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

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    SOLUTIONS MANUAL FOR by Measurement and Detection of Radiation, Third Edition Nicholas Tsoulfanidis

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    PROBLEMS 2.1 What 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, the probability to have one 4 in 3 throws is: P(only one 4) =   3472 0 216 75 6 5 6 1 ! ! 1 3 ! 3 2 1 .                 The same result can be obtained by enumerating all possible events. Throw 1 2 3 Outcome 4 No – 4 No – 4 No – 4 4 No – 4 No – 4 No – 4 4 To throw only one 4: ( ) Note: The probability to have at least one four is larger P (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: 4 4 No – 4  4 No – 4 4  No – 4 4 4  4 4 4  Total Probability = 216 91 216 1 216 5 3 216 75     2.2 What is the probability when drawing one card from each of three decks of cards that all three cards will be diamonds? Answer: The p robability of drawing a diamond from one deck is . The p robability that this event happens three times is ( using Equation 2.15): ( ) ( ) f (4) = 1 6 13 52

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    2.3 A box contains 2000 computer cards. If five faulty cards are expected to be found in the box, what is the probability of finding two faulty cards in a sample of 250? Answer: The probability of finding a faulty card is P(faulty card) = . The probability of finding two faulty cards in a sample of 250 is (using Binomial distribution, Equation 2.45): ( ) ( ) ( ) ( ) ( ) ( ) 2.4 Calculate the average and the standard deviation of the probability density function     a b / x f   1 when: 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 r 2 : 2.5 The energy distribution of thermal (slow) neutrons in a light - wave reactor follows very closely the Maxwell - Boltzmann distribution:   dE e E A dE E N kT / E   W here : N(E) dE = number of neutrons with kinetic energy between E and E + dE k = Boltzmann constant = 1.380662 × 10 − 23 J/° K T = temperature, K A = constant Show that : (a) The mode of this distribution is : kT E 2 1  (b) The mean is : kT E 2 3  Answer: (a) Use Equation 2.24                   kT E E Ae e kT AE e E A dE E dN kT E kT E kT E 2 2 1 2 1 1 1 2 1 0 Or: kT E e 2 1 mod  (b) Use Equation 2.27 kT dE Ee A dE Ee EA E kT E kT E 2 3 0 / 0 /        2.6 If the average for a large number of counting measurements is 15, what is the probability that a single measurement will produce the result 20? Answer: Use Poisson statistics, Equation 2.50:   4.18% or 10 18 . 4 10 433 . 2 10 159 . 3 10 325 . 3 ! 20 15 20 2 18 7 23 15 20            e P 2.7 For the binomial distribution, prove : 5 2000 = 0.0025     12 , 12 1 2 2 2 a b a b dx a b x x x b a           2 1 a b dx a b x x b a     

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    ) p ( m (c) pN n ) b ( p ) a ( 2 N n n N        1 1 0 Answer: (a)          N N n N n N n n N ) p ( p ! n )! n N ( ! N p 0 0 1 1 1 Because the binomial distribution represents the n th term of the binomial expansion (x+y) N . The n th term is:           n n N y x n n N n N N ! ) 1 ( ) 2 ( (multiply and divide by (N-n)!) = n n N y x n n N N     ! )! ( ! In the present case: x=p , y = 1- p   1 1 ) p 1 ( p p N N N 0 n ) n ( N        (b)                    N n N n N y n N n n ) n ( N Np ) p p ( Np )! n ( )! n n ( ) N ( Np ) p ( p ! n )! n N ( ! N n np n 0 0 1 1 1 1 1 (c) m p ) 1 N ( N m ) p 1 p ( p ) 1 N ( N m ) p 1 ( p )! 2 n ( )! n N ( )! 2 n ( p ) 1 N ( N m ) p 1 ( p ! n )! n N ( ! N ) 1 n ( n 2 2 N 2 n 2 N 2 n 2 n N n n                           Combining terms ) p 1 ( m ) p 1 ( m ) p 1 ( Np p N Np Np p N m m p ) 1 N ( N 2 2 2 2 2 2 2 2                 2.8 For the Poisson distribution, prove        0 2 1 n n m ) c ( m x ) b ( P ) a ( Answer: (a) (b)               n m m n m m n n n m e me ! n m e e ! n m n nPn m 1 (c)                                n ) n ( N ) n ( N ) n ( N n ) n ( N n ) n ( N n ) n ( N n ) n ( N n ) n ( N N n ) n ( N np p ) n ( n p n ) n ( n p n n m n p n m m m p n np m p m p ) n m ( 1 1 2 2 2 2 2 2 2 2 2 2 0 2 2 1 ! ! 0              m m n n n m m n n e e n m e e n m Pn 2 2 2 2 2 2 2 2 2 2 2 m n n m m P n nP m P m Pn ) n m ( n n n n                                          n n m n n n n n m n n n m m m ! n m e m nP e ! n m ) n ( n P n n n P n n 2 2 2 2 2 2 1 1

