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

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PROBLEM SOLUTION MANUAL FORFundamentals of NuclearScience and EngineeringThird EditionbyJ. Kenneth ShultisandRichard E. FawDept. of Mechanical and Nuclear EngineeringKansas State UniversityManhattan, KS 66506

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Chapter 1Fundamental ConceptsPROBLEMS1.Both the hertz and the curie have dimensions of s−1.Explain the differencebetween these two units.Solution:Thehertzis used for periodic phenomena and equals the number of “cyclesper second.” Thecurieis used for the random or stochastic rate at which aradioactive source decays, specifically, 1 Ci = 3.7×1010decays/second.2.Advantages of SI units are apparent when one is presented with units ofbarrels,ounces,tons, and many others.(a) Compare the British and U.S. units for the gallon and barrel (liquid anddry measure) in SI units of liters (L).(b) Compare thelong ton,short ton, andmetric tonin SI units of kg.Solution:Unit conversions are taken from the handbookConversion Factors and Tables,3d ed., by O.T. Zimmerman and I. Lavine, published by Industrial ResearchService, Inc., 1961.(a) In both British and U.S. units, the gallon is equivalent to 4 quarts, eightpints, etc. However, the quart and pint units differ in the two systems. TheU.S. gallon measures 3.7853 L, while the British measures 4.546 L. Notethat 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 in3, 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 shorthundredweight (cwt). The metric ton is exactly 1000 kg, and the long tonis 20 long cwt, 22.4 short cwt, 2240 lb, or 1016 kg.1-1

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1-2Fundamental ConceptsChap. 13.Compare the U.S. and British units ofounce(fluid), (apoth), (troy), and(avdp).Solution:The U.S. and British fluid ounces are, respectively, 1/32 U.S. quarts (0.02957L) and 1/40 British quarts (0.02841 L). The oz (avdp.)is exactly 1/16 lb(avdp), i.e., 0.02834 kg. Avdp., abbreviation foravoirdupoisrefers to a systemof weights with 16 oz to the pound.The apoth.apothecaryor troy ounce isexactly 480 grains, 0.03110 kg.4.Explain the SI errors (if any) in and give the correct equivalent units for thefollowing units: (a) mgrams/kiloL, (b) megaohms/nm, (c) N·m/s/s, (d) gramcm/(s−1/mL), and (e) Bq/milli-Curie.Solution:(a) Don’t mix unit abbreviations and names; SI prefixes only in numerator:correct form isμg/L.(b) Don’t mix names and abbreviations and don’t use SI prefixes in denomi-nator: correct formnohm/m.(c) Don’t use hyphen and don’t use multiple solidi: correct formN m s−2.(d) Don’t mix names and abbreviations, don’t use multiple solidi, and don’tuse parentheses: correct form g cm s mL or better10μg m s L.(e) Don’t mix names with abbreviations, and SI prefix should be in numerator:correct formkBq/Ci.5.Consider H2, D2, and H2O, treated as ideal gases at pressures of 1 atm andtemperatures 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 gasesunder the same conditions, i.e.,ρm=p/RT.From Table 1.5,R= 8.314472Pa m3/K. Forp= 0.101325 MPa= 1 atm., andT= 293.2◦K ,ρm= 41.56mol/m3. Multiplication by molecular weights yield, respectively, 83.78 , 167.4,and 749.0 g/m3for the three gases.6.In vacuum, how far does light move in 1 ns?Solution:∆x=c∆t= (3×108m/s)×(10−9s) = 3×10−4m =30 cm.

