Solution Manual for Measurement and Detection of Radiation, 3rd Edition

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SOLUTIONS MANUAL FORbyMeasurement andDetection of Radiation,Third EditionNicholas Tsoulfanidis

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PROBLEMS2.1What is the probability when throwing a die three times of getting a four in any of the throws?Answer:The probability of getting a four in one throw is:. Using the binomial distribution, theprobability to have one 4 in 3 throws is:P(only one 4) =34720216756561!!13!321.The same result can be obtained by enumerating all possible events.Throw123Outcome4No–4No–4No–44No–4No–4No–44To throw only one 4:()Note: The probability to have at least one four is largerP(at least one 4) = P(one 4) + P(two 4’s) + P(three 4’s) =()( )( )()( )( )Using possible outcomes, one should extend the table given above, as follows:44No–44No–44No–444444Total Probability =21691216121653216752.2What is the probability when drawing one card from each of three decks of cards that all three cards will bediamonds?Answer:The probability of drawing a diamond from one deck is.The probability thatthis eventhappens three timesis(usingEquation 2.15):()()f(4)=161352

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2.3A box contains 2000 computer cards.If five faulty cards are expected to be foundin the box, what is theprobability of finding two faulty cards in a sample of 250?Answer:The probability of finding a faulty card is P(faulty card) =. The probability offinding two faulty cards in a sample of 250 is (using Binomial distribution, Equation 2.45):()()()()()()2.4Calculate the average and the standard deviation ofthe probability density function ab/xf1when:a ≤ × ≤ b.(This probability distribution function is used for the calculation to round off errors.)Answer:Use Equation 2.27, or in this case Equation 2.26, for. Use Equation 2.33 for r2:2.5The energy distribution of thermal (slow) neutrons in a light-wave reactor follows very closely theMaxwell-Boltzmann distribution:dEeEAdEENkT/EWhere:N(E) dE=number of neutrons with kinetic energy betweenEandE+dEk=Boltzmann constant = 1.380662 × 10−23J/° KT=temperature,KA= constantShow that:(a)The mode of thisdistribution is:kTE21(b)The mean is:kTE23Answer:(a)Use Equation 2.24kTEEAeekTAEeEAdEEdNkTEkTEkTE2212111210Or:kTEe21mod(b) Use Equation 2.27kTdEEeAdEEeEAEkTEkTE230/0/2.6If theaverage for a large number of counting measurements is 15, what is the probability that a singlemeasurement will produce the result 20?Answer:Use Poissonstatistics, Equation 2.50:4.18%or1018.410433.210159.310325.3!2015202187231520eP2.7For the binomial distribution, prove:52000=0.002512,121222ababdxabxxxba21abdxabxxba

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)p(m(c)pNn)b(p)a(2NnnN110Answer:(a)NNnNnNnnN)p(p!n)!nN(!Np00111Because the binomial distribution represents the nthterm of the binomial expansion (x+y)N. The nthterm is:nnNyxnnNnNN!)1()2((multiply and divide by (N-n)!) =nnNyxnnNN!)!(!In the present case:x=p,y =1-p11)p1(ppNNN0n)n(N(b)NnNnNynNnn)n(NNp)pp(Np)!n()!nn()N(Np)p(p!n)!nN(!Nnnpn0011111(c)mp)1N(Nm)p1p(p)1N(Nm)p1(p)!2n()!nN()!2n(p)1N(Nm)p1(p!n)!nN(!N)1n(n22N2n2N2n2nNnnCombining terms)p1(m)p1(m)p1(NppNNpNppNmmp)1N(N222222222.8For the Poisson distribution, prove021nnm)c(mx)b(P)a(Answer:(a)(b)nmmnmmnnnmeme!nmee!nmnnPnm1(c)n)n(N)n(N)n(Nn)n(Nn)n(Nn)n(Nn)n(Nn)n(NNn)n(Nnpp)n(npn)n(npnnmnpnmmmpnnpmpmp)nm(1122222222220221!!0mmnnnmmnneenmeenmPn22222222222mnnmmPnnPmPmPn)nm(nnnnnnmnnnnnmnnnmmm!nmemnPe!nm)n(nPnnnPnn22222211

