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Solution Manual for Fundamentals of Nuclear Science and Engineering, 3rd Edition

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    PROBLEM SOLUTION MANUAL FOR Fundamentals of Nuclear Science and Engineering Third Edition by J. Kenneth Shultis and Richard E. Faw Dept. of Mechanical and Nuclear Engineering Kansas State University Manhattan, KS 66506

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    Chapter 1 Fundamental Concepts PROBLEMS 1. Both the hertz and the curie have dimensions of s − 1 . Explain the difference between these two units. Solution: The hertz is used for periodic phenomena and equals the number of “cycles per second.” The curie is used for the random or stochastic rate at which a radioactive source decays, specifically, 1 Ci = 3 . 7 × 10 10 decays/second. 2. Advantages of SI units are apparent when one is presented with units of barrels , ounces , tons , and many others. (a) Compare the British and U.S. units for the gallon and barrel (liquid and dry measure) in SI units of liters (L). (b) Compare the long ton , short ton , and metric ton in SI units of kg. Solution: Unit conversions are taken from the handbook Conversion Factors and Tables , 3d ed., by O.T. Zimmerman and I. Lavine, published by Industrial Research Service, Inc., 1961. (a) In both British and U.S. units, the gallon is equivalent to 4 quarts, eight pints, etc. However, the quart and pint units differ in the two systems. The U.S. gallon measures 3.7853 L, while the British measures 4.546 L. Note that the gallon is sometimes used for dry measure, 4.405 L U.S. measure. The barrel in British units is the same for liquid and dry measure, namely, 163.65 L. The U.S. barrel (dry) is exactly 7056 in 3 , 115.62 L. The U.S. barrel (liq) is 42 gallons (158.98 L) for petroleum measure, but otherwise (usually) is 31.5 gallons (119.24 L). (b) The common U.S. unit is the short ton of 2000 lb, 907.185 kg, 20 short hundredweight (cwt). The metric ton is exactly 1000 kg, and the long ton is 20 long cwt, 22.4 short cwt, 2240 lb, or 1016 kg. 1-1

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    1-2 Fundamental Concepts Chap. 1 3. Compare the U.S. and British units of ounce (fluid), (apoth), (troy), and (avdp). Solution: The U.S. and British fluid ounces are, respectively, 1/32 U.S. quarts (0.02957 L) and 1/40 British quarts (0.02841 L). The oz (avdp.) is exactly 1/16 lb (avdp), i.e., 0.02834 kg. Avdp., abbreviation for avoirdupois refers to a system of weights with 16 oz to the pound. The apoth. apothecary or troy ounce is exactly 480 grains, 0.03110 kg. 4. Explain the SI errors (if any) in and give the correct equivalent units for the following units: (a) mgrams/kiloL, (b) megaohms/nm, (c) N · m/s/s, (d) gram cm/(s − 1 /mL), and (e) Bq/milli-Curie. Solution: (a) Don’t mix unit abbreviations and names; SI prefixes only in numerator: correct form is μ g/L . (b) Don’t mix names and abbreviations and don’t use SI prefixes in denomi- nator: correct form nohm/m . (c) Don’t use hyphen and don’t use multiple solidi: correct form N m s − 2 . (d) Don’t mix names and abbreviations, don’t use multiple solidi, and don’t use parentheses: correct form g cm s mL or better 10 μ g m s L . (e) Don’t mix names with abbreviations, and SI prefix should be in numerator: correct form kBq/Ci . 5. Consider H 2 , D 2 , and H 2 O, treated as ideal gases at pressures of 1 atm and temperatures of 293.2 ◦ K . What are the molecular and mass densities of each. Solution: According to the ideal gas law, molar densities are identical for ideal gases under the same conditions, i.e., ρ m = p/RT . From Table 1.5, R = 8.314472 Pa m 3 /K. For p = 0.101325 MPa= 1 atm., and T = 293.2 ◦ K , ρ m = 41.56 mol/m 3 . Multiplication by molecular weights yield, respectively, 83.78 , 167.4, and 749.0 g/m 3 for the three gases. 6. In vacuum, how far does light move in 1 ns? Solution: ∆ x = c ∆ t = (3 × 10 8 m/s) × (10 − 9 s) = 3 × 10 − 4 m = 30 cm .