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Fundamental ConceptsChap. 11-37.In a medical test for a certain molecule, the concentration in the blood isreported as 57 mcg/dL. What is the concentration in proper SI notation?Solution:123 mcg/dL = 10−310−2g/10−1L = 1.23×10−4g/L =57μg/L.8.How many neutrons and protons are there in each of the following nuclides:(a)11B, (b)24Na, (c)60Co, (d)207Pb, and (e)238U?Solution:Nuclideneutronsprotons11B6524Na131160Co3327207Pb12582238U146929.Consider the nuclide71Ge. 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 anisomer.Solution:(a)71As, (b)59Ga, and (c)71mGe10.Examine the Chart of the Nuclides to find any elements, withZless that thatof lead (Z= 82), that have no stable nuclides. Such an element can have nostandard relative atomic mass.Solution:Promethium (Z= 61) and Technetium (Z= 43)11.What are the molecular weights of (a) H2gas, (b) H2O, and (c) HDO?Solution:From Table A.3,A(O) = 15.9994 g/mol; from Table B.1A(H) = 1.007825g/mol andA(D) = 2.014102 g/mol.(a)A(H2) = 2A(H) = 2×1.007825 =2.01565 g/mol(b)A(H2O) = 2A(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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1-4Fundamental ConceptsChap. 112.What is the mass in kg of a molecule of uranyl sulfate UO2SO4?Solution:From Table A.3,A(U) = 238.0289 g/mol,A(O) = 15.9994 g/mol, andA(S) =32.066 g/mol.The molecular weight of UO2SO4is thusA(UO2SO4) =A(U) + 6A(O) +A(S) = 238.0289 + 6(15.994) + 32.066 = 366.091 g/mol = 0.336091 kg/mol.Since one mol containsNa= 6.022×1023molecules, the mass of one moleculeof UO2SO4=A(UO2SO4)/Na= 0.366091/6.002×1023=6.079×10−25kg/molecule.13.Show by argument that the reciprocal of Avogadro’s constant is the gramequivalent of 1 atomic mass unit.Solution:By definition one gram atomic weight of12C is 12 g/mol.Thus the mass ofone atom of12C isM(126C) =12 g/molNaatoms/mol = 12Nag/atom.But by definition, one atom of12C has a mass of 12 u. Therefore,1 u =1 u12 u/(12C atom)(12Nag/(12C atom))=1Nag.14.Prior to 1961 the physical standard for atomic masses was 1/16 the mass of the168O atom. The new standard is 1/12 the mass of the126C atom. The change ledto advantages in mass spectrometry. Determine the conversion factor neededto convert from old to new atomic mass units. How did this change affect thevalue of the Avogadro constant?SolutionFrom Table B.1, the168O atom has a mass of 15.9949146 amu. Thus, the pre-1961 atomic mass unit was 15.9949146/16 post-1961 units, and the conversionfactor is thus 1 amu (16O) = 0.99968216 amu (12C).The Avogadro constant is defined as the number of atoms in 12 g of unboundcarbon-12 in its rest-energy electronic state, i.e., the number of atomic massunits per gram. Using data from Table 1.5, one finds thatNais given by thereciprocal of the atomic mass unit, namely, [1.6605387×10−24]−1= 6.0221420×1023mol−1. Pre-1961, the Avogadro constant was more loosely defined as thenumber of atoms per mol of any element, and had the best value 6.02486×1023.

