Atomic and Nuclear Physics

Dr. Zee, Sung-Kyun(zee@kaeri.re.kr, spoymz@gmail.com)

Session 1 : Nuclear Physics Basics

• Introduction and Nuclear Physics Basics

Atom and Nucleus

• In this session we will discuss the building blocks of the atom including the Neutron, Proton and Electron.

• We will also discuss how the Atomic Number specifies the elements and how they are arranged in the Periodic Table.

• And we will discuss how Isotopes of an element have different Atomic Mass.

• Finally, nuclear stability and unstable nuclei will be discussed.

Discovery of X-ray

• In 1895, Roentgen was working with cathode ray tube. During experiment, when the tube was shielded with heavy black paper, a previously unknown invisible light or ray that was being emitted from the tube.

• He named the new ray, X-ray.

W. Roentgen

Nuclear Evolution

• 1895 : Roentgen, X-ray Discovery • 1896 : Becquerel, Radioactivity of U • 1898 : Curie , Radioactivity of Ra, Po• 1902 : Rutherford, α, β, γ Discovery • 1905 : Einstein, The Theory of Relativity• 1932 : Chadwick, Neutron Discovery• 1939 : Hahn & Strassman, Fission Reaction• 1941 : Manhattan Project • 1942 : Fermi, The 1st Nuclear Reactor Development

Enrico Fermi Curie Einstein

Atomic Nature of Matter

• Atomic Structure• Energy Level of Atom• Excitation and Ionization of Atom

Atomic Nature of Matter

Atomic Structure• Atom = Nucleus + Electron

- Radius of Atom : 10-10 m- Radius of Nucleus : 10-15 m

• Electron: Orbital Electron

SymbolMass Electric

Chargeamu kg MeV

Electron e 5.4858E-4 9.11E-31 0.511 -e

Proton p 1.007277 1.673E-27 938.28 +e

Neutron n 1.0086665 1.675E-27 939.57 0

Models of Atom - Chronicle

Atom Electron Atomic Nucleus Atomic Model Neutron

Daltonin 1808

Thomsonin 1897

Rutherfordin 1911

Bohrin 1913

Chadwickin 1932

• Billiards Model• Atoms are solid,

invisible, undestroyablesmall balls

• Plum pudding model

• Negative electrons in a positive framework

• Atoms are mostlyempty space

• Negative electrons orbit a positive nucleus

• Electrons orbit the nucleus in “shells”

• Electrons can be bumped up to a higher shell if hit by an electron or a photon of light

• Electron Clouds surround nucleus

Nucleus and Electrons

Protons and neutrons together form the nucleus of the atom.

The nucleus determines the identity of the element and its atomic mass.

Protons and neutrons have essentially the same mass. Only proton is positively charged while the neutron has

no charge, so the nucleus has a positive electric charge.

Electrons are negatively charged particles that surround the nucleus in “orbits” similar to moons orbiting a planet.

Atomic Structure

Electrons circle the atom in atomic orbitals which represent different energy states.

Electrons closest to the nucleus are the most tightly bound (the K-shell).

Going outward from the nucleus, the other orbitals are called the L, M, N, O and P-shells.

The atom is electrically neutral atom, the number of electrons is equal to the number of protons in the nucleus.

Nucleus

Electron

Bohr’s Model of the Hydrogen Atom

hmvr 2hmvr 22

2 3r mv h

34 2: Planck's constant ( 6.626 10 m / ) wave length, : speed of light

m ncE E h h

h kg sc

Discrete Energy Level- n = 1: ground state- n = 2,3,4 .. excited state

nhmvr 2

Energy of Electromagnetic Wave

Electron Shell of an Atom

Example : Total # of Electrons up to M shell ?- K (n=1): # of electrons = 2- L (n=2): # of electrons = 8- M (n=3): # of electrons = 18

Total 28 electrons

M (n=3)

L (n=2)

K (n=1)Ze

Pauli’s Exclusive Law

Allowable # of Electrons in an Electron shell = 2 x n2

n = Principal Quantum Number K shell : n = 1

L shell : n = 2M shell : n = 3

Energy Level : K < L <M < N…

Atomic Structure

Mass Charge(atomic mass units)