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    Fundamental Concepts Chap. 1 1-3 7. In a medical test for a certain molecule, the concentration in the blood is reported as 57 mcg/dL. What is the concentration in proper SI notation? Solution: 123 mcg/dL = 10 − 3 10 − 2 g/10 − 1 L = 1 . 23 × 10 − 4 g/L = 57 μ g/L . 8. How many neutrons and protons are there in each of the following nuclides: (a) 11 B, (b) 24 Na, (c) 60 Co, (d) 207 Pb, and (e) 238 U? Solution: Nuclide neutrons protons 11 B 6 5 24 Na 13 11 60 Co 33 27 207 Pb 125 82 238 U 146 92 9. Consider the nuclide 71 Ge. Use the Chart of the Nuclides to find a nuclide (a) that is in the same isobar, (b) that is in the same isotone, and (c) that is an isomer. Solution: (a) 71 As, (b) 59 Ga, and (c) 71 m Ge 10. Examine the Chart of the Nuclides to find any elements, with Z less that that of lead ( Z = 82), that have no stable nuclides. Such an element can have no standard relative atomic mass. Solution: Promethium ( Z = 61) and Technetium ( Z = 43) 11. What are the molecular weights of (a) H 2 gas, (b) H 2 O, and (c) HDO? Solution: From Table A.3, A (O) = 15 . 9994 g/mol; from Table B.1 A (H) = 1 . 007825 g/mol and A (D) = 2 . 014102 g/mol. (a) A (H 2 ) = 2 A (H) = 2 × 1 . 007825 = 2 . 01565 g/mol (b) A (H 2 O) = 2 A (H) + A (O) = 2 × 1 . 007825 + 15 . 9994 = 18 . 0151 g/mol (c) A (HDO) = A (H) + A (D) + A (O) = 1 . 007825 + 2 . 014102 + 15 . 9994 = 19 . 0213 g/mol

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    1-4 Fundamental Concepts Chap. 1 12. What is the mass in kg of a molecule of uranyl sulfate UO 2 SO 4 ? Solution: From Table A.3, A (U) = 238.0289 g/mol, A (O) = 15.9994 g/mol, and A (S) = 32.066 g/mol. The molecular weight of UO 2 SO 4 is thus A (UO 2 SO 4 ) = A (U) + 6 A (O) + A (S) = 238 . 0289 + 6(15 . 994) + 32 . 066 = 366 . 091 g/mol = 0 . 336091 kg/mol. Since one mol contains N a = 6 . 022 × 10 23 molecules, the mass of one molecule of UO 2 SO 4 = A (UO 2 SO 4 ) /N a = 0 . 366091 / 6 . 002 × 10 23 = 6 . 079 × 10 − 25 kg/molecule . 13. Show by argument that the reciprocal of Avogadro’s constant is the gram equivalent of 1 atomic mass unit. Solution: By definition one gram atomic weight of 12 C is 12 g/mol. Thus the mass of one atom of 12 C is M ( 12 6 C) = 12 g/mol N a atoms/mol = 12 N a g/atom . But by definition, one atom of 12 C has a mass of 12 u. Therefore, 1 u = 1 u 12 u/( 12 C atom) ( 12 N a g/( 12 C atom) ) = 1 N a g . 14. Prior to 1961 the physical standard for atomic masses was 1/16 the mass of the 16 8 O atom. The new standard is 1/12 the mass of the 12 6 C atom. The change led to advantages in mass spectrometry. Determine the conversion factor needed to convert from old to new atomic mass units. How did this change affect the value of the Avogadro constant? Solution From Table B.1, the 16 8 O atom has a mass of 15.9949146 amu. Thus, the pre- 1961 atomic mass unit was 15.9949146/16 post-1961 units, and the conversion factor is thus 1 amu ( 16 O) = 0.99968216 amu ( 12 C). The Avogadro constant is defined as the number of atoms in 12 g of unbound carbon-12 in its rest-energy electronic state, i.e., the number of atomic mass units per gram. Using data from Table 1.5, one finds that N a is given by the reciprocal of the atomic mass unit, namely, [1 . 6605387 × 10 − 24 ] − 1 = 6 . 0221420 × 10 23 mol − 1 . Pre-1961, the Avogadro constant was more loosely defined as the number of atoms per mol of any element, and had the best value 6 . 02486 × 10 23 .