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Fundamental ConceptsChap. 11-515.How many atoms of234U are there in 1 kg of natural uranium?Solution:From Table A.3, the natural abundance of234U in uranium is found to bef(234U) = 0.0055 atom-%. A massmof uranium contains [m/A(U)]Naura-nium atoms. Thus, the number of234U atoms in the massm= 1000 g areN(234U) =f(234U)mNaA(U)= 0.000055 1000×(6.022×1023)238.0289=1.392×1020atoms.16.A bucket contains 1 L of water at 4◦C where water has a denisty of 1 g cm3.(a) How many moles of H2O are there in the bucket? (b) How many atoms of11H and21D are there in the bucket?Solution:(a) The relative atomic weight of waterA(H2O) = 2A(H)+A(O) = 2(1.00794)+(15.9994) = 18.01528.Then the number of water moleculesmols of H2O = mass(H2O)A(H2O)=1000 g18.01258 g/mol =55.5 mol.(b) Number of molecules of H2O = 55.5 mol×Namol−1= 55.5×6.60221×1023= 3.343×1025molecules.Then the number of atoms of both11Hand21D atoms = 2×no. of H2O molecules = 6.6856×1025atoms. FromTable A.4, the isoptopic abundances are found to beγ(11H) = 0.999885andγ(21D) = 0.000115. ThenN(11H) = (0.999885)(6.6856×1025) =6.69×1025atomsandN(21D) = (0.000115)(6.6856×1025) =7.69×1021atoms.17.How many atoms of deuterium are there in 2 kg of water?Solution:Water is mostly H2O, and so we first calculate the number of atoms of hydrogenN(H) in a massm= 2000 g of H2O isN(H) = 2N(H2O) = 2mNaA(H2O)'2mNaA(H2O)= 2 2000×(6.022×1023)18= 1.34×1026atoms of H.

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1-6Fundamental ConceptsChap. 1From Table A.4, the natural isotopic abundance of deuterium (D) is 0.015atom-% in elemental hydrogen. Thus, the number of deuterium atoms in 2 kgof water isN(D) = 0.00015×N(H) =2.01×1022atoms.18.Estimate the number of atoms in a 3000 pound automobile. State any assump-tions you make.Solution:The car massm= 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 aboutmequiv= 1500 kg. Thusthe number of atoms in the car isN=mequivNaA(Fe)= (1.5×106)(6.022×1023)56=1.6×1028atoms.19.Calculate the relative atomic weight of oxygen.SolutionFrom Table A.4, oxygen has three stable isotopes:16O,17O, and18O withpercent abundances of 99.757, 0.038, and 0.205, respectively.Their atomicmasses, in u, are found from Table B.1 and equal their relative atomic weights.Then from Eq. (1.2)A(O) =γ(16O)100A(16O) +γ(17O)100A(17O) +γ(18O)100A(18O)= 99.75710015.994915 + 0.038100 16.999132 + 0.205100 17.999160 =15.999405.20.Natural uranium contains the isotopes234U,235U and238U. Calculate therelative atomic weight of natural uranium.SolutionFrom Table A.4, the three isotopes234U,235U, and238U 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. Thenfrom Eq. (1.2)A(O) =γ(234U)100A(234U) +γ(235U)100A(235U) +γ(238U)100A(238U)= 0.0055100234.040945 + 0.720100 235.043923 + 99.2745100238.050783=238.02891.

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Fundamental ConceptsChap. 11-721.Does a sample of carbon extracted from coal have the same relative atomicweight as a sample of carbon extracted from a plant? Explain.SolutionThe carbon extracted from coal has only two isotopes, namely12C and13Cwith with abundances of 98.93% and 1.07%, respectively. The relative atomicweight is thus is slightly larger than 12 that would result if there were no13C,namely 12.0107. Carbon extracted from plant material, however, also containsthe radioactive isotope14C produced in the atmosphere by cosmic rays. Thus,the relative atomic weight isconceptuallygreater than that of carbon from coalin which all the14C has radioactively decayed away.However, as discussed in Section 5.8.1, the amount of14C in plant material isextremely small (1.23×10−12atoms per atom of stable carbon).Thus,14Cwould 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/cm3with a mass fraction of oxygen of 0.23. What is the atom density (atom/cm3)of18O?Solution:From Eq. (1.5), the atom density of oxygen isN(O) =woρNaA(O)= 0.23×0.0012×(6.022×1023)15.9994= 1.04×1019atoms/cm3.From Table A.4 isotopic abundance of18O in elemental oxygen isf18= 0.2atom-% of all oxygen atoms. Thus, the atom density of18O isN(18O) =f18N(O) = 0.002×1.04×1019=2.08×1016atoms/cm3.23.A reactor is fueled with 4 kg uranium enriched to 20 atom-percent in235U.The remainder of the fuel is238U. The fuel has a mass density of 19.2 g/cm3.(a) What is the mass of235U in the reactor? (b) What are the atom densitiesof235U and238U in the fuel?Solution:(a) Letm5andm8be the mass in kg of an atom of235U and238U, and letn5andn8be the total number of atoms of235U and238U in the uraniummassMU= 4 kg. For 20% enrichment,n8= 4n5, so thatMU=n5m5+n8m8=n5m5+ 4n5m8=n5m5(1 + 4m8m5).Heren5m5=M5is the mass of235U in the uranium massMU. From thisresult we obtain usingm5/m8'235/238M5=MU[1 + 4m8m5]−1= 4 kg[1 + 4(238235)]−1=0.7919 kg.

