Tài liệu Chapter XXIII Nuclear Physics pdf

• We have for a nucleus: A = Z + NThe mass number = the total number of protons and neutrons The atomic number =the ordinal numeral of theelement in the Periodic Table = the charge numbe

GENERAL PHYSICS III

Optics

&

Quantum Physics

Chapter XXIII Nuclear Physics

§ 1 The structure and properties of nuclei

§ 2 Nuclear binding energy and nuclear force

§ 3 Nuclear stability and radioactivity

It is well known for you that every atom contains as its center

a nucleus that is:

positively charged

much smaller in size than the atom

carrying almost the total mass of the atom

This chapter provides deeper knowledge about NUCLEI

§1 The structure and properties of nuclei:

1.1 The structure of nuclei:

• Nucleus consist of two sorts of particles: protons (p) and neutrons (n)

(the common name: nucleons)

• The mass of nuleons:

mp = 1.007276xu; mn = 1.008665xu; (Note: me = 0.000548580 x u).where 1u = 1.66053873x10-27 kg → the atomic mass units

In nuclear physics, the masses of particles are usually expressed

in the energy units – MeV (It is based on the formula E = mc2 ) Then,

1u = 931 MeV  mp = 938.2 MeV; mn = 939.5 MeV; (me = 0.511 MeV)

1.2 The spin and magnetic moment:

• Protons and neutrons have the same spin s = ½

• The magnetic moment: p = +2.49 N; n = -1.91 N

T

J m

(It is 1386 times smaller than the Bohr magneton)

• An application of the proton spin: Magnetic Resonance Imaging

In order to image tissue of various types, Magnetic ResonanceImaging detects the small difference in the numbers of “up” and

“down” hydrogen proton spins generated when the object studied isplaced in a magnetic field

Magnetic resonance imaging (MRI) depends

on the absorption of electromagnetic radiation

by the nuclear spin of the hydrogen atoms

in our bodies (in water molecules)

The nucleus is a proton with

spin ½, so in a magnetic field B there are only

two possible spin directions with definite energy

The energy difference between these states is

E=2pB, with p = 1.41 x 10-26 J /Tesla

• We have for a nucleus: A = Z + N

The mass number =

the total number of

protons and neutrons

The atomic number =the ordinal numeral of theelement in the Periodic Table

= the charge number

= the number of protons

The number

of neutrons

Denote a nucleus by the symbol: ZXA ( 1H1, 2He4, 8O16,… )

• The isotopes of an element: the nuclei that have the same Z (that is

belong to the same element), but different AExamples:

The isotopes of hydrogen: 1H1 , 1H2 (deuterium), 1H3 (tritium)

The isotopes of carbon: 6C12 (carbon-12), 6C13 (carbon-13); etc

1.3 Mass and radius of nuclei:

It is known today about 2500 different nuclei (including different

isotopes) 300 nuclei are stable, the rest nuclei are unstable (radioactive)

• The radius of nuclei: The rather exact results of measuments give

fermi 3

1 m

10 3

.

1 15 A1/3 A1/3

m) 10

(1fermi  15

§2 Nuclear binding energy and nuclear force:

The component paricles (protons, neutrons) of a nucleus bind stronglytogether inside it The evidence for this is that we must spend an amount

of energy (to bombard the nucleus by other particles) to separate a

nucleus to individual protons and neutrons

Thus we have for the nuclear binding energy EB the following equation:

EB = { [Z.mp + (A-Z).mn] – M } .c2

The total mass of separate nucleons > The mass of nucleus

2.1 Nuclear binding energy:

• Note: The last equation can be written in terms of atomic masses:

EB = { [Z.mH + (A-Z).mn] – Matom }.c2

This equation is more convenient, because there are usually in Tables

not nuclear masses, but the atomic masses

• Example: For 2He4:

Matom = 4.0026 u = 3728.0 MeV

EB = [ 2 x 938.7 + 2 x 939.5 ] – 3728.0 = 28.4 MeV

The difference [the total mass of separate nucleons - the mass of nucleus] is called

the mass defect of the nucleus

The binding energy per nucleon: It is the ratio

An important observation is that

the binding energy per nucleon is nearly the same for almost nuclei !

the binding energy EBthe number of nucleons A

The dependence of EB /A on A for all nuclei is as follows

The graph represents the dependence of the binding energy per nucleon(EB/A) as a function of mass number A.

Remarks:

• The maximum of EB/A is for the values A ~ 50 ÷ 60 (Cr ÷ Zn)

It is about 8.7 – 8.8 MeV

• Light nuclei (with few nucleons) have small EB/A (for example,

EB/A ~ 1.1 MeV for 1H2 (deuterium); EB/A ~ 7.1 MeV for 2He4)

• EB/A for heavy nuclei decreases as increasing A

( EB/A ~ 7.5 MeV for the isotopes of uranium)

This dependence of EB/A on A leads to important consequences

Two possibilities of giving off energy:

Nuclear Fission: Division of a heavy nucleus into two fragments of

comparable mass

Nuclear Fusion: Synthesis of some light nuclei to form a larger nucleus

So after division the binding energy of the system increases about

240 MeV Since binding energy is negative → the total energy of the

system decreases → the remainder (240 MeV) converts into the kineticenergy of the fragments The division of isotopes of Uranium, Plutonium,…gives large sources of energy – called the atomic energy

