Unit-III: Nuclear Energy (unit-3)15872160375e9afea59b659.pdfS. N. Ghoshal, Nuclear Physics, S. Chand...

Unit-III: Nuclear Energy

Dr. Prashanta Kumar Khandai

Department of PhysicsEwing Christian College

Allahabad

Email: [email protected] Number: +91-9122008433, +91-7717756468

April 12, 2020

Dr. Prashanta Kumar Khandai (ECC) Lecture April 12, 2020 1 / 56

Reference (Text) Books

• Arthur Beiser, Concept of Modern Physics, Mc Graw Hill Publication.

• S. N. Ghoshal, Nuclear Physics, S. Chand Publication.


Outline

1 Introduction

2 Binding EnergyMass defect and Packing fraction

3 Nuclear ModelsLiquid drop modelShell model

4 Yukawa’s theory of Nuclear force

5 Nuclear Reactions


Introduction

Beginning of Nuclear Physics

• The era of Nuclear Physics started in the year 1909, when Rutherfordproposed his famous atomic model on the basis of alpha scatteringexperiment.• In 1909 Hans Geiger and Ernest Marsden performed the alpha scatteringexperiment in collaboration with Rutherford. They fired a beam of alphaparticles (He nuclei) at gold foils.


Introduction

Continued.....

• They observed that fraction of alpha (α−) particles back scattered wasabout 1 in 8000. That means, 7999 α particles were entered in to themetal foil and went in forward direction.

• In 1911, Rutherford sorted out the reason of back scattering of αparticles in 180o. He proposed that, there must be some strong positivecharge particles concentered in a small region creating a large electric field.He named it as nucleus and then there is a beginning of Nuclear Physics.

• 2009 was celebrated as 100 years of Nuclear Physics.


Introduction

About Nucleus

• Atomic nuclei are made up of two different types of particles: protonsand neutrons. Neutrons and protons are commonly called as nucleons.

• proton carries one electronic unit of positive charge (+e) and has a massabout 1836 times the electronic mass (me).

• The neutron is electrically neutral and is slightly heavier than the proton.

• The protons and neutrons are held together inside the nucleus by verystrong short range force, called as nuclear force.

• The nuclear force is different from gravitational and electric forces.


Introduction

Continued......

• The sum of the numbers of neutrons (N) and protons (Z) inside thenucleus is known as its mass number (A), so that A = N + Z.

• A nucleus of an atom (X) of atomic number Z and mass number A issymbollically written as A

ZX.

• The nuclei with same Z but different A are called isotopes. Here thenuclei contain the same number of protons but different number ofneutrons. Ex: Hydrogen has three isotopes: 1

1H,21H,

31H.

• The nuclei with same mass number (A) but different atomic number Zare called isobars, while the nuclei with same number of neutrons arecalled isotones.


Introduction

Nuclear mass

• Atomic mass refers to the mass of neutral atoms, not of bare nuclei.Thus the atomic mass always includes the mass of its Z electrons.

• Atomic masses are expressed in a.m.u (atomic mass unit) or u, which areso defined that the the mass of a 12

6 C is 12 u.1a.m.u = 1u = 1

12 . Mass of 126 C = 1

12 .12gNA

= 1g6.023×1023

⇒ 1u = 1.67× 10−24g = 1.67× 10−27kg. The energy equivalent of a massunit is 931.5MeV.

• So the nuclear mass (Mnuc) is obtained from the atomic mass (M) bysubtracting the masses of Z orbital electrons, i.e.,Mnuc = M − [Zme −Be(Z)]. where Be(Z) the binding energy of the Zelectrons in the atom which is very small.


Introduction

Table of particle mass


Introduction

Table of particle mass


Introduction

Nuclear Size

• When we talk about nuclear radius, we assume that nucleus has aspherical shape. This is expected because of the short range character ofthe nuclear force.