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Combining terms:2.9For the normal distribution, show(a) 1dxxP(b)mx(c) the variance is σ2Answer:(a)121222dxe))mx((Look up tables; integral is of the form0222adxexa)(b)Change variable toy = x - mdy2ye2dxemdy2ye2)my(x22y22y222()(c))()(2)()()(2222xPmdxxxPmdxxPxdxxPmxdxexxmx222)(2221set()√,√Combining terms:σ2= σ2+ m2-m2=σ22.10Ifn1,n2,...,nNare mutually uncorrelated random variables with a common varianceσ2, show that221NNnniAnswer:Consider the expression̅as being of the formNjjjna1(Equation 2.36a) with:mmmmm222dxexx)mx(2222dyemdyyedyeydyemyyyyy2222222222222122222212)0(422mm222222mxmmx

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jiNaj1,Nai11,Using Equation 2.41:2.11Show that in a series ofNmeasurements, the resultRthat minimizes the quantityNiinRQ12is,Rnwherenis given by Equation 2.31.Answer:To find the minimum of Q, solve0R.NiNiiiinNR,nRnRRQ1102NiinNR12.12Prove Equation2.62 using tables of the error function.The table of error functions can be foundhttp://www-dsp.elet.polimi.it/fondstiol/comperrfnc.pdf.Answer:Equation 2.62can be written as:mm)mx(mmedxdx)x(GA2222Call:myx,ymx1021222222dyyFyeedyAIn tables,one finds the integral:22121022terfdyety102683.06826.03413.02222dyedyy2.13As part of a quality control experiment, the lengths of 10 nuclear fuel rods have been measured with thefollowing results in meters:2.602.622.652.582.612.622.592.592.602.63What is the average length?What is the standard deviation of this series of measurements?Answer:Average length:m.ii60921010110122021000045400040909111iim.,..N2.14The average number ofcalls in a 911 switchboard is 4calls/hour. What is the probability to receive 6 calls inthe next hour?222211122222211111111)(NNNNNNNannNjNiiji  

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Answer:%1010.0e!64e!nmp46mnn2.15At a uranium pellet fabrication plant the average pellet density is 17 × 103kg/m3with a standard deviationequal to 103kg/m3.What is the probability that a given pellet has a density less than 14 × 103kg/m3?Answer:We have to find the probability that the density331014mkgwith331017mkgand3310mkg(See section 2.14).310310171014173333222133dttetGP(Equation 2.78)Fromtable 2.2:%..P1300013032.16A radioactive sample was counted once and gave 500 counts in 1 min.The corresponding number for thebackground is 480 counts. Is the sample radioactive or not?What should one report based on thismeasurement alone?Answer:The answer depends on the standard error of the net counting rate using Equation 2.93:mincbgr20480500Using Equation 2.96:minc.r3311480150022%.rr15620331error.Based on this measurement alone, one cannot tell whether or not the sample is radioactive.Note:The counting rate is so low that dead time need not be considered.2.17A radioactive sample gave 750 counts in 5 min. When the sample was removed, the sealer recorded 1000counts in 10 min. What is the net counting rate and its standard percent error?Answer:Use Equation 2.93:mincr501001501010005750And Equation 2.96:minc.r3261030100100025750%13126.0503.6rrstandard error(Again, dead time need not be considered)2.18Calculate the average net counting rate and its standard error from thedata given below:GtG(min)BtB(min)355512010355513010355513210Answer:Use Equation 2.97:minc..r619602382213110132130120536538535531

