Class 11 Nuclear Physics Notes

Unit 5 – Modern Physics

Chapter 24 -Nuclear Physics

Nuclear Physics:

The branch of physics dealing with the study of atomic nucleus is called nuclear physics. It includes the study of properties of nucleus, nuclear phenomena, interaction of nuclei, nuclear transmutation and their application.

Nucleus:

Nucleus is highly dense central core of the atom where almost all of the mass and all the positive charge of the atom are concentrated in a very small region as compared to the atom.

Atomic nucleus : Its constituents

1. According to Rutherford’s planateary model of atom, the entire positive charge and most of the mass of the atom are concentrated in a small volume called the nucleus and a suitable number of electrons revolve around it just as planets revolve around the sun.
2. From the results of Rutherford scattering experiments, nuclear size is found to be of the order of 10^-14 m. Hence most of the atom is empty.

Proton:

Proton was discoverd by Rutherford in 1911. It has positive charge of 1.6 × 10^-19C. It has rest mass of 1.6726 × 10−27 kg, which is about 1837 times the rest mass of an electron.A proton has an intrinsic(spin) angular momentum equal to 1/2 It also possesses a magnetic moment much smaller than that of an electron.

Neutron:

It is chargeless particle having mass slightly greater than that of proton. Its rest mass is 1838 times mass of electron. It has intrinsic angular momentum equal to that of proton. Inspite of being neutral, a neutron also possesses a small magnetic moment.

The following terms are used to describe the composition of an atomic nucleus:

1. Nucleons. Protons and neutrons which are present in the nuclei of atoms are collectively known as nucleons.
2. Atomic Number. The no of protons in the nucleus is called the atomic number of the element. It is denoted by Z.
3. Mass number. The total no of protons and neutrons present in a nucleus is called the mass no of element. It is denoted by A. Hence for the neutral atom, we have the following relations:
   Number of protons in an atom =Z
   Number of electrons in an atom =Z
   Number of nucleons in an atom =A
   Number of Neutrons in an atom =A-Z=N
4. Nuclear mass. The total mass of protons and neutrons present in a nucleus is called the nuclear mass.
   

Nuclide.

When an atom is talked of with particular reference to its nuclear composition, it is called nuclide. Thus a nuclide is a specific nucleus of an atom characterised by its atomic number Z and mass number A.

It is symbolically represented as ${}_Z{}^{A}X$ Where X= chemical symbol of the element,
Z= atomic number, and A= mass number

For example
Gold nucleus is represented as ${}_{79}{}^{197}Au$ . It contains 197 nucleons, of which 79 are protons and 118 neutrons.

Isotopes, Isobars, Isotones and Isomers

Isotopes:-

The atoms of an element which have same atomic number but different mass number are called isotopes .
For example ${}_2{}^{1}H ,{}_{ 1}{}^{2}H$ are isotops.

Isobars:-

The atoms which have same mass number but different atomic numbers are called isobars.

For example${}_1{}^{3}H,{}_{ 2}{}^{3}He$

Isotones:-

The nuclides having the same no of neutrons are called isotones .
For example${}_{17}{}^{37}Cl,{}_{ 19}{}^{39}K$

Isomers:-

There are the nuclei with same atomic number and same mass number but existing different ground states.

General properties of nucleus

1) Nuclear size:

The size of nucleus is very small. According to Rutherford, the size of nucleus (i.e radius) is 10^-14m to10^-15m while that of atom is about 10^-10m. The radius of nucleus is almost 10, 000 times smaller than the size of atom. The empirical formula for nuclear radius is R = R₀A^1/3 , whereR₀ = 1.2 × 10^-15m= 1.2fm.

2) Charge :

The nuclei consists of protons and neutrons. Protons are positively charged and neutrons are neutral. So the nuclei are positively charge. The charge of nuclei Ze.

3) Mass :

Since the nucleus consists of protons and neutrons. So, the mass of nucleus is the sum of mass of the protons and mass of neutrons. Assumed nuclear mass=Zm_p + Nm_n Where Z is atomic number,m_p is mass of proton,N is number of neutron and m_n is mass of neutron.

Nuclear density:

The mass per unit volume of a nucleus is called nuclear density.

∴ Nuclear density(ρ) = $\frac{Nuclear mass}{Nuclear volume} = \frac{A{m}_N}{\frac{4}{3}\Pi R^3}=\frac{A{m}_N}{\frac{4}{3}\Pi R {\left(R_0{A}^{\frac{1}{3}}\right)}^3}=2.30 \times {10}^{17}kg{m}^{–3}$

The density of nucleus is independent of mass number A. All nuclei have approximately same density.

