that a nucleus survives n such intervals is

Qn = (1 − ∆t)n

Consider a fininte time interval: t = n∆t, the probability Qn will be

Qn =

(1 −

λt

n

)n

using Binomial series:

(1 + x)m = 1 + mx +m(m − 1)

2!x2 +

m(m − 1)(m − 2)

3!x3 + . . .

it follows

Qn = 1 − nλt

n+

n − (n − 1)

2!λ2t2

n2 −n(n − 1)(n − 2)

3!λ3t3

n3

= 1 − λt +

(1 −

1n

)λ2t2

2!−

(1 −

1n

)(1 −

2n

)λ3t3

3!+ . . .

∴ limn→∞

Qn = 1 − λt +λ2t2

2!−

λ3t3

3!+ . . .

= e−λt =N

N◦

⇒ N = N◦e−λt

Half life

A quantity that is of interest in connection with the radioactive decay of nuclide is what is referred toas the half-life t1/2 of a nuclide. It is the time it takes a given number of nuclide to decrease by half innumber via radioactive decay. The half life of a nuclide is related to the decay constant of the nuclide.

If t1/2 is the half life of a nuclide, then equation (2.2) implies:

N(t1/2) =N(0)

2= N(0)e−λt1/2

⇒ e−λt1/2 =12⇒ −λt1/2 = ln

12

= −0.693

⇒ −t 12

=−0.693

λ⇔ λ =

0.693t1/2

(2.3)

Mean Life (τ)

The mean or average life is the average life time for the decay of radioactive atoms. The mean life τ iscalculated as follows.

The number of atoms dN which decay between t and t + dt is given by:

dN = λNdt ⇒ dN = λN◦e−λtdt by equation (2.2) (2.4)

since the decay process is statistical, any single atom may have a life sapn from 0 to ∞

∴ τ =1

N◦

∫∞

0N◦λe−λtdt = λ

∫∞

0te−λtdt =

1λ

(2.5)




2.1. BASIC RELATIONS OF RADIOACTIVITY 2-3

Activity

The rate of decay of a radioactive material is known as its activity. i.e.

A =dN

dt⇒ A = −λN

Equation (2.2) can be expressed interms of activity as:

A = A◦e−λt

The unit of activity is the Curie (Ci).

1 Ci = 3.7× 1010 disintegrations per second (dps)

Note:-

� The definition of Curie is based on the rate of decay of 1 g of radium which was originally measuredto be 3.7 × 1010 dps. Although recent measurements have established that the decay rate is 3.6 ×1010 dps/g of radium, the original definition of Curie remain unchanged.

The SI unit for activity is becquerel (Bq). The becquerel is smaller but more basic than the Curie

1 Bq = 1 dps = 2.70× 10−11 Ci

Another unit of radioactivity is Rutherford (rd)

1 rd = 106 dps

I EXAMPLE 2.1: Estimating half-life from activity

One gram of radium is reduced by 2.1 mg in 5 years by α-decay. Calculate the half-life of radium.

Solution: Initial mass is 1 g. Therefore mass left behind is 0.9979 g

N

N◦= e−λt ⇒ 0.9979

1= e−λt ⇒ ln(5λ) = ln 0.9979 − ln 1 = −2.102× 10−3

⇒ T = 1672 years

notice that the number of nuclei is proportional to the mass in a given sample J

I EXAMPLE 2.2: Carbon-14 Dating

A carbon specimen found in a cave contained 18 as much 14C as an equal amount of Carbon in living

matter. Calculate the approximate age of the specimen. (14C has a half life of 5568 years.)

Solution:

λ =0.693t1/2

=0.6935568

per year

N

N◦= e−λt =

18

⇒ t =ln 1

8−λ

≈ 16708 years

Notice that it is the ratio of the two isotops of carbon that is used, not the actual amount of carbon in aspecimen, that is used to determine the age of organic materials. J





2.1.2 Successive Radioactive Transformations

In nature there are three long chains, or radioactive series, of radio elements stretching through the lastpart of the periodic system of elements. These are the Uranium, Actinium and Thorium series. In theUranium series the mass number (A) of each member can be expressed as A = 4n + 2, where n is aninteger. Similarly A = 4n + 3 in the Actinium and A = 4n in the Thorium series.

The 4n + 1 series (Neptunium series) is missing in nature because its longest-lived radionuclide 237Np(2.20 × 106 yr) has a relatively short half-life compared with the approximate age of the earth’s crust,5 × 109 yr. The Uranium series, Thorium series, Actinium series and the Neptunium series. The seriesare as given in tables 2.1 to 2.4.

Table 2.1: The Uranium Series (4n + 2)

Nuclides Type of Half lifeDecay

23892U α 4.5× 109 yrs

23490Th β 24.1 days

23491Pa β 6.7 hrs

23492U α 2.5× 105 yrs

23090Th α 8.0× 104 yrs

22688Ra α 1620 yr

22286Rn α 3.82 d

21884Po α 3.05 min

21482Pb β 26.8 min

21483Bi β 19.7 min

21484Po α 1.64× 10−4 yrs

21082Pb β 19.4 yr

21083Bi β 5.0 d

21084Po α 138.3 d

20682Pb – stable

Table 2.2: The Actinium Series (4n + 3)


23592U α 7.1× 108 yrs

23190Th β 25.6 hrs

23191Pa α 3.43× 104 yrs

22789Ac β 21.6 yrs

22790Th α 18.17 d

22388Ra α 11.68 d

21986Rn α 3.92 sec

21584Po β 1.83× 10−3 sec

21585At α 10−4 sec

21183Bi β 2.15 min

21184Po α 0.52 sec

20782Pb — stable

Apart from the members of these natural chain of radio elements, there are stand alone radioactiveelements in nature. Some of them are 40K (t1/2 = 1.27 × 109 yr), 48Ca (t1/2 > 7 × 1018 yr), 129I (t1/2 =1.6× 107 yr), 14C (t1/2 = 5568 yr) etc.





Table 2.3: The Thorium Series (4n)


23290Th α 1.39× 1010 yrs

22888Ra β 6.7 hrs

22889Ac β 6.13 yrs

22890Th α 1.91 yrs

22488Ra α 3.64 d

22086Rn α 51.5 d

21684Po β 0.16 sec

21685At α 3× 10−4 sec

21283Bi β 60.5 sec

21284Po α 3× 10−7 sec

20882Pb α stable

Table 2.4: The Neptunium Series (4n + 1)


24194Pu α 13 yrs

23792U β 6.75 days

23793Np α 2.2× 106 yrs

23391Pa β 27 d

23392U α 1.62× 105 yrs

22990Th α 7340 yr

22588Ra β 14.8 d

22589Ac α 10 d

22187Fr α 4.8 min

21785At α 0.018 sec

21383Bi α 47 min

20981Tl β 2.2 min

20982Pb β 3.3 h

20983Bi — stable

Radioactive Growth and Decay

Equation (2.2) is the solution for the case only when a nuclide decays into another stable nuclide. For achain of decay, let us assume nuclide 1 decay into nuclide 2 and nuclide 2 in turn decay into nuclide 3 andso on. For such a case, assuming nuclide 3 is stable, the number of nuclei of nuclides (N1(t), N2(t), N3(t),)obey the following relations:

dN1(t)

dt= −λ1N1(t) (2.6)

dN2(t)

dt= −λ2N2(t) + λ1N1(t) (2.7)

dN3(t)

dt= λ2N2(t) (2.8)

The solution for N1(t) is

N1(t) = N1(0)e−λ1t





To get the solution for N2(t), substitute the solution for N1(t) into equation 2.7. Thus

dN2(t)

dt= −λ2N2(t) + λ1N1(0)e−λ1t

⇒ dN2(t)

dt+ λ2N2(t) = λ1N1(0)e−λ1t

From the general form of the solution of first order differential equation the solution for N2(t) is:

N2(t) = e−∫

λ2t

∫λ1N1(0)e−λ1(t)e

∫λ2tdt + Ce−

∫λ2t

⇒ N2(t) = e−λ2t

∫λ1N1(0)e−λ1teλ2tdt + Ce−λ2t

⇒ N2(t) =λ1

(λ2 − λ1)N1(0)e0 + Ce0

⇒ C = N2(0) −λ1

λ2 − λ1N1(0)

∴ N2(t) =λ1

(λ2 − λ1)N1(0)e−λ1t −

λ1

λ2 − λ1N1(0)e−λ2t + N2(0)e−λ2t (2.9)

To get the solution for N3(t) substituting the solution for N2(t) into the equation for N3(t) gives:

dN3(t)

dt= λ2N2(t)

=λ2λ1

(λ2 − λ1)N1(0)e−λ1t −

λ1λ2

(λ2 − λ1)N1(0)e−λ2t + λ2N2(0)e−λ2t

Integrating this gives:

N3(t) =λ2λ1

(λ2 − λ1)N1(0)

∫e−λ1tdt −

λ2λ1

(λ2 − λ1)N1(0)

∫e−λ2tdt+

+ λ2N2(0)

∫e−λ2tdt

⇒ N3(t) =−λ2

(λ2 − λ1)N1(0)e−λ1t +

λ1

(λ2 − λ1)N1(0)e−λ2t − N2(0)e−λ2t + C1

N3(t) =−λ2

(λ2 − λ1)N1(0)e0 +

λ1

(λ2 − λ1)N1(0)e0 − N2(0)e0 + C1

⇒ C1 = N3(0) +λ2

(λ2 − λ1)N1(0) −

λ1

(λ2 − λ1)N1(0) + N2(0)

⇒ N3(t) = N3(0) + N2(0)[1 − e−λ2t

]+

N1(0)λ2

(λ2 − λ1)

[1 − e−λ1t

]

− N1(0)λ1

(λ2 − λ1)

{1 − e−λ2t

}

(2.10)

These are then, the solution for N1(t), N2(t)andN1(t)

N1(t) = N1(0)e−λ1(t) (2.11a)

N2(t) = N1(0)

[λ1

(λ2 − λ1)e−λ1t −

λ1

(λ2 − λ1)e−λ2t

](2.11b)

N3(t) = N1(0)

[1 +

λ1

(λ2 − λ1)e−λ1t −

λ2

(λ2 − λ1)e−λ2t

](2.11c)

The time evolution of the concentrations N1(t), N2(t)andN3(t) looks as shown in figure 2.1





N(t)

N1(t)

N2(t)

N3(t)

t

Fig. 2.1: N1,2,3(t) vs t

The above analysis is for a series that consists of two unstable nuclides and one final stable nuclide. Inthe general case there can be K nuclides involved in a series of radioactive decays where K is any integerand the Kth nuclide is a stable nuclide. For this general case the equations for the number of the nuclidesare

dN1(t)

dt= −λ1N1(t) (2.12a)

dN2(t)

dt= λ1N1(t) − λ2N2(t) (2.12b)

dN3(t)

dt= λ2N2(t) − λ3N3(t) (2.12c)

...

dNi(t)

dt= λi−1Ni−1(t) − λiNi(t) (2.12d)

...

dNk−1(t)

dt= λk−2Nk−2(t) − λk−1Nk−1(t) (2.12e)

dNk(t)

dt= λk−1Nk−1(t) (2.12f)

The solutions to these equations would be obtained in the same way

dN1(t)

dt= −λ1N1(t) ⇒ N1(t) = N1(0)e−λ1t (2.13a)

dN2(t)

dt= λ1N1 − λ2N2(t) ⇒ dN2(t)

dt+ λ2N2(t) = λ1N1 = λ1N1(0)e−λ1t

⇒ N2(t) = e−λ2t

∫eλ2t · λ1N1(0)e−λ1t + Ce−λ2t

⇒ N2(t) = e−λ2t λ1N1(0)

λ2 − λ1

∫e(λ2−λ1)t · (λ2 − λ1)dt + Ce−λ2t

⇒ N2(t) =λ1N1(0)

(λ2 − λ1)e−λ2t · e(λ2−λ1)t + Ce−λ2t

⇒ N2(t) =λ1

(λ2 − λ1)N1(0)e−λ1t + Ce−λ2t

from N2(0) = 0, it follows C = −λ1

(λ2 − λ1)N1(0)

(2.13b)





∴ N2(t) =λ1

(λ2 − λ1)N1(0)e−λ1t −

λ1

λ2 − λ1N1(0)e−λ2t

⇒ N2(t) =λ1

(λ2 − λ1)N1(0)e−λ1t −

λ1

λ2 − λ1N1(0)e−λ2t (2.13c)

dN3(t)

dt= λ2N2 − λ3N3

⇒ dN3

dt+λ3N3 = λ2N2 =

λ2λ1

(λ2 − λ1)N1(0)e−λ1t +

λ2λ1

(λ1 − λ2)N1(0)e−λ2t

⇒ N3(t) = e−λ3t ·∫ {

λ2λ1

(λ2 − λ1)N1(0)e−λ1t +

λ2λ1

(λ1 − λ2)N1(0)e−λ2t

}eλ3tdt

+ Ce−λ3t

⇒ N3(t) =λ2λ1N1(0)

(λ2 − λ1)(λ3 − λ1)eλ1t +

λ2λ1N1(0)

(λ1 − λ2)(λ3 − λ2)eλ2t + Ce−λ3t

N3(0) = 0 follows from the boundary condition. Hence

⇒ 0 =λ2λ1

(λ2 − λ1)(λ3 − λ1)N1(0) +

λ2λ1

(λ1 − λ2)(λ3 − λ2)N1(0) + C

⇒ C =

{λ1λ2

(λ3 − λ1)(λ3 − λ2)

}N1(0) =

λ1λ2

(λ1 − λ3)(λ2 − λ3)N1(0)

