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 9

Curve of Binding Energy per Nucleon

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.