Spin angular momentum: Both the proton and neutron have intrinsic spin. The spin angular momentum is S =$\sqrt{s(s + 1)} \frac{h}{2\Pi } where s = \frac{1}{2}$ is spin quantum number.

Resultant angular momentum:

The proton and neutron in a nucleus have orbital angular momentum. The resultant angular momentum of nucleus is obtained by adding spin and orbital angular momentum of all nucleons within the nucleus. This total angular momentum is called nuclear spin.

Nuclear magnetic moment:

A proton has a positive charge and due to its spin, it should have a magnetic dipole moment. However, neutron is electrically neutral but it can have magnetic moment.

Atomic mass:

One atomic mass unit is defined as th of actual mass $\frac{1}{12}$th of carbon-12 atom.
Atomic mass unit is denoted by amu or just u.
$\therefore$Thus
1amu =$\frac{1}{12}$ X Mass of carbon − 12atom
=$\frac{1}{12}$ X $1.992678 \times {10}^{–26}Kg$
1 amu = $1.660565\times {10}^{–27}Kg$

Nuclear force :

Nuclear force is a strong attractive force that binds the protons and Neutrons together inside the tiny nucleus.
Properties of nuclear force
1 Strongest interaction
2.Short range force
3.Charge independent character
4.Spin independent character
5 Exchange Forces

Einstein mass energy relation:

Before 20th century, it was assumed that the mass and energy are two distinct physical quantities. In 1905, Einstein explained the inter-relationship between mass and energy in his special theory of relativity. According to this theory, mass and energy are inter-changeable i.e. mass can be converted into energy and viceversa.

According to Einstein, the energy equivalent of a mass ∆m is
-E = ∆mc²

where c = 3 × 10⁸ ms^-1 is speed of light in vacuum.

This equation represents the Einstein’s mass energy relation. According to this
relation 1 kg of mass of any matter is equivalent to 9 × 10¹⁶J. The energy equivalent of mass of an electron, proton and neutron are respectively given by m_e = 0.511 MeV, m_p= 938.279 MeV and m _n =939.573 Mev.

Electron Volt

It is defined as the energy acquired by an electron when it is accelerated through a potential difference of 1 volt and it is denoted by eV.
1eV = 1.602 × 10^-19J
1 MeV=10⁶eV = 1.602 × 10^-13J

Relation between amu and MeV: Energy equivalent of amu

The Einstein’s mass-energy equivalence relation is $E = m{c}^2$
To determine the energy equivalent of one atomic mass unit, we take
$m=1amu=1.66 \times {10}^{–27}$
$c=3 \times {10}^8 m{s}^{–1}$
$E=1.66\times {10}^{–27}\times (3\times {10}^8{)}^{2}J$
=$\frac{1.66\times {10}^{–27}\times (3\times {10}^8{)}^2}{1.602\times {10}^{–19}}ev$
=931MeV
$\therefore 1amu=931Mev$

Express 16 mg mass into equivalent energy in eV.

solution here,
$m=16mg=16\times {10}^{–6}kg$
$c=3 \times {10}^{8}m{s}^{–1}$
$E = m{c}^2$
$= 16\times {10}^{–6}kg \times (3 \times {10}^{8}m{s}^{–1}{)}^2$
$= \frac{16\times {10}^{–6}kg \times (3 \times {10}^{8}m{s}^{–1}{)}^2}{1.602 \times {10}^{–19}}ev$
$=9\times {10}^{30}eV$

Taking one atomic mass unit equal to 931 MeV.Calculate the mass of ${}_6{}^{12}c$atom.
given,
$1amu=931\times 1.602\times {10}^{–13}J$
using mass energy relationship
$E = m{c}^2 or m=\frac{E}{c^2}$
$1amu=\frac{931\times 1.602\times {10}^{–13}J}{(3 \times {10}^8{)}^2}$
$=1.66 \times {10}^{–27}kg$
${}_6{}^{12}c =12atm$
$= 12\times 1.66\times {10}^{–27}\mathrm{kg}$
$= 1.99\times {10}^{–26}\mathrm{kg}$

Mass Defect

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

Consider the nucleus${}_X{}^{Z}A$It has Z protons and (A-Z) neutons.Therefore, its mass defect is
$∆m=Z{m}_p+(A - Z)m_n-m$
Where mp,mn and m are the rest masses of a proton, neutron and the nucleus${}_X{}^{Z}A$respectively.