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    Fundamental Concepts Chap. 1 1-5 15. How many atoms of 234 U are there in 1 kg of natural uranium? Solution: From Table A.3, the natural abundance of 234 U in uranium is found to be f ( 234 U) = 0 . 0055 atom-%. A mass m of uranium contains [ m/ A (U)] N a ura- nium atoms. Thus, the number of 234 U atoms in the mass m = 1000 g are N ( 234 U) = f ( 234 U) mN a A (U) = 0 . 000055 1000 × (6 . 022 × 10 23 ) 238 . 0289 = 1 . 392 × 10 20 atoms . 16. A bucket contains 1 L of water at 4 ◦ C where water has a denisty of 1 g cm 3 . (a) How many moles of H 2 O are there in the bucket? (b) How many atoms of 1 1 H and 2 1 D are there in the bucket? Solution: (a) The relative atomic weight of water A (H 2 O) = 2 A (H)+ A (O) = 2(1 . 00794)+ (15 . 9994) = 18 . 01528 . Then the number of water molecules mols of H 2 O = mass(H 2 O) A (H 2 O) = 1000 g 18 . 01258 g/mol = 55 . 5 mol . (b) Number of molecules of H 2 O = 55 . 5 mol × N a mol − 1 = 55 . 5 × 6 . 60221 × 10 23 = 3 . 343 × 10 25 molecules. Then the number of atoms of both 1 1 H and 2 1 D atoms = 2 × no. of H 2 O molecules = 6 . 6856 × 10 25 atoms. From Table A.4, the isoptopic abundances are found to be γ ( 1 1 H) = 0 . 999885 and γ ( 2 1 D) = 0 . 000115. Then N ( 1 1 H) = (0 . 999885)(6 . 6856 × 10 25 ) = 6 . 69 × 10 25 atoms and N ( 2 1 D) = (0 . 000115)(6 . 6856 × 10 25 ) = 7 . 69 × 10 21 atoms . 17. How many atoms of deuterium are there in 2 kg of water? Solution: Water is mostly H 2 O, and so we first calculate the number of atoms of hydrogen N (H) in a mass m = 2000 g of H 2 O is N (H) = 2 N (H 2 O) = 2 mN a A (H 2 O) ' 2 mN a A (H 2 O) = 2 2000 × (6 . 022 × 10 23 ) 18 = 1 . 34 × 10 26 atoms of H .

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    1-6 Fundamental Concepts Chap. 1 From Table A.4, the natural isotopic abundance of deuterium (D) is 0.015 atom-% in elemental hydrogen. Thus, the number of deuterium atoms in 2 kg of water is N (D) = 0 . 00015 × N (H) = 2 . 01 × 10 22 atoms . 18. Estimate the number of atoms in a 3000 pound automobile. State any assump- tions you make. Solution: The car mass m = 3000 / 2 . 2 = 1365 kg. Assume most the this mass is iron. If the atoms in non-iron materials (e.g., glass, plastic, rubber, etc.) were con- verted to iron, the car mass would increase to about m equiv = 1500 kg. Thus the number of atoms in the car is N = m equiv N a A (Fe) = (1 . 5 × 10 6 )(6 . 022 × 1023) 56 = 1 . 6 × 10 28 atoms . 19. Calculate the relative atomic weight of oxygen. Solution From Table A.4, oxygen has three stable isotopes: 16 O, 17 O, and 18 O with percent abundances of 99.757, 0.038, and 0.205, respectively. Their atomic masses, in u, are found from Table B.1 and equal their relative atomic weights. Then from Eq. (1.2) A (O) = γ ( 16 O) 100 A ( 16 O) + γ ( 17 O) 100 A ( 17 O) + γ ( 18 O) 100 A ( 18 O) = 99 . 757 100 15 . 994915 + 0 . 038 100 16 . 999132 + 0 . 205 100 17 . 999160 = 15 . 999405 . 20. Natural uranium contains the isotopes 234 U, 235 U and 238 U. Calculate the relative atomic weight of natural uranium. Solution From Table A.4, the three isotopes 234 U, 235 U, and 238 U have isotopic abun- dances of 0.0055%, 0.720%, and 99.2745%, respectively. Their atomic masses, in u, are found from Table B.1 and equal their relative atomic weights. Then from Eq. (1.2) A (O) = γ ( 234 U) 100 A ( 234 U) + γ ( 235 U) 100 A ( 235 U) + γ ( 238 U) 100 A ( 238 U) = 0 . 0055 100 234 . 040945 + 0 . 720 100 235 . 043923 + 99 . 2745 100 238 . 050783 = 238 . 02891 .