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1-8Fundamental ConceptsChap. 1The mass of238UM8=MU−M5= 3.208 kg.(b) The volumeVof the uranium isV=MU/ρU= (4000 g)/(19.2 g/cm3) =208.3 cm3. Hence the atom densities areN5=M5NaA5V= (791.9 g)(6.022×1023atoms/mol)(235 g/mol)(208.3 cm3)=9.740×1021cm−3N8=M8NaA8V= (3208 g)(6.022×1023atoms/mol)(238 g/mol)(208.3 cm3)=3.896×1022cm−324.A sample of uranium is enriched to 3.2 atom-percent in235U with the remainderbeing238U. What is the enrichment of235U in weight-percent?Solution:Let the subscripts 5, 8 and U refer to235U,238U, and uranium, respectively.For the given atom-% enrichment, The number of atoms in a sample of theuranium areN5= 0.0320NUandN8= 0.9680NU.The massM5andM8of235U and238U in the sample isM5= 0.0320NUm5andM8= 0.9680NUm8,wherem5andm8is the mass of an atom of235U and238U, respectively.The enrichment in weight-% is thuse(wt-%) = 100×M5M5+M8= 100×0.0320m50.0320m5+ 0.9680m8=100×0.03200.0320 + 0.9680(m8/m5)'100×0.03200.0320 + 0.9680(238/235)=3.16 wt-%.25.A crystal of NaCl has a density of 2.17 g/cm3.What is the atom density ofsodium in the crystal?Solution:Atomic weights for Na and Cl are obtained from Table A.3, so thatA(NaCl)=A(Na) +A(Cl) = 22.990 + 35.453 = 58.443 g/mol. Thus the atom densityof Na isN(Na) =N(NaCl) =ρNaClNaA(NaCl) = 2.17×6.022×102358.443=2.24×1022cm−3.

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Fundamental ConceptsChap. 11-926.A concrete with a density of 2.35 g/cm3has a hydrogen content of 0.0085weight fraction. What is the atom density of hydrogen in the concrete?Solution:From Eq. (1.5), the atom density of hydrogen isN(H) =wHρMaA(H)= (0.0085)(2.35 g/cm3)(6.022×1023atoms/mol)1 g/mol=1.20×1022atoms/cm3.27.How much larger in diameter is a uranium nucleus compared to an iron nucleus?Solution:From Eq. (1.7) the nuclear diameter isD= 2RoA1/3so thatDUDF e=(AUAF e)1/3'(23856)1/3= 1.62.Thus,DU'1.62 DFe.28.By inspecting the chart of the nuclides, determine which element has the moststable isotopes?Solution:The elementtin (Sn)has 10 stable isotopes.29.Find an internet site where the isotopic abundances of mercury may be found.Solution:http://www.nndc.bnl.gov30.The earth has a radius of about 6.35×106m and a mass of 5.98×1024kg.What would be the radius if the earth had the same mass density as matter ina nucleus?Solution:From the text, the density of matter in a nucleus isρn'2.4×1014g/cm3. Themass of the earthM=ρ×Vwhere the volumeV= (4/3)πR3.Combiningthese results and solving for the radius givesR=(3M4πρ)1/3=(3(5.98×1027g)4π(2.4×1014g/cm3))1/3= 1.81×104cm =181 m.