Proton 1 1Neutron 1 0Electron 1/1840 -1

1 atomic mass unit = 1.66 x 10-27 kg (~ 1g/mol)One unit of charge = 1.6 x 10-19 CoulombEnergy Units (1 eV ~ 1.6 x 10-19J)

• Binding energy of outermost electrons ~ order of eV(energy involved in chem. reactions))

• Binding energy of nucleons (constituents of nucleus) ~ order of MeV= 1.66 x 10-27 kg

One unit of charge = 1.6 x 10-19 Coulomb atomic mass unit = 1.66 x 10-27 kg(note) 1 amu: 1/12 of the mass of unbound neutral atom of C-12

Size of the atom : 10-10 m Size of the nucleus : 10-15 m

Atomic Number

The number of protons determines the element. Atomic number (Z)

= number of protons Elements with the same number of protons

but different numbers of neutrons are called isotopes.

Some isotopes are radioactive.Element Symbol ZHydrogen H 1Helium He 2Cobalt Co 27Iodine I 53Iridium Ir 77Uranium U 92

Atomic Numbers of some element

Elements

The number of protons in an atom dictate the element.

For an uncharged atom, the number of electrons equals the number of protons.

Sample of Element - Neon

10Ne20

Periodic Table of the Elements

All known elements are shown on the Periodic Table

Isotope

The following is a convention used to simplify the identification of isotopes.

A

ZX Where “X” is the chemical symbol, “A” the mass number and “Z” is the number of protons (atomic number). Hydrogen, deuterium and tritium are represented as:

1

1H2

1H3

1HThe subscript is sometimes omitted because the chemical symbol gives same information. Other forms of the notation include: H-3 or 3H, Co-60 or 60Co, U-238 or 238U

Isotopes

An isotope of an element has:• the same number of protons• but different number of neutrons

Some isotopes are stable and some are unstable (radioactive).

1H 2H 3H

Chart of Nuclides

There are many isotopes. Most have more neutrons than protons. Some are stable but most are unstable.

equal number of protons and neutrons

Nuclear Stability

A stable or non-radioactive nuclide is one whose atoms do not decay

Line of stability

N > Z

Too manyneutrons

for stability

Too manyprotons

for stability

Nuclear Stability

For the heaviest stable nuclei, N is about 1.5 times Z

The presence of the extra neutrons overcomes the positively charged protons’ tendency to repel each other and disrupt the nucleus

The nucleus is held together by the Nuclear Force (attractive force) btwn. nucleons compensating repulsive coulombian force btwn. protons

Unstable Nuclei

• Nuclei which do not fall on the line of stability tend to be unstable or “radioactive” They are called “radionuclides”

• A few radionuclides do fall on the line of stability, but their rate of decay is so slow that for all practical purposes they are stable

– K-40, V-50, Rb-87, In-115, Te-130, La-138, Nd-144, Sm-147, Lu-176, Re-187, Th-232, U-235, and U-238

• Radionuclides undergo a process called radioactive transformation or disintegration

• In this process, the nucleus emits particles to adjust its neutron (N) to proton (Z) ratio

• This change in the N to Z ratio tends to move the radionuclide toward the line of stability

Some Common Radionuclides

Naturally occurring 235U and 238U 60Co, 137Cs, 90Sr found in nuclear power plants 192Ir used in radiography 99mTc used in nuclear medicine 131I used in treatment of thyroid conditions

