Chapter 31 Nuclear Physics GOALS

When you have mastered the contents of this chapter, you should be able to achieve the following goals: Definitions Define each of the following terms, and use each term in an operational definition: atomic mass number alpha, beta, and gamma radiation isotope radioactivity isobar half-life isotone nuclear fission nuclear binding energy nuclear fusion Carbon-14 Dating Describe and explain the physical basis of carbon-14 dating. Radioactivity Problems Solve problems involving radioactivity. Binding-Energy Problems Solve nuclear binding-energy problems.

PREREQUISITES

Before beginning this chapter you should have achieved the goals of Chapter 5, Energy, and Chapter 28, Atomic Physics.

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Chapter 31 Nuclear Physics OVERVIEW

Almost all of the mass of the atom is located in the nucleus. Investigations into the nature of nuclear matter has led to several models which help us understand many observable nuclear properties. These observations include nuclear binding energies, natural radioactivity, fission, and fusion. In this chapter you will be introduced to the basis of nuclear physics and several phenomenalogical applications which result.

SUGGESTED STUDY PROCEDURE When you begin your study of this chapter, you should be familiar with the following Chapter Goals: Definitions, Carbon-14 Dating, Radioactivity Problems, and Binding Energy Problems. A further discussion of each term listed under the Goal of Definitions section of this Study Guide chapter. Next, read text sections 31.1-31.6 and 31.9-31.11. As you read, take special note of Table 31.1 which outlines the relative strength of the four interactional forces. Be sure to look carefully at the Example on page 692 and 693 and on page 698. Answers to questions you encounter during your reading will be discussed in the second section of this Study Guide chapter. Now turn to the end of the chapter and read the Chapter Summary and do Summary Exercises 1-6. Then do Algorithmic Problems 1-4 and Exercises and Problems 1-3, 13, 20, 22, and 23. After you complete each of these exercises, check your answer against those that are given. For more work with the concepts introduced in this chapter, turn to the Examples section of this Study Guide chapter and work through the problems discussed. Now you should be prepared to attempt the Practice Test given at the end of this Study Guide chapter. Work completely through the test before checking your answers. For assistance with any part, see specific text sections or refer again to this Study Guide chapter. This study procedure is outlined below. --------------------------------------------------------------------------------------------------------------------Chapter Goals Suggested Summary Algorithmic Exercises Text Readings Exercises Problems & Problems --------------------------------------------------------------------------------------------------------------------Definitions 31.1,31.2,31.3, 1 1,2 1 31.5,31.9-31.10 Carbon-14 31.6 2 23 Dating Radioactivity 31.6 3,4 3,4 20,22 Problems Binding-Energy 31.4 5,6 2,3,13 Problems

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DEFINITIONS ATOMIC MASS NUMBER (A) - The number of nucleons (protons and neutrons) in an atomic nucleus. The atomic mass number of elements ranges from 1 for hydrogen to 260 for hahnium. ISOTOPE - Refers to atoms that have the same atomic number but different numbers of neutrons. Since the chemical properties of materials are determined by their electronic structure, hence their atomic number, isotopes have the same chemical properties. ISOBAR - Nuclei that have the same atomic mass number, A, are called isobars. Since the atomic number, rather than the atomic mass number, determines the chemical properties of a material isobars will have quite different chemical properties, e.g. neon, sodium, and magnesium. ISOTONE - Nuclei with equal numbers of neutrons are called isotones. The chemical properties of isotones is quite different, consider boron, carbon and nitrogen. NUCLEAR BINDING ENERGY - The energy that binds neutrons and protons in a nucleus is equal to the mass-energy difference between separate nucleons and the particular nucleus. The Einstein equation that relates mass to energy is used to relate the mass difference between the separate masses to the combined mass to the binding energy, see Eq. 31.4. RADIOACTIVITY - The name given to the process occurring when unstable decay with characteristic radiation. This was discovered in 1896, barely one generation ago. Modern human life has probably changed more since the discovery of radioactivity than all previous changes combined. Has radioactivity been the cause of this tremendous change? ALPHA RADIATION - The radiation that consists of helium nuclei emitted by heavy radioactive nuclei. The alpha particle was used in the scattering experiments conducted by Rutherford, et. al. which lead to the nuclear model for matter. BETA RADIATION - The radiation that results in electron or positron emission from a radioactive nucleus. The beta emission process which did not seem to conserve energy lead Fermi to propose a new particle, not experimentally verified for nearly 20 years! Such confidence in conservation laws! GAMMA RADIATION - Radiation of high energy electromagnetic waves. Gamma radiation has a high penetration ability. It is difficult to shield persons from gamma emission processes.