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Chapter 2Modern Physics ConceptsPROBLEMS1.An accelerator increases the kinetic energy of electrons uniformly to 10 GeVover a 3000 m path.That means that at 30 m, 300 m, and 3000 m, thekinetic energy is 108, 109, and 1010eV, 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 textT=mc2−moc2we obtainm=T /c2+mo.(P2.1)From Eq. (2.5) in the textm=mo/√1−v2/c2, which can be solved forv/cto givevc=√1−m2om2'1−12m2om2,ifmom<<1.(P2.2)(a) For an electron (mo=me) withT= 108eV = 100 MeV, Eq. (P2.1) givesm=100 MeV931.5 MeV/u +me= 0.1074 u + 0.0005486 u =0.1079 u.Thenm2e/m2= (0.0005486/0.1079)2= 2.59×10−5. Finally, from Eq. (P2.2)above, we obtainvc'1−12m2om2= 1−1.29×10−5=0.999987.(b) For an electron withT= 109eV = 1000 MeV, we similarly obtainm=1.0741u andv/c=0.99999987.(c) For an electron withT= 1010eV = 104MeV, we similarly obtainm=10.736u andv/c=0.9999999987.Alternative solution:Use Eq. (P2.4) developed in Problem 2-3, namelyvc={1−[mec2T+mec2]2}1/2.2-1

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2-2Modern Physics ConceptsChap. 22.Consider a fast moving particle whose relativistic massmis 100percent greaterthan its rest massmo, i.e.,m=mo(1 +). (a) Show that the particle’s speedv, relative to that of light, isvc=√1−1(1 +)2.(b) Forv/c <<1, show that this exact result reduces tov/c' √2.Solution:(a) We are givenm−momo=mo((1 +)−1)mo=.But we also havem−momo=1mo[mo√1−v2/c2−mo].Equating these two results yields=1√1−v2/c2−1.Solving this result forv/cgivesvc=√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 givesvc'√1−(1−2) =√2.3.In fission reactors one deals with neutrons having kinetic energies as high as10 MeV. How much error is incurred in computing the speed of 10-MeV neu-trons by using the classical expression rather than the relativistic expressionfor kinetic energy?Solution:A neutron with rest massmn= 1.6749288×10−27kg has a kinetic energyT= (107eV)(1.602177×10−19J/eV) = 1.602177×10−12J. For the neutronmnc2= 939.56536 MeV.Classically:vc=√2T /mn=[2×1.602177×10−121.6749288×10−27]1/2= 4.373993×107m/s.

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Modern Physics ConceptsChap. 22-3Relativistically: From the text we haveT=mc2−moc2=moc2√1−v2/c2−moc2.Solving this equation forvyields the relativistic speedvrvr=c{1−[moc2T+moc2]2}1/2.(P2.4)Substitution then givesvr=c{1−[939.5653610 + 939.56536]2}1/2= 0.1447459c= 4.339373×107m/s.Thus the percent error in the classical speed is = 100(vc−vr)/vr=0.798%.4.What speed (m s−1) and kinetic energy (MeV) would a neutron have if itsrelativistic mass were 10% greater than its rest mass?Solution:We are given (m−mo)/mo≡= 0.1. From Problem 2-2vc=√1−1(1 +)2=√1−11.12= 0.4167.Thus the neutron’s speed isv= 0.4167c=1.25×108m/s.The kinetic energy can be calculated fromT=mc2−moc2=moc2[1√1−v2/c2−1].Formoc2= 939.6 MeV andv/c= 0.4167 we obtainT= 939.6[1√1−0.41672−1]=94.0 MeV.5.Show that for a relativistic particle the kinetic energy is given in terms of theparticl’s momentum byT=√p2c2+m2oc4−mcc2.Solution:Squaring Eq. (2.17) and rearranging the terms one obtainsT2+ 2T moc2−p2c2= 0

