Chapter 31- Nuclear Physics

1. Chapter 31- Nuclear Physics

3. 31.1 Nuclear Structure The atomic nucleus consists of positively charged protons and neutral neutrons.

4. 31.1 Nuclear Structure atomic mass number atomic number

5. 31.1 Nuclear Structure Nuclei that contain the same number of protons but a different number of neutrons are known as isotopes.

6. 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.

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

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

9. 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. • Recall that E0 = mc2 • If the separated nucleons are also at rest, that energy is in the form of greater mass. • Binding Energy (B) = mseparated c2 - mnuc c2

10. 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.

11. 31.3 The Mass Deficit of the Nucleus and Nuclear Binding Energy • B = mseparated c2 - matom c2 • 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 1hydrogen (1.0078 u) for all protons (ZmH ). Note that this is slightly different from the 1.00794 value given in the periodic table!

12. 31.3 The Mass Deficit of the Nucleus and Nuclear Binding Energy 1u of mass is equivalent to 931.5 MeV/c2 c2 = 931.5 MeV/u B = mseparated c2 - matom c2

13. 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 4.0026u and the atomic mass of 1hydrogen isotope is 1.0078u. Using atomic mass units, instead of kilograms, obtain the binding energy of the helium nucleus.

14. 31.3 The Mass Deficit of the Nucleus and Nuclear Binding Energy

15. QUESTION The atomic mass of this particular isotope of iron is 55.9349u. Recall that the neutron has a mass of 1.0087u and the 1hydrogen atom a mass of 1.0078u

17. Binding Energy per Nucleon Curve

18. 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.

19. α DECAY The general form for α decay is: where P is the parent nucleus, D is the daughter nucleus and The process whereby one element becomes another is called transmutation. This reaction occurs when the mass of the parent nucleus is greater than the mass of the daughter nucleus plus the mass of an alpha particle. This is the case for some high-Z nuclei located beyond Bi on the binding energy curve. It is energetically favorable for these nuclei to eject an alpha particle, because the daughter is bound more tightly than the parent.

20. α DECAY of Uranium

21. α DECAY and the release of energy The atomic mass of 238U is 238.0508 u, the atomic mass of 234Th is 234.0436 u and the atomic mass of a helium atom is 4.0026. Determine the energy released when uranium 238 undergoes α decay. 238.0508u {234.0436u + 4.0026u} 238.0462 Since energy is released, the combined mass of thorium and the alpha particle should be less than that of the uranium .

22. α DECAY of Uranium 31.4 Radioactivity 238.0508u {234.0436u + 4.0026u} 238.0462 ∆E = ∆mc2 = (.0046u)(931.5)MeV/u = 4.3 MeV. Most of this energy is in the form of kinetic energy of the recoiling particles.

23. βDECAY The general form for β decay is: β decay, like α decay, results in transmutation of the parent nucleus. In this case a neutron within the nucleus decays into a proton and a negative particle that is indistinguishable from an orbital electron. 14C becomes 14N due to this mechanism. β plus decay also occurs. In this case a proton decays into a neutron and positron, which has the same mass as an electron, but a positive charge.

24. 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

25. 31.4 Radioactivity Gamma knife – I hope I never see you here!

26. Stop to think The cobalt isotope 60Co (Z = 27) decays to the nickel isotope 60Ni (Z = 28). The decay process is: • Alpha decay. • Beta-plus decay. • Beta-minus decay. • Gamma decay.

27. Stop to think The cobalt isotope 60Co (Z = 27) decays to the nickel isotope 60Ni (Z = 28). The decay process is: • Alpha decay. • Beta-plus decay. • Beta-minus decay. • Gamma decay.

28. 31.5 The Neutrino During beta decay, energy is released. However, it is found that most beta particles do not have enough kinetic energy to account for all of the energy released. The additional energy is carried away by a neutrino.

29. 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 remaining vs. number of tosses would result in an exponential graph, like the one on the right. Tossing a single coin is a random process, but doing so repeatedly shows a definite pattern. Radioactive decay shows this same pattern.

30. How often does alpha, beta, or gamma decay occur within a sample of radioactive material? • We can defineλ as the decay constant, the probability that a nucleus will decay in the next second. 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.

31. 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.

32. 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, N0 is the number of parent nuclei at time t = 0, and λ is the decay constant. Since A = λN we can also say:

33. It is often convenient to know how long it will take for one-half the sample to decay. Half- Life, T1/2 Taking the natural log of both sides: where T1/2 the half-life, is defined as the time in which ½ of the radioactive nuclei disintegrate.

34. 31.6 Radioactive Decay and Activity

35. A sample starts with 1000 radioactive atoms. How many half-lives have elapsed when 750 atoms have decayed? • 2.5 • 2.0 • 1.5 • 0.25

36. A sample starts with 1000 radioactive atoms. How many half-lives have elapsed when 750 atoms have decayed? • 2.5 • 2.0 • 1.5 • 0.25

37. Cesium activity • The isotope 137Cs is a standard source of gamma rays. The half-life is 30.0 years. • How many 137Cs atoms are in a source that has an activity of 1.85 x 105Bq? • What is the activity of the source 10 years later?

38. Cesium activity • The isotope 137Cs is a standard source of gamma rays. The half-life is 30.0 years. • How many 137Cs atoms are in a source that has an activity of 1.85 x 105Bq? A = λ N and we are looking for N. We can use the relationship between T1/2 and λ :

39. Cesium activity First, express T1/2 in terms of seconds: T1/2 = 30.0 y(3.15 x 107 s/y) = 9.45 x 108 s λ = (ln 2)/T1/2 λ = .693/9.45 x 108 s λ = 7.33 x 10-10 s-1 Thus the number of 137 Cs is: A/ λ = N 1.85 x 105 Bq/ 7.33 x 10-10 s-1 = 2.5 x 1014 atoms.

40. Cesium activity What is the activity of the source 10 years later? A= 1.47 x 105 Bq

41. 31.7 RadiocarbonDating – 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. • 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. • T1/2 = 5730 years for 14 C. It undergoes beta decay to 14 N. • Other isotopes with longer half-lives are used to date geological materials. • . In particular Uranium was used to obtain an date for the age of the Earth (5.4 billion years)

42. Carbon dating • Archeologists excavating a site have found a piece of charcoal from a fireplace. Lab measurements find the 14 C activity to be .07 Bq per gram. What is the radiocarbon age of the charcoal? Known Find A= .07Bq/g t A0 = .23 Bq/g T1/2 = 5730 years λ = (ln 2)/T1/2

43. Carbon dating • First, find λ, the decay constant Known Find A= .07Bq/g t A0 = .23 Bq/g T1/2 = 5730 years Tip: If we leave λ in terms of the T1/2 , we can find t in years : λ = (ln 2)/T1/2 = .693/5730 λ = 1.209 x 10-4 y-1

44. Carbon dating Known Find A= .07Bq/g t A0 = .23 Bq/g T1/2 = 5730 years = 1.807 x 1011 s λ = 1.209 x 10-4 y-1 Ln (A/A0) = - λt Ln (A/A0)/- λ = t T = 9839 or 9800 years

45. 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

46. 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

47. 31.8 Radioactive Decay Series The sequential decay of one nucleus after another is called a radioactive decay series.

48. 31.8 Radioactive Decay Series

49. 31.9 Radiation Detectors A Geiger counter

50. 31.9 Radiation Detectors A scintillation counter