∴ N3(t) =λ1λ2

(λ2 − λ1)(λ3 − λ1)N1(0)e−λ1t +

λ1λ2

(λ1 − λ2)(λ3 − λ2)N1(0)e−λ2t

+λ1λ2

(λ1 − λ3)(λ2 − λ3)N1(0)e−λ3t dN4(t)

dt= λ3N3 − λ4N4

⇒ dN4

dt+ λ4N4 = λ3N3

=λ1λ2λ3

(λ2 − λ1)(λ3 − λ1)N1(0)e−λ1t +

λ1λ2λ3

(λ1 − λ2)(λ3 − λ2)N1(0)e−λ2t

+λ1λ2λ3

(λ1 − λ3)(λ2 − λ3)N1(0)e−λ3t

⇒ N4(t) = e−λ4t ·∫

λ3N3(t)eλ4tdt + Ce−λ4t

⇒ N4(t) =λ1λ2λ3N1(0)

(λ2 − λ1)(λ3 − λ1)(λ4 − λ1)e−λ1t

+λ1λ2λ3N1(0)

(λ1 − λ2)(λ3 − λ2)(λ4 − λ2)e−λ2t

+λ1λ2λ3N1(0)

(λ1 − λ3)(λ2 − λ3)(λ4 − λ3)e−λ3t + Ce−λ4t

N4(0) = 0

⇒ λ1λ2λ3N1(0)

(λ2 − λ1)(λ3 − λ1)(λ4 − λ1)+

λ1λ2λ3N1(0)

(λ1 − λ2)(λ3 − λ2)(λ4 − λ2)

+λ1λ2λ3N1(0)

(λ1 − λ3)(λ2 − λ3)(λ4 − λ3)+ C = 0

⇒C =

{λ1λ2λ3

(λ2 − λ1)(λ3 − λ1)(λ4 − λ1)+

λ1λ2λ3

(λ1 − λ2)(λ3 − λ2)(λ4 − λ2)

+λ1λ2λ3

(λ1 − λ3)(λ2 − λ3)(λ4 − λ3)

}N1(0)

C =λ1λ2λ3

(λ1 − λ4)(λ2 − λ4)(λ3 − λ4)N1(0)





∴ N4(t) =λ1λ2λ3

(λ2 − λ1)(λ3 − λ1)(λ4 − λ1)e−λ1t

+λ1λ2λ3

(λ1 − λ2)(λ3 − λ2)(λ4 − λ2)e−λ2t

+λ1λ2λ3

(λ1 − λ3)(λ2 − λ3)(λ4 − λ3)e−λ3t

+λ1λ2λ3

(λ1 − λ4)(λ2 − λ4)(λ3 − λ4)e−λ4t (2.13d)

Thus from the above solutions it is seen that for all 1 6 i 6 k − 1

Ni(t) = c1ie−λ1t + c2ie

−λ2t + . . . + cjie−λit (2.13e)

where

c1i =1 · λ1 · λ2 · · · λi−1N1(0)

1 · (λ2 − λ1) · (λ3 − λ1) . . . (λi − λ1)(2.13f)

c2i =1 · λ1 · λ2 · · · λi−1N1(0)

1 · (λ1 − λ2) · (λ3 − λ2) . . . (λi − λ2)(2.13g)

...

cji =1 · λ1 · λ2 · · · λi−1N1(0)

1 · (λ1 − λj) · (λ2 − λj) . . . (λi − λj)(2.13h)

i.e. Ni(t) =

i∑

j=1

Cjie−λjt where cji =

1 ·∏i−1

j=1 λj

1 ·∏i

m=1m 6=j

(λm − λj)(2.13i)

For the last nuclide of the series, ie, for the stable nuclide

dNk

dt= λk−1Nk−1

⇒ Nk(t) =

∫λk−1Nk−1(t)dt ⇒ Nk(t) =

∫ k−1∑

j=1

cj(k−1)eλjtdt

⇒ Nk(t) =

k−1∑

j=1

∫cj(k−1)e

λjtdt ⇒ Nk(t) =

k−1∑

j=1

∫ ∏(k−2)l=1 λlN1(0)eλjt

∏k−1m=1

for m 6=j(λm − λj)

dt

=

(k−1)∑

j=1

(− 1

λj

) ∏(k−2)l=1 λlN1(0)e−λjt

∏k−1m=1


+ k1 (k1 is a constant)

⇒ Nk(t) = 0 ⇒(k−1)∑

j=1

(−

1λj

) ∏(k−2)l=1 λl∏k−1

m=1for m 6=j

(λm − λj)N1(0) + k1 = 0

⇒ Nk(t) =

(k−1)∑

j=1

(− 1

λj

) ∏(k−2)l=1 λlN1(0)∏k−1m=1


(1 − e−λjt

)

So, for the general case of a decay series containing k nuclides, with only the first nuclide being present





at initial time and the kth nuclide being a stable nuclide, the general solution is

N1(t) = N1(0)e−λ1t (2.14a)

Ni(t) =

i∑

j=1

∏i−1l=1 λlN1(0)∏j

m=1for m 6=j

(λm − λj)e−λjt for 2 6 i 6 (k − 1) (2.14b)

...

Nk(t) =

k−1∑

j=1

(1

λk−1

) ∏k−2l=1 λlN1(0)∏(k−1)

m=1for m 6=j

(λm − λj)N1(0)

{1 − e−λjt

}(2.14c)

Radioactive Equilibrium

For any process equilibrium is the state when the time of change of quantities zero. So strictly for equi-librium to exist in a radio active decay series λ1 must be zero but this of course is not the case.

Equation (2.14b) shows that N2 is zero both at t = 0 and t = ∞, Hence the activity of the daughter passesthrough a maximum value at some intermediate time tm. tm can be obtained by putting

dN2

dt= 0 ⇒ λ1e

−λ1tm = λ2e−λ2tm

⇒ tm =ln(λ1/λ2)

λ2 − λ1=

ln(T1/T2)

T2 − T1T is the half life

This relation shows tm is positive and real for either T1 > T2 or T1 < T2. The activities of parent anddaughter are equal, the situation is called ideal equilibrium. This situation exists only at the momentwhen, time= tm. This implies that parent activity exceeds daughter activity in the time range 0 < t < tm.Conversely between t = ∞ and tm dNB/dt is negative and daughter activity exceeds the activity of itsparent.

There is no known instance of decay series for which T1 = T2. Due to the differences in half lives andhence the magnitudes of the decay constants, several cases of equilibrium arise depending on the relativehalf lives of the parent the daughter.

Extremely long lived Parent (T1 À T2)This is the case of λ1 ¿ λi for i > 2 from the above solution:

Ni(t) = Ci1e−λ1t + Ci2e

−λ2t + · · ·+ Ciie−λit

Let τi be the half life of the ith nuclide. Then τi À 1 ⇒ λ1 ¿ 1

e−λ1t ≈ 1 ⇒ τi = 0(1) (for i > 2)

⇒ λi = 0(1) = e−λit −→ 0 as t → ∞ (for i > 2)

Therefore after a short time

Ni ≈ ci (for i 6 1 6 (k − 1))

⇒dNi

dt≈ 0 (for 1 6 1 6 (k − 1))





This means then that N1(t) ≈ N1(0)

dN2(t)

dt= λ1N1 − λ2N2 = 0 ⇒ λ1N1 = λ2N2

dN3(t)

dt= λ2N2 − λ3N3 = 0 ⇒ λ2N2 = λ3N3

...

dNi(t)

dt= λi−1Ni−1 − λiNi = 0 ⇒ λi−1Ni−1 = λiNi

...

dNk−1(t)

dt= λk−2Nk−2 − λk−1Nk−1 = 0 ⇒ λk−2Nk−2 = λk−1Nk−1

This equilibrium is known as secular (longterm) equilibrium. So for secular equilibrium

N1(t) ≈ N1(0)

Ni(t) ≈λ(i−1)

λiN(i−1) (for 2 6 i 6 k − 1)

This for example applies for the natural radioactive decay series. Uranium series, the Actinium seriesand the Thorium series.

Relatively Long-lived parent (λ1 < λ2)The second type of equilibrium transient equilibrium is obtained when the half life of the mother nuclideis larger than the half life of the daughter nuclide but the different is not as large as in the case of secularequilibrium. In cases where there is transient equilibrium.

N1(t) = N1(0)e−λ1t (For all i > 2, Ni(t) ≈ ci1e−λ1t)

The reason is that from the general solution

Ni(t) = ci1e−λ1t + ci2e

−λ2t + · · ·+ ciie−λit

but because λi > λ1 for i > 2 after a short time cije−λjt is negligible compared with ci1e

−λ1t for j > 2, and

thus for transient equilibrium

N1(t) = N1(0)e−λ1t

Ni(t) = cie−λit (for i > 2)

For example 227Pa decays as shown

Table 2.5: Decay of 227Pa# Nuclide Type of decay half life1 227

91Pa α 38.3 min2 223

89Ac α 2.2 min3 219

87Fr α 0.025 sec4 215

85At α 10−4 sec5 211

83Bi α 2.16 min6 209

81Tl β 4.79 min7 209

82Pb – Stable

In this series because τ1 is larger than τi, i > 2 there is transient equilibrium.





???????????????? Exercises: 2.1

1. What is the mass of one Rutherford of 234U

2. 1 gram of caesium-137(137Cs ) decays by β-emission with a half life of 30 years. If the initial activityof caesium is 1.0 mCi and atomic mass of caesium is 137, what is

(a) the resulting isotope

(b) the number of atoms left after 5 years

(c) the activity of caesium after 5 yaears

3. The normal activiy of a living matter containing carbon is found to be 15 decays per minute pergram of carbon. An archeological specimen gives 6 decays per minute per gram of carbon. If the halflife of carbon is 5730 years, estimate the approximate age of the specimen.

4. For the Uranium, Actinium and Thorium radioactive decay series, write down the solution for num-bers of nuclei for the 3rd and 6th nuclides in the decay series.

5. If 1 gm of thorium 232 is allowed to decay by natural radioactivity determine the mass of the decayseries after 10 yrs.

i) with out assuming secular equilibrium

ii) by assuming secular equilibrium.

2.2 Alpha Decay

As it has been pointed out in the last section one of the ways in which nuclides decay is by emitting α-particles. α-particles are generally emitted by very heavy nuclei containing to many nucleons to remainstable. The emission of such a nucleon cluster as a whole rather than the emission of single nucleon isenergetically more advantageous because of the particularly high binding energy of alpha-particles. Theparent nucleus (Z,A) is transformed as

AZ X → A-4

Z-2 Y + α + Qα (2.15)

where Qα is the total energy released in the reaction known as disintegration energy.

2.2.1 Energy

One of the quantities of interest in connection with α-decay is the energy of α-particles emitted in thedecay process. This energy is measured by what is called a magnetic spectrograph.

photographic

plateSource

paths of alphas

chamber where

there is magnetic field

2ri

Fig. 2.2: Magnetic Spectrograph




2.2. ALPHA DECAY 2-13

In a magnetic spectrograph a magnetic field is used to make the α-particles move in a circular path. Bymeasuring the radius of the circular path, the velocity of the α-particles and hence their kinetic energy isdetermined

In their circular path, the α-particles satisfy the relation

Mαv2

r= Hqv ⇒ v =

( q

M

)Hr

q is the charge of two protons, Mα is the mass of a Helium nucleus. From this the energy is given by

E =12

Mαv2 =12

q2

MαH2r2 (2.16)

The energy is usually expressed in MeV. Examples of measurements obtained are given in table 2.6

Table 2.6: Energy of α-particles

Source Energy(MeV)230

90Th 4.685226

88Ra 4.80212

84Po 8.78211

83Bi 6.62

In spontaneous α-decay reaction given by equation (2.15), it follows

Myvy = mαvα conservation of momentum

Qα =

(12

Myv2y +

12

mαv2α

)− 0 conservation of KE

=12

My

(mαvα

My

)2

+12

mαv2α =

12

mαv2α

(mα

My+ 1

)

=12

mαv2α

(4

A − 4+ 1

)∵ mass ratio ≈ mass number ratio

=12

Kα

(A

A − 4

)(2.17)

Since A is large Qα ≈ Kα = 12mαv2

α. i.e. the alpha particle carries away most of the disintegration energy.

Theoretically, the α-particle emission process was first explained by Gamow and others as the tunnelingof the α-particle through the potential barrier of the nucleus. Alpha particles therefore show a mono ener-getic energy spectrum. Since barrier transmission is independent of energy, most α-sources are generallylimited to the range ≈ 4 − 8 MeV with the high energy sources having higher transmission probabilityand thus shorter half life.

I EXAMPLE 2.3:

In the α decay: 23892 U → 234

90 Th + 42He + Q, calculate

(a) the Q of the α decay by 238U and,

(b) the approximate speeds of the daughter and the α particle.

Solution(a) Using masses from standard tables

Q = (MU + MTh − Mα)× 931.49 MeVamu

= (238.04955 + 234.043593 − 4.002602) ' 3.13 MeV





(b) The energy released (3.13 MeV) in the decay is much less than the rest energy of both products. Wecan use the non-relativistic form of the energy and momentum. This gives

vTh = 1.21× 107 m/s

vTh = 2.11× 105 m/s

Notice that the Lorentz factor is approximately 1.0006, which is close to unity. Hence, our non-relativisticcalculation is valid. J

I EXAMPLE 2.4:21283 Bi decays with a half life of 60.5 min by emitting five groups of α-particles with energy 6.08 MeV,6.04 MeV, 5.76 MeV, 5.62 MeV and 5.60 MeV. Calculate the α disintegration energies. What is the daugh-ter nucleus. Sketch its level scheme.