• Here we can say that the distribution of protons and neutrons inside thenucleus is uniform and so the nuclear charge density ρc is approximatelyconstant.

• Since the nuclear mass is almost linearly proportional to the massnumber A, this means that ρm ∼ A/V = constant ⇒ V ∝ A. Assuming anucleus with radius R we can get V = 4

3πR3 ∝ A⇒ R ∝ A1/3

R = R0A1/3. Where R0 = 1.2× 10−15m = 1.2fm is called the nucleus

radius parameter.


Introduction

Problems


Binding Energy

Binding energy

• If we want to break a nucleus of Z protons and N neutrons completely, acertain minimum amount of energy is to be supplied to the nucleus. Thisenergy is known as the binding energy of the nucleus.

• Conversely if we start with Z protons and N neutrons at rest, and thenbring them together to form the nucleus of mass number A = N + Z andnuclear charge Z, then an amount of energy equal to the binding energy ofthe nucleus will be evolved.

• This evolution energy of the nucleus takes place due to the disappearanceof a fraction of the total mass of the Z protons and N neutrons, out whichthe nucleus is formed. If the quantity of mass disappearing is ∆M , thenthe binding energy is EB = ∆Mc2. Here ∆M is called as the mass defect.


Binding Energy

Continued.....

• It is clear that the mass of the nucleus must be less than the sum of themasses of the constituent neutrons and protons. That is∆M = ZMp +NMn −Mnucl(A,Z) = ZMH +NMn −M(A,Z). WhereM(A,Z) is the mass of the atom of atomic number Z and mass numberA, MH is the mass of hydrogen atom. Hence the binding eneergy of thenucleus isEb = [ZMH +NMn −M(A,Z)]c2 = [ZMp +NMn −Mnucl(A,Z)]c2

• Now let us see what the source of this energy is ?According to the special theory of relativity, mass and energy areequivalent. The mass of a body can be transformed into energy in certainphysical and chemical processes and vice versa.

• If a body of mass m completely converted into energy, then it yieldsE = mc2. Thus 1g, when completely converted into energy, gives9× 1013joules of energy.


Binding Energy Mass defect and Packing fraction

Mass defect

• The departure of the measured atomic mass M(A,Z) from the massnumber (A) is quite significant. The difference between M and A is knownas mass defect (∆M): ∆M = M(A,Z)−A.

• For example, since the atomic mass of 4He (4.00263 u) is slightly greaterthan the mass number 4, its mass defect is +0.002603. On the other hand75As has the atomic mass 74.9215967 u, which is less than the massnumber 75, its mass defect is -0.078403 u.

• Thus mass defect can be both positive and negative. For very light andvery heavy atoms, the mass defct is positive while for the intermediateatoms it is negative.



Packing fraction

• The mass defect of an atom divided by its mass number is known as thepacking fraction (f), a term introduced by F.W. Aston. Thus

f = ∆MA =

M(A,Z)A − 1⇒M(A,Z) = A(1 + f)

• f has the same sign as ∆M and is positive for very light and very heavyatoms. It is negative for atoms in the intermediate region.

• It is found that the packing fraction f varies in a systematic manner withthe mass number A. It is seen that for very light nuclei the f is +ve anddecreases rapidly with incresing A and then becomes −ve for A > 20. fattains a minimum (negative) at A ≈ 60 and then rises slowly for higher Abecomes positive for A > 180.

• The systematic variation of f with A can be understood from thebinding energy per nucleon.



Packing fraction curve



Binding energy per nucleon

• If the binding energy EB of a nucleus AZX is divided by the mass number

A, we get the binding energy per nucleon in the nucleus, which is knownas the binding fraction (fB) and is given by

fB = EBA =

[ZMH+NMn−M(A,Z)]c2

A .

⇒ fB = EBA =

[ZMH+NMn−M(A,Z)]A .

• Here we have assumed that the masses are expressed in energy unit sothat c2 has been omitted.

