Chapter 31 Nuclear Physics 31 1 Nuclear Structure

Chapter 31 - Nuclear Physics

31. 1 Nuclear Structure The atomic nucleus consists of positively charged protons and neutral neutrons.

31. 1 Nuclear Structure atomic mass number atomic number

31. 1 Nuclear Structure Nuclei that contain the same number of protons but a different number of neutrons are known as isotopes.

The Strong Force The strong force has four important properties: 1. It is an attractive force between any two nucleons. 2. It does not act on electrons. 3. It is a short-range force, acting only over nuclear distances. 4. Over the range where it acts, it is stronger than the electrostatic force that tries to push two protons apart.

As the number of protons increase, the number of neutrons must increase even more for stability All elements with >83 protons are unstable

Radioactivity – the spontaneous disintegration or rearrangement of internal nuclear structure.

• A nucleus is a bound system. You need to supply energy to separate a stable nucleus into separated nucleons. • This energy is called the binding energy. • Experimental evidence shows that the mass of any nucleus is less than the mass of the same number and type of nucleons when they are separated, i. e. at rest and out of the range of the forces from other nucleons. Why is this?

• According to Einstein, mass and energy are related. The energy of an object at rest is: E 0 = mc 2 • The binding energy is transformed into the greater mass of the separated nucleons. • Binding Energy (B) = mseparated c 2 - mnuc c 2

31. 3 The Mass Deficit of the Nucleus and Nuclear Binding Energy Where Δm, the mass defect, is the difference between the nuclear mass and the mass of the separated nucleons.

31. 3 The Mass Deficit of the Nucleus and Nuclear Binding Energy • Values for binding energy are very small, when using SI units of Joules. • Often the unit used is called the electron volt, where: 1 e. V = 1. 60 x 10 -19 J 1 Me. V = 1 x 106 e. V

31. 3 The Mass Deficit of the Nucleus and Nuclear Binding Energy 1 u of mass is equivalent to 931. 5 Me. V/c 2 = 931. 5 Me. V/u B = mseparated c 2 - matom c 2

31. 3 The Mass Deficit of the Nucleus and Nuclear Binding Energy B = mseparated c 2 - matom c 2 • Use atomic mass units (amu) for the mass of the atom (matom ) when calculating the resting energy of the atom. Values in periodic table. • Use amu value for neutron (1. 0087 u) for all neutrons (Nmn ). • Use atomic mass of 1 hydrogen (1. 0078 u) for all protons (Zm. H ). This accounts for the mass of the electrons. Note that this is slightly different from the 1. 00794 value given in the periodic table!

31. 3 The Mass Deficit of the Nucleus and Nuclear Binding Energy Example 3 The Binding Energy of the Helium Nucleus Revisited The atomic mass of helium is 4. 0026 u and the atomic mass of 1 hydrogen isotope is 1. 0078 u. Using atomic mass units, instead of kilograms, obtain the binding energy of the helium nucleus.

31. 3 The Mass Deficit of the Nucleus and Nuclear Binding Energy

QUESTION The atomic mass of this particular isotope of iron is 55. 9349 u. Recall that the neutron has a mass of 1. 0087 u and the 1 hydrogen atom a mass of 1. 0078 u

Binding Energy per Nucleon Curve

31. 4 Radioactivity When radioactive material disintegrates spontaneously, certain kinds of particles and/or high energy photons are released. These are called, respectively alpha (α) rays, beta (β) rays, and gamma(γ) rays. A magnetic field can separate these three types of particles emitted by radioactive nuclei.

Alpha (α) DECAY The general form for α decay is: • P is the parent nucleus, • D is the daughter nucleus and • He is an alpha particle, which is a helium nucleus (2 protons and 2 neutrons) • Since the daughter product has 2 less protons than the parent nucleus, it becomes another element • The process whereby one element becomes another is called transmutation.

α DECAY • When alpha decay occurs, the daughter product and the alpha particle have less mass than the parent nucleus. • This is in marked contrast to the separation of a stable nucleus, where the separated nucleons has a greater mass. • The fact that the daughter nucleus plus alpha particle have less mass implies that alpha decay releases energy. • Alpha decay occurs in some high-Z nuclei located beyond bismuth (Bi) on the binding energy curve.

α DECAY and the release of energy Plutonium has a mass of 239. 052 u, and undergoes α decay to become with a mass of 235. 043 u. The mass of a helium atom (used for the mass of the alpha particle) is 4. 0026. Determine the energy released (in Me. V) when Plutonium 239 undergoes α decay. 239. 052 u (235. 043 u + 4. 0026 u + E) E =. 0064 u (931. 5 Me. V/u) = 5. 96 Me. V This energy manifests itself as kinetic energy of the alpha particle and daughter particle.

Beta Decay and Quarks • Nucleons are composed of elementary particles called quarks. • There are 6 known types of quarks, but we will only consider 2 of these types: – Up quarks have a charge of +2/3 e – Down quarks have a charge of - 1/3 e

Beta Decay and Quarks • In the most common type of beta decay, ( known as β- decay), one of the down quarks in the neutron decays into an up quark, by emitting a beta particle with a total charge of -1. This particle is indistinguishable from an orbital electron.