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    Combining terms: 2.9 For the normal distribution, show (a)        1 dx x P (b) m x  (c) the variance is σ 2 Answer: (a) 1 2 1 2 2 2          dx e )) m x ( (Look up tables; integral is of the form      0 2 2 2 a dx e x a ) (b) Change variable to y = x - m                             dy 2 ye 2 dx e m dy 2 y e 2 ) m y ( x 2 2 y 2 2 y 2 2 2 ( ) (c)                      ) ( ) ( 2 ) ( ) ( ) ( 2 2 2 2 x P m dx x xP m dx x P x dx x P m x         dx e x x m x 2 2 2 ) ( 2 2 2 1    set ( ) √ , √ Combining terms: σ 2 = σ 2 + m 2 - m 2 = σ 2 2.10 If n 1 , n 2 ,..., n N are mutually uncorrelated random variables with a common variance σ 2 , show that   2 2 1     N N n n i Answer: Consider the expression ̅ as being of the form   N j j j n a 1 (Equation 2.36 a ) with : m m m m m        2 2 2 dx e x x ) m x (           2 2 2 2                            dy e m dy ye dy e y dy e m y y y y y 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2          2 2 2 2 2 1 2 ) 0 ( 4 2 2 m m            2 2 2 2 2 2 m x m m x     

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    j i N a j   1 , N a i 1 1   , Using Equation 2.41 : 2.11 Show that in a series of N measurements, the result R that minimizes the quantity       N i i n R Q 1 2 is , R n  where n is given by Equation 2.31. Answer: To find the minimum of Q, solve 0     R .                 N i N i i i i n NR , n R n R R Q 1 1 0 2   N i i n N R 1 2.12 Prove Eq uation 2.62 using tables of the error function. The table of error functions can be found http://www- dsp.elet.polimi.it/fondstiol/comperrfnc.pdf . Answer: Equation 2.62 can be written as:                 m m ) m x ( m m e dx dx ) x ( G A 2 2 2 2 C all : m y x , y m x                  1 0 2 1 2 2 2 2 2 2 dy y F y e e dy A In tables, one finds the integral :            2 2 1 2 1 0 2 2 t erf dy e t y        1 0 2 683 . 0 6826 . 0 3413 . 0 2 2 2 2 dy e dy y  2.13 As part of a quality control experiment, the lengths of 10 nuclear fuel rods have been measured with the following results in meters: 2.60 2.62 2.65 2.58 2.61 2.62 2.59 2.59 2.60 2.63 What is the average length? What is the standard deviation of this series of measurements? Answer: Average length : m . i i 609 2 10 10 1                     10 1 2 2 021 0 000454 0 00409 0 9 1 1 1 i i m . , . . N   2.14 The average number of calls in a 911 switchboard is 4 calls/hour. What is the probability to receive 6 calls in the next hour?   2 2 2 2 1 1 1 2 2 2 2 2 2 1 1 1 1 1 1 1 1 ) (     N N N N N N N a n n N j N i i j i                                            