• For deuteron (21H) atom, since Z = 1, N = 1,⇒ EB = [MH +Mn −Md] =

(1.007825 + 1.008665− 2.014102)× 931.5 = 2.224MeV. SoEBA = fB = 2.224

2 = 1.112MeV per nucleon.



Continued......

• For 42He atom, since Z = 2, N = 2,

EB = (2× 1.007825 + 2× 1.008665− 4.002603)× 931.5 = 28.3 MeVSo EB

A = fB = 28.34 = 7.08MeV per nucleon.

• For 168 O atom, since Z = 8, N = 8, EB = 127.62 MeV, So

EBA = fB = 127.62


• For 5626Fe atom, since Z = 26, N = 30, EB = 492.275 MeV, So

EBA = fB = 492.275




Binding energy (per nucleon) curve

• The binding fraction of different nuclei represent the relative strengths oftheir binding. So 2H is very weakly bound, compared to 4He or 16O. Thenature of variation of fB or EB

A for different nuclei is shown graphically.



Binding energy (per nucleon) curve.....

• EBA (fB) for very light nuclei is very small and rises rapidly with A

attaining a value of ≈ 8MeV/nucleon for A ∼ 20. It then rises slowly withA and attains a maximum of 8.7MeV/nucleon at A ∼ 56. For higher A, itdecreases slowly.

• For 20 < A < 180, the variation of EBA (fB) is very slight, so that it may

be taken to be approximately constant in this region having mean value of∼ 8.5MeV/ nucleon.

• For very heavy nuclei (A > 180), fB decreases monotonically with theincrease of A. For the heaviest nuclei EBA (fB) ∼ 7.5 MeV/ nucleon.

• For very light nuclei, there are rapid fluctuations in the values of fB. Inparticular, peaks are observed for even-even nuclei 4He, 8Be, 12C, 16O

etc. Similar, but less prominent peaks are observed at the values of Z orN equal to 20, 28, 50, 82 and 126. These are called as magic numbers.Dr. Prashanta Kumar Khandai (ECC) Lecture April 12, 2020 21 / 56


Continued.....

• The appearance of peak shows greater stability of the correspondingnuclei relative to its neighbourhood.

• The nature of the binding energy per nucleon (binding fraction) curve iscomplementary to the nature of the packing fraction curve.

• fB increases or decreases as f decreases or increases respectively.



Continued.....


Nuclear Models

Nuclear models

• To understand the observed properties of the nucleus of an atom it isnecessary to have an adequate knowledge about the nature of the nuclearforce. Unfortunately the nuclear force is not as well understood as theelectromagnetic and gravitational force.• We know that a very strong short range force acts between the nucleonsand its mathematical form is not known to us.• Although Yukawa’s theory as well as some other theories give us someidea about the force, but none of these theories gives us a fullunderstanding of the nature of the force.• It is extremely difficult to develop a satisfactory theory or model of thestructure of the nucleus made up of a large number neutrons and protons,as it is impossible to solve the Schrodinger equation for many body system.


Nuclear Models Liquid drop model

Liquid drop model

• The liquid drop model treats the nucleus as a drop of incompressiblenuclear fluid of very high density.The nucleus is made of nucleons (protonsand neutrons), which are held together by the nuclear force (a residualeffect of the strong force). This is very similar to the structure of aspherical liquid drop made of microscopic molecules.

• It was first proposed by George Gamow and then developed by NielsBohr and John Archibald Wheeler. The model was later applied by C. F.von Weizsacker and H. A. Bethe to develop a semi-emperical formula forthe binding energy of the nucleus.• We know that the binding energy of a nucleus is proportional to thenumber of nucleons within it, so the B.E per nucleon is nearly a constantfor most nuclei. Thus we can make a conclusion that the nuclear forceattains a saturation value, so that each nucleon can interact only with itsnearest neighbors.