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RADIOACTIVE HALF-LIFE - The time required for one-half the radioactive nuclei in a sample to decay. The half- lives of materials vary widely from a small fraction of a second for very unstable nuclei to millions of years for the more stable radioactive nuclei. NUCLEAR FISSION - The process occurring when a heavy unstable nucleus splits to form two lighter nuclei with the release of considerable energy and a few neutrons. The process is exploited in nuclear reactors. This was the first man made nuclear process. It seems to not occur anywhere without human intention. NUCLEAR FUSION - The process that occurs when two light nuclei get close enough together that the nuclear force causes them to fuse to form a heavier nucleus with a release of energy. The basic source of energy in stars. This is the basic natural nuclear process and has so far not been duplicated in a scientific laboratory for any extended period of time.

ANSWERS TO QUESTIONS FOUND IN THE TEXT SECTION 31.1 Introduction The basic physics of this chapter is contained in the explanation of two facts. If a heavy nucleus is divided into two smaller nuclei by a fission process, the final mass of the nuclei is less than the initial mass of the heavy nucleus. If two light nuclei are fused together to form a larger nucleus the mass of the final large nucleus is less than the total mass of the two initial light nuclei. That is, in both processes, splitting heavy nuclei and joining light nuclei, mass is lost. That mass loss appears as its equivalence in energy according to Einstein's E=mc2 equation. SECTION 31.2 The Nucleus The six carbon isotopes contain six protons and four, five, six, seven, eight, and nine neutrons respectively. SECTION 31.3 Properties of the Nucleus The great density of nuclear matter seems much greater than we could predict from the forces of interaction. The forces of interaction for atoms and bulk materials are primarily electromagnetic. The nuclear force of interaction is about 100 times the electromagnetic force yet the nuclear matter is a thousand billion times as dense as an atom. SECTION 31.4 Nuclear Binding Energy The binding energy per nucleon is seven times greater for helium than it is for deuterium. It seems safe to predict that helium is a much more stable nucleus than deuterium. SECTION 31.6 Stable and Unstable Nuclei The neutrino prediction of Fermi ranks as a great triumph for the conservation of energy principle. Today in studying the increasing maze of "elementary" particles basic conservation principles are used to guide experimental studies.

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SECTION 31.10 Fusion The binding energy per nucleon graph (Figure 31.1) gives us a way of predicting which nucleons would be appropriate for fission or for fusion. The most stable nuclei are near an atomic mass number of 55. Clearly only nuclei with mass numbers much larger than 55 are suitable candidates for fission. In a similar way only nuclei with atomic mass numbers much less than 55 are suitable candidates for fusion. Look at the location of 36Li. What do you predict about that nucleus?

EXAMPLES NUCLEAR REACTION PROBLEMS 1. The energy released in the radioactive decay 92

238

U → 24He + 90234Th

is 4.27MeV. Calculate the mass of 234Th. What data are given? 238 4

U mass = 238.0498 amu

He mass = 4.00260 amu

Energy Released = 4.27MeV What data are implied? This is an isolated system with no external energy input. What physics principles are involved? Conservation of energy using the Einstein relation. What equation is to be used? E = mc2 Solution

(31.4)

Initial Energy = 238.0498 amu Final Energy = 4.0026 amu + 4.27 MeV + mass of Th Since 1 amu = 931 MeV; 4.27 MeV = .0046 amu. Final Energy = 4.0072 amu + mass of the Th mass of Th = 238.0498 - 4.0072 234

Th = 234.0428 amu.

