Post on 28-Dec-2015
Ch 30 1
Chapter 30
Nuclear Physics and Radioactivity
© 2006, B.J. LiebSome figures electronically reproduced by permission of Pearson Education, Inc., Upper Saddle River, New Jersey Giancoli, PHYSICS,6/E © 2004.
Ch 30 2
Definitions•nucleons: collective name for protons and neutrons (very similar except for charge)
•nuclides: different types of nuclei
•atomic number (Z): number of protons
•atomic mass number (A): number of protons and neutrons
•neutron number (N): N = A – Z
•notation: AZX where X is the chemical symbol
•example: 146C has A =14, Z = 6 ( note that C conveys the
same information) and N = A – Z = 14 – 6 = 8, so 6 protons and 8 neutrons and 6 atomic electrons.
Ch 30 3
Definitions•isotopes: same Z but different N. They behave the same way chemically CCCCCC 16
6156
146
136
126
116 ,,,,,
•natural abundance: % with which an isotope is found in nature (12C = 98.9 %, 13C = 1.1%)
Ch 30 4
Properties of the NucleusRadius: experiments show that radius is
3115 )102.1( Amr
•would expect this if each nucleon occupied the same volume, so nuclear volume r3 A
•edge of nucleus is somewhat fuzzy
•density of all nuclei is approximately the same
constant
34 3
r
mAdensity N
Ch 30 5
Properties of the Nucleus
Mass Units:
1 u = 1.6605 X 10-27 kg = 931.5 MeV / c2
Nuclear Spin I: results from nucleon spin and orbital motion; is either integer or half-integer
Ch 30 6
Binding Energy
total binding energy : energy that would be required to separate nucleus into protons and neutrons
average binding energy : total binding energy divided by A.
binding energy of last nucleon : energy to free one neutron or proton.
Thus, energy is released when a nucleus is formed. This is the source of energy of the sun.
Ch 30 7
Example 30-1. Use data from the book to calculate the radius of the 16O nucleus, its binding energy, binding energy per nucleon and the biding energy of the last neutron added.
3115 )102.1( Vmr
uumm
06932.8)008665.1(8)(8 uuHm 06260.8)007825.1(8)(8
16O has 8 protons and 8 neutrons.
Mass of 16O = 15.994915 u
Δm = 0.13701 u
16.13192 u
3115 16)102.1( mr m15100.3
Binding energy
Binding energy per nucleon
)5.931)(13701.0(2
uMeVucm MeV7.127
MeVMeV
98.716
7.127
Ch 30 8
Example 30-1 (Continued). Use data from the book to calculate the radius of the 16O nucleus, its binding energy, binding energy per nucleon and the biding energy of the last neutron added.
Binding energy of the last neutron added
Mass of 16O = 15.994915 u
Mass of 15O = 15.003065 u
mn = 1.008665 u
sum = 16.01173 u
um 016819.0
MeVmc 7.152
)5.931)(016819.0(2
uMeVucm
Ch 30 10
Curve of Binding Energy per Nucleon
•certain nuclei are exceptionally stable ( 4He and 12C )
•for small A, energy can be released if nucleons are added (fusion)
•for large A, energy can be released if nucleus is split in half (fission)
Ch 30 11
Two Forces Acting in NucleusElectromagnetic: protons have + charge and repel each other with a force that is long range
Strong Nuclear Force:
•attractive
•strong
•short range ~ 10-15 m, so nucleons of opposites sides of nucleus do not feel it
•Stability of a Nucleus : results from balance of these two forces
Ch 30 12
Two Forces Acting in NucleusCurve of Binding Energy:
•rises for small A because strong nuclear force dominates
•decreases for large A because long- range electromagnetic force dominates
Ch 30 13
Quantum Mechanics in NucleusProtons and neutrons fill quantum states separately, because they are different particles, a proton and neutron can have the same set of quantum numbers. (Pauli exclusion principal)
• this favors Z ≈ N for stability
•electromagnetic force favors N > Z for stability
•Curve shows line of stability•Heavier nuclei have more neutrons than protons
•Nuclei that are not on line of stability undergo radioactive decay
Ch 30 14
Radioactivity•In 1896 Becquerel discovered uranium could darken photographic film without light.