Similarities between Liquid drop and Nucleus of Atom

• The nuclear force near the nuclear surface is similar to the surfacetension force on the surface of the liquid drop.

• Each nucleon in a nucleus interacts only with its nearest neighbors, like amolecule in a liquid .

• The density of nuclear matter is independent of its volume. Similarly thedensity of liquid is independent of the size of the drop/ volume of liquid.

• The emission of nuclear radiations from nucleus is analogous to theemission of molecules from the liquid drop during evaporation.

• The constant B.E. per nucleon (Eb ∼ 8MeV) is analogous to theconstant latent heat of vaporization of the liquid.



Bethe-Weizsacker formula

• This semi-emperical formula explains the relation between the nuclearbinding energies with the mass number of the nucleus and also shows howthe B.E of a nucleus is affected by various parameters.

• If M(A,Z) be the atomic mass of an element AZX, then we can write:

M(A,Z) = ZMH +NMn − EB.Where the binding energy EB of the nucleus can be expressed as the sumof number of terms as given below:• Volume energy: Nuclear volume increases linearly with the numbe ofnucleons, i.e., V ∝ A .So Ev = avA , where av is called the volume energyparameter.• Surface energy: Since the nucleons on the surface of a nucleus are lesstightly bound, so there is a decrease in B.E proportional to the number ofnucleons which depends on the surface area, i.e., A = 4πR2 = 4πR2

0A2/3.

Then Es = −asA2/3 , where as is called the surface energy parameter.



Continued....

• Coulomb energy: The Coulomb repulsion between the protons in thenucleus also tends to weaken the nuclear binding energy. The Coulombenergy EC of a nucleus is the work that must be done to bring together Zprotons from infinity into a spherical aggregate the size of the nucleus.The P.E of a pair of protons r apart is equal to: V = − e2

4πε0r. Since there

are Z(Z − 1)/2 pairs of protons, so Energy ∝ Z(Z−1)e2

8πε0R.

So EC = −ac Z(Z−1)A1/3 , where ac is another proportionality constant.

• Asymmetry energy: we have seen that except for Coulomb repulsionbetween protons, stable nuclei prefer to have N ≈ Z. But in case ofheavier nuclei, there are more number of neutrons than protons to weakenthe Coulomb repulsion. Thus due to this asymmetry, there will be a

reduction in the B.E which is: Ea = −aa (N−Z)2

A = −aa (A−2Z)2

A .



Continued....

• Pairing energy: As we know that binding energy per nucleon is more foreven-even nuclei. So neutrons and protons always tend to be in pairingform inside the nucleus. Such pairing tends to increase the strength ofbinding energy. That means, the pairing energy is positive for even-evennuclei, 0 for even-odd and odd-even nuclei and negative for odd-odd

nuclei. Ep = (±, 0)apA3/4 .

• The semi-emperical B.E formula or the Bethe-Weizsacker formula is:

EB = avA− asA2/3 − ac Z(Z−1)A1/3 − aa (A−2Z)2

A (±, 0)apA3/4 (1)

Then the formula for B.E per nucleon is:

EBA = av − asA−1/3 − ac Z(Z−1)

A4/3 − aa (A−2Z)2

A2 (±, 0)apA7/4 (2)



Continued....

• A set of coefficients that gives a good fit with the data is as follows:av = 14.1 MeV, as = 13.0 MeV, ac = 0.6 MeV, aa = 19.0 MeV andap = 33.5 MeV.• The atomic mass of an isotope can be written as:

M(A,Z) = ZMH+NMn−(avA−asA2/3−ac Z(Z−1)

A1/3 −aa(A−2Z)2

A (±, 0)apA3/4

).



Problem from Beiser book


Nuclear Models Shell model

Shell model

• The nuclear shell model is a model of the atomic nucleus which uses thePauli exclusion principle to describe the structure of the nucleus in termsof energy levels.The first shell model was proposed by Dmitry Ivanenko (together with E.Gapon) in 1932. The model was developed in 1949 following independentwork by several physicists, most notably Eugene Paul Wigner, MariaGoeppert Mayer and J. Hans D. Jensen, who shared the 1963 Nobel Prizein Physics for their contributions.