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    Fundamental Concepts Chap. 1 1-7 21. Does a sample of carbon extracted from coal have the same relative atomic weight as a sample of carbon extracted from a plant? Explain. Solution The carbon extracted from coal has only two isotopes, namely 12 C and 13 C with with abundances of 98.93% and 1.07%, respectively. The relative atomic weight is thus is slightly larger than 12 that would result if there were no 13 C, namely 12.0107. Carbon extracted from plant material, however, also contains the radioactive isotope 14 C produced in the atmosphere by cosmic rays. Thus, the relative atomic weight is conceptually greater than that of carbon from coal in which all the 14 C has radioactively decayed away. However, as discussed in Section 5.8.1, the amount of 14 C in plant material is extremely small (1 . 23 × 10 − 12 atoms per atom of stable carbon). Thus, 14 C would increase the atomic weight only in the 12th significant figure! 22. Dry air at normal temperature and pressure has a mass density of 0.0012 g/cm 3 with a mass fraction of oxygen of 0.23. What is the atom density (atom/cm 3 ) of 18 O? Solution: From Eq. (1.5), the atom density of oxygen is N (O) = w o ρN a A (O) = 0 . 23 × 0 . 0012 × (6 . 022 × 10 23 ) 15 . 9994 = 1 . 04 × 10 19 atoms/cm 3 . From Table A.4 isotopic abundance of 18 O in elemental oxygen is f 18 = 0 . 2 atom-% of all oxygen atoms. Thus, the atom density of 18 O is N ( 18 O) = f 18 N (O) = 0 . 002 × 1 . 04 × 10 19 = 2 . 08 × 10 16 atoms/cm 3 . 23. A reactor is fueled with 4 kg uranium enriched to 20 atom-percent in 235 U. The remainder of the fuel is 238 U. The fuel has a mass density of 19.2 g/cm 3 . (a) What is the mass of 235 U in the reactor? (b) What are the atom densities of 235 U and 238 U in the fuel? Solution: (a) Let m 5 and m 8 be the mass in kg of an atom of 235 U and 238 U, and let n 5 and n 8 be the total number of atoms of 235 U and 238 U in the uranium mass M U = 4 kg. For 20% enrichment, n 8 = 4 n 5 , so that M U = n 5 m 5 + n 8 m 8 = n 5 m 5 + 4 n 5 m 8 = n 5 m 5 ( 1 + 4 m 8 m 5 ) . Here n 5 m 5 = M 5 is the mass of 235 U in the uranium mass M U . From this result we obtain using m 5 /m 8 ' 235 / 238 M 5 = M U [ 1 + 4 m 8 m 5 ] − 1 = 4 kg [ 1 + 4 ( 238 235 )] − 1 = 0 . 7919 kg .

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    1-8 Fundamental Concepts Chap. 1 The mass of 238 U M 8 = M U − M 5 = 3 . 208 kg. (b) The volume V of the uranium is V = M U /ρ U = (4000 g) / (19 . 2 g/cm 3 ) = 208 . 3 cm 3 . Hence the atom densities are N 5 = M 5 N a A 5 V = (791 . 9 g)(6 . 022 × 10 23 atoms/mol) (235 g/mol)(208 . 3 cm 3 ) = 9 . 740 × 10 21 cm − 3 N 8 = M 8 N a A 8 V = (3208 g)(6 . 022 × 10 23 atoms/mol) (238 g/mol)(208 . 3 cm 3 ) = 3 . 896 × 10 22 cm − 3 24. A sample of uranium is enriched to 3.2 atom-percent in 235 U with the remainder being 238 U. What is the enrichment of 235 U in weight-percent? Solution: Let the subscripts 5, 8 and U refer to 235 U, 238 U, and uranium, respectively. For the given atom-% enrichment, The number of atoms in a sample of the uranium are N 5 = 0 . 0320 N U and N 8 = 0 . 9680 N U . The mass M 5 and M 8 of 235 U and 238 U in the sample is M 5 = 0 . 0320 N U m 5 and M 8 = 0 . 9680 N U m 8 , where m 5 and m 8 is the mass of an atom of 235 U and 238 U, respectively. The enrichment in weight-% is thus e (wt-%) = 100 × M 5 M 5 + M 8 = 100 × 0 . 0320 m 5 0 . 0320 m 5 + 0 . 9680 m 8 = 100 × 0 . 0320 0 . 0320 + 0 . 9680( m 8 /m 5 ) ' 100 × 0 . 0320 0 . 0320 + 0 . 9680(238 / 235) = 3 . 16 wt- % . 25. A crystal of NaCl has a density of 2.17 g/cm 3 . What is the atom density of sodium in the crystal? Solution: Atomic weights for Na and Cl are obtained from Table A.3, so that A (NaCl) = A (Na) + A (Cl) = 22 . 990 + 35 . 453 = 58 . 443 g/mol. Thus the atom density of Na is N (Na) = N (NaCl) = ρ NaCl N a A (NaCl) = 2 . 17 × 6 . 022 × 10 23 58 . 443 = 2 . 24 × 10 22 cm − 3 .