Beta Decay and Quarks • In the most common type of beta decay, ( known as βdecay), one of the down quarks in the neutron decays into an up quark, by emitting a beta particle with a total charge of -1. This particle is indistinguishable from an orbital electron. • In the process, an extremely small particle (0. 0004% of the mass of the electron) called an antineutrino( ) is emitted.

β- DECAY The general form for β- decay is: β- decay, like α decay, results in transmutation of the parent nucleus into a different element and releases energy. This type of radioactivity occurs in unstable nuclei with more neutrons than protons. 14 C decays into 14 N due to this mechanism.

β- DECAY The radioactive isotope of carbon, Carbon 14, has an atomic mass of 14. 003241 u. It is converted by β- decay into Nitrogen 14 (atomic mass= 14. 003074 u). What is the energy (in Me. V) released in this process?

β- DECAY The radioactive isotope of carbon, Carbon 14, has an atomic mass of 14. 003241 u. It is converted by β- decay into Nitrogen 14 (atomic mass= 14. 003074 u). C 14 N 14 + E E = (14. 003241 -14. 003074) u x 931. 5 Me. V/u E =. 1556 Me. V Note that the mass of the β- particle and the antineutrino is included in the atomic mass of the nitrogen.

β+ DECAY β plus decay also occurs, but is a much less common phenomenon. In this case a proton decays into a neutron, a neutrino, and positron, which has the same mass as an electron, but a positive charge.

β+ DECAY β plus decay requires energy, it does not release it. Therefore it does not happen spontaneously

31. 4 Radioactivity γ DECAY γ decay occurs when a nucleus in a higher energy state spontaneously jumps to a lower energy state. The result is emission of a very high frequency photon. excited energy state lower energy state Note that γ decay does not result in transmutation.

31. 4 Radioactivity Gamma knife surgery – I hope I never see you here!

Stop to think The cobalt isotope 60 Co (Z = 27) decays to the nickel isotope 60 Ni (Z = 28). The number in the rh corner is A, atomic mass. The decay process is: A. Alpha decay. B. Beta-plus decay. C. Beta-minus decay. D. Gamma decay.

31. 6 Radioactive Decay and Activity Imagine tossing a coin. Will it be heads or tails? If you tossed 1000 coins, you’d very likely find 500 heads and 500 tails. If you tossed the 500 heads, you’d get about 250 heads, 250 tails. Tossing the 250 heads you’d get about 125 heads and so on. A graph of number of heads (N) remaining vs. number of tosses (T) would result in an exponential graph, like the on the right.

31. 6 Radioactive Decay and Activity Tossing a single coin is a random process, but tossing many coins results in a definite pattern. Each time you toss, about ½ the coins will be heads, and ½ will be tails. Radioactive decay shows this exact same pattern.

How often does the radioactive decay occur within a sample of radioactive material? • We can define λ as the decay constant, the probability that a nucleus will decay in the next second. λ has units of inverse seconds (s-1) • For example if λ=. 01 s-1, it means that a nucleus has a 1% chance of decay in the next second. • The probability that a nucleus will decay in a small time period, Δt, is: λ Δt • If there are N independent nuclei in a sample, the number of nuclei expected to decay in the time period, Δt: Δ N = -N λ Δt • And therefore Δ N/ Δt = - λN • The negative sign indicates that the ΔN is a loss of radioactive nuclei.

How often does alpha, beta, or gamma decay occur within a sample of radioactive material? • The activity (A) of a radioactive sample is defined as the number of disintegrations per second: A = |Δ N/ Δt| We have shown that the activity is equal to λ N. • The SI unit of activity is the becquerel (one Bq = 1 disintegration per second) • A non-SI unit of activity often used in the medical industry is the Curie, where 1 Ci = 3. 7 x 1010 Bq. • Activity can be determined by direct measurement.

This is a graph of N, the number of radioactive nuclei in a sample, vs. time. It can be shown that the exponential graph, shown on the left, has the following equation: where N is the number of radioactive or parent nuclei at a given time t, N 0 is the number of parent nuclei at time t = 0, and λ is the decay constant. Since A = λN we can also say:

Half- Life, T 1/2 It is often convenient to know how long it will take for one-half the sample to decay. Taking the natural log of both sides: where T 1/2 the half-life, is defined as the time in which ½ of the radioactive nuclei disintegrate.

31. 6 Radioactive Decay and Activity

A sample starts with 1000 radioactive atoms. How many half-lives have elapsed when 750 atoms have decayed? A. 2. 5 B. 2. 0 C. 1. 5 D. 0. 25

Cesium activity • The isotope 137 Cs is a standard source of gamma rays. The half-life is 30. 0 years. a. How many 137 Cs atoms are in a source that has an current activity of 1. 85 x 105 Bq? b. What is the activity of the source 10 years later?