• The shell model is partly analogous to the atomic shell model whichdescribes the arrangement of electrons in an atom. When adding nucleons(protons or neutrons) to a nucleus, there are certain points where thebinding energy of the nucleus is significantly higher than the nearest one.This observation, that there are certain magic numbers of nucleons: 2, 8,20, 28, 50, 82, 126 which are more tightly bound than the next highernumber, is the origin of the shell model.• Some nuclei contain magic numbers of protons and neutrons both.Examples are 4He (Z = 2, N = 2), 16O (Z=8, N=8), 40Ca(Z = 20, N = 20), 208Pb (Z = 82, N = 126). These are called doubly magicnuclei and they show exceptionally high stability.



Features of shell model

• The single particle energy levels have the characteristics of occuring ingroups separated from each other by energy intervals that are largecompared to the typical energy difference between the levels of eachgroup. The levels belonging to each group are called shell.

In a single particle shell model

1 It is assumed that nucleons in the nucleus move independently in amean potential or common potential.

2 Nucleons do not interact with each other.

3 The properties of nucleus are attributed to the single unpairednucleon.

4 Most of the nucleons are paired to contribute a net zero spin and zeroorbital angular momentum.



Continued.....

• To obtain the nucleonic configuration and to get the exact magicnumbers, one has to solve the Schrodinger equation for a harmonicpotential with an extra spin-orbit interaction term adding to it.

That is Hψ = Eψ ⇒(−P 2

2m + V (r) + Vso(r)~l.~s)ψ = Eψ.

⇒(−~2

2m ∇2 + 1

2mω2r2 + Vso(r)~l.~s

)ψ = Eψ

where ~j = ~l + ~s⇒ ~l.~s = ~2

2

[j(j + 1)− l(l + 1)− s(s+ 1)

].

• For a single value of l: j splits into two values, i.e., j = l + 1/2 andj = l − 1/2.For j = l + 1/2⇒ ~l.~s = ~2

2 l and the total potential: V = V (r) + Vsol~2

2 .For j = l − 1/2⇒ ~l.~s = −(l + 1)~

2

2 and the total potential:V = V (r)− Vso(l + 1)~

2

2 .As Vso is negative, the state with j = l + 1/2 is lowered in energy than thestate j = l − 1/2. For a particular j we have total 2j + 1 states.



Continued....

The solution of Schrodinger equation is E = N + 32~ω, here N = 2n+ l− 2,

n = 1, 2, 3, ...... represents the radial quantum number andl = 0, 1, 2, .....(n− 1) is the azimuthal quantum number. So N can takevalues 0, 1, 2, .........The sequence of energy levels, taking spin-orbit interaction in to account,is shown below.



Continued....



Application of shell model

• Using shell model, one can find out the ground state nucleonicconfiguration of any nucleus. Here the unpaired nucleon will decide thephysical and chemical properties of nucleus. Because nucleons always tendto form pair to result zero spin and zero orbital angular momentum.

For example: 178 O9. Here as the neutron number is odd in oxygen nucleus,

so we can find its configuration as: (1s1/2)2, (1p3/2)4(1p1/2)2(1d5/2)1.

• We can also find out the ground state spin-parity (Jp) of the nucleus.Here parity is: p = (−1)l. For the 17

8 O9 nucleus, the ground statespin-parity of the nucleus is: Jp = J(−1)l = (5

2 )+, as J = 5/2.• For all even-even nuclei JP = 0+.


Yukawa’s theory of Nuclear force

Meson Theory of Nuclear forces

• Around 1935, Hideki Yukawa proposed a fundamental theory of nuclearforces involving the exchange of massive charged particles betweenneutrons and protons called heavy quanta, (now known as pions).