SolutionBy equation (2.17)

Qα =A

A − 4Kα

∴ Qα1 =212208

× 6.08 = 6.20 MeV

similarlyQα2 = 6.16 MeV, Qα3 = 5.87 MeV

Qα4 = 5.73 MeV, Qα5 = 5.71 MeV

The daughter is 208Tl. Energy levels (excited states) of 208Tl which are fed by the α-ray groups are shownin figure 2.3

Tl (ground state)208

α4

0.49

0.47

0.33

0.04

α3

α2

α1

α0

Bi (ground state)212

{Excited

states of

208Tl

Fig. 2.3: Decay scheme of 212Bi

Equation (2.17) gives a good approximation to the experimental results:α0 (6.203 MeV, 27.2 %)α1 (6.160 MeV, 69.9 %)α2 (5.874 MeV, 1.70 %)α3 (5.730 MeV, 0.15 %)α4 (5.711 MeV, 1.10 %)α5 (5.584 MeV, 0.016 %) J





2.2.2 Absorption of α-particles

As α-particles pass through matter, they lose energy almost entirely by collisions. Bremsstrahlung isnegligible while radiative loss may not always be negligible. The number of collisions and the amount ofenergy loss in the α’s is subject to statistical fluctuations and so all members of initially mono-energeticbeam may not have identical ranges as they pass through an absorber.

Number of alphas

N0

thicknesstraversed

N0

2

R0

at R0

half the particles

get stopped.

A

B

count rate and range relationship for monoenergetic alpha'sA=

B=A normal probability distribution centered on the mean range R

0

RE

R

Fig. 2.4: Path length of monoenergetic α s from 210Po in a cloud chamber Fig. 2.5: straggling

Range RThe range of α-particles is the distance from the source where the number α-particles sharply goes tozero. The actual relation between the number of particles and the distance from the source is as shownin figure 2.4.

Table 2.7 shows the range of α-particles for some substances

Table 2.7: Range of α-particles

Substance Range (cms)air 6.953Aluminium 0.00406Mica 0.0036Copper 0.00183Gold 0.00140

An abrupt change in the trajectory near the end of the track is the result of a collision between the slowlymoving particle and a massive nucleus. The variabilities in the latter stages of the motion produce arange broadening known as straggling. To take the variables due straggling into account, two types ofrange are defined:

mean range (R0) is the distance at which the count rate is reduced to half its value.

extrapolated range (RE) tangent drawn to the count rate at the mean range will intersect the rangeaxis at the RE

The two are related by

RE = Rm +π

2α = Rm + 0.866α α is staggering parameter (2.18)

The connection between α-particle energy as measured by range and decay constant was studies for thethree naturally occurring radioactive series by Geiger and Nuttal in 1911. Their observations were





1. the range and the energy are related as:

R ∝ E3/2 (2.19)

2. longer lived nuclide emit less energetic α-particles

3. Table 2.8 illustrates the pattern of the energies and ranges of α-particles and the decay constants ofα-emitters .

4. As it is seen from table 2.8, the longer lived nuclide emit less energetic α-particles. As a resultthe range of the α-particles emitted by the longer lived nuclides is smaller than the range of theα-particles emitted by the shorter lived nuclides. In other words the larger the decay constant of theα-radioactive decay is then the larger is the energy of the α-particle and the range of the α-particles.It is also seen that for small change in the energy of the α-particles or the range of the α-particlesthe change in the decay constant is large. This pattern is put together by an equation that is referredto as the Geiger-Nuttall equation which is given by equation (2.20)

log λ = A log R + B or log λ = A ′ log E + B ′ (2.20)

Table 2.8: Energy and Range of α-particles

Nuclide Mean Range Decay Constant Half life α-Energy232

90Th 2.49 cm 1.58× 10−18 sec−1 1.39× 1010 yrs 4.06 MeV226

88Ra 3.30 cm 1.36× 10−11 sec−1 1.62× 108 yrs 4.86 MeV228

90Th 3.98 cm 1.16× 10−8 sec−1 1.9 yrs 5.52 MeV222

86Rn 4.05 cm 2.10× 10−6 sec−1 3.83 days 5.59 MeV218

84Po 4.66 cm 3.78× 10−3 sec−1 3.05 min 6.11 MeV216

84Po 5.64 cm 4.33× 100 sec−1 0.16 min 6.90 MeV214

84Po 6.91 cm 4.26× 10+3 sec−1 1.64× 10−4 sec 7.83 MeV212

84Po 8.57 cm 2.31× 10+6 sec−1 3× 10−7 yrs 8.95 MeV

5. The plot of log λ vs log R gives a straight line. This is the relation that has been arrived from experi-mental measurements. It was also the first result that was obtained theoretically.

Theoretically the relation between λ and E or R is obtained by solving Schrödinger’s equation for α-particlein a nucleus. The equation that is solved is Schrodinger equation with the potential

V(r) =2Ze2

r2 (Coulomb’s Potential)

This is a potential barrier and so α-particle penetrate or leave a nucleus by the tunneling effect. Fromsuch a theoretical analysis the relation that is obtained for λ, the decay constant is

λ =Vα

r0exp

{−8(Z − 2)e2

ÃhVα(δ0 − sin δ0 cos δ0)

}

where δ0 = cos−1{

MαV2αr

4e2(Z − 2)

} 12

and Vα is the velocity of the emitted α particles relative to the nucleus. Mα is the mass of an α particle,r0 is the effective radius of the nucleus emitting the α-particle.

Vα is related to the measured velocity of the emitted α particle by

Vα = V

{1 +

Mα

Mr

}





where V is the measured velocity of the α particle and Mr is the mass of the recoil nucleus. The fact thatthis result is in agreement with the experimental result (i.e. the Geiger-Nuttall equation) can be seen bytaking the log of both sides of the equation

log λ = log(

Vα

r0

)−

8(Z − 2)e2

ÃhVα(δ0 − sin δ0 cos δ0)

When approximations are made this simplifies to

log λ ≈ a − b(Z − 2)

(1

Vα

)

so when a plot of log λ versus 1Vα

is made the result is a straight line. This is in agreement with theGeiger-Nuttall equation.

log λ

1/v

Fig. 2.6: plot of log λ versus 1Vα

Stopping power, S(E)In addition to range, the stopping power of a substance is also a point of interest. The two are of courserelated. Stopping power is the decrease in the energy of the α particle when it moves a distance in thesubstance of one unit of length. i.e.

S(E) =−dE

dx(2.21)

Thus the relation between range and stopping power is given by:

R(E) =

∫R

0dx ⇒ R(E) =

∫E

0

dE

S(E)

⇒ dR(E)

dE=

1S(E)

(2.22)

A quantity that is defined in connection with stopping power is the relative stopping power:

(Relative Stopping Power of a Substance) =(stopping power in air)

(Stopping Power in Substance)(2.23)

I EXAMPLE 2.5: Range of α-particles

(i) Calculate the range of a 7 MeV α particle in aluminum if the relative stopping power of Al 1700(ii) Calculate the thickness of Al that is equivalent in stopping power to 1 meter air.

Solution:Range in air of an α-particle is given by the relation

R = 0.00318E3/2 = 0.0031873/2 = 0.05889 m

∴ Range in aluminium (RAl) =Rair

S=

0.058891700

= 3.46411× 10−5 meter.





If the density of aluminium is ρAl, then the equiv. thickness in Al will be

RAlρAl =

(Rair

S

)ρAl = 1.58 kg/m3

J

???????????????? Exercises: 2.2

1. Show that 236Pu is unstable against α-decay. Use mass values 236.04607 u for 236Pu, 232.03717 ufor 234U and 4.00260 u for 4He.

2. The Q value for the α-decay of 214Po is 7.83 MeV. What is the energy of the α-particles emitted.

3. The highest energy of α particle emitted in the decay of 238U to 234Th is 4196 ± 4 keV. From thisinformation and the known mass of 238U, compute the mass of 234Th.

4. Use the uncertainty principle to estimate the minimum speed and kinetic energy of an α particleconfined to the interior of a heavy nucleus.

2.3 Beta Decay

Another way in which nuclides decay radioactively is by the emission of β particles. There are three typesof beta decays.

a) Negatron (β−) emission ⇒ n → p + β− + ν

Examples of nuclides that decay by the emission of β− particles are 64He, 27

12Mg, 5020Ca, 130

50 Sn, 20980 Hg,

23792 U, 245

94 Pu.

b) Positron (β+) emission ⇒ p → n + β+ + ν

Examples of nuclides which decay by the emission of β+ particles are 84B, 30

15P, 5427Cu, 100

47 Ag, 19280 Hg.

c) Electron capture (EC) ⇒ p + e− → n + ν

When unstable nuclides have a high Z-value (proton number) the coulomb barrier tends to preventthe emission of positron. However, transforming proton into a neutron is effected by the capture oforbital electron by the nucleus.

ν is the antineutrino and ν is the neutrino. The neutrino a particle which has zero rest mass and nocharge like a photon. It does not interact much with matter and so it is difficult to detect. It was firstpostulated that it is emitted during the emission of β+. Because there would not be conservation of energywith out the emission of a particle of the mass of a neutrino just as with the emission of α-particles theenergy of the particles and the range of the particles are two quantities of importance associated with β

decay.

2.3.1 β− emission

The process of β− emission can be expressed by the generalized equation

AZ X → A

Z+1Y + β− + ν

Just as with α-particles, the velocity and thus the energy of β particles is measured by using a magneticfield to make the β particle through a circular path. Basically the relation for the velocity is then the




2.3. BETA DECAY 2-19

same, except that, because the β particles move with high velocity (≈ 0.99 c). The mass of the β particlesis not taken to be constant. Thus from the relation

Hqv =mv2

r⇒ v =

Hqr

m=

Hqr

m0e/√

1 − v2/c2=

Hqer

m0e

√1 −

v2

c2

From the measurement of r the velocity then is determined from this equation. From the velocity theenergy is given by E = (m − m0)c

2

⇒ E =

{m0e√

1 − v2/c2− m0e

}⇒ E = m0ec2

{1√

1 − v2/c2− 1

}

In this way the energy of β particles from a β-emitter that is measured is found to be of the type shownin figure 2.7

n(E)

Ε

observedspectrum

Expected

Em

Fig. 2.7: β-particles emitted by radioactive nuclei show a continuous energy spectrum with an upper limit which is

characteristic of the concerned nuclide

where n(E) is the number of particles with the energy E. Thus the energy of the β-particles from a givenemitter forms a continuous spectrum. The maximum energy of particles is referred to as the endpointenergy. The endpoint energies of β-particles emitters is listed in table 2.9

Table 2.9: Endpoint Energies

Nuclide End Point Energy (MeV)228

88Ra 0.0536027Co 0.31

19879Au 0.972411Na 1.39

21283Bi 2.255625Mn 2.867633As 3.12

For example 32P decays to 32S as shown in figure 2.8. Considering the law of conservation of energy oneexpects all β particles to have the same E = 1.71 MeV kinetic energy. Explaining this observation has notbeen easy. Fraction of the energy ∆E = 1.71 MeV is associated with the emitted beta particle. But thedisposition of the balance of the energy posed an enigma.

The suggestion was made by Pauli and developed by Fermi (1934) that the emission of β particle isaccompanied by another particle, the neutrino. Its charge is zero but it does possess momentum andenergy. The energy





Q=1.71 MeV

P32

S32

β-

E2

E1

Fig. 2.8: Energy level diagram for the decay of 32P

In many instances of β decay, decay does not occur with a direct transition to the ground state of thedaughter. The case of 198Au is an example:

1.33 MeV

Q=1.37 MeV

Au198

2.7 day

β1=0.28 (0.98%)

Hg198

β2=0.96 (99%)

β3=1.37 (0.02%)

stable

1.09

0.41

0.68

Fig. 2.9: 99% of the decay process in 198Au, Em = 0.96 MeV, is associated with the emission of 0.41 MeV γ-photons

Energetics of β− particles

If the mass-energy conservation law is applied to β− decay where a nucleus of mass 1ZM to 1

Z+1M oneobtains:

1Z M = 1

Z+1M + m◦ + mν + Tβ + Tν + TM1 + Tγ

1Z M = 1

Z+1M + m◦ + (Tβ + Tν) + Tγ

1Z M = 1

Z+1M + m◦ + Emax + Tγ

Add the mass of Z electrons in both sides of the above can be written interms of atomic mass as

ZM = Z+1M + Emax + Tγ

2.3.2 Absorption of β-particles

The range of β-particles is larger than the range of α-particles. While α-particles have a range in air ofonly a few centimeters, the range of β-particles in air is about 1000 cms. Thus β-particles would have to beabsorbed by a medium like aluminum. The reason for this is because when they pass through a medium.β-particles create less ionization than α-particles. But while α-particles get absorbed or loose their energyby only ionization, β-particles loose their energy by X-ray emission (bremsstrahlung, braking radiation).The loss of energy by bremsstrahlung is more significant for absorbers of the heavier elements. In generalthe ration between the loss of energy by ionization and the loss of energy by radiation is given by

(dE/dX)rad

(dE/dX)ion=

EZ

800

Where E is the energy of the β-particles in MeV and Z is the atomic number of the absorber. In generalthe absorption of β-particles is exponential i.e.

n(x) = n(0)e−µx




2.4. γ-DECAY 2-21

where n(x) is the number of particles at a distance of x in the absorber. n(0) is the number of particleswhen there is no absorber, and µ is a quantity referred to as the absorption coefficient.

The distance in the absorber where the radiation of β-particles equals the background radiation is therange of the β-particles in the medium, Rβ. as would be expected the range is related to the energy. Theempirical relation between the range and energy for β-particles is

R =

{412E

(1.265−0.0954 ln E0)0 , for E < 2.5 MeV;

530E0 − 106, for E > 2.5 MeV.

Where E0 is the endpoint energy of the β-particles. (Here R is in Mg/cm2)

These two curves (the two straight lines) are referred to as the Sargent Curves. The upper curve is thecurve for what are referred to as allowed transitions. The lower curve is the curve for what are referred toas forbidden transitions. The terms allowed and forbiden are to indicate that one type of transition muchmore probable than the other.