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σr(Equation 2.98)minc..r32393610038225110531%..rr403806132standard error2.19A countingexperiment has to be performed in 5 min. The approximate gross and background rates are 200counts/min. and 50 counts/min., respectively.(a)Determine the optimum gross and background count times.(b)Based on the times obtained in (a), what is the standardpercent error of the net counting rate?Answer:(a)From Equation2.100:5020050.ttGB,Since:tG+ tB= 5 , 0.5 tG+ tB=5, tG=333 min., tB= 1.67 min.r= 220–50 = 150minc(b)minc...tbtgBGr49967150333220%...rr3606301504992.20The strength of a radioactive source was measured with a 2%standard error by taking a gross count for timetmin and a background for time2tmin. Calculate the timetif it is given that the background is 300counts/min and the gross count 45,000 counts/min.Answer:Since no information is given about dead time, consider it negligible (e.g. detector is scintillator).Using Equations2.93 and 2.96:300450002300450003004500022202.022tbtgtBtGtBtGrr894t=212.48,st39.3min056.0Check:minctr8981230045000,%.rr2020447008982.21The strength of radioactive source is to be measured with a detector that has a background of 120 ± 8counts/min. The approximate gross counting rate is 360 counts/min.How long should one count if the netcounting rate is to be measured with an error of 2%?Answer:Use Equation 2.102:0642336081002120360360222GtThere is no way to get 2 % accuracy under these conditions. The error in background alone is 3.3%2408With the data given, the best one can do is:22228100x120360Which gives:

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%3.376.564x2.22The bucklingB2of a cylindrical reactor is given by:2224052 HR.BWhere:R=reactor radiusandH= reactor heightIf the radius changes by2% and the height by 8%, by what percent willB2change? TakeR= 1 m,H= 2 m.Answer:24424424424422322232222222208024020140524440524440524...HHHRRR.HHRR.HHBRRBB   2454.0156.005.0m,2225.8mB,054025845022...BB2.23UsingChauvenet’s criterion, should any of the sealer readings listed below be rejected?1151211031511211057510310510710010811311010197110109103101Answer:N=20,97504011.. A reading should be rejected if it deviates from the average by more thanthe 97.5% error. By interpolation from Table 2.4, number of standard deviation is 2.23.141937411910891072021582)nni(.NninSince23114232.., the data 75 and 151 should be rejected.Note:To obtain number ofσ’sexactly, one may usetables of error function(http://www-dsp.elet.polimi.it/fondstiol/comperrfnc.pdf). Proceed as follows:12N=140=0.025,012502121.NArea listed in table:0.5–0.0125 = 0.4875, number of242.s'2.24As a quality control test in a nuclear fuel fabrication plant, the diameter of 10 fuel pellets has been measuredwith the following results (inmm): 9.50,9.80, 9.75, 9.82, 9.93, 9.79, 9.81, 9.65, 9.99, and 9.57. Calculate(a)the average diameter(b)the standard deviation of this set of measurements(c)the standard error of the average diameter(d)should any of the results be rejected based on theChauvenet criterion?Answer:(a)mm76.9109.57+9.99+9.65+9.81+9.79+9.93+9.82+9.75+9.80+9.50dav(b)mm15.09205.01Ndavdi1012