Cesium activity, part A • The isotope 137 Cs is a standard source of gamma rays. The half-life is 30. 0 years. a. How many 137 Cs atoms are in a source that has a current activity of 1. 85 x 105 Bq? Known Find T 1/2 = 30. 0 years N, number of Cs atoms A = λ N = 1. 85 x 105 Bq To find λ, the decay : constant, we can use the relationship between it and half-life

Cesium activity, part A a. How many 137 Cs atoms are in a source that has a current activity of 1. 85 x 105 Bq? A = λ N = 1. 85 x 105 Bq λ = 7. 33 x 10 -10 s-1 N = A/ λ = 2. 53 x 1014 atoms

Cesium activity b. What is the activity of the source 10 years later? A 0 = 1. 85 x 105 Bq λ = 7. 33 x 10 -10 s-1 t = 10 yrs converted to seconds t= 3. 15 x 108 s A= 1. 47 x 105 Bq Activity decreases over time.

Cesium activity 1. Express T 1/2 in terms of seconds: T 1/2 = 30. 0 y(3. 15 x 107 s/y) = 9. 45 x 108 s 2. Find decay constant, λ: λ = (ln 2)/T 1/2 λ =. 693/9. 45 x 108 s λ = 7. 33 x 10 -10 s-1 3. Thus the number of 137 Cs atoms is: A/ λ = N 1. 85 x 105 Bq/ 7. 33 x 10 -10 s-1 = 2. 5 x 1014 atoms.

Formation of C-14 • Carbon 14 is produced in the upper atmosphere when energized neutrons form from interactions between high energy cosmic radiation and atmospheric gases. • The energetic neutron is absorbed by a nitrogen 14 nucleus and a proton is ejected. • Transmutation occurs and an unstable isotope of carbon (C-14) is created.

31. 7 Radiocarbon. Dating – visiting with Oetzi In 1949, Willard Libby developed a method of using the radioactive isotope 14 C to determine the age of organic materials up to about 50, 000 years old. Libby won a Nobel Prize for his work. .

31. 7 Radiocarbon. Dating – visiting with Oetzi • The concentration of 14 C in the is about 1 part per trillion (1 atom of 14 C for 8. 3 x 1011 atoms of 12 C. This seems small, but is measurable by modern chemical techniques. • Living organisms have an activity of 0. 23 Bq per gram carbon. After death, this activity decreases. • materials.

31. 7 Radiocarbon. Dating – visiting with Oetzi • T 1/2 = 5730 years for 14 C. It undergoes beta decay to 14 N. • Other isotopes with longer half-lives are used to date geological materials. • Uranium-Lead isotope dating was used to obtain a value for the age of the Earth (4. 5 billion years).

Carbon dating • Archeologists excavating a site have found a piece of charcoal from a fireplace. Lab measurements find the 14 C activity of the charcoal to be. 07 Bq per gram. What is the radiocarbon age of the charcoal? Known Find A=. 07 Bq/g t A 0 =. 23 Bq/g T 1/2 = 5730 years λ = (ln 2)/T 1/2

Carbon dating • First, find λ, the decay constant Known Find A=. 07 Bq/g t A 0 =. 23 Bq/g T 1/2 = 5730 years Tip: If we leave λ in terms of the T 1/2 , we can find t in years : λ = (ln 2)/T 1/2 =. 693/5730 λ = 1. 209 x 10 -4 y-1

Carbon dating Known Find A=. 07 Bq/g t A 0 =. 23 Bq/g T 1/2 = 5730 years λ = 1. 209 x 10 -4 y-1 Ln (A/A 0) = - λt Ln (A/A 0)/- λ = t T = 9839 or 9800 years

Conceptual Example 12 Dating a Bottle of Wine A bottle of red wine is thought to have been sealed about 5 years ago. The wine contains a number of different atoms, including carbon, oxygen, and hydrogen. The radioactive isotope of carbon is the familiar C-14 with ½ life 5730 yr. The radioactive isotope of oxygen is O-15 with a ½ life of 122. 2 s. The radioactive isotope of hydrogen is called tritium and has a ½ life of 12. 33 yr. The activity of each of these isotopes is known at the time the bottle was sealed. However, only one of the isotopes is useful for determining the age of the wine. Which is it? A. C-14 B. O-15 C. H-3

EOC 47 The practical limit to ages that can be determined by radiocarbon dating is about 41, 000 years. In a sample of this age, what percentage of C-14 remains?

EOC 47 The practical limit to ages that can be determined by radiocarbon dating is about 41, 000 years. In as sample of this age, whatj percentage of C-14 remains? 0. 70%

EOC #50 When a sample from a meteorite is analyzed, it is determined that 93. 8% of the original mass of a certain radioactive isotope is still present. The age of the meteorite is calculated to be 4. 51 x 109 yrs. What is T 1/2 in years of this isotope?

EOC #50 When a sample from a meteorite is analyzed, it is determined that 93. 8% of the original mass of a certain radioactive isotope is still present. What is T 1/2 in years of this isotope? Answer: 4. 88 x 1010 years

31. 8 Radioactive Decay Series The sequential decay of one nucleus after another is called a radioactive decay series.

31. 8 Radioactive Decay Series

31. 9 Radiation Detectors A Geiger counter

31. 9 Radiation Detectors A scintillation counter