The theoretical approach to the study of β-decay is a theory that is referred to as the fermi theory ofβ-decay. In this theory the emission of β-particles is supposed to be (as pointed above) due to the reaction

n → p + β+ + ν

p → n + β− + ν

To change a proton to a neutron or a neutron to a proton, a force, called the fermi force is assumed to existthen by following the same methods as in wave (quantum) mechanics the probability that β-particles within a certain energy will be emitted is determined. The result is that the probability that β-particles withenergy between E − dE and E + dE will be emitted in unit time, φ(E) is given by:

φ(E)dE = G2|M|2F(Z, E)(E + m0c2)(E2 + 2m0c

2E)× (E0 − E)2dE

Where G is a constant M is the matrix element, which is a measure of the probability that a nucleuswould undergo a decay by β emission.

F is a function which represents of a coulomb force m0 is the rest mass of an electron. From this the decayconstant is given by

λ =

∫E0

0φ(E)dE (2.24)

with assumptions that M is a constant and F is unity, which is valid for many situations. Equation (2.24)for φ(E) reduces to

φ(E)dE = A2(E + m0c2)(E2 + 2m0cE)(E0 − E)2dE (2.25)

When (2.25) is substituted in (2.24), the result is λ ≈ kE60 to a first approximation. From this it is seen

that:

logλ10 = logk

10 +6 logE010

This gives a straight line for a graph of logλ10 versus logE0

10 in agreement with the Sargent curves. So, fromthis and other more accurate plots that referred to as kurie plots or fermi plots where plots of (φ(E)/p2)1/2

versus E are made with

p =

{(E

C

)2

+ 2m0E

}1/2

the fermi theory of β-decay is taken to in agreement with experimental observation.

Some times during β-decay of nuclides γ-particles are emitted. This happens when the transition fromthe mother nuclide to the daughter nuclide is a transition to an excited state of the daughter nuclide andas the daughter nuclide falls to the ground state it emits a γ-particle. For example in the transition207 F → 20

8 Ne + β−





2.4 γ-Decay

A nucleus that decays by α or β -emission is usually left in excited state (Ei). Upon de-excitation therewill be transition to lower state of energy (Ef) by emitting ∆E = Ei − Ef. ∆E is released by one of thefollowing three em interactions:

i.) γ-ray emission

ii.) Internal Conversion

iii.) Internal pair-production

2.4.1 γ-ray emission

Like the electron shell structure of the atom, the nucleus is also characterized by discret energy levels.Transitions between these levels can be made by the emission (or absorption) of electromagnetic energyof the correct energy.

The energies of these photons, from a few hundred keV to few Mev, characterizing, the high bindingenergy of the nucleus.

Energy level diagrams illustrating the specific energy structure of some typical γ-ray sources are shownbelow.

0

1.227MeV

3+

Na22

2.6yr

β+

γ0+

2+

0.662 MeV

Cs137

27yr

92%

β- γ0

7/2+

0

1.33 MeV

2.505 MeV

Co60

Ni60

5.2yr

2+

4+

5+

β-

0

0.0144 MeV

0.136 MeV

7/2-

EC

5/2-

270d

Co57

Fig. 2.10: Nuclear level diagrams of a few common gamma sources

Most γ-sources are “placed” in their excited states as the result of a β - disintegration. Since electronsand positrons are more easily absorbed in matter, the β-particles in such sources can be “filtered” out byenveloping them with sufficient absorbing material leaving only the more penetrating γ-rays.

Nuclear IsomerismIf two half-lives are observed in a given sample is it necessarily due to two different isotopes?

23490Th → 234

91Pa + 0-1e + v

234pa (protactinium) was found to decay by emission of β-particles of two different hay-lives of 1.18 minand the other of 6.66hr.




2.4. γ-DECAY 2-23

The two hay-lives are due to two different energy states. (excited)of the same nucleus such long-livedexcited states of the same nucleus are referred to as isomeric states or nuclear isomers or metastablestates and denoted by “m” next to the mass number in their formula 234mpa

Isomeric statesMost excited states in nuclei make almost immediate transitions to a lower state some nuclear statesmay live very much longer. Their de-excitation is usually hindered by a large spin difference betweenlevels. (i.e. for bidden transitions) resulting in lifetimes ranging from seconds to years. A nuclrde whichis “trapped” in one of these metastable states will thus show radiactive properties different from those inmore normal states. Such nuclei are called isomers and are denoted by a m next to the mass number intheir formula.eg 60mCo or 69mZn

Annihilation RadiationAnother source of high-energy photons is the annihilation of positions. The γ-spectrum from a thickposition source will thus show a peak at 511 kev (corresponding to the detection of one of the annhilationphotons) in addition to the peaks characteristic of transitions in the daughter nucleus

Gamma-ray spectrum of 22Nasource as observed with NaI. detector position annhilation may take placein the detector or the source itself

2.4.2 Internal Conversion

γ-emission is the most common mode of nuclear de-excilation, transitions may also occur through internalconversion. In this process, the nuclear excitation energy is directly transferred to bound such as electrongets knocked out of the atom. atomic electron ratherthan emitted as a photon.

The electrons ejected in this process (known as internal conversion)

• have K.E equal to the excitaion energy minus atomic binding energy

• is called internal conversion electrons

• are monoenergetic unlike electrons from β-decay

While the k-shell electrons are the most likely electrons to be ejected, electrons in other orbitals may alsoreceive the excitation energy. Thus an internal may also receive the excitation energy. Thus an internalconversion lines, with their energy difference being the differences in the b.e. of respective orbitals thekinetic energy of conversion electron

ke = ∆E − Be

where Ke = kinetic energy of the converted electron∆E = Nuclear excitation energy between initial state and final state

Note

(i) The continuous β-spectra are super imposed by discret lines due to conversion electrons.

(ii) Internal conversion was wrongly interpreted as internal photoelectric effect. Expelimental evidencesindicate the pricence of conversion electrons without γ-emission

(iii) If a given radioactive source emits Nγγ-rays and Ne conversion electgrons during the same intervalof then the internal conversion coefficient α is defined as

α =Ne

Nγ





2.4.3 Internal Pair production

This process occurs much less frequently than others to crete an electron-position pair, ∆E must be atleas1 − 02 Mev

I EXAMPLE 2.6:

203Te atoms resulting from β-decay or 203Hg atoms emit 4 groups of conversion electrons with kineticenergies of 266.2, 264.2, 263.6 and 193.3 KeV. To what shell of Tl atom K, LI, LII, LIII does each groupcorrespond? The electron Binding energies in the shells are 87.7, 15.4, 14.8, and 12.7KeV respectively.Calculate the energies of γ-quanta concurrent with that decay

SolutionKe = ∆E − Be

The maximum K.E. of the group corresponds to LIII conversion electrons as they are least tightly bound

266.3 KeV = ∆E − 12.7 =⇒ ∆E = 279 KeV

using the same ∆E279 Kev for all

γ corresponding to Lii is 264.2 = 279 − 14.8 KeV

LI 263.6 = 279 − 15.4 KeV.

K 193.3 = 279 − 87.7

The γ energy concurrent with β -decay ∼ 279 mev J

Radioactive decay may also be accompanied by gamma emission. This is how the nucleus rids itselfof excess energy if it is in an excited state after emitting an alpha or beta particle. Actually as it hasbeen pointed the last section in the emission of β-particles is accompanied by γ-emission. This is alsotrue, although less frequently for α-emission. So in many cases the emission of γ-particles by nuclides isassociated not with the transformation of nuclides to other types of nuclides but rather with the changeof energy levels of nuclides. But the transformation of nuclides from one isomeric state to another canbe brought about only by γ-emission. This is because isomers are both isobaric and isotopic (same A andsame Z) and to change in between isomers can not be brought about by γ or α-emission.

Just as with α and β particles energy of γ particles and the absorption of γ particles are two things ofinterest in connection with γ decay or emission of nuclides. Since the emission of γ-particles is one typeof electromagnetic radiation the most general way of measuring the energy of γ-particles is by measuringthe wave length of γ-particles using diffraction grating. from the wavelength of course, the frequency andthus the energy are determined. For γ-particles of medium energy the energy of electrons (by using amass spectrometer) scattered by the γ-particles by Compton scattering.

In general γ-particles are more penetrating than both α and β particles. The absorption of γ particles byabsorbing media is exponential. i.e. the intensity of γ-radiation at a distance of x in an absorbing mediumis given by

I(x) = I0e−µx (2.26)

where I0 is the intensity of the source at x = 0 and µ is the absorption coefficient.

The absorption of γ-particles in an absorbing medium is due to three types of interactions of γ-particles.These are:

1. due to photoelectric effect (photo electricity)

2. due to Compton scattering

3. due to electron-positron formation (pair production)




2.4. γ-DECAY 2-25

Thus the above absorption coefficient µ has three components:

µ = µp + µc + µB (2.27)

These different components are of varying significant for varying ranges of energy. For example, µB

doesn’t exit for energies less than 2m0ec2 where m0e is the rest mass of electron while as of high energies

it is the dominant component.

As with α and β decay there is a logarithmic relation between the decay constant λ and the energies ofγ-particles. Two general relations are a

λ = C1A(2L/3)E(2L+1) (for electron multiple transition)

λ = C2A(2L/3)E(2L+1) (for magnetic multipole transition)

where C1 and C2 are constants and A is a parameter that varies from one nuclide to another nuclide.Thus in general

logλ10 = a + b log E (2.28)

where a and b are constants. Thus a plot of logλ10 versus logE

10 is a straight line as it is for α and β-decay.As it has been pointed out above γ decay is the type of decay that exists between isometric states. Forexample, for 69

30Zn, 7732Ge, and 124

51 Sb the decay schemes are as shown below.

???????????????? Exercises: 2.3

1. For the α-emission for 224Th log λ = 56.13−105.05/Uα and the energy of the α particles is 7.33 MeV.What is the half life of thorium 224?

2. 211Rn emits three groups of α particles with energies of 5.847 MeV, 5.779 MeV and 5.612 MeV as-sociated with the α-emission is the emission of three groups of γ-particles of energies 0.0687 MeV,0.169 MeV and 0.238 MeV. What is the decay scheme of 211Rn?

3. From the following table of β+ emitters make a plot of logλ10 versus logE0

10 . From your plot (A SargentCurve) determine E0 for the β+ particles emitted by 25Al (t 1

2= 7.3 sec) and 33Cl(t 1

2= 2.0 sec)

Nuclide Half live End point Energy E0

11C 1200 sec 0.98 MeV13N 606 sec 1.24 MeV15O 122 sec 1.68 MeV17F 70 sec 1.72 MeV19Ne 18.4 sec 2.18 MeV21Na 22.8 sec 2.52 MeV23Mg 11.9 sec 2.99 MeV27Si 4.9 sec 3.48 MeV29P 4.6 sec 3.60 MeV31Si 3.2 sec 3.86 MeV35A 1.86 sec 4.4 MeV41Sc 0.87 sec 4.94 MeV

4. 76Si emits four groups of β-particles with end point energies 2.97 MeV, 2.41 MeV, 1.76 MeV, and0.36 MeV. Associated with the β emission is the emission of 5 groups of γ-particles with energies of0.561 MeV, 0.643 MeV, 1.2 MeV, 1.4 MeV and 2.05 MeV. What is the decay scheme of 76As?





5. What is α-radiation? why is it deflected less in magnetic field than β-radiation?

6. Which of the three radiations (α, β and γ) is not diflected by either magnetic or electric fields?

7. Determine the half-life of radon if 1.75× 105 out of 106 atoms decay per day.

8. 23892U (1g-mass) emits 1.24× 104 α-particles per second. Determine the decay constant.

9. What is the difference between the nuclei of the chlorine isotopes 3517Cl and 37

17Cl? How can you explainthe fact that Cl has relative atomic mass of 35.5 in the periodic table.

10. What are the resulting nuclei after the α and β -decay of Xenon.

11. What must the energy of a γ-photon be for it to be convertible into an electron-position pair.

12. Give one possible explanation for.

(a) wide energy spectrum of β-rays.(b) very narrow energy range of α-particles.(c) isomeric state of a given nuclei.

13. Determine the charges of the lithium, copper, and uranium−238 nuclei in Coulombs. Estimate theelectrostatic repulsion energy and compare with nuclear B.E in each case

14. One gram of radium emits 3.7 × 1010α-particles per sec. Determine the charge of this radiation incoulombs

15. Analyze the following reactions and determine whether energy is liberated evolved:

42He + 4

2He → 73Li + 1

1H,→94Be + 2

1H → 105B + 1

0n

63Li + 2

1H → 242He + 1n




RREEAADDIINNGG ##33

UNIT 3

INTERACTION OF IONIZING RADIATIONS WITH MATTER

Contents3.1 Interaction of X and γ-rays with matter. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-13.2 Interaction of Charged Particle with Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-73.3 Interaction of Neutrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-10

3.1 Interaction of X and γ-rays with matter.

In order to understand the physical basis for radiation protection, it is necessary to know the mechanismsby which radiations interact with matter. In most instances, interactions involve a transfer of energy fromthe radiation to the matter with which it interacts. Matter consists of atomic nuclei and extranuclearelectrons. Radiation may interact with either or both of these constituents of matter. The probability ofoccurrence of any particular category of interaction depends on the type and energy of the radiation aswell as on the nature of the absorbing medium.

In all instances, excitation and ionization of the absorber atoms results from their interaction with theradiation. Ultimately, the energy transferred either to tissue or to a radiation shield is dissipated as heat.The interaction of photons is independent of their nature of origin.

3.1.1 Interaction Mechanisms

There are 12 possible processes by which photons (X-rays and γ-rays) may interact with matter. Theseare classified in table 3.1.