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(c)%5.0049.0761.9048.0d,mm048.010avdavdav(d)SinceN = 10, check whether any measurements lie away from average by more than the 95% error (seesection 2.16) which is 1.96 σ = 0.294 mm. No measurements should be rejected based on that criterion.(Check the largest value first, 9.99–9.76 = 0.230 < 0.294).2.25Using the data ofProblem2.13, what is the value of accepted lengthxaif the confidence limit is 99.4%?Answer:From Problem 2.13,= 2.609 m and σ = 0.21 m. If the confidence limit is 99.4%,(xa) shouldnot deviate fromby more than 2.5 σ’s (see Table 2.2), or:m661.2021.0*5.2609.25.22.26An environmental sample has been collected for determination of210Po content.The sample is chemicallyseparated and counted in an instrument with the following results 60 days after sampling.Chemical Yield80%Counting Efficiency20%Sample Counts (gross)20 countsSample Count Time30 minutesBackground Counts10 countsBackground Count Time30 minutesHalf-life of Po-210138 days(a)What was the sample210Po net counting rate at the time of the sampling?(b)What is the standard error of value determined in part (a)?(c)The lower limit of detection (LLD) at the 95% confidence level has been defined as: LLD =1.645(2√2)Sbwhere Sbis the standard deviation. Calculate the LLD for this determination.(d)Does the activity level of this sample exceed the LLD for this determination?Answer:(a)Let An, Ag, Aband Sn, Sg, Sb,represent net, gross and background count rates and standard deviations atthe time of analysis: Ac, and Screpresent net activity and standard deviation, corrected for chemical yieldand counting efficiency at the time of sampling. The net count rate attime of analysis is:whichmust be converted for radiological decay, counting efficiency, and chemical yield to determine thedisintegration rate at the time of sampling. Corrections for radiological decay are made by using thedecay constant λ (determined from the half-life; λ = ln2/t1/2) and the time t from sampling to analysis.Aoverall efficiency term Y, can be defined as the product of individual efficiencies (such as the chemicalyield and counting efficiency) and applied to thelaboratory result.The chemical yield is defined, as theamount if the end-product recovered in a chemical process, is sometimes expressed as a percentage ofexpected recovery to indicate the efficiency of the process.If a correction factor C is defined as:()()()()()()Then the rate in disintegrations per minute at the time of sampling is:()()(b)The distribution of results when observing radioactive decay is described by Poisson statistics when thecounting interval is short compared with the half-life.The standard deviation of the distribution is thesquare root of the mean for a particular sampling interval.In this case, observations are made of thenumber of background or gross counts, nbor ng, registered in a corresponding time interval, tbor tg.Thetime intervals are assumed to have no associated measurement error, and the standard deviation in theobserved number of counts nis√. The standarddeviation associated with a count rate, therefore is

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√The standard deviation of the sum or the difference of two measurements is the square root ofthe sum of the squares of the standard deviations associated with the measurements. The standarddeviation of the net count rate obtained in the laboratory is:√()()The relative uncertainty inAcis the same as that in An. Their associated standard deviations, Scand Sn.differ by the correction factor C which accounts for yield, chemical efficiency and radioactive decay. Thestandard deviation in the original activity in the sample is:() (8.45) =(c)The lower limit of detection for thecounting portion of the analysis is:(√)(√)~4.653[()=Correcting for yield, efficiency, and radioactive decay as in part)b), gives an LLD for thedetermination of the original activity of 4.1 min-1(d)The activity level of the sample does not exceed the LLD since An= 0.33 min-1which is less than 0.49min-12.27Prove that for radioactivity measurements the value of MDA is given by the equation MDA =k2+ 2CDL, ifkα=kβ=k.Hint: whenn=MDA, the variance202MDA.Answer:Starting with Equation 2.104:222kCDLMDAkCDLMDAkCDLkCDLMDADD2222kCDLMDAkCDLMDA(Equation 2.103:CDL=kασ0)Now use:kα= kβ:CDLkMDAkCDLMDAMDACDLMDAkCDLMDACDLMDA20222222222.28A sample was counted for 5 min and gave 2250 counts; the background, also recorded for 5 min, gave 2050counts. Is this sample radioactive?Assume confidence limits of both95% and 90%.Answer:(a)95% confidence limit:k=1.645, G=2250, B=2050,mincB952050Using Equation 2.106:MDA=1.6452+4.053σβ=44.8≈45mincn40520505225040<45, not radioactive with 95% confidence level(b)90% confidence limitk=1.2853.3463.395.19285.122285.12BMDA40>34.2, yes it is, with 90% confidence limit2.29Determine the dead time of adetectorbased on the following data obtained with the two-source method:g2= 14,000 counts/ming12=26,000 counts/min