• According to Yukuwa’s theory, every nucleon continuously emits andreabsorbs pions. If another nucleon is nearby, an emitted pion may shiftacross to it instead of returning to its parent nucleon. The associatetransfer of momentum is equivalent to the action of a force.

• Nuclear forces are repulsive at very short range as well as beingattractive at greater nucleon-nucleon distances; otherwise the nucleons ina nucleus would mesh together.• The mathematical form of nuclear potential given by Yukawa is given byVY ukawa = −g2 e−µr

r . where g is a magnitude scaling constant, µ is theYukawa particle mass, r is the radial distance to the particle.



Continued ......

• If nucleons constantly emit and absorb pions, why are neutrons andprotons never found with other than their usual masses. So it should be aclear violation of the law of conservation of mass.

• The answer is based on the Heisenberg uncertainty principle and it limitsthe accuracy with which certain combinations of measurements can bemade.

• The emission of a pion by a nucleon, which does not change in mass,can take place provided that the nucleon reabsorbs it or absorbs anotherpion emitted by a neighboring nucleon so soon afterward that even inprinciple it is impossible to determine whether or not any mass change hasactually been involved.



Continued ......

• From the uncertainty principle in the form ∆E.∆t ≥ ~2 an event in which

an amount of energy ∆E is not conserved is not prohibited so long as theduration of the event does not exceed ~

2∆E . Using this condition, let usestimate the pion mass.

• Let us assume that a pion travels between nucleons at a speed of v ∼ c;that the emission of a pion of mass mπ represents a temporary energydiscrepancy of ∆E. ∼ mπc

2 (it neglects the KE of pion); and that∆E.∆t ∼ ~. Nuclear forces have a maximimum range r of about 1.7fm andthe time ∆t needed for the pion to travel this far is ∆t = r

v ∼rc .

• So we have ∆E.∆t ∼ ~⇒ (mπc2)( rc ) ∼ ~⇒ mπ = ~

rc = 2× 10−28kg. Thisrough figure is about 220 times the rest mass of electron.



Discovery of pion

• A dozen years after the Yukawa’s proposal, particles with the propertieshe had predicted were actually discovered. The rest mass of charged pionsis 273me and that of neutral pions is 264me, not far from the aboveestimate.

• Two factors contributed to the belated discovery of pion. First, enoughenergy must be supplied to a nucleon so that its emission of a pionconserves energy. Thus atleast mπc

2 ∼ 140 MeV of energy is required.To furnish a stationary nucleon with this much of energy in a collision, theincident particle must have more K.E than mπc

2 so that the momentum aswell as the energy be conserved.

Particles with K.Es of several hundred MeV are required to produce freepions and such particles are found in cosmic radiation that bombards theearth. Recently high energy accelerators also able to produce pions bycolliding two nuclei with very high energy.Dr. Prashanta Kumar Khandai (ECC) Lecture April 12, 2020 41 / 56


Continued.....

• The second reason for the lag between the prediction and theexperimental discovery of the pion is its instability; the mean lifetime ofthe charged pion is only 2.6× 10−8s and that of the neutral pion is8.4× 10−17s. The lifetime of π0 is so short that its existence was notestablished until 1950.


Nuclear Reactions

Nuclear Reactions

• The first and foremost point of nuclear reaction is to understand whethera nuclear reaction yields energy or just has a reverse effect. Therefore, it isimportant to calculate energy gain or loss in a nuclear reaction in terms ofits Q-value.• When a target nucleus is bombarded with an energetic projectile adifferent nucleus forms along with another particle. In this process eitherenergy gets released or absorbed depending upon how masses of thereactants change, and is known as nuclear reaction.• Nuclear physicists understood this process on the basis of formation of a”compound nucleus” when a projectile interacts with a target nucleus. Thecompound nucleus can decay in different channels forming various productnuclei and other particles and does not remember the way it was formed.• The application of nuclear reaction is immense. The two most importantnuclear reactions, i.e. fission and fusion are the most efficient ways ofelectrical power production and hence are part and parcel of humancivilization.Dr. Prashanta Kumar Khandai (ECC) Lecture April 12, 2020 43 / 56

Nuclear Reactions

Nuclear Reactions.....