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    Fundamental Concepts Chap. 1 1-9 26. A concrete with a density of 2.35 g/cm 3 has a hydrogen content of 0.0085 weight fraction. What is the atom density of hydrogen in the concrete? Solution: From Eq. (1.5), the atom density of hydrogen is N (H) = w H ρM a A (H) = (0 . 0085)(2 . 35 g/cm 3 )(6 . 022 × 10 23 atoms/mol) 1 g/mol = 1 . 20 × 10 22 atoms/cm 3 . 27. How much larger in diameter is a uranium nucleus compared to an iron nucleus? Solution: From Eq. (1.7) the nuclear diameter is D = 2 R o A 1 / 3 so that D U D F e = ( A U A F e ) 1 / 3 ' ( 238 56 ) 1 / 3 = 1 . 62 . Thus, D U ' 1 . 62 D Fe . 28. By inspecting the chart of the nuclides, determine which element has the most stable isotopes? Solution: The element tin (Sn) has 10 stable isotopes. 29. Find an internet site where the isotopic abundances of mercury may be found. Solution: http://www.nndc.bnl.gov 30. The earth has a radius of about 6 . 35 × 10 6 m and a mass of 5 . 98 × 10 24 kg. What would be the radius if the earth had the same mass density as matter in a nucleus? Solution: From the text, the density of matter in a nucleus is ρ n ' 2 . 4 × 10 14 g/cm 3 . The mass of the earth M = ρ × V where the volume V = (4 / 3) πR 3 . Combining these results and solving for the radius gives R = ( 3 M 4 πρ ) 1 / 3 = ( 3(5 . 98 × 10 27 g) 4 π (2 . 4 × 10 14 g/cm 3 ) ) 1 / 3 = 1 . 81 × 10 4 cm = 181 m .

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    Chapter 2 Modern Physics Concepts PROBLEMS 1. An accelerator increases the kinetic energy of electrons uniformly to 10 GeV over a 3000 m path. That means that at 30 m, 300 m, and 3000 m, the kinetic energy is 10 8 , 10 9 , and 10 10 eV, respectively. At each of these distances, compute the velocity, relative to light (v/c), and the mass in atomic mass units. Solution: From Eq. (2.10) in the text T = mc 2 − m o c 2 we obtain m = T /c 2 + m o . (P2.1) From Eq. (2.5) in the text m = m o / √ 1 − v 2 /c 2 , which can be solved for v/c to give v c = √ 1 − m 2 o m 2 ' 1 − 1 2 m 2 o m 2 , if m o m << 1 . (P2.2) (a) For an electron ( m o = m e ) with T = 10 8 eV = 100 MeV, Eq. (P2.1) gives m = 100 MeV 931 . 5 MeV/u + m e = 0 . 1074 u + 0 . 0005486 u = 0 . 1079 u . Then m 2 e /m 2 = (0 . 0005486 / 0 . 1079) 2 = 2 . 59 × 10 − 5 . Finally, from Eq. (P2.2) above, we obtain v c ' 1 − 1 2 m 2 o m 2 = 1 − 1 . 29 × 10 − 5 = 0 . 999987 . (b) For an electron with T = 10 9 eV = 1000 MeV, we similarly obtain m = 1 . 0741 u and v/c = 0 . 99999987 . (c) For an electron with T = 10 10 eV = 10 4 MeV, we similarly obtain m = 10 . 736 u and v/c = 0 . 9999999987 . Alternative solution: Use Eq. (P2.4) developed in Problem 2-3, namely v c = { 1 − [ m e c 2 T + m e c 2 ] 2 } 1 / 2 . 2-1