Packing Fraction:-

The packing fraction of a nucleus is its massdefect per nucleon.

$P.F of nucleus =\frac{ mass defect}{mass number}=\frac{∆m}{A}$
Paking fraction is directly related to the availability of nuclear energy and the stability of the nucleus.

Binding Energy

The binding energy of a nucleus is defined as the energy required to break up a nucleus into its constituent protons and neutrons and to separate them to such a larger distance that they may not interact with each other.
The binding energy of the nucleus is given by
B.E = ∆mc²
B.E = [ Zm_p + (A − Z) m_n − m ] c²
If these mass is measured in atomic mass unit then
B.E =[Zm_p + (A − Z) m_n − m] × 931Mev

Binding energy per nucleon or average binding energy :-

It is the total energy of the nucleus divided by its mass number A.

$\therefore B ̄.E =\frac{Total binding Energy}{Mass number} =\frac{B.E}{A}$
It determine the measure of the stability of the nucleus against disintegration.The greater value of the average binding energy, higher will be the stability of nucleus and vice versa.

Binding energy per Nucleon curve

Above shows the graph of binding energy per nucleon drawn against mass number A.
The binding energy curve reveals the following important features:
1. Except for some nuclei like${}_2{}^{4}He, {}_6{}^{12}C and {}_8{}^{18}O$the values of binding energy per nucleon lie on or near a smooth curve.
2. B.E per nucleon is small for light nuclei like$1H^1,1H^{2}and 1H^3$
3. In mass number range 2 to 20, there are well defined maximum and minimum on the curve. The maxima occur for${}_2{}^{4}He, {}_6{}^{12}C and {}_8{}^{18}O$indicating the higher stability of these nuclei than the neighbouring ones. The minima, corresponding to low stability, occure for ${}_3{}^{7}Li ,{}_5{}^{10}Band {}_7{}^{14}N.$
The curve has broad maximum close to value 8.5 MeV per nucleon in the mass number range from about 40-120. It has peak value of 8.8 MeV per nucleon for${}_{26}{}^{56}Fe.$
5.As the mass number increases further, the$\frac{B.E}{nucleon}$shows a gradual decrease and drops to 7.6 mev per nucleon for${}_{92}{}^{238}U$This decrease is due to coulomb repulsion between the protons which makes the heavier nuclei stable.

Soddy-Fajan’s Displacement law

1. When a radioactive nucleus emits an alpha particle, its atomis
   number decreases by 2 and mass number decreases by 4.
2. When a radioactive nucleus emits a beta particle its atomic no
   increased by 1 but mass remains same.
3. .The emission of a gamma particle does not change the mass
   number or the atomic number of the atomic nucleus.
   Multiplication Factor;

The ratio of secondary neutrons produced to the initial number of neutrons is called multiplication factor. It is denoted by K.
$K =\frac{number of Neutrons in any generation}{No of neutrons in previous generation}$
If K=1,The reaction is steady or critical
If K>1, the reaction is supercritical
If K<1, the reaction is dying down or sub-critical

Nuclear reaction vs Chemical reaction

Nuclear Reaction:

A reaction which involves the change of stable nucleus of one element into the nucleus of another element is called nuclear reaction.
A nuclear reaction differs markedly from a chemical reaction. In chemical reaction, only the electron revolving around the nucleus take part in the reaction and no chance occurs inside the nucleus where as in a nuclear reaction, the nucleus itself undergoes a transformation.
The energy changes involved in chemical reactions are much smaller than the energy change involved in nuclear reaction.
A + a → B + b + Q

Nuclear fission:-

It is the process in which a heavy nucleus (A>230) when excited gets split up into two smaller nuclei of nearly comparable masses.

For Example : ${}_{92}{}^{235}U + {}_0{}^{1}n \to {}_{56}{}^{141}Ba +{}_{36}{}^{92}Kr + 3_0^1 n+Q$

Nuclear Fusion

It is the process of fusion of two smaller nuclei into heavier nucleus with the liberation of large amount of energy.

For example:${}_1{}^{2}H + {}_1{}^{2}H \to {}_2{}^{4}He + 24MeV$
These reactions require the extreme conditions of temperature andpressure so that the reacting nuclei can overcome their electrostatic repulsion.