Let us first discuss what the Q-value of a nuclear reaction is and how tocalculate it.In general, any nuclear reaction can be represented asa+A→ B + b.........(1).This can also be written like: A(a, b)B.From conservation of energy principle, we can write(K.E +R.M.E)LHS = (K.E +R.M.E)RHS ........(2).(ka +mac

2) +MAc2 = (kB +MBc

2) + (kb +mbc2).

Assuming the target nucleus at rest i.e., kA = 0⇒kB + kb − ka = (MA +ma −MB −mb)c

2.......(3).The difference in K.E of reaction products and projectile is known as theenergy balance or Q-value of the reaction.So Q = kB + kb − ka = (MA +ma −MB −mb)c

2............(4).


Nuclear Reactions

Continued.....

• Depending on whether Q-value is positive or negative, the nuclearreactions can be classified into two categories; namely, exoergic andendoergic.

• In exoergic reaction,(MA +ma) > (MB +mb), i.e. (kB + kb) > ka andhence Q > 0. When the input mass is greater than the output mass, somemass is lost in the form of energy at the expense of lost mass.

• On the other hand, in endoergic reaction Q < 0, i.e. the output massbeing larger than the input mass, some mass has been created at theexpense or ”loss” of the output kinetic energy.

• The negative sign of the Q-value indicates that the reaction is endoergic.


Nuclear Reactions

Threshold Energy of Nuclear reaction

• The minimum energy required to have a nuclear reaction is called”Threshold energy” for the said reaction. The threshold energy is obtainedby the relation: Tthres = |Q| × (ma+mA)

mA

The energy of the outgoing particle is given by:Tb = Tthres ×mb × ma

(ma+mA)2.

• Consider a reaction as: H21 + C12

6 ⇒ N137 + n1

0 − 0.28MeV

Here the threshold energy is: Tthresh = 0.28× 2+1212 = 0.33MeV.

The K.E. of outgoing particle (n10) is: Tb = Tn = 0.33× 1×2

(2+12)2= 0.0033

MeV.


Nuclear Reactions

Nuclear Transmutation

• Nuclear transmutation is the conversion of one chemical element or anisotope into another chemical element. Because any element (or isotope ofone) is defined by its number of protons (and neutrons) in its atoms, i.e.in the atomic nucleus, nuclear transmutation occurs in any process wherethe number of protons or neutrons in the nucleus is changed.

• A transmutation can be achieved either by nuclear reactions (in which anoutside particle reacts with a nucleus) or by radioactive decay, where nooutside cause is needed.• Most stars carry out transmutation through fusion reactions involvingHydrogen and helium, while much larger stars are also capable of fusingheavier elements up to iron late in their evolution.

• One type of natural transmutation observable in the present occurs whencertain radioactive elements present in nature spontaneously decay by aprocess that causes transmutation, such as alpha or beta decay.An example is the natural decay of potassium-40 to argon-40, which formsmost of the argon in the air. Also on Earth, natural transmutations fromthe different mechanisms of natural nuclear reactions occur, due to cosmicray bombardment of elements (for example, to form carbon-14), and alsooccasionally from natural neutron bombardment.


Nuclear Reactions

Nuclear Transmutation


Nuclear Reactions

Types of Nuclear reactions

In a reaction a+A→ B + b, if a = b and A = B, then it is called elasticscattering. Here ka + kA = kb + kB.

In inelastic scattering, there exists: (a) Compound nuclear reaction (lowenergy): a+A→ c∗ → b+B and(b) Direct reaction (high energy) a+A→ c+D.