Table 3.1: Photon Interactions with matterType of Interaction ⇒Interaction with ↓

Complet Absorption of Photons Elastic Scattering(Coherent)

Inelastic Scattering(Incoherent)

I. Atomic Electrons Photoelectric effectσpe ∝ Z4 (low energy)

∝ Z5 (high energy)

Rayleigh ScatteringσR ∝ Z2 (low energy limit)

Compton Scatteringσ ∝ Z

II. Nucleons Photonuclear reactions (hν > 10 MeV)(γ, n), (γ, p), (γ, f) etc ∝ Z

Elastic Nuclear Scattering Nuclear ResonanceScattering

III. Electric Field ofCharged Particles

Pair productiona) Field of nucleus

kn ∝ Z2(hν) > 1.02 MeVb) Field of electron

Ke ∝ Z(hν) > 2.04 MeV

Delbrük scattering

IV. Mesons Photomeson productionhν > 140 MeV

Among these large number of possible interaction mechanisms, only four major types play an importantrole in radiation measurements:

3-1

3-2 3. INTERACTION OF IONIZING RADIATIONS WITH MATTER

• photoelectric absorption• Compton Scattering• pair production• Rayleigh scattering

All these processes lead to partial or complete transfer of photon energy to electron.

Photoelectric Absorption

In the photoelectric absorption process, a photon undergoes an interaction with an absorber atom in whichthe photon completely disappears. In its place, an energetic photoelectron is ejected by the atom from oneof its bound shells. The interactions with the atom as a whole and can not take place with free electrons.For γ-rays of sufficient energy, the most probable origin of the photoelectron is the most tightly bound ork-shell of the atom. The photoelectron appears with an energy given by

Ee− = hν − Eb (3.1)

where hν is the energy of incident photon and Eb is the binding energy of the photoelectron in its originalshell

The photoelectric cross section strongly depends upon the atomic number of the absorber and the energyof the photon. It is approximately given by

σPE =Zn

E3.5γ

(3.2)

where the exponent n varies between 4 and 5 over the gamma-ray energy region of interest.

Compton Scattering

As the energy of photon increases above the k-edge, cross section for the photoelectric absorption of photonbecomes rapidly insignificant. The important energy loss mechanism beyond this point and upto around1.5 MeV photons, is Compton scattering.

Compton scattering occurs when a photon of medium energy undergoes an elastic collision with a free, ornearly free electron. In matter, of course, the electrons are bound, however, if the photon energy is highwith respect to the binding energy, the binding energy can be ignored and the electrons can be essentiallyfree.

Recoililing

electron

Scattered

PhotonIncident

Photonφ

θ

Fig. 3.1: Kinematics of Compton Scattering

The incoming gamma-ray photon is deflected through an angle θ with respect to its original direction.The photon transfers a portion of its energy to the electron (assumed to be initially at rest), which is thenknown as a recoil electron. Because all angles of scattering are possible, the energy transferred to theelectron can vary from zero to a large fraction of the gamma-ray energy.

By conservation of Energy

hν = hν ′ + T = hν ′ + m◦C2

[1√

1 − v2/c2− 1

](1)




3.1. INTERACTION OF X AND γ-RAYS WITH MATTER. 3-3

By Conservation of Momentum

hν

c=

hν ′

ccos φ +

m◦v√1 − v2/c2

cos θ x-component (2)

0 =hν ′

csin φ −

m◦v√1 − v2/c2

sin θ y-component (3)

From simultaneous equations for conservation of energy and momentum the expression that relates theenergy transfer and the scattering angle for any given interaction can be shown to be:

hν ′ =hν

1 + hνm◦c2 (1 − cos θ)

(3.3)

where m◦c2 isthe rest mass energy of the electron (0.511 MeV)

The probability that the photon will be Compton scattered falls off steadily with increasing energy ofphotons. Its dependence on the atomic number of absorber has been found to be very little.

σcs =Z

Eγ(3.4)

I EXAMPLE 3.1: Energy lost by photons

What fraction of their energies do 1 MeV and 0.1 MeV photons lose if they are scattered through an angleof 90◦.

Solution

(i) For 1 MeV photon

hν ′ =hν

1 + hνm◦c2 (1 − cos θ)

=1 MeV

1 + 1 MeV0.511 MeV (1 − cos 90◦)

= 0.338 MeV

The fractional energy loss is thus

1 −hν

hν ′=

1 − 0.3881

= 0.612 = 61.2%

(ii) In the case of 0.1 MeV gamma rays, the energy of the scattered photon is 0.0835 MeV, and the frac-tional energy loss is only 16.5%.

Notice that the photon of higher energy loses greater fraction of its energy in the energy range considered.J

I EXAMPLE 3.2: Maximum Energy of Compton Recoil Electron

Compute the maximum energy of the Compton Recoil electrons resulting from the absorption in Al of2.19 MeV γ-rays.

The recoil electron maximum energy (Emax) corresponds to the minimum energy of scattered photon(hν ′min)

hν ′ =hν

1 + hνm◦c2 (1 − cos θ)

⇒ hν ′min =hν

1 + 2hνm◦c2

cos θ = −1 at the minimum value

⇒ Emax = hν − hν ′ = hν

(1 −

11 + 2hν/m◦c2

)=

hν

1 + m◦c2

2hν





Substituting hν = 2.19 MeV we obtain

= 1.961 MeV

J

Pair Production

If the gamma ray energy exceeds twice the rest mass energy of an electron (1.02 MeV), the process of pairproduction is energetically possible.

In pair production process the photon disappears and its materialization into an electron-positron pairtakes place. The energy of the photon partly appears as the rest masses of the two particles and partly asthe kinetic energies of the electron (Ee) and positron (Ep).

electron

positron

511 keV 511 keV

two annihilation

photons

Fig. 3.2: Pair production and pair Annihilation

hν = Ee + Ep + 2m◦c2 (3.5)

Note:-

� the available energy (hν − 2m◦c2) can be divided in any proportion between electron and positron,but in general it is almost divided equally.

� pair production can take place in the presence of a nucleus so as to conserve momentum by recoilof nucleus.

� after production of a pair, the positron and electron are projected in a forward direction and losetheir kinetic energy by excitation, ionization and bremsstrahlung, as with any other high-energyelectron. When the positron has expended all of its kinetic energy, it combines with an electron toproduce two quanta of 0.511 MeV each of annihilation photons.

� the cross section of the pair production process increases both as energy of photon and atomicnumber Z of the absorber.

σpp ∝ Z2 ln E (3.6)

This means pair production is most important process with photons of high energy in heavy ele-ments.

Rayleigh Scattering

If the photon energy is smaller than the binding energy of the electron, then the electron is not removedby photon from its shell. It only vibrates after absorbing the photon. The vibrating photon acts as a dipoleand thus emit photons of same energy int eh forward direction. Because the outgoing photon energyremain unaltered and electron of the medium is neither removed nor excited the process is also calledcoherent scattering.

Rayleigh scattering generally occurs at higher Z materials.




3.1. INTERACTION OF X AND γ-RAYS WITH MATTER. 3-5

Compton effect

dominant

Photoelectric effect

dominant

Pair production

dominant

Z o

f ab

sorb

er

hν in MeV

Fig. 3.3: The relative importance of the three major types of photon interaction. The lines show the values of Z and

hν for which the two neighboring effects are just equal

Photo Nuclear Reactions

When the photon energy approaches the binding energy of the nucleon in the nucleus it is possible for aphoton to initiate a nuclear reaction such as (γ,n), (γ,p), (γ,d), (γ,f) etc. The cross section of such reactionsis generally very low compared to total cross section for interaction with atomic electrons. Hence energylost by these photo nuclear reactions is not that important upto photon energy of nearly 10 MeV.

3.1.2 Gamma Ray Attenuation

Consider a beam of photons of intensity I◦ incident upon a plane of absorbing material as shown

Collimator

X

dX

Absorbing material

IIo

Fig. 3.4:

At a depth x with in the material this intensity is this intensity is reduced to I due to interaction alongits way. within an incremental thickness dx there is a further reduction in I by dI. the probability forinteraction with in dx is dI/I. Thus the probability for interaction per unit thickness is :

µ =dI

I

(1dx

)(3.7a)

Interms of µ we may write the decrease in intensity with in dx

−dI = µIdx (3.7b)





This simple differential equation, when solved by using the boundary condition that I = I◦ at x = 0, yields

I = I◦e−µx for good geometry (3.8)

I = B× I◦e−µx for broad beam geometry (3.9)

where B is buildup factor, which accounts for scattered radiation reaching the reference point. B is afunction of shield material and its thickness and energy of the radiation and also a particular quantitybeing observed.

The linear attenuation coefficient µ is related to the total microscopic cross section σt by

µ = Nσt (3.10)

where N is the number of targets per unit volume. For this reason µ is sometimes referred to as macro-scopic cross section. If we neglect minor interaction processes, the linear attenuation coefficient is thesum of the individual attenuation coefficients for the individual interactions

µ = µPE + µCS + µpp (3.11)

Half Value Thickness (HVT)is the thickness of an absorber at which the intensity of the incident beam falls to its half value.

µ =0.693HVT

(3.12)

Mean free-pathis a quantity that describes the average distance travelled by a photon before absorption. The mean freepath is given by

Xm =1µ

(3.13)

Mass Attenuation Coefficient (µm)

Mass attenuation coefficient is the probability of interaction per unit of path length expressed in terms ofmass/area. It is obtained by dividing the linear attenuation coefficient by the density (ρ) of the material.

µm =µ

ρ=

1ρ

[µPE + µCS + µpp] (3.14)

At a given γ-ray energy, the mass attenuation coefficient is independent of the physical state of a givenabsorber. For example it is the same for water whether present in vapour or liquid form.

The mass attenuation coefficient of a compound or mixture of elements can be calculated from(

µ

ρ

)

compuond=

∑

i

Wi

(µ

ρ

)

i(3.15)

where the wi factors represent the weight fraction of element i in the compound or mixture.

In terms of the mass attenuation coefficient, the attenuation law for gamma rays now takes the form

I = I◦e(−µm)ρx (3.16)

A unit very often used for expressing thickness of absorbers is the surface density or mass thickness. Thisis given by the mass density of the material times its thickness in normal units of length. i.e.

mass thickness = mass density× thickness

t = ρx (3.17)




3.2. INTERACTION OF CHARGED PARTICLE WITH MATTER 3-7

mass thickness units are more convenient than normal length units because they are more closely relatedto the density of interaction centers.

They thus have the effect of normalizing materials of different mass densities.

Note:-

� Equal mass thickness of different materials will have roughly the same effect on the same radiation.

In terms of mass thickness and mass attenuation coefficient, the attenuation law thus becomes

I = I◦e(−µm)ρx = I◦e(−µm)t (3.18)

I EXAMPLE 3.3:

The radioisotope 24Na emits γ-rays of energies 1.378 MeV and 2.754 MeV in succession. After passingthrough 27.5 gm/cm2 of lead (density=11 gm/cm3) calculate their relative intensities. The linear absorp-tion coefficients are 48 and 62 respectively for the compounds.

γ1=1.378 MeV µ1 =48 m−1 ρb=11 gm/cm3

γ2=2.754 MeV µ2 =62 m−1 t = 27.5 gm/cm2 I◦ is the same before attenuation. Therefore:

I1 = I◦e−µ1x = I◦e−µmt = I◦

[−0.48× 27.5

11

]

I2 = I◦e−µ2x = I◦e−µmt = I◦

[−0.68× 27.5

11

]

⇒ I1

I2= 1.42

J

3.2 Interaction of Charged Particle with Matter

Studies of the interactions of charged particles with the matter through which they pass have been veryinformative from the earliest scattering experiments of Rutherford till the sophisticated high energy ex-periments of today. An understanding of these interactions has led to a more detailed knowledge of atomicand nuclear structure, to a better insight into the nature of the radiations themselves, and to their effectson living systems.

A charged particle passing through matter loses energy as a result of electromagnetic interactions withthe atoms and molecules of the surrounding medium. The character of these interactions and the mech-anism of the energy loss depends on the charge and velocity of the particle and on the characteristics ofthe medium.


Two principal features characterize the passage of charged particles in matter. These are loss of energy bythe particle and deflection of the particle from its incident direction. These are the results of the followinginteractions.

1. Inelastic collisions with orbital electrons (excitation and ionization of atoms),

2. Radiative losses in the field of nuclei (Bremsstrahlung emission),

3. Elastic scattering with nuclei and

4. Elastic scattering with orbital electrons.





Which of these interactions actually take place is a matter of chance. However energetic electrons lose en-ergy mainly by inelastic collisions which produce ionization and excitation, and also by radiation. Chargedparticles in general lose energy mainly by the coulomb interactions with the atomic electrons. If the en-ergy transferred to the electrons in an atom is sufficient to raise it to higher energy state in the atom, thisprocess is called excitation. If the energy transferred is more, the electron is ejected out of this system.This process is called ionization. These two processes are closely associated and together constitute theenergy loss by inelastic collision. The ejected electron will lose its kinetic energy and finally attach itselfto another atom thereby making it a negative ion. These together constitute an ion pair. Some of theelectrons ejected may have sufficient energy to produce further ionization. Such electrons are called delta(d) rays. In any case, the energy for these processes comes from the kinetic energy of the incident particle,which is slowed down.

Charged particles are classified mainly into two groups: heavy charged particles of mass comparable withthe nuclear mass (protons, alpha particles, mesons, and atomic and molecular ions), and electrons.