Excitation and Ionization

In this session we will discuss : Electrons Binding Energy which holds the electrons to the nucleus Excitation and Ionization which are processes by which

electrons are removed from their stable location

Electron Shells

• Pauli’s Exclusive Law : Allowable # of ‘e’ in a shell

2n2

KL

MNO

91

40 Zr

Shell nMaximum

No. of Electrons

K 1 2 2

L 2 8 8

M 3 18 18

N 4 32 10

O 5 50 2

P 6 72

9140 Zr

Bohr’s Model of Hydrogen Atom

hmvr 2hmvr 22

2 3r mv h

nhmvr 2

Discrete Energy Level- n = 1: ground state- n = 2,3,4 .. excited state

chhEE nm

Energy of Electromagnetic Wave

Energy Level of Hydrogenby Bohr’s Atomic Model

2 2 40

2 2 2

2 13.6n

Z m eE eVn h n

n = ∞

n = 5

n = 4

n = 3

n = 2

n = 1

E∞ = 0

E5 = -0.54

Eφ = -0.85

E3 = -1.51

E2 = -3.4

E1 = -13.6

En > 0 Free Electron

En < 0 Electronsin Atom

Energy Level of Nucleus

Energy Level of Atom– Discrete Energy Level

in Electron Orbit– Radiation emitted due to

transition (L K): X ray– X-ray Energy ~ eV,

Energy Level of Nucleus– Discrete Energy Level– Radiation emitted: Gamma ray– Gamma ray energy ~MeV

Free Electron

Excited State

Energy

Ground State

Ener

gy

Excitation and Ionization of Atom

Ground State : Lowest energy level of orbital electron

Excitation- Movement of orbital electrons to higher energy level due

to excitation of atom by radiation (α , β , Gamma, X ray, )

Transition - Movement of electrons at an excited state to lower energy

level - Radiation (X ray) is emitted with the energy difference

Ionization - If orbital electron receives larger energy than binding

energy of electron and nucleus from outside, orbital electron becomes free electron. Then the atom is ionized.

- Ionization energy is required to make orbital electron to free electron.

(Ex) Ionization Energy of Hydrogen: 13.6 eV

Ionization of a K-shell electron

Ek > EL

EL > Ek

E > 0

Excitation

Ionization

Radiation(α, β)

Transition

Nucleus

Nucleus Structure• Nucleus = Proton + Neutron

- Z = Number of Protons, Atomic Number - N = Number of Neutrons- Mass number, A = Z + N

Nuclides

Nuclide Catalogued with Z, N, A - Isotope : Elements of same number of protons Z, different mass number A

Similar chemical nature, different nuclear characteristics - Isobar : Elements of same mass number A, different number of protons Z- Isotone : Elements of same number of neutrons N- Isomer : Nucleus that remains in excited state for a measurable period of

time, before dropping down to ground state by the emission of gamma radiation

Z N A Examples

Isotope = ≠ ≠

Isobar ≠ ≠ =

Isotone ≠ = ≠

Isomer = = = 811375681

13756

32602832

5927

51903952

9038

1210

11

,

,

,

,

BaBa

NiCo

YSr

HH

m

Nuclear Forces

Forces present in the nucleus- Electrostatic Forces between Charged Particles- Gravitational Forces between two Objects that have Mass- Nuclear Forces

• Strong attractive force that is independent of charge• Very short range

Forces Acting in the Nucleus

Force Interaction Range

Gravitational Very week attractive force between all nucleons

Relatively long

ElectrostaticStrong repulsive force between like charged particles (protons)

Relatively long

Nuclear Force

Strong attractive force between all nucleons

Extremely short

Mass Defect

Einstein’s Theory of Relativity- Mass and energy are equivalent and convertible, one to the other.

Mass Defect (△m ) △m = Z(mp + me ) + (A-Z)mn – matom (amu)

where mp = mass of proton (1.007277 amu)mn = mass of neutron (1.008665 amu)me = mass of electron (0.000548597 amu)matom = mass of nuclide (amu)Z = Atomic Number (number of protons)A = mass number (number of nucleons)

Atomic Mass Unit (amu) : One-twelfth the mass of the neutral 12C atom

2 8 mc (J) , where c = 3x10 m/s, m = mass (kg)E

Atomic Mass Units (amu)

Atomic weight : Unitless- Mass of neutral atom relative to the mass of a neutral 12C atom

Gram atomic weight : gm- Amount of substance having a mass in grams equal to the atomic weight.

Same as one mole for monatomic substances.

