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Atomic and Nuclear Physics
Dr. Zee, Sung-Kyun([email protected], [email protected])
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