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2-4Modern Physics ConceptsChap. 2The solution of this quadratic equation givesT= 12{−2moc2±√4m2oc4+ 4p2c2}Only the + sign gives a physically meaningful result. Rearrangement gives thedesired realtion.6.For a relativistic particle show that Eq. (2.21) is valid.Solution:From the definition ofηone hasη2+1 =P2(moc)2+1 =p2c2(moc2)2+1 = (mc2)2−(moc2)2(moc2)2+1 = (W2−1)+1 =W2.7.Prove the relationships given in (a) Eq. (2.19), (b) Eq. (2.20), and (c) Eq. (2.21).Solution:(a) From the definition ofηandWone immediately hasβ=vc=pmc=ηW .(b) BecauseW2= 1 +η2, thenβ2=(vc)2=η2W2=η21 +η2.(c) Becauseβ=η/WandW2= 1 +η2, one hasβ21−β2=η2/W21−η2/W2=η2/(1 +η2)1−η2/(1 +η2) =η2(1 +η2)−η2=η2.From this result we seeβ21−β2=p2m2oc2=c2p2(moc2)2,but we knowp2c2=T2+ 2T moc2, soβ21−β2=T2+ 2T moc2(moc2)2=(Tmoc2)2+2Tmoc2=(Tmoc2)2(1 + 2moc2T).

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Modern Physics ConceptsChap. 22-58.In the Relativistic Heavy Ion Collider, nuclei of gold are accelerated to speedsof 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 theapparent diameter of such a gold nucleus in its direction of motion relative tothat perpendicular to the motion.Solution:The relativistically contracted diameterDto the uncontracted di-ameterDowhenv/c= 0.9995 isD/Do=√1−v2/c2=√1−0.99952=√1−(1−0.0005)2'√1−(1−2×0.0005) =√0.001 = 0.031.Hence the gold nucleus appears to flatten to3.1%of its at-rest width.9.Muons are subatomic particles that have the negative charge of an electronbut are 206.77 times more massive. They are produced high in the atmosphereby cosmic rays colliding with nuclei of oxygen or nitrogen, and muons arethe dominant cosmic-ray contribution to background radiation at the earth’ssurface. A muon, however, rapidly decays into an energetic electron, existing,from its point of view, for only 2.20μs, on the average. Cosmic-ray generatedmuons typically have speeds of about 0.998cand thus should travel only afew hundred meters in air before decaying. Yet muons travel through severalkilometers of air to reach the earth’s surface.Using the results of specialrelativity explain how this is possible. HINT: consider the atmospheric traveldistance as it appears to a muon, and the muon lifetime as it appears to anobserver on the earth’s surface.Solution:Muon’s Point of View:A muon, with a lifetimeto= 2.20×10−6s andtraveling with a speedv= 0.998c, travels on the average a distanced=vto=0.998(3.00×108m/s)(2.29×10−6s) = 660 m.If the muon is created at an altitudeLo, from the muon’s point of view thedistance to the surface (approaching with speedv= 0.998c) is relativisticallynarrowed or contracted to a distanceL=Lo√1−v2/c2=Lo√1−0.9982= 0.063Lo.For example, ifLo= 10 km,L= 630 m, so that, on the average, almost halfof the muons will reach the surface.Surface Observer’s Point of View:An observer on the earth’s surfaceobserves the muon approaching at a speedv= 0.998cand the muon’s lifetimeappears to expand (the muon’s internal clock appears to slow) ast=to√1−v2/c2=to√1−0.9982= 15.9to= 3.49×10−5s.In such a lifetime, the muon can traveld= 0.998c×t= 10,500 m so that itcan reach the surface from an altitude of 10 km before decaying.

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