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    2-2 Modern Physics Concepts Chap. 2 2. Consider a fast moving particle whose relativistic mass m is 100  percent greater than its rest mass m o , i.e., m = m o (1 +  ). (a) Show that the particle’s speed v , relative to that of light, is v c = √ 1 − 1 (1 +  ) 2 . (b) For v/c << 1, show that this exact result reduces to v/c ' √ 2  . Solution: (a) We are given m − m o m o = m o ((1 +  ) − 1) m o = . But we also have m − m o m o = 1 m o [ m o √ 1 − v 2 /c 2 − m o ] . Equating these two results yields  = 1 √ 1 − v 2 /c 2 − 1 . Solving this result for v/c gives v c = √ 1 − 1 (1 +  ) 2 . (P2.3) (b) For  << 1 we have (1 +  ) − 2 ' 1 − 2  + · · · . Substitution of the approxi- mation into Eq. (P2.3) above gives v c ' √ 1 − (1 − 2  ) = √ 2 . 3. In fission reactors one deals with neutrons having kinetic energies as high as 10 MeV. How much error is incurred in computing the speed of 10-MeV neu- trons by using the classical expression rather than the relativistic expression for kinetic energy? Solution: A neutron with rest mass m n = 1 . 6749288 × 10 − 27 kg has a kinetic energy T = (10 7 eV)(1 . 602177 × 10 − 19 J/eV) = 1 . 602177 × 10 − 12 J. For the neutron m n c 2 = 939 . 56536 MeV. Classically : v c = √ 2 T /m n = [ 2 × 1 . 602177 × 10 − 12 1 . 6749288 × 10 − 27 ] 1 / 2 = 4 . 373993 × 10 7 m/s .

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    Modern Physics Concepts Chap. 2 2-3 Relativistically : From the text we have T = mc 2 − m o c 2 = m o c 2 √ 1 − v 2 /c 2 − m o c 2 . Solving this equation for v yields the relativistic speed v r v r = c { 1 − [ m o c 2 T + m o c 2 ] 2 } 1 / 2 . (P2.4) Substitution then gives v r = c { 1 − [ 939 . 56536 10 + 939 . 56536 ] 2 } 1 / 2 = 0 . 1447459 c = 4 . 339373 × 10 7 m/s . Thus the percent error in the classical speed is = 100( v c − v r ) /v r = 0 . 798 % . 4. What speed (m s − 1 ) and kinetic energy (MeV) would a neutron have if its relativistic mass were 10% greater than its rest mass? Solution: We are given ( m − m o ) /m o ≡  = 0 . 1. From Problem 2-2 v c = √ 1 − 1 (1 +  ) 2 = √ 1 − 1 1 . 1 2 = 0 . 4167 . Thus the neutron’s speed is v = 0 . 4167 c = 1 . 25 × 10 8 m/s . The kinetic energy can be calculated from T = mc 2 − m o c 2 = m o c 2 [ 1 √ 1 − v 2 /c 2 − 1 ] . For m o c 2 = 939 . 6 MeV and v/c = 0 . 4167 we obtain T = 939 . 6 [ 1 √ 1 − 0 . 4167 2 − 1 ] = 94 . 0 MeV . 5. Show that for a relativistic particle the kinetic energy is given in terms of the particl’s momentum by T = √ p 2 c 2 + m 2 o c 4 − m c c 2 . Solution: Squaring Eq. (2.17) and rearranging the terms one obtains T 2 + 2 T m o c 2 − p 2 c 2 = 0