•Marie Curie isolated radioactive radium and polonium.
•products were not understood, so named after Greek alphabet , and .
Ch 30 15
Alpha Decay
•Alpha is 4He nucleus
•“Daughter” nucleus has A - 4 and Z – 2
•disintegration energy Q: 2cMMMQ daughterparent )(
Ch 30 16
Alpha Decay
•Alpha is emitted because it has a high binding energy (see Curve of Binding Energy).
•Parent nucleus has high A
•Alpha is not very penetrating, could not pass through a piece of paper
Ch 30 17
Beta Decay
•Emission of electron (e-) and an antineutrino
•or emission of a positron (e+) and a neutrino ()
•the positron is the antiparticle of the electron
•thus each Beta decay results in emission of “particle” and “antiparticle”
•this decay process occurs because of the weak nuclear force.
•the electron is not an orbital electron, it is created in the decay
•Sometimes called - or +.
Ch 30 18
Beta Decay eNC 14
7146
•A is constant and a neutron changes into a proton in the nucleus. (Z’ = Z + 1)
•The atomic mass of 14N includes an extra orbital electron so it is not necessary to subtract an electron mass when calculating Q
•Antineutrino mass is treated as approximately zero.
Ch 30 19
Positron Decay eFNe 19
91910
•positron is the antiparticle of the electron; it has same mass and opposite charge
•A is constant and proton changes into a neutron in the nucleus (Z’= Z -1)
•atomic mass of 19F includes one fewer orbital electron than 19Ne and thus is necessary to subtract two electron masses when calculating Q.
•Electron Capture: nucleus captures an orbital electron and emits a neutrino. This has same effect as positron decay.
Ch 30 20
Beta Decay and the Lepton Family
eNC 147
146
•Electrons, positrons, neutrinos and antineutrinos belong to a family of particles called “leptons”
•Electrons and neutrinos are “particles” with lepton number L = +1
•Positrons and antineutrinos are antiparticles with lepton number L = -1
•Since L = 0 on the left side of the arrow the lepton numbers on the right side of the arrow must add up to 0 and thus an antineutrino must be emitted.
L 0 → 0 +1 -1
Ch 30 21
Example 30-2. Calculate the energy released in the beta decay of 3H to 3He.
uHm 016049.33 uHem 016029.33
uuuHemHmm 00020.0016029.3016049.333
These calculations are done with atomic masses. The atomic mass of 3H includes one electron and the atomic mass of 3He includes two electrons. Thus we do not have to account for the mass of the electron emitted because the 3He mass already has includes an “extra” electron.
This energy is shared by the electron, the neutrino and the nucleus in such a way as to conserve energy and momentum.
)/5.931)(00020.0(2 uMeVucm keVMeV 6.180186.0 18.6 KeV
KE of Electron
Probability
eHeH 32
31
We don’t have to include the neutrino mass– its mass is very small.
Ch 30 22
Gamma Decay
•Gamma is high energy photon
•Just like atoms, nuclei have excited states and photon are emitted when nucleus changes from higher to lower state
•Nucleus remains the same, just looses energy
•Energy differences between states are typically in MeV range.