A striking difference, in the absorption of heavy charged particles, electrons and photons, is that onlyheavy charged particles (and electrons only in a limited sense) have a range. A heavy charged particleusually loses a relatively small fraction of its energy in a single collision with an atomic electron. Conse-quently, a monoenergetic beam of heavy charged particles, in traversing a certain amount of matter, losesenergy without changing the number of particles in the beam. Ultimately they will all be stopped afterhaving crossed practically the same thickness of absorber. This minimum amount of absorber that stops aparticle is its range. Electrons exhibit a more complicated behavior. They radiate electromagnetic energyeasily because they have a large value of e/m and hence are subject to violent accelerations under theaction of electric forces. Moreover, they undergo scattering to such an extent that they follow irregulartrajectories. For electromagnetic radiation, on the other hand, the absorption is exponential.

3.2.2 Interaction of Heavy Charged Particles

Bethe Formula of Stopping Power

The stopping power (−dE/dx) of a material for a fast moving heavy charged particle is given by the Betheformula:

S = −

(dE

dx

)

coll=

4πe4Z2

m◦v2 NZB (3.19)

Where B =

[ln

2m◦v2

I− ln

(1 −

v2

c2

)−

v2

c2

]

where ze is the charge of the incident particle, v its velocity, N the number density of atoms (numberof atoms per unit volume) of the material having atomic number Z, m◦ the electron rest mass and e

the electron charge. The quantity I is a material property called the mean excitation energy, which isa logarithmic average of the excitation energies of the medium weighted by the corresponding oscillatorstrengths. Except for elements with very low atomic number Z, the mean excitation energies in eV areapproximately to 10Z.

At very low velocities, i.e., when v is comparable with the velocity of the atomic electrons around the heavyparticles (in the case of hydrogen, v = c/137), the heavy ion neutralizes itself by capturing electrons forpart of the time. This results in a rapid fall off of ionization at the very end of the range. On the otherhand, at extremely high energies, with v ≈ c, ionization increases for several reasons. The relativisticcontraction of the coulomb field of the ion is one of them. Part of the energy is carried away as light.

Bragg’s Rule

It is generally assumed that chemical and atomic aggregation phenomena affects stopping power to a verylimited extent. This is embodied in the Bragg rule for the evaluation of the mean excitation potential:

n ln I =∑

ln Ii (3.20)




3.2. INTERACTION OF CHARGED PARTICLE WITH MATTER 3-9

where ni is the number density of electrons associated with element and Ii is the mean excitation potentialfor that element. The implication of this relation is that the stopping power in a compound is the sum ofthe stopping powers of the individual elements.

It is observed that chemical binding does affect the mean excitation potential but the effect decreasesrapidly with increasing atomic charge. The increased validity of Bragg’s law with increasing atomiccharge relies on the increased dependence of I on inner-shell electrons, which are insensitive to chemicaleffects. Deviation from Bragg’s rule should be more apparent at low energies for which the logarithmicterm in the Bethe formulae becomes a sensitive function of I.

Scaling Laws for Stopping Power and Range

To draw some conclusions of considerable practical importance, the Bethe stopping power formula can bewritten as

dE

dx= z2λ(v) (3.21)

or, remembering that the kinetic energy of a particle of mass M is E = Mε(v), where ε is a function of the

velocity only, we have, using E as a variable

−dE

dx(E) = z2λE

E

M(3.22)

or using v as a variable

−dE

dx(v) =

z2

Mλv (3.23)

The last two relations allow us to write the energy loss as a function of energy for any particle, once theenergy loss as a function of energy is known for protons. In particular, protons, deuterons, and tritium ofthe same velocity have the same stopping power.

Similar scaling relations obtain for the range. Using the velocity as variable, one has

Rv(v) =

∫v

0

dv

(dv/dx)=

M

z2

∫v

v

dv

λv(v)=

M

z2 ρv(v) (3.24)

or using energy as a variable

RE(E/M) =M

z2 ρE(E/M) (3.25)

The above mentioned equations for range are not exact, for the neutralisation phenomena occurring atthe end of the range and other corrections are neglected; but it is sufficiently accurate for most cases,excluding very low energies. As an example of the application of the last equation mentioned above, wecan verify that a deuteron of energy E has twice the range of a proton of energy E/2.

A semiempirical power law valid from a few MeV to 200 MeV for the proton range-energy relation is

R(E) = (E/9.3)1.8 (3.26)

where E is in MeV and R in meters of air.

3.2.3 Interaction of Light Charged Particles

Light charged particles are electrons and positrons. As all forms of ionizing radiation eventually result ina distribution of low-energy electrons, the interaction of light charged particles are of central importancein radiation biology. The large difference in mass betwween electrons and heavy charged particles hasimprotan consequences for interactions.

Light charged particles deposit energy through two mechanisms: Collisional losses and Radiative losses





Hig

h K

inet

ic e

ner

gy e

-

(bremstrahlung)

(bre

mst

rahl

ung)

(bre

mst

rahl

ung)

e- delta ray

e- delta ray

Fig. 3.5: Life history of a fast electron

Collision Losses

• Electrons lose energy via interactions with orbital electrons in the medium.• This leads to excitation of the atom or ionization.• Energy loss via these mechanisms is called “collisional loss".• Maximum energy transfer occurs in a “head-on" collision between two particles of masses m and M:

and can be expressed as

Qmax =4mME

(M + m)2 (3.27)

where E is the kinetic energy of the incident particle.• The electron collides with a particle of identical mass and thus large scattering angles are possible.• This results in a track that is very tortuous instead of the straight path of a heavy charged particle.

Radiative Losses: bremsstrahlung

A second mechanism of energy loss is possible because of the small mass of the light charged particle(negligible with HCPs).

A charged particle undergoing a change in acceleration always emits “radiative” electromagnetic radiationcalled bremsstrahlung.

The larger the change in acceleration, the more energetic the bremsstrahlung photon.

For electrons, the bremsstrahlung photons have a continuous energy distribution that ranges downwardfrom a maximum equal to the kinetic energy of the incoming electron.

The efficiency of bremsstrahlung in elements of different atomic number Z varies nearly as Z2.

Notice that bremsstrahlung increases with electron kinetic energy and atomic number Z.

3.3 Interaction of Neutrons

The behavior of neutrons in matter is quite different from that of either charged particles or gamma rays.Since neutrons are uncharged, no coulomb force comes into play and a neutron possesses free access tothe nucleus of all atoms.

For neutrons to interact with matter, i.e. for the nuclear force to act, they must either enter the nucleus orcome sufficiently close to it. The type of the interaction taking place between a neutron and the nucleusdiffers depending upon the kinetic energy of the incident neutron.




3.3. INTERACTION OF NEUTRONS 3-11

3.3.1 Energy Classification

For the purpose of study of neutron interactions; neutrons are classified as below:

• Thermal neutrons Energy below 0.5 eV• Intermediate neutrons 0.5 eV – 100 keV• Fast neutrons 100 keV – 20 MeV• High energy neutrons above 20 MeV.

Study of the neutron interaction with matter requires the knowledge of neutron energy spectrum. Formany applications the spectrum is poorly known. All neutrons are fast by birth and lose energy by col-liding elastically with atoms in their environment and then after being slowed down to thermal energiesthey are captured by the nuclei of the absorbing medium.

3.3.2 Neutron Sources

The most prolific source of neutrons is the nuclear reactor. The splitting of a uranium or a plutoniumnucleus in a nuclear reactor is accompanied by the emission of several neutrons. These fission neutronshave a wide range of energies, peak at ≈ 0.7 MeV and have a mean value of ≈ 2 MeV.

Copious neutron beams may be produced in accelerators by many different reactions. For example, bom-bardment of beryllium by high-energy deuterons in a cyclotron produces neutrons according to the reac-tion:94Be + 2

1D → (115Be)* → 10

5B + 10n

Another commonly used neutron source depends on the bombardment of beryllium with alpha particles.The reaction, in this case, is

94Be + 4

2He → (136C)* → 12

6C + 10n

Radium, polonium, and plutonium are used as the source of the alpha particles. Photoneutrons are theother important source of neutrons. In this process gamma radiations from 24Na, 226Ra, 124Sb, 72Ga and140La bombarding 9

4Be and 21H give neutrons (having an energy distribution).


There are a number of processes by which a neutron can interact with matter.

1. Elastic scattering (n,n) In elastic collision both the momentum and kinetic energy of the system ofneutron and interacting nucleus are conserved. The process may be regarded as essentially a Sbil-liard ballT type of collision. In each collision with a stationary nucleus, the neutron transfers partof its kinetic energy to the nucleus depending on the angle through which the neutron is scattered.

2. Inelastic scattering In the range of energy above 0.5 MeV inelastic scattering begins to occur [(n,n), (n, 2n), (n,γ) type of reactions]. In this case a part of kinetic energy of the incident neutron isgiven off in the form of one or more gammas. This process always takes place through the formationof compound nucleus. This type of interaction is not possible unless the neutron energy exceeds acertain threshold

3. Nonelastic scattering (Nuclear Reactions Involving Emission of Charged Particles) Nonelasticreactions occur at high neutron energies [(n,α), (n,p) type]. These are the reactions with energythresholds in which the neutron causes the emission of charged particles (protons or other heavierparticles) from the target nucleus. An example of particular importance in biological tissue is the (n,p) reaction of 14N with slow neutrons: 14

7N + 10n → 16

6C + 11H

This reaction produces a proton of 0.58 MeV energy.

4. Radiative Capture (n,γ) Nuclear process in which a neutron is captured by the target nucleusand the excess energy emitted as radiation, is called radiative capture process. Conditions for such





reactions are especially favorable during slow neutron (with energy < 1 ev) interaction with medium.Cross section for these processes usually decreases with the inverse of the neutron velocity. It is avery common process, for it occurs with a wide variety of nuclides from low to high mass numbers.This process is extensively used for the production of isotopes by exposing stable nuclides to slowneutrons in a nuclear reactor.

5. Spallation reaction: In this process the target nucleus is fragmented with the emission of severalparticles often including neutrons. The process becomes significant only at neutron energies of about100 MeV or greater.

6. Nuclear fission( n,f): In certain reactions involving heavy atomic nuclei, the capture of neutronresults in the formation of an excited state of a compound nucleus so unstable that it splits up intotwo smaller nuclei. This process is of fundamental importance for the operation of nuclear reactors.

3.3.4 Attenuation of neutrons

All neutrons, at the time of their birth, are fast. Generally, fast neutrons lose energy by colliding elas-tically with atoms in their environment, and then, after being slowed down to thermal or near thermalenergies, they are captured by nuclei of the absorbing material. When absorbers are placed in a col-limated beam of neutrons and the transmitted neutron intensity is measured, as was done for gammarays, it is found that neutrons, too, are removed exponentially from the beam. Instead of using linear ormass absorption coefficients to describe the ability of a given absorber material to remove neutrons fromthe beam, it is customary to designate only the microscopic cross section Σ for the absorbing material.The product σN, where N is the number of absorber atoms per cm3, is the macroscopic cross section Σ.The removal of neutrons from the beam is thus given by:

I = I◦e−σNt = I◦e−Σt

where I◦ = Initial intensity of incident neutrons, I = Transmitted intensity of neutrons after passingthrough t cm and t thickness of the target.

I EXAMPLE 3.4:

In an experiment designed to measure the total cross section of lead for 10 MeV neutrons, it was foundthat a 1-cm-thick lead absorber attenuated the neutron flux to 84.5% of its initial value. The atomicweight of lead is 207.21, and its specific gravity is 11.3. Calculate the total cross section from these data.Solution

N =6.023× 1023 atoms

mole207.21 g

mole× 11.3

gcm3 = 3.29× 1022 atoms

cm3

I

Io= 0.845 = exp[−σ× 3.29× 1022atoms/cm2]

⇒ ln(0.845) = −3.29× 1022σ ⇒ σ = 5.11× 10−24cm2

σ = 5.1barns, and the macroscopic crossection is 0.168 cm−1

J




Reading 1: STRUCTURE AND STATIC PROPERTIES OF NUCLEI. Complete reference :Nuclear Physics Lecture notes

From Department of Physic Addis Ababa University, Ethiopia.

Abstract : This is lecture notes given by the author of the Module at Ababa University. Rationale: This section has a well illustrated content on Structure and Static properties of the atomic nucleus. There are illustrative examples and exercises in the topic.

Reading 2: Radio Activity. Complete reference :Nuclear Physics Lecture notes From Department of Physic Addis Ababa University, Ethiopia.

Abstract :This is lecture notes given by the author of the Module at Addis Ababa University. Rationale: This section has a well illustrated content on Radioactivity. There are several illustrative numerical examples and exercises relevant to the second activity of this module.

Reading 3: INTERACTION OF IONIZING RADIATIONS WITH MATTER. Complete reference :Nuclear Physics Lecture notes From Department of Physic Addis Ababa University, Ethiopia. Abstract :This is lecture notes given by the author of the Module at Addis Ababa University. Rationale: This section has a well illustrated content on Interaction of Ionizing Radiation With Matter. There are several illustrative numerical examples and exercises relevant to the second activity of this module.

Reading 4: Nuclear Reacations Complete reference :Nuclear Physics Lecture notes From Department of Physic Addis Ababa University, Ethiopia. Abstract :This is lecture notes given by the author of the Module at Addis Ababa University. Rationale: This section has a well illustrated content on Nuclear Reaction There are several illustrative numerical examples and exercises relevant to the second activity of this module.

RREEAADDIINNGG ##44

UNIT 4

NUCLEAR REACTIONS

4.1 Nuclear Reactions In GeneralNuclear reactions are generally taken to be reactions in which particles interact with nuclides and endup giving rise to nuclides and particles

p + T → x + R (4.1)

where T is the target nuclide, p is the particle (projectile) that interacts with T; R is the product nuclideand x is the particle that is emitted.

Table 4.1: Examples of nuclear reactions

Reaction Shorthand form11H + 7

3Li → 74Be + n 7

3Li( 11H,n) 7

4Be21H + 12

6C → 137N + n

4He + 93Be → 12

4C + nn + 238

92U → 23992U + γ

n + 23994Pu → 240

94Pu + γ

n + 24094Pu → 241

94Pu + γ

n + 23892U → 237

92U + n + n

A shorthand way to denote the above reactions is:

T(p, x)R (4.2)

Equation (4.1)can be generalizes as

T + p → R +∑

xi (4.3)

where xi are the product particles.