Avogadro’s Law- The number of atoms or molecules in a mole of any substance is

a constant, which is called, Avogadro’s number ; NA =0.602217 x 1024

Mass of one atom of 12C = 12/ NA =1.99264 x 10-23 g

Atomic Mass Unit (amu) : One-twelfth the mass of the neutral 12C atom

Energy equivalent of 1 amuEnergy of the rest electron

Atomic Mass Unit (u) : kg

Energy of 1 amu

Energy of the rest electron

Electron Volt (eV)1eV = kinetic energy of electron accelerating 1 electron with 1 Volt

-2 71 21 = 1 .6 6 0 5 x 1 0 k g1 2

Ca m u

2 8 mc (J) , where c = 3x10 m/s, m = mass (kg)E 2 -27 8 2 -10 E mc = 1.6605 x 10 x (3 x 10 ) = 1.429242 x 10 Joule = 931.5 MeV

2 -31 8 2rest 0

-14-13

E m c = 9.1095 x 10 x (2.9979 x 10 )1MeV = 8.1871 x 10 Joule x = 0.5110 MeV

1.6022 x 10 Joule

-19 -19 1.6022 x 10 (C) x 1 Volt(V) = 1.6022 x 10 Joule

Binding Energy

Binding Energy- the amount of energy that must be supplied to a nucleus to completely

separate its nucleons- the energy equivalent of the Mass Defect : △mc2

Binding Energy2 2

2

{ ( ) ( ) }

{1.007826 1.008665 } Joule {1.007826 1.008665 } x 931.5 Mev

B p e n atom

atom

atom

E mc Z m m A Z m m c

Z N m cZ N m

△

< Example > Binding Energy of Nucleus?matom = Atom Mass Helium = 4.001502

• BE = (2x1.007826 + 2x 1.008665 – 4.001502) x 931.5 MeV = 29.3 MeV

42 eH

Example

Calculate the mass defect and the binding energy for uranium-235. One uranium-235 atom has a mass of 235.043924 amu.

Solution:

Step 1: Calculate the mass defect

Step 2: Calculate the binding energy.

p e n atomm = [Z(m + m ) + (A- Z)m ]- m

m = [92x(1.007826 amu) (235 92)x1.008665 amu] 235.043924 amum 1.91517 amu

931.5 MeVB.E = m1 amu

931.5 MeV = 1.91517 amu = 1,784 MeV1 amu

Binding Energy per Nucleon

AEf B

B

42 eH

Binding Energy per Nucleon

: measure of force btwn. nucleons

If is large, nucleons have strong binding, and consequently a large energy is required to separate nucleus into nucleons.

Sharp increase at low a value, broad maximum at ~ A=50

The heaviest nuclei are easily fissionable compared to lighter nuclei.

AEf B

B

Bf

Bin

ding

Ene

rgy

per N

ucle

on (M

eV)

atomic mass number

Bf

Nuclear Fission Reaction

• Nuclear Fission Process• Nuclear Fission Cross Section• Fission Energy• Delayed Neutron• Fission Products• Decay Heat of Fission Products

Nuclear Fission Reaction

Nuclear Fission- A heavy nucleus absorbs a neutron and splits into

two fragments with the ejection of several high-velocity (fast) neurons.

U235 Fission Reaction235 92 142 192 38 54 02.5 (200 )U n Sr Xe n Q MeV

- Most fission products of Uranium have either mass number (90-100) or

mass number (130-140)

- Energy released per fission is approximately 200MeV

- Average neutrons emitted per fission= 2.43

In the Chemical Reaction, 2 2+ O + 4.2 eVC CO

Nuclear Fission Yield and Neutron Energy

Nuclear Fission Yield

- Most fission products of Uranium haveeither mass number (90-100) or mass number (130-140)

0 2 4 6 8 10 12Fission Neutron Energy, MeV

01

0.2

0.3

0.4

n(E

)

Fission Neutron Energy- Energies of neutrons released from fission:

0.1 MeV ~ 15 MeV- Most probable Energy ~ 0.7 Mev- Average energy of fission neutrons : 2MeV

Nuclear Fission Process

Liquid Drop Model– Excitation Energy: the energy level of a nucleus

above its ground state– Critical energy for fission :

• the minimum excitation energy required for fission reaction occur

• The minimum energy for compound nucleus to havein order to deform into the state C

A

B

C

D

Nuclei Critical Energy (CE)(MeV)

Binding Energy (BE) of Compoundnucleus (MeV)

BE - CE

Fission *Cross

Section(barn)

U233 6.0 7.0 +1.0 MeV 531

U235 6.5 6.8 +0.3 MeV 584

U238 7.0 5.5 -1.5 MeV 11 m

Pu239 5.0 6.6 +1.6 Mev 747

• For U238, BE – CE = -1.5 MeV. The kinetic energy of an incident neutron must be above1.5MeV for fission to be possible.

Nuclear Fission Cross Section

• Thermal(0.0253 eV) data for the fissile nuclides

Totalfission

Scattering

U-235 Pu-239Total

fission

Scattering

Puγn,Puγn,Pu

Fissionfn,PuPuNpUγn,U

UPaThγn,Th

24194

24094

23994

23994

23994

2.3d,β23993

23m,β23992

23892

23392

27.4d,β23391

22m,β23390

23290

233U 578.8 531.1 0.0899 2.287 2.492

235U 680.8 582.2 0.169 2.068 2.418

239Pu 1011.3 742.5 0.362 2.108 2.871

241Pu 1377 1009 0.365 2.145 2.927

†a f

* From Neutron Cross Sections, Brookhaven National Laboratory report BNL-325, 3rd ed., 1973.

fa †f

γ

σσ

α

Fission Energy

MWhjMeV 2313 1045.4106021.1

jMWh 9106.3

Form Emitted(MeV)

Recoverable(MeV)

Fission fragments

FF decay

-ray

-ray

neutrino

Prompt

Fission Neutron

Captured -ray

168

8

7

12

7

5

-

168

8

7

-

7

5

3-12

Total 207 198-207

Fission Energy Produced from 1g of U-235

2 32 3

F i s s io n E n e rg y p ro d u c e d f ro m 1 2 3 5 6 .0 2 1 0 #( ) 2 0 0 ( ) 4 .4 5 1 0 ( )#2 3 5

2 2 .7 8 ( ) 0 .9 5 /

g U

M e V M W hg M e V

M W h M W D gg

7

7

( ) x 6 ,0 0 0 k cal/k g 2 2 .8 M w h /g = 2 2 .8 x 1 0 0 0 k w h /g x 8 6 0 k cal/k w h

1 .9 6 0 8 x 1 0 k cal/g1 .9 6 0 8 x 1 0 k cal/g = = 3 ,2 6 8 k g

6 ,0 0 0 k cal/k g

c

c

T kg

T

Calculate fission energy for 1g of U-235.

Calculate amount of coal (Tc) to produce same energy as the fission energy from 1g of U-235. If the heat generation of coal is 6,000 kcal/kg

Calculate amount of petroleum (Q) to produce same energy as the fission energy from 1g of U-235. If the heat generation of petroleum is 9,700 kcal/liter 7

7

x 9 ,7 0 0 k cal/lite r 1 .9 6 0 8 x 1 0 k cal/lite r1 .9 6 0 8 x 1 0 k cal/lite r = = 2 ,0 0 0 lite r

9 ,7 0 0 k cal/lite r

Q

Q

Example: Calculation of Fission Energy

Calculate fission energy for a typical fission reaction listed below.

Solution

This mass difference can be converted to an energy equivalent.