• Synthesis of two nuclei 1H2 into the nucleus 2He4 (fusion):

(EB)i = 2x1.1x2 MeV

= 4.4 MeV

(EB)f = 7.1x4 MeV

= 28.4 MeVThe remainder about 24 MeV is given off

This type of energy is called the fusion energy

• Comparision: With the same mass of material the synthesis reaction

give off much more energy than the division reaction

2.2 Nuclear force:

The large binding energy of nucleons inside nucleus is the evidence ofthe strong interaction between nucleons  there exists forces betweenthem, and this kind of forces is called nuclear forces

Some charisteristics of nuclear forces:

They must be attractive, and much stronger the repulsive electricalforces between positive-charged protons

They are short-range forces: The range is of the order 10-15 m

They become negligible at distances d > 2.10-15 m

At d < 10-15 m the attractive forces are replaced by the repulsive

They do not depend on the eletrical charge of nucleons The bindingforces have the same magnitude for two protons, or two neutrons,

or proton and neutron This is an important feature and called

the charge-independence of nuclear forces

They depend on the mutual orientation of nucleons The binding

forces favour paires of protons or neutrons with opposite spins

§3 Nuclear stability and radioactivity:

They have the property of saturation: It means that one nucleoncannot interact simultaneously with all other nucleons in the nucleus,but only with those few in its vicinity (otherwise the binding energyper nucleon will not be nearly constant)

To give a theoretical basis for these characteristics of nuclear forcesone introduced models of nuclei There is no completely perfect model,every model has successfulness and disvantage

Among 2500 known nuclei, there are about 300 stable The others areunstable, and they decay to convert into other nuclei with emitting

particles and electromagnetic radiation This kind of processes is called

radioactivity

We consider the following processes:

(i) -decay; (ii) -decay;

3.1 Alpha decay:

In this process a nucleus emits the nuclei 2He4 (-particles):

ZXA → Z-2YA-4 + 2He4

Example: 92U238 → 90Th234 + 2He4

• The velocity of emmited a-particles is about

107 m/s, the kinetic energy ~ few MeV

• When -particles travel through matter, they

ionize molecules of matter, loss their energy

and stop The race in normal air: several

centimeters, in solid: several tens of

micrometers (-particles cannot pass

through an usual paper sheet)

• Alpha decay is the tunnel effect of -particles

through a potential-energy barrier as shown in the picture

• Alpha decay is usually accompained by -rays (electromagnetic radiation)

234 1

234

90Th ZPa e 

(protactinium)(thorium)

• Another example:

• The energy of emitted electrons

in this decay is not definte, but

has a continuous spectrum from 0

to Emax.

• The speeds of electrons are up to

0.9995 c  the electron’s motion

is relativistic

• Anti-neutrino:

It is the anti-particle of neutrino Both neutrino and its anti-particlehave the rest mass 0

It is neutral  the electric charge of the system is conserved

Its spin is ½  the total spin of the system is conserved

It shares the given off energy in the decay process with the

electron The given off energy is distributed between the electronand the anti-neutrino in different ways, holding the energy

p

• Note: Since the mass of proton is smaller than that of neutron,

the process is impossible for free protons

(But it is possible for protons inside nuclei by taking energy from

other nucleons)

3.2.3 Electron capture (K-capture):

• It is the process in which the nucleus absorbs one of the electrons inK-shell of the atom (or in L-, M-shells, but with small probabilities) Inthe result, one of protons converts to neutron:

40

and the corresponding transformation of the nucleus is

• The place of the captured electron will be filled by an electron fromother shells  this process is accompaned by emitting electromagneticwaves in the band of X-rays

• Example:

3.3 Decay rate and half-lives:

Suppose that we have an amount of radioactive matter Due to

radioactive processes (decays), the initial nuclei transform into other

nuclei  the number of the initial matter decreases in time

• Denote by N(t) the number of nuclei (and also atoms) at the moment t.

During an interval dt the number of nuclei (atoms) decreases dN :

Ndt

const t

where N0 = N(t=0) is the number of nuclei of the initial matter at t=0;

is a constant which characterizes the radioactivity of the matter It

is different for different elements is called the decay constant

• The half-life T1/2: By definition, it is the time required for the number

of radioactive nuclei to decrease to one-half the original number N0

2 / 1





693

0 2

ln

2 /

T

• We can derive the relation between the half-life and the decay constant:

• Unites for T1/2 → s, for → 1/s

• The half-lives are very different for different elements For the knownelements the range of half-lives is from 3x10-7seconds to 5x1015 years

Any nucleus consists of the component particles: protons (p) and

neutrons (n)

Proton has the electrical charge +1, neutron has no charge

Spin of proton and neutron is ½

The nucleus ZXA consists of Z protons and (A-Z) neutrons

The binding energy of the nucleons inside a nucleus is determined by

the mass defect of the nucleus (the difference between the total mass

of separate nucleons and the mass of nucleus)

The dependence of the binding energy per nucleon on the mass number Agives predictions about characteristics of nuclear forces between

nucleons inside nucleus It allows also to explain two ways of having

energy: fission and fusion

Radioactivity: The most of nuclei are unstable under decay processes.After decay processes the nucleus transforms into other nucleus.

The rate of radioactive process is described by the equation:

t

e N t

The half-life of an radioactive element T1/2: It is the required time

to decrease to one-half the initial amount of radioactive matter