Nuclear reactions may be divided roughly into two groups called compound nucleus reactions and directreactions.

4.1.1 Compound Nucleus

When nuclides are bombarded by particles the way the reaction occurs is as follows, first a compoundnucleus is formed with the absorption of the incident particle by the target nucleus. This compoundnucleus exists for a short time (10−14–10−18 sec) in what is referred to as a virtual state, then it decaysinto the excited state or ground state of the product nucleus. For example in the reaction

21H + 238

92U → 23893Np + n + n

what happens first is the formation of a compound nucleus via the process

21H + 238

92U → 24093Np (comopound nucleus)

4-1

4-2 4. NUCLEAR REACTIONS

Then the compound nucleus that is formed decays via the process

24093Np → 238

93Np + 10n + 1

0n

So the general process in which nuclear reactions occur is:

First: (Incident particle) + (Target nucleus) → (Compound nucleus)

Second: (Compound nucleus) → (Product nucleus) + (emitted particle)

During the formation of a compound nucleus some of the energy of the incident particle goes into theformation of the compound nucleus and some of the energy goes into raising the energy level of thenucleus. Which position goes to which is determined by the use of conservation of energy and momentum.If m is the mass of the incident particle, MCN is the mass of the compound nucleus, M is the mass of thetarget nucleus, v is the initial speed of the incident particle and V is the speed of the compound nucleus,then from conservation of momentum

mv = MCNV =⇒ V =m

MCNv

Let E be the energy that goes to raising the energy level of the nucleus and let E0 be the energy of theincident particle, then,

Energy that goesto raise the energylevel of the particles

=

(Energy ofincident particle

)−

(k.E of compoundnucleus

)

i.e. E =12

mv2 −12

MCNV2 =12

mv2 −12

MCN

(m

MCNv

)2

⇒E =12

mv2{

1 −m

MCN

}⇒ E =

12

mv2{

1 −m

M + m

}

⇔E = E0

{M

M + m

}

Note:- maximum K.E. that can be converted to excitation energy a compound nucleus is more likely to beformed when the excitation energy provided exactly matches one of its energy levels than the excitationenergy has some other value.

4.1.2 Direct reactions

In a compound nucleus reaction it is expected that all sense of direction of the incident particle would belost and that the produced particles would boil off with an essentially isotropic distribution in the centreof mass frame. The discovery that processes such as inelastic proton scattering are strongly peaked in thedirection which is forward with respect to the incident particle and further that these processes exhibitonly very gradual changes in cross-section with energy pointed to some other form of reaction than thecompound nucleus.

We will just summarise a few such processes -

Single collisions between the incoming particle and a nucleon in the target nucleus in which the incidentparticle emerges with reduced energy (inelastic scattering).

As in a but with the struck nucleon carrying off most of the energy.

The incoming particle does not really enter the nucleus but excites it in passing.

A proton enters the nucleus and exchanges charge with one of the neutrons - leaving as a neutron (chargeexchange reactions).




4.2. NUCLEAR CROSS-SECTION 4-3

The incident nucleon collects a second one from the nucleus to form a deuteron (pick up reactions eg (n,d))

An incident deuteron or other light nucleus leaves one or more of its nucleons in the target nucleus(stripping reactions).

These reactions can be thought of as surface reactions but in this context it should be remembered thatthe surface forms a large part of even quite heavy nuclei.

4.1.3 Conservation Laws

In ordinary nuclear Reaction the following conservation laws are valid. Conservation of

1. Energy 2. Momentum 3. Angular Momentum

4. Charge 5. Nucleons 6. Spin

7. Parity 8. Isotopic Spin

???????????????? Exercises: 4.1

1. What is the product in the reaction 59Co(p,n)?

2. Write the nuclear reaction in table 4.1 in shorthand form.

3. In the reaction

42He + 14

7N → 178O + 1

1H

(a) Show that charge is conserved

(b) determine the energy shared, as kinetic energy, by the Oxygen and the proton

4. Compare the time required by a projectile particle to cross a target nucleus to the life time of acompound nucleus.

4.2 Nuclear Cross-section

In the last section it has been pointed out that compound nuclei are formed during nuclear reaction.A measure of the probability that a compound nucleus would decay in a certain manner is given by aquantity that is referred to as the level width of the compound nucleus for a certain decay manner. Bydefinition the level width is given by:

Γ =h

2πτ(4.4)

Where Γ is the level with, h is Plank’s constant and τ is the mean life time of the compound nucleus.

The origin of the definition of Γ is Heisenberg’s uncertainty principle. By the uncertainty principle

(∆E∆t) ≈ h

2π= Ãh

If ∆t is taken to be τ then it follows that

∆E ≈ h

2πτ





There are different possible manners of decay. For each manner of decay there is a level width. The totallevel width for a given compound nucleus is the sum of all the partial level widths i.e.

Γt = Γp + Γα + Γγ + · · ·where Γt = is the total level width

Γp = is the level width for decay by emitting a proton

Γα = is the level width for decay by emitting an α-particle etc.

The fact that Γ is a measure for the probability of decay in a certain manner is seen from the fact

Γ =h

2πτ=

hλ

2π

and

N(t) = N(0)e−λt

and thus the larger λ is the faster or more probable would the decay be.

The level width is a measure of the probability of certain manner of decay after the compound nucleushas been formed. The probability of the formation of the compound nucleus, i.e., the probability of theinteraction of the incident particles and the target nuclei, is given by the nuclear cross-section. The cross-section is represented by σ, and it can be defined as the rate of interaction per target nucleus per unitintensity beam of the incident particles.

Rate of interaction ∝ (number of particles arriving in a unit time)Rate of interaction ∝ (number of density of target nuclide).

Thus

Rate of interaction = σ× I× n×A× dx

where σ is a proportionality factor; I is the number of particles hitting target per unit area in unit time;n is number density of target nuclides; A area of target and dx is thickness of the target. Thus

σ =Rate of Interaction

I×N×A× dx(4.5)

The nuclear cross-section is of the order of the cross-sectional area of a nucleus (πR2 ≈ 10−24 cm2). So theunit that is used for the measurement of cross section is the barn (b)

1 b = 10−24 cm2 (4.6)

Like the level width there are cross-sections for different types of interaction like σe for elastic scattering;σi for inelastic scattering; σa for absorption; σ(p, α), σ(α, p) etc. The total cross-section σt is the sum ofall cross-sections

σt = σe + σi + σa + σ(α, p) + σ(p, α) + · · · (4.7)

During the formation of a compound nucleus, there would be a cross-section for the formation of a com-pound nucleus σC. Then, for interaction of the type

X + x → Compound nucleus → Y + y

the cross-section for the reaction is given by

σ(x, y) = σcΓy

Γ

where Γy level width for decay by emission of particle y and Γ is total level width.




4.2. NUCLEAR CROSS-SECTION 4-5

In general cross-sections depend on the energy of the incident particle and the physical properties of thetarget nuclide.

The fact that cross-sections are measures of the probability of interaction is seen from the relations forthe intensity of the incident beam as function of position in the target. From the definition of cross-sectioni.e.equation (4.5). Therefore in any distance dx the differential decrease in intensity, which is equal to theinteraction with in dx is given by:

σt =−dI(x)

I(x)NAdx⇒ −dI(x)

I(x)NAdx= σNAdx

per unit area this becomes

−dI(x)

I(x)= σtNdx ⇒ ln I(x) = −σtNx + C

⇒ I(x) = I(0)e−Nσtx

let I(x) be I0, then

I(x) = I0e−Nσtx (4.8)

So it is seen that the larger σt is, the higher is the rate of decrease of I(x), the higher is the probability of

interaction. With Σt = Nσt equation (4.8) can be written as:

I(x) = I0e−Σtx (4.9)

Since at x = 0 the intensity is I0 and at x = x the intensity is I(x) the intensity is I(x) is the fraction of theincident particles that have not interacted up to a distance of x. So,I(x)/I0 = eΣtx is the probability that aparticle would reach a distance of x without interacting.

I(x) = I0e−Σtx ⇒ dI(x) = −Σt I0e

−Σtx

︸︷︷︸dx ⇒ dI(x) = −ΣtI(x)dx

⇒−dI(x)

I(x)= Σtdx

Since −dI(x) is the decrease in the intensity within a distance of dx, −dI(x)I(x) is the probability that a

particle which has survived without interacting up to a distance of x will interact with the distance dx.But | −dI

I(x) |= Σtdx. Therefore Σtdx is the probability that a particle which has survived without up to a

distance x will interact with in a distance of dx. Therefore (Σtdx)dx = σt is the probability of interaction per

unit length. So, Σt = Nσt and is referred to as the macroscopic cross-section. Macroscopic cross-sectionis the probability of interaction per unit length. The unit of Σt is cm−1. Therefore the probability that aparticle would interact at a distance of x. P(x)dx is

P(x)dx =I(x)

I0Σtdx ⇒ P(x)dx = eΣtxΣtdx

⇒P(x)dx = ΣteΣtxdx

This is the probability that a particle will interact at a distance of x inside the target. From this it isobtained that the probability that a particle would interact between the distance x1andx2 is given

∫x2

x1

P(x)dx =

∫x2

x1

ΣteΣtxdx = e−Σtx1 − e−Σtx2

and the probability that the particle would interact within an infineite thickness is∫∞

0P(x)dx =

∫∞

0Σte

Σtxdx =[−e−Σtx

]∞0 = 1





The average distance that particle would move without interacting which is referred to as the mean freepath is given by

l =

∫∞

0xp(x)dx =

∫∞

0Σte

−Σtxdx

⇒ l =

∫∞

0xe−Σtx(Σt)dx =

[−e−Σtxx

]∞0 −

∫∞

0−e−Σtx(Σt)dx

⇒ l = 0 −1Σt

[0 − 1] =1Σt

∴ l =1Σt

Thus the macroscopic cross-section is the reciprocal of the mean free path. This could also have been ar-rived at from the above interpretation of Σt. Σt is the probability of interaction per unit length. Therefore

1Σt

is the average distance, that a particle would move without interacting, which is the mean free path.

I EXAMPLE 4.1:

There are approximately 6 × 1023 atoms/m3 in solid aluminium. A beam of 0.5 meV neutrons is directedat an aluminium foil 0.1 mm thick. If the capture cross-section for neutrons of this energy in Aluminiumis 2× 10−31 m2 find the fraction of incident neutrons that are captured.

SolutionN = 6× 1028 atoms/m3; E = 0.5 MeV; dx = 0.1 mm; δt = 2× 10−31 m2

from the relation

I(x) = I0eΣtx where,Σt = Nδt∣∣∣∣−

dI

I(x)

∣∣∣∣ = Σtdx

(Σtdx)/dx is the probability of interaction per unit length

⇒Σt = Nδt/l = 1.6× 10−6cm−1

J

I EXAMPLE 4.2:

The density of iron is about 8×103 kg/m3. The neutron-capture cross-section is about 2.5 b. What fractionof an incident beam of neutrons is absorbed by a sheet or iron 1 cm thick ?

Solution

δt = 2.5 b = 2.5× 10−28m2

The number of atoms per m3 of iron is

mFe = 5.59× 1.66× 10−2710−27 kg/amu = 9.3× 10−26 kg/atom

⇒ n =8× 103 kg/m3

9.3× 10−26 kg/atom

Σt = nδt = 0.215

J




4.3. CLASSIFICATION OF NUCLEAR REACTIONS 4-7

???????????????? Exercises: 4.2

1. The total cross-section of Nickel for 1 MeV neutrons is 3.5 barns. What is the fractional attenuationof a beam of neutrons on passing through a sheet of nickel 0.01 cm in thickness. Given density ofnickel 8.9 gram/cm3

4.3 Classification of Nuclear Reactions

Nuclear Reactions can be classified in different ways. In terms of the mechanism of interaction nuclearreactions can be categorized as follows:

Elastic Scattering: The incident particle strikes the target nucleus and leaves without energy loss. α

particle scattering in gold foil is a good example of the elastic scattering.

Inelastic Scattering: The scattered particle may loss K.E. in excess of that required for an elastic colli-sion with the nucleus. Formation of compound nucleus is an example of such scattering.

Disintegration: On striking the target nucleus the incident particle is absorbed and a different particleis ejected. The product nucleus differs from target nucleus.

147 N + 4

2He → 178 O + 1

1H

Photo disintegration Very energetic γ rays interact with the nucleus of an atom and may be absorbed21H + γ → 1

1H + 10n

Radiative Capture: A particle may combine with a nucleus to produce a new or compound nucleus inan excited state. The excess energy is emitted in the form of γ-rays

Direct Reactions: This pickup reaction in which the nucleus immediately emit a nucleon or the incidentparticle is stripped of one of its particle in the nucleus.

Heavy Ion Reactions: Nuclear reaction induced by heavy ions.

Spontaneous Decay: β and α emission processes are in this class of nuclear reactions

Spallation Reactions: Heavy nuclei gain sufficient energy, from a single incident particle, to emit sev-eral particles. The process of Fission is an example of this process.

High Energy Reactions: This class of reactions are taking place in the energy range about 150 MeV,in which new kinds of particles (mesons, strange particles) are produced along with neutrons andprotons.