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g2= 15,000 counts/minb= 50counts/minAnswer:Use Equation 2.11005026000150001400050260001500014000250260001500050260001400050150001400026000150001400020109521017441043538212...1231228810435210752104354101744101744.....τ1=6.9 × 10-5min or 1.15 × 10-6sec or 1.15 μ secτ2= 7.87 × 10-6min or 0.13 × 10-6sec or 0.13 μ sec(τ2is the correct one since τ1g>> 1)Note: One could use the simpler Equation 2.111, sinceb<<giwhich gives the same results.2.30If the dead time of a detector is 100μs, what is the observed counting rate if the loss of counts due to deadtime is equal to 5%?Answer:sc.g.g500101000500506Proof:f.ngn050but211gnggcgntf.ng0501950.ngfng1gfg.f195010509501..gfg2.31Calculatethe true net activity and its standard percent error for a sample that gave 70,000 counts in 2 min.Thedead time of the detector is 200μs.The background is known to be 100 ± 1 counts/min.Answer:Use Equation 2.113:mincr39523100102001027000012700006Forσruse Equation 2.114:%..rminc.rr4010343952317051691287431270000102006027000011322462.32Calculate the true net activity and its standard error based on the following data:G= 100,000 countsobtained in 10 minB=10,000 countsobtained in 100 minThe dead time of thedetectoris 150μs.

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Answer:Use Equation 2.113:mincr10156100102561001000010150601010000011010000064610521015060100.bForr, use Equation 2.114:%..rmincgrr301023101563333111061001000010100000113224

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PROBLEMS3.1What is the speed of a 10-MeV electron?What is its total mass, relative to its rest mass?Answer:Use Equation 3.1. Solve for:21*MM99882025111051101...Total Mass = T + mc2= 10.511 MeV,51920511051110...times larger than its rest mass.3.2What is the speed of a proton with a total mass equal to 2Mc2?(M is the proton rest mass).Answer:Same as above:86602112.smC810598.23.3What is the kinetic energy ofa neutron that will result in 1%error difference between relativistic and classicalcalculation of its speed?Answer:Let that kinetic energy be T. The classical equation for speed is:CMTcl2.The relativistic equation is:CMTMrel21Solve for T the equation:cMTMcMTMcMT.relrelcl22112010Mtakes into account the speed of light, c2Set:221101.01112:MMTOr:110111011122..102121011222..01331.MeV...M)(T51255939013301(OrT = 1.33% of its rest mass)3.4What is the mass of an astronaut traveling with speedυ= 0.8c? Mass at rest is 70 kg.sm.C81099642

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Answer:kg..MM*M661168011223.5What is the kinetic energyof an alpha particle with a total mass 10%greater than its rest mass?Answer:MMM10.1*,MeV.MeV..MT5937247893141013.6What would the density of graphite be if the atomic radius were 10−13m?[Atomic radius (now) 10−10m;density of graphite(now) 1600 kg/m3.]Answer:VMLet331061mkg.NandρC= density of compact graphite3323434CNNatomCatomNCRRRMRMWhere:mRN1010€RC=10−13m931310101010NC3121061mkg.C3.7Calculate the binding energy of the deuteron. [M(1H)=1.007825 u;M(2H) = 2.01410 u.]Answer:Use Equation 3.24:MeV.MeV.u.u...,MMM,BnH2242481931103882014102200866510078251121233.8One of the most stable nuclei is55Mn. Its nuclidic mass is 54.938 amu. Determine its total binding energyandaverage binding energy per nucleon.Answer::Mn552525 protons + 25 electrons25(1.00783 u = 25.19575 u30 neutrons30(1.00867 u) = 30.26010 uTotal mass = 55.45585 uMn5525mass = 54.938 uDifference = 0.518 uBE= (0.158 u) (931.5MeV/u) = 483MeVand:nucleonMev.nucleonsMeVnucleonBE788554833.9Calculate the separation energy of the last neutron of241Pu [M(240Pu)=240.053809 u;M(241Pu)=241.056847 u].Answer:UsingEquation3.25:MeVMeVuuMMMBnn241.5481.931005627.0)056847.241008665.1053809.240(94,24194,2403.10Assume that the average binding energy per nucleon (in some new galaxy) changes withAas shown in thefollowing figure:(a)Would fission or fusion or both release energy in such a world?(b)How much energy would be released if a tritium (3H) nucleus and a helium (4He) nucleus combined toform a lithium nucleus? [M(3H)=3.016050 u;M(4He)=4.002603 u;M(7Li)=7.016004 u.]Answer:

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