1 235 140 93 10 92 55 37 0+ + + 3 n U Cs Rb n

23592

10

U 235.043924 amu

n 1.008665 amu

9 33 7

1 4 05 5

10

R b 9 2 .9 1 6 9 9 a m u

C s 1 3 9 .9 0 9 1 0 a m u

3 ( n ) 3 .0 2 5 9 9 a m u

236.052589 amu 235.85208 amu

Mass difference = Mass of Reactants - Mass of Products = 236.052589 amu - 235.85208 amu = 0.200509 amu

931.5 MeVFission Energy = 0.020059 amu = 186.8 MeVamu

Prompt and Delayed Neutrons

Prompt Neutron– great majority (over 99%) of the neutrons produced

within 10-13 seconds of actual fission event

Delayed Neutron– Small portion (< 1%) of the neutrons produced some time after the

fission process– Emits immediately following the first beta decay of fission fragment

(delayed neutron precursor)

Delayed Neutron FractionNucleus Fraction

U233 0.0026

U235 0.0065

Pu239 0.0021

Delayed Neutrons

Delayed neutrons – Neutron-rich fission products undergoing beta decay – Daughter is produced in an excited state with sufficient

energy to emit a neutron

Delayed neutron fraction : β– Total fraction of the delayed neutron among

fission neutrons

Delayed Neutron Data for Thermal Fission in 235U

Group Half-lifesec

Decay constantλi, sec-1

EnergykeV

Yield, neutrons

per fission

Fractionβi

123456

55.7222.726.222.300.6100.230

0.01240.03050.1110.3011.143.01

250560405450

--

0.000520.003460.003100.006240.001820.00066

0.0002150.0014240.0012740.0025680.0007480.000273

Total yield : 0.0158Total delayed fraction (β) : 0.0065

8735 Bromine Br

8736 Krypton Kr

8737 Rubidum Rb

8738 StrontiumSr

nKr 8637

Fission Product

Distribution of Fission Product– Compound nucleus splits up more

than 40 different ways yielding 80 fission products

– Range of Mass numberLight group : A= 80-100 ( A0=95) Heavy group : A= 125-155(A0=139)

– More symmetrical fissions induced by fast neutrons

9638Sr 140

54Xe

Continuous Decay due to neutron rich

14058

14057

14056

14055

14054 CeLaBaCsXe

Decay Heat of Fission Products

Fission products produce energy by emitting radiations.• Fission products are transited to stable nuclides.

• Decay Heat After Reactor Shutdown– About 6~8% of the reactor power right after reactor shutdown

135135135135135 BCXeITe as

Decay Heat Generation after Reactor Shutdown

Reactor Safety Issues due to Decay Heat of Fission Products– One of the major safety concerns in reactor safety– Major Reactor Safety System

• Residual Heat Removal System (RHRS) • Shutdown Cooling System(SDC)• Decay Heat Removal Circuit(DRC)• PRACS, DRACS, IRACS, PDRC etc.

Fission Product Decay Power for a Reactor• Consider a reactor that has been operating at a constant power P0

for the finite time t0

Where P(ts) = the power(rate of energy release) emanating from the fission products at the time ts after shut down

P(t0+ts) = the power emanating from the fission products at t0 + ts

0 0

0 0 0

( , ) ( ) ( ) = s s sP t t P t P t tP P P

Nuclear Fusion Reaction

MeVnHeTD 58.17432

MeVpTDD 02.4322 Thermofusion

Plasmaconfinement

International Thermonuclear Experimental Reactor

21

8

O For sustainable Fusion Reaction , We need to confine D ( ) in plasma state

n > T (1x10 ) where n = plasma density, = confinement time, T = absolute tem

H

K

o

perature

Example

Calculate Q value (the difference in the rest mass energies before and after reaction) for the following fusion reaction.

For atomic masses, = 3.016049 amu, = 2.014102 amu ,

= 1.008665 amu and = 4.002604 amu

3 41 2 ( , ) H d n He

21 H42He1

0 n

Solution – Before reaction = 3.016049 amu , = 2.014102 amu

Total mass before reaction = 5.030151 amu

– After reaction = 1.008665 amu, = 4.002604 amuTotal mass after reaction = 5.0011269 amu

– Difference in mass = 5.030151 amu - 5.0011269 amu = 0.0018882amu

– Q = 0.0018882 x 931 MeV = 17.6 MeV > 0

Therefore, this fusion reaction is an exothermic reaction.

31 H

31 H 2

1 H

10 n 4

2He

Thank you