In terms of the incident particles nuclear reactions can be classified into

i) Photon induced reaction

ii) neutron induced reactions

iii) reactions induced by alpha particles

iv) deuteron induced reactions

v) reactions induced by γ-particles





Each type of reaction can also be classified in terms of the product particle. For proton induced reactionsa possible reaction is a reaction of the form

AZ X + 1

1H → (A-3)(Z-1) Y + 4

2He

Examples of such reactions are:

63Li + 1

1H → 32He + 4

2He

94Be + 1

1H → 63Li + 4

2He

Another possible reaction is a reaction in which the result is a neutron. The general form of such a reac-tion is:

AZ X + 1

1H → (A)(Z+1)Y + 1

0n


115 Bi + 1

1H → 116 C + 1

0n

188 O + 1

1H → 189 F + 1

0n

Still another possible reaction is a reaction of the form:AZ X + 1

1H → (A+1)(Z+1) Y + γ


73Li + 1

1H → 84Be + γ

126 C + 1

1H → 137 N + γ

Still another possible reaction is a reaction of the form:AZ X + 1

1H → (A-1)(Z) Y + 2

1H


94Be + 1

1H → 84Be + 2

1H

73Li + 1

1H → 63Li + 2

1H

For reactions induced by neutrons there are different possible forms. Possible forms of reactions arereactions of the form:

AZ X + 1

0n → (A-3)(Z-2) Y + 4

2He (emission of α)

AZ X + 1

0n → (A)(Z-1)Y + 1

1H (emission of p)

AZ X + 1

0n → (A-1)(Z) Y + 1

0n + 10n (emission of n)

AZ X + 1

0n → (A+1)(Z) Y + 1

0n + γ (emission of photon)

Examples for these reactions are:

2713Al + 1

0n → 2411Na + 4

2He (emission of α)

147 N + 1

0n → 146 C + 1

1H (emission of p)

6329Cu + 1

0n → 6229Cu + 1

0n + 10n (emission of n)

23892 U + 1

0n → 23992 Y + 1

0n + γ (emission of photon)




4.3. CLASSIFICATION OF NUCLEAR REACTIONS 4-9

For reactions induced by α-particles possible forms of reaction are:

AZ X + 4

2He → (A+3)(Z+1) Y + 1

1H (emission of photon)

AZ X + 4

2He → (A+3)(Z+2) Y + 1

0n (emission of neutron)

AZ X + 4

2He → (A+4)(Z+2) Y + γ


147 N + 4

2He → 178 O + 1

1H (emission of photon)

2713Al + 4

2He → 3015P + 1

0n (emission of neutron)

94Be + 4

2He → 136 C + γ

Reactions which are induced by deuterons have the following possible forms.

AZ X + 2

1H → (A+1)Z Y + 1

1H

AZ X + 2

1H → (A+1)(Z+1) Y + 1

0n

AZ X + 2

1H → (A-2)(Z-1) Y + 4

2He


126 C + 2

1H → 136 C + 1

1H

73Li + 2

1H → 84Be + 1

0n

A168 O + 2

1H → 147 N + 4

2He

For γ-particles to induce reaction, the particles must be of high energy. The more common type of reactionsthat are induced by γ particles are of the form:

AZ X + γ → (A-1)

Z Y + 10n

An example of such reaction is

3115P + γ → 30

15P + 10n

Nuclear reactions are classified in terms of the target nuclei as reactions involving

i) Light nuclei 1 6 A 6 25

ii) Intermediate nuclei 26 6 A 6 80

iii) Heavy nuclei A > 80

Nuclear reactions are also classified in terms of the energy of the incident particles as

i) Low energy, 0 6 E 6 1000 eV

ii) Intermediate energy, 1 keV < E 6 500 keV

iii) High energy, 0.5 MeV < E 6 10 MeV

iv) Very high energy, 10 MeV < E 6 50 MeV

v) Ultra high energy, E > 50 MeV





Table 4.2: Reaction classification on the basis of energy

Intermediate nuclei Heavy nuclide

n p α d n p α d

Low 0–1 keV n γ

γ n

n n n p n

Intermediate γ γ γ n

1 keV–0.5 MeV α p

n n n p n n n p

High α p p n p p p n

0.5–10 MeV p α α p,n γ γ γ p,n

2n 2n

Very High 2n 2n 2n p 2n 2n 2n p

10–50 MeV n n n 2n n n n 2n

p p p p,n p p p n,p

n,p p,p n,p 3n p,n p,n n,p 3n

2p 2p 2p d 2p 2p 2p d

α α α > 3 α α α > 3

> 3 > 3 > 3 > 3 > 3 > 3 > 3 > 3

This can be summarized for reactions involving intermediate and heavy nuclei by the following table:

All nuclear reactions are studied or analyzed by using several conservation principles. These are conser-vation of nucleons; charge; linear and angular momentum and conservation of energy.

Of these the more important is the conservation of energy. For the general form of a nuclear reaction

X + x = Y + y

from the conservation of energy(Ex + mxc2) + MXc2 =

(Ey + myc2) + EY + MYc2

where mx, my, MX, MY are the rest masses of x, y, X, and Y.

Much analysis involving nuclear reaction is done using this equation.

4.4 Fusion and Fission Reactions

As it is remembered, the binding energy of a nucleus is the the energy that is needed to break up a nucleoninto its constituent nucleons. Thus the higher the binding energy per nucleon of a nuclide the more stableit is.

So it is seen that nuclide of mass numbers 30 ∼ 80 the more stable nuclides. Whenever a more stableconfiguration can be achieved nucleons would try to achieve that configuration. For this reason some-times nuclides of small mass number combine to form nuclides of higher mass number. In this way theB.E/nucleon would be increased and thus there will be a more stable configuration such a reaction is re-ferred to as fusion reaction. Thus fusion reaction occurs when nuclides of A 6 15 combine to form heaviernuclides.

In another way heavy nuclides i.e. nuclides with A > 80 break up into nuclides of smaller mass number.




4.4. FUSION AND FISSION REACTIONS 4-11

-1

-2

-3

-4

-5

-6

-7

-8

-90 20 40 60 12080 100 140 160 180 200 220 240

FUSION FISSION (A~200)

Nucleon Number A∆

Mass

/nu

cleo

n (

MeV

/C2)

7Li

55Fe

238U

1H

27Al

22NaXeMo

Fig. 4.1: Mass defect per nucleon vs mass number

As it is seen from the above diagram B.E/nucleon will increase and thus the nucleons will be in a morestable configuration. Such reactions are referred to as fission reaction.

In both fusion and fission reactions energy is given off because the nucleons go from a state of lowerB.E/nucleon. The energy that is given off is 4(B.E/nucleon)×N, where 4(B.E/nucleon) is the change inB.E/nucleon and N is the number of nucleons taking part in the reaction.

4.4.1 Fusion in Stars

Fusion reactions are the source of the energy that is produced by stars. Among the fusion reactions thatoccur in stars are:

11H + 1

1H −→ 21H + β+ + ν

11H + 2

1H −→ 32H + γ

32He + 3

2He −→ 42He + 1

1H + 11H

The result reaction of these reactions is

411H −→ 4

2He + 2β+ + 2γ + 2ν

Other fusion reactions that occur in stars are

126 C + 1

1H −→ 137 N + γ

147 N + 1

1H −→ 158 O + γ

74Be + 1

1H −→ 85B + γ

32He + 4

2He −→ 74B + γ

4.4.2 Fission reactions

Fission reactions are the source of energy in nuclear power reactors. For some nuclides, when the nuclidesare bombarded by neutrons, the nuclides break up in to lighter nuclides. Such nuclides are referred to asfissile nuclides. The probability that a fission reaction will occur during the interaction of a fission reaction





Table 4.3: Fission cross sections of some fissile nuclides

Nuclide σf in barns22790Th 1500

23091Pa 1500

23392U 525

23592U 582

will occur during the interaction of a fissile nuclide and a neutron is given by the fission cross-section σf.Examples of some fissile nuclides and their cross-sections are as given in the table 4.3.

When heavy nuclide break up in to lighter nuclides, they break up into many lighter nuclides ranging inmass number from about 70 to about 160.

4.5 Reactor Basics

Nuclear reactors are probably the most important application of nuclear physics. There are many typesof reactors. In terms of purpose they can be classified as:

i) Research Reactors

ii) Reactors for the production of isotopes

iii) Nuclear Power reactors

The function of reactors is based on fission nuclear reactions. During a fission reaction, not only does anuclide disintegrate into other nuclide but also neutrons are given off during fission. These neutrons thatare given off create more reactions and thus more neutrons. In this way a chain reaction is created. Thisis the basis of the function of reactors. A fission reaction is started, with the bombardment of a fissilenuclide by neutrons. the fission reactions themselves create more neutrons and more fission reaction andso once it is started the fission reaction sustain itself under the appropriate conditions. The ratio betweenthe number of neutrons (or fissions) at any one generation (of reactions) to the number of neutrons in thepreceding generation is referred to as the multiplication factor η.

In any reactor, when the value of η is unity, the reactor is said to be critical. When η is less than unity thereactor is said to be subcritical. When η is greater than unity the reactor is said to be supercritical. Whenneutrons are produced during fission some of them react with the nuclei of the fissile material and somediffuse out of the reactor (or the given body of fissile material).

Consider for example; a spherical reactor , of radius r,. Reactions would be occurring with in the volumeof the reactor and neutrons would be escaping through the surface of the reactor so:

(number of neutrons reacting) ∼43

πr3

(number of neutrons escaping) ∼43

πr2

(number of neutrons escaping)(number of neutrons reacting)

∼1r

so for small r, the ratio of the neutrons that escape is larger than the ratio of neutron reactions and soη < 1, and thus there is sub-criticality for a certain value r. The ratio of neutrons escaping will be smallenough that a chain reaction can be maintained, i.e. η = 1 and thus there will be criticality. For large r,the ratio of the neutrons that escape will be so small that η > 1 and thus there would be supercriticality.The size of reactor (or in general the size of the fissile material) for which a chain reaction can be main-tained is referred to as the critical size. The mass of the fissile material of critical size is called the critical




4.5. REACTOR BASICS 4-13

mass. For example for 23892 U the critical size is rc = 8.7 cms and the critical mass is mc = 5.2 kg. So basi-

cally in nuclear reactor a body of fissile material of critical size is used to create a self sustaining fissionreactions. During the fission reaction, heat is given off. The heat from the fission reactions boil water andheats steam directly or indirectly and the steam runs the turbines of an electric power generator. Withthis being the basic idea behind nuclear power reactors, there are three types of nuclear reactors. Theseare:

i) Thermal reactors (low energy reactors)ii) Intermediate reactors

iii) Fast reactors (high energy reactors)

The more widely built types of reactors are thermal reactors. In thermal reactors the energy of theneutrons is < 0.3 eV. This is because as mentioned in the last chapter the fission cross section varies withenergy of the neutrons, and for many fissile materials the fission cross-section is large for small energyand decrease with increasing energy of neutrons. For example for 235

92 U for neutron energy of 0.025 eVσf = 577 b where as for neutron energy of 2 MeV, σf = 13. b.

So in thermal reactors some suitable material is used to slow down the neutrons produced during fissionto the thermal range, ie < 0.3 eV. Such a material is referred to as a moderator. Materials that are used asmoderators are water, heavy water, graphite, beryllium and beryllium oxide. In general, light materialsare good moderators.

Intermediate reactors are not used that much. In intermediate reactors, through neutrons of energy up to10 eV are used. The advantage of intermediate reactors over thermal reactors is that less moderation isneeded. As a result they can be more compact than thermal reactors. Fast reactors used neutrons of highenergy. In fast reactors no moderator is used. Because of this the size of the fissile material that is neededto achieve criticality is large. For example with 235

92 U the critical mass that is needed for a fast reactor 22.2times the critical mass for a thermal reactor of the same size. An advantage of the fast reactor though isthat it can be used as a breeder reactor i.e. it produces power while at the same time it produces nuclearfuel. For example in one reactor the core of the reactor is 235

92 U enriched to 23.5% and surrounding thisuranium depleted of 235

92 U to < 0.35%. This region of reactor where where the breading, i.e., the productionof 239

94 Pu, occur is referred to as the blanket.

At least one neutron produced during each fission must, on the average, initiate another fission. If toofew neutrons initiate fissions, the reaction will slow down and stop. If precisely one neutron per fissioncauses another fission, energy will be released at a constant rate (which is the case in a nuclear reactor);and if the frequency of fission increases, the energy release will be so rapid that and explosion will occur(which is the case in an atomic bomb). These situations are respectively called subcritical, critical andsupercritical.

???????????????? Exercises: 4.3

1. Enumerate the possible modes of decay of the compound nucleus 2010Ne

2. In the 9Be(α,p)12B reaction, using 21.7 MeV α-particles, proton groups of 6.96 MeV, 6.08 MeV, and5.45 MeV were observed at right angles to the incident α particles. Calculate the Q-values for thesegroups and the corresponding energy levels. To what nucleus do these energy levels refer?

3. What is the physical meaning of the multiplication factor in a chain fission reaction? How manyneutrons will there be in the hundredth generation if the fission process starts from 1000 neutronsand k = 1.05?

4. Calculate the binding energy of thermal neutrons added to the following nuclei: 227Th, 233U, 235U,239Pu,238U,242Pu. Which of these nuclei are fissionable under thermal neutrons?

5. Explain why a self-sustaining chain reaction can not be obtained with 238U (natural) as fuel (exceptwith heavy water as moderator in a homogeneous assembly)




BIBLIOGRAPHY

Burcham, W. and Jobes, M. (1995). Nuclear and Particle Physics. Addison Wesley.

Evans, R. (1955). The Atomic Nucleus. McGraw-Hill Book Company, New York.

Fenyves, E. and Haiman, O. (1969). The Physical Principles of Nuclear Radiation Measurement.AKADÉMIAI KIADÓ, Budapest.

H., C. (1989). Introduction to Health Physics. Pergamon Press, 2nd edition.

Kaplan, I. (1962). Nuclear Physics. Addison-Wesley Publ.Co., 2nd edition.

Tayal, D. (1997). Nuclear Physics. Himalaya Publishing House, New Delhi.

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