When the wavelength (λ) of the incident particle is of the order of full sizeof the target, then it belongs to compound nuclear reaction. If λ is of theorder of size of a nucleon in the nucleus,then it is direct reaction.


Nuclear Reactions

CoM frame and lab frame

In center of mass (CoM) frame, total linear momentum = 0 i.e, bothprojectile and target particle are approaching towards each other.In labouratory (lab) frame, target is at rest and the projectile is in motion.Generally all experimental (measurements) are done in lab frame andanalysis are done in CoM frame.


Nuclear Reactions



Nuclear Reactions



Nuclear Reactions

About Nuclear fission

• Nuclear fission was discovered by two German chemists Otto Hahn andF. Strassmann in 1939. They did a very careful chemical analysis bybombarding slow neutrons to uranium nucleus and as a result, the uraniumnucleus broke up in to two medium sized fragments such as barium andkrypton. This reaction is symbollically written as:n1

0 + U23592 ⇒ U236∗

92 ⇒ Ba14156 +Kr92

36 + 3n10.

• We can determine the energy release in fission quantitatively as follows:From the mass-energy equivalence principle, we can writeQ = M(U235) +Mn −M(Ba141) +M(Kr92)− 3Mn

Q = 235.04278u+1.00866u−140.9129u−91.89719u−3×1.00866u = 0.21537u

Q = 200.6MeV .

• So nuclear fission is a highly exoergic reaction. Because of this enormousenergy release during fission, it is possible to obtain very large quantity ofenergy from a small amount of uranium.Dr. Prashanta Kumar Khandai (ECC) Lecture April 12, 2020 53 / 56

Nuclear Reactions

About Nuclear fission.......

• For example, if 1g of uranium is completely fissioned, then we cancalculate the energy released from Q-value:We know that 235 g of U235

92 contains 6.023× 1023 uranium atoms.Then 1 g of U235

92 contains 6.023×1023

235 = 2.564× 1021 atoms.Hence the energy release per gram of U235 isE = 2.564× 1021 × 200.6× 1.6× 10−13J

So the value of energy is 8.229× 1010J = 2.29× 104kWh.

• A thermal power generator having a capacity of 1 MW (heat) wouldhave to be run for 229 hours to generate this amount of energy.

• The energy released in burning of a coal (C +O2 ⇒ CO2) is 4 eV peratom of carbon. The quantity of energy released when 1 kg of carbon iscompletely burnt isε = 6.023×1023×4×1.6×10−19×103

12 J = 3.213× 107J = 8.926kWh.So mass of carbon required for generating 2.29× 104kWh energy is2.29×104

8.926 = 2.56× 103 kg.Dr. Prashanta Kumar Khandai (ECC) Lecture April 12, 2020 54 / 56

Nuclear Reactions

About Nuclear fission

• In this type of reaction, two (or more) light nuclei fuse together toproduce a heavier nucleus.• For very light nuclei, such reactions are usually exoergic, which can beunderstood from the binding fraction (fB) curve.• For very light nuclei, fB


Nuclear Reactions

Bohr- Wheeler Model

• N. Bohr and J. A. Wheeler put forward the theory of nuclear fissionbased on the liquid drop model of the nucleus in 1939.• They provided an explanation of the fission process and predicted whichnuclei should be most fissionable. Generally two forces are at work in aheavy nucleus: the nuclear force, which holds the nucleus together, andthe Coulomb electrical force, tends to blow the nucleus apart.•The mechanism of the fission process is given by the theory based onliquid drop theory by assuming the nucleus to be nearly spherical drop andthen before breaking, the nucleus should go some intermediatedeformation process.


Date post:	18-Mar-2021
Category:	Documents
Upload:	others
View:	3 times
Download:	0 times

Unit-III: Nuclear Energy (unit-3)15872160375e9afea59b659.pdfS. N. Ghoshal, Nuclear Physics, S. Chand...

Documents