•Often Gamma decay follows Alpha or Beta decay
Ch 30 23
Conservation Laws
Radioactive Decay obeys a number of conservation laws
•Conservation of Mass-Energy
•Conservation of Momentum: The neutrino was first proposed to account for “missing” energy and momentum
•Conservation of Electric Charge
•Conservation of Nucleon Number (A)
•Conservation of Lepton Number: anti-lepton is created at same time as lepton
Ch 30 24
Exponential DecayRadioactive decay is similar to the decay of charge on a capacitor in an RC circuit
• N0 is the initial number of nuclei
• N is number of nuclei at a later time
• N is number of decays in time t
• is called the decay constant
teNN 0
Ch 30 25
Activity
tet
N
t
N
0
Activity which is defined as the number of decays per time (N / t) also decreases exponentially
Ch 30 26
Half-Life
As with RC decay, we define the half-life T1/2
6930
21
.T
Exponential decay always results when the number of decays is proportional to the number of un-decayed nuclei present and thus
Nt
N
Ch 30 27
Example 30-3 (43) The iodine isotope 53131I (half-life = 8.02 days) is used for
diagnosis of thyroid function. If 682 g are ingested by a patient, determine the activity (a) immediately, (b) 1.0 h later when the thyroid is being tested, and (c) 6 months later.The decay constant
is:
81.15)/3600)(/24)(/5.30)(6)(10000.1( 16 hsdayhmodaysmost
The initial number of nuclei is
(a) When t=0, we get
(b) When t=1.0 h, the exponent is
So we get
(c) When t=6 months, the exponent is
So we get
2
1
693.0
T
nucleimolatomsmolg
gN 1823
6
010134.3)/1002.6(
/131
10682
teNN 0
316 106.3)/3600)(0.1)(10000.1( hshst
teNN 0
teNN 0
1610000.1)/3600)(/24)(02.8(
693.0 shsdayhrdays
sdecayses /1013.3)10134.3)(10000.1( 1201816
sdecayses /1013.3)10134.3)(10000.1( 12003600.01816
sdecayses /1026.4)10134.3)(10000.1( 581.151816
Ch 30 28
Decay Series238U decays with a half-life of 4.5 billion years. After each decay a series of reactions occur relatively quickly. The approximate age of solar system is 5x109year so we now have about ½ of the 238U that originally existed. 234U (2.5x105 year) exists in nature only because it is a daughter of 238U.
Ch 30 29
Radioactive Dating•Radioactivity has been used to determine the age of many objects that range in age from several hundred to several billion years.
•Carbon Dating: 14C is produced in the atmosphere by the reaction
pCNn 146
147
• 14C behaves like 12C and is taken in by living plants. After the plant dies it no longer takes in 14C.
• 14C decays with a half-life of 5700 years
•ratio of 14C/ 12C can determine age of plant or object plant was made from.
•older objects can be dated by other radioactive nuclei
Ch 30 30
Example 30-4. (30-77) The practical limit for carbon-14 dating is about 60,000 years. If a bone contains 1.0 kg of carbon, and the animal died 60,000 years ago, what is the activity today? The ratio of 14C to 12C is ≈ 1.3x10-12.
The mass of carbon 60,000 years ago was essentially 1.0 kg, for which the corresponding number of atoms would be
However, a small fraction will in fact be 14C atoms, namely
The decay constant is
Activity today is given by
2
1
693.0
T
atomsmolg
molatomsgN 25
233
121002.5
/12
)/1002.6)(100.1(
atomsmolatomsN 131225
141052.6)/103.1)(1002.5(
teNActivity 0)1016.3)(000,60)(1083.3(13112
7112
)1052.6)(1083.3( yrsyrs
es
1127
1083.3)/1016.3)(5730(
693.0
syrsyr
1/18.0 sdecays
Ch 30 31
Coulomb Barrier and Tunneling
•In alpha decay, the combined effect of the strong nuclear force and the Coulomb force produces the potential energy diagram shown above.
•In classical physics, the alpha particle does not have enough kinetic energy to surmount the barrier. A small part of its wave function extends beyond the barrier and this permits the decay, but with a long half-life.
Ch 30 32
Detection of Radiation
•Nuclear detectors are sensitive to the ions created when a high energy charged particle passes through them or photons given off by the ions.
•If the radiation is not charged (gamma ray or neutron) it must first interact and make a charged particle that can be detected
•Example: A gamma ray can interact via photoelectric effect, Compton effect or pair production.
•The resulting charged particles usually ionize many atoms producing a signal that can be detected.
Ch 30 33
Geiger Counter•Consists of a metal can with a low pressure gas
•A thin wire down the center is charged to approximately +500 V
•A high energy charged particle will ionize a few of the atoms in the gas•The electrons from these ionizations are accelerated toward the + 500 V wire. They collide with other atoms producing additional electrons
•This avalanche of electrons creates a current in the wire that is detected and counted by a circuit
Ch 30 34
Scintillation Counter•Atoms of scintillation materials are easily raised to excited states by charged particles
•Drop back to ground state by emitting photons.
•Single high energy particle or gamma ray can cause many atoms to scintillate
•Example: NaI crystal1.0 MeV gamma ray results in many photons
•Photons create ~1000 photoelectrons in photomultiplier tube
•Tube multiplies these electrons by factor of 1 x106 which is a sufficient signal for instruments.