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    2-4 Modern Physics Concepts Chap. 2 The solution of this quadratic equation gives T = 1 2 { − 2 m o c 2 ± √ 4 m 2 o c 4 + 4 p 2 c 2 } Only the + sign gives a physically meaningful result. Rearrangement gives the desired realtion. 6. For a relativistic particle show that Eq. (2.21) is valid. Solution: From the definition of η one has η 2 +1 = P 2 ( m o c ) 2 +1 = p 2 c 2 ( m o c 2 ) 2 +1 = ( mc 2 ) 2 − ( m o c 2 ) 2 ( m o c 2 ) 2 +1 = ( W 2 − 1)+1 = W 2 . 7. Prove the relationships given in (a) Eq. (2.19), (b) Eq. (2.20), and (c) Eq. (2.21). Solution: (a) From the definition of η and W one immediately has β = v c = p mc = η W . (b) Because W 2 = 1 + η 2 , then β 2 = ( v c ) 2 = η 2 W 2 = η 2 1 + η 2 . (c) Because β = η/W and W 2 = 1 + η 2 , one has β 2 1 − β 2 = η 2 /W 2 1 − η 2 /W 2 = η 2 / (1 + η 2 ) 1 − η 2 / (1 + η 2 ) = η 2 (1 + η 2 ) − η 2 = η 2 . From this result we see β 2 1 − β 2 = p 2 m 2 o c 2 = c 2 p 2 ( m o c 2 ) 2 , but we know p 2 c 2 = T 2 + 2 T m o c 2 , so β 2 1 − β 2 = T 2 + 2 T m o c 2 ( m o c 2 ) 2 = ( T m o c 2 ) 2 + 2 T m o c 2 = ( T m o c 2 ) 2 ( 1 + 2 m o c 2 T ) .

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    Modern Physics Concepts Chap. 2 2-5 8. In the Relativistic Heavy Ion Collider, nuclei of gold are accelerated to speeds of 99.95% the speed of light. These nuclei are almost spherical when at rest; however, as they move past the experimenters they appear considerably flat- tened in the direction of motion because of relativistic effects. Calculate the apparent diameter of such a gold nucleus in its direction of motion relative to that perpendicular to the motion. Solution: The relativistically contracted diameter D to the uncontracted di- ameter D o when v/c = 0 . 9995 is D/D o = √ 1 − v 2 /c 2 = √ 1 − 0 . 9995 2 = √ 1 − (1 − 0 . 0005) 2 ' √ 1 − (1 − 2 × 0 . 0005) = √ 0 . 001 = 0 . 031 . Hence the gold nucleus appears to flatten to 3.1 % of its at-rest width. 9. Muons are subatomic particles that have the negative charge of an electron but are 206.77 times more massive. They are produced high in the atmosphere by cosmic rays colliding with nuclei of oxygen or nitrogen, and muons are the dominant cosmic-ray contribution to background radiation at the earth’s surface. A muon, however, rapidly decays into an energetic electron, existing, from its point of view, for only 2.20 μ s, on the average. Cosmic-ray generated muons typically have speeds of about 0 . 998 c and thus should travel only a few hundred meters in air before decaying. Yet muons travel through several kilometers of air to reach the earth’s surface. Using the results of special relativity explain how this is possible. HINT: consider the atmospheric travel distance as it appears to a muon, and the muon lifetime as it appears to an observer on the earth’s surface. Solution: Muon’s Point of View: A muon, with a lifetime t o = 2 . 20 × 10 − 6 s and traveling with a speed v = 0 . 998 c , travels on the average a distance d = vt o = 0 . 998(3 . 00 × 10 8 m/s)(2 . 29 × 10 − 6 s) = 660 m. If the muon is created at an altitude L o , from the muon’s point of view the distance to the surface (approaching with speed v = 0 . 998 c ) is relativistically narrowed or contracted to a distance L = L o √ 1 − v 2 /c 2 = L o √ 1 − 0 . 998 2 = 0 . 063 L o . For example, if L o = 10 km, L = 630 m, so that, on the average, almost half of the muons will reach the surface. Surface Observer’s Point of View: An observer on the earth’s surface observes the muon approaching at a speed v = 0 . 998 c and the muon’s lifetime appears to expand (the muon’s internal clock appears to slow) as t = t o √ 1 − v 2 /c 2 = t o √ 1 − 0 . 998 2 = 15 . 9 t o = 3 . 49 × 10 − 5 s . In such a lifetime, the muon can travel d = 0 . 998 c × t = 10 , 500 m so that it can reach the surface from an altitude of 10 km before decaying.

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