2. In the slow neutron reaction 0

1

n + 510B →

3

7

Li + ? + Q

(a) Determine the unknown component of the reaction and its energy released. What data are given? The basic nuclear equation is given. What data are implied? There is one component missing from the equation.

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What physics principles are involved? Conservation of atomic number and atomic mass number can be used to determine the unknown component. Then conservation of energy using the Einstein equation can be used to find the energy released. What equations are to be used? E = mc2 (31.4) Solution There are 11 amu's shown on the left hand side of the equation and 5 units of atomic number. There are 7 and 3 respectively on the right hand side of the equation. That leaves 4 amu and 2 units of atomic number not counted. Thus the particle must be 24X, where X must be helium. So the missing component is an alpha particle 24He. mass of 01n = 1.00867 amu mass of 510B = 10.01295 --------------total = 11.02162 amu mass of 37Li = -7.01601 -------------4.00561 mass of 24He = -4.00260 amu Q = .00301 amu = 2.80 MeV RADIOACTIVITY PROBLEMS 3. The half life of Rn222 is 3.82 days. (a) What is the decay constant of this isotope in sec-1? (b) If a sample of Rn222 contains 1016 atoms at time t=0, how many Rn222 atoms will be left after 25 days? What data are given? t1/2 = 3.82 days = 3.30 x 105 seconds No = 1016 atoms What physics principles are involved? The exponential decay of radioactive nuclear is basic to this problem. What equations are to be used? N = No exp(-λt) (31.12) t1/2 = 0.693/λ (31.13) Solutions (a) The decay constant can be found from the half-life using Equation 31.13. λ = 0.693/t1/2 = 0.693/(3.30 x 105 sec) λRn222 = 2.10 x 10-6 sec-1 (b) The number of atoms at time t is given by Equation 31.12 N = (1016)exp - (25 days = 24 hr/d x 60 min/hr = 60 sec/min)λ N = 1016 exp - (2.16 x 106)(2.10 x 10-6) n = 1016 exp(-4.54) = 1.07 x1014 atoms

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BINDING-ENERGY PROBLEMS 4. What is the binding energy of 37Li whose atomic mass is 7.01822 amu ? What data are given? Atomic number = 3 (number of protons) Atomic mass number = 7 (number of nucleons) Atomic mass = 7.01822 amu (includes 3 electrons) What data are implied? The 37Li nucleus consists of 3 protons and 4 neutrons. What physics principles are involved? The Einstein equation is combined with conservation of mass equations to compute nuclear binding energies. What equations are to be used? BE = (Zmp + Nmn)c2 - Mc2 (31.7) Solution The 37Li atom is equivalent to 3 11H atoms and 4 neutrons. Their separate masses are 3 x 11H = 3(1.007825) = 3.023475 amu 4 x n = 4(1.008665) = 4.034660 amu -----------------7.058135 amu 7 Li -7.01822 3 ----------------------------BE = .03992 amu BE = 37.19 MeV There are 7 nucleons in 37Li so the binding energy per nucleon is 5.312 MeV/nucleon.

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PRACTICE TEST 1. a) The half-life of C-14 is 5730 years. The activity of C-14 from a living sample is 16.0 counts per minute. If the activity of carbon from an ancient wood sample is found to be 1.88 counts/min, what is the age of the sample? b) Comment on the "accuracy" of this dating method.

2. Given the following nuclear reaction: H + 37Li → 2He + ? a) Complete the reaction by filling in the blanks with proper numbers and/or symbols. b) Calculate the amount of energy released in this reaction. 1

1

1

1

H = 1.00813 amu,

3

7

Li = 7.01613 amu,

ANSWERS:

2

4

He = 4.00386 amu, and 1 amu = 931 MeV.

1. 17,800 yr. In actual practice, the 1.88 counts/min may be less than the normal "background count" due to natural occurring disintegration. Since normal variations in background occur, it becomes difficult to distinguish between background and radiation attributed to the sample itself. Thus the possibility of error is great. 2. 2 2He4 + Q,

Q = 15.4 MeV

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