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    Answer: % 10 10 . 0 e ! 6 4 e ! n m p 4 6 m n n                       2.15 At a uranium pellet fabrication plant the average pellet density is 17 × 10 3 kg/m 3 with a standard deviation equal to 10 3 kg/m 3 . What is the probability that a given pellet has a density less than 14 × 10 3 kg/m 3 ? Answer: We have to find the probability that the density 3 3 10 14 m kg    with 3 3 10 17 m kg    and 3 3 10 m kg   (See section 2.14).                 3 10 3 10 17 10 14 17 3 3 3                 3 2 2 2 1 3 3 dt t e t G P (Equation 2.78) From table 2.2:   % . . P 13 0 0013 0 3        2.16 A radioactive sample was counted once and gave 500 counts in 1 min. The corresponding number for the background is 480 counts. Is the sample radioactive or not? What should one report based on this measurement alone? Answer: The answer depends on the standard error of the net counting rate using Equation 2.93: min c b g r 20 480 500      Using Equation 2.96: min c . r 3 31 1 480 1 500 2 2     % . r r 156 20 3 31    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. 2.17 A radioactive sample gave 750 counts in 5 min. When the sample was removed, the sealer recorded 1000 counts in 10 min. What is the net counting rate and its standard percent error? Answer: Use Equation 2.93 : min c r 50 100 150 10 1000 5 750      And Equation 2.96 : min c . r 32 6 10 30 100 1000 25 750       % 13 126 . 0 50 3 . 6    r r  standard error (A gain, dead time need not be considered) 2.18 Calculate the average net counting rate and its standard error from the data given below: G t G (min) B t B(min) 355 5 120 10 355 5 130 10 355 5 132 10 Answer: Use Equation 2.97 :   min c . . r 61 9 60 2 38 221 3 1 10 132 130 120 5 365 385 355 3 1                

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    σ r ( Equation 2.98)  min c . . r 3 2 3 93 6 100 382 25 1105 3 1      % . . r r 4 038 0 61 3 2     standard error 2.19 A counting experiment has to be performed in 5 min. The approximate gross and background rates are 200 counts/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 standard percent error of the net counting rate? Answer: (a) From Equation 2.100 : 5 0 200 50 . t t G B   , Since: t G + t B = 5 , 0.5 t G + t B =5, t G =333 min., t B = 1.67 min . r = 220 – 50 = 150 min c (b) min c . . . t b t g B G r 49 9 67 1 50 33 3 220       % . . . r r 3 6 063 0 150 49 9     2.20 The strength of a radioactive source was measured with a 2 % standard error by taking a gross count for time t min and a background for time 2 t min. Calculate the time t if it is given that the background i s 300 counts/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 Eq uation s 2.93 and 2.96: 300 45000 2 300 45000 300 45000 2 2 2 02 . 0 2 2           t b t g t B t G t B t G r r  894 t = 212.48 , s t 39 . 3 min 056 . 0   Check : min c t r 898 1 2 300 45000      , % . r r 2 02 0 44700 898     2.21 The strength of radioactive source is to be measured with a detector that has a background of 120 ± 8 counts/min. The approximate gross counting rate is 360 counts/min. How long should one count if the net counting rate is to be measured with an error of 2 % ? Answer: Use Equation 2.102:   0 64 23 360 8 100 2 120 360 360 2 2 2             G t T here is no way to get 2 % accuracy under these conditions. The error in background alone is 3.3%       240 8 With the data given, the best one can do is:   2 2 2 2 8 100 x 120 360            Which gives:

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    % 3 . 3 76 . 5 64 x   2.22 The buckling B 2 of a cylindrical reactor is given by : 2 2 2 405 2                H R . B W here : R = reactor radius and H = reactor height If the radius changes by 2% and the height by 8% , by what percent will B 2 change? Take R = 1 m, H = 2 m. Answer:             2 4 4 2 4 4 2 4 4 2 4 4 2 2 3 2 2 2 3 2 2 2 2 2 2 2 2 08 0 2 4 02 0 1 405 2 4 4 405 2 4 4 405 2 4 . . . H H H R R R . H H R R . H H B R R B B                                                                      2 454 . 0 156 . 0 05 . 0     m , 2 2 25 . 8   m B , 054 0 25 8 45 0 2 2 . . . B B    2.23 Using Chauvenet’s criterion, should any of the sealer readings listed below be rejected? 115 121 103 151 121 105 75 103 105 107 100 108 113 110 101 97 110 109 103 101 Answer: N=20, 975 0 40 1 1 .   . A reading should be rejected if it deviates from the average by more than the 97.5% error. By interpolation from Table 2.4, number of standard deviation is 2.23. 14 19 3741 19 108 9 107 20 2158 2            ) n ni ( . N ni n Since 2 31 14 23 2 . .   , the data 75 and 151 should be rejected. Note: To obtain number of σ’s exactly, one may use tables of error function ( http://www- dsp.elet.polimi.it/fondstiol/comperrfnc.pdf ). Proceed as follows: 1 2 N = 1 40 = 0.025 , 0125 0 2 1 2 1 . N        A rea listed in table : 0.5 – 0.0125 = 0.4875, number of 24 2 . s '   2.24 As a quality control test in a nuclear fuel fabrication plant, the diameter of 10 fuel pellets has been measured with the following resu l ts (in mm): 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 the Chauvenet criterion? Answer: (a) mm 76 . 9 10 9.57 + 9.99 + 9.65 + 9.81 + 9.79 + 9.93 + 9.82 + 9.75 + 9.80 + 9.50 d av         (b)   mm 15 . 0 9 205 . 0 1 N dav di 10 1 2       

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    (c) % 5 . 0 049 . 0 761 . 9 048 . 0 d , mm 048 . 0 10 av dav dav         (d) Since N = 10, check whether any measurements lie away from average by more than the 95% error (see section 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 ). 2.25 Using the data of Problem 2.13, what is the value of accepted length x a if the confidence limit is 99.4 % ? Answer: From P roblem 2.13,  = 2.609 m and σ = 0.21 m . If the confidence limit is 99.4%,  (x a ) should not deviate from  by more than 2.5 σ’s (see Table 2.2), or: m 661 . 2 021 . 0 * 5 . 2 609 . 2 5 . 2         2.26 An environmental sample has been collected for determination of 210 Po content. The sample is chemically separated and counted in an instrument with the following results 60 days after sampling. Chemical Yield 80% Counting Efficiency 20% Sample Counts (gross) 20 counts Sample Count Time 30 minutes Background Counts 10 counts Background Count Time 30 minutes Half-life of Po-210 138 days (a) What was the sample 210 Po 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)S b where S b is 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 A n , A g , A b and S n , S g , S b , represent net, gross and background count rates and standard deviations at the time of analysis: A c , and S c represent net activity and standard deviation, corrected for chemical yield and counting efficiency at the time of sampling. The net count rate at time of analysis is : which must be converted for radiological decay, counting efficiency, and chemical yield to determine the disintegration rate at the time of sampling. Corrections for radiological decay a re made by using the decay constant λ (determined from the half - life; λ = l n2/t 1/2 ) and the time t from sampling to analysis. A overall efficiency term Y, can be defined as the product of individual efficiencies (such as the chemical yield and counting efficiency) and applied to the laboratory result . The chemical yield is def ined, as the amount if the end - product recovered in a chemical process, is sometimes expressed as a percentage of expected 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 the counting interval is short compared with the half - life. The standard deviation of the distribution is the square root of the mean for a particular sam pling interval. In this case, observations are made of the number of background or gross counts, n b or n g , registered in a corresponding time interval, t b or t g . The time intervals are assumed to have no associated measurement error, and the standard dev iation in the observed number of counts n is √ . The standard deviation 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 of the sum of the squares of the sta ndard deviations associated with the measurements. The standard deviation of the net count rate obtained in the laboratory is : √ ( ) ( ) The relative uncertainty in A c is the same as that in A n . Their associated standard deviations, S c and S n . differ by the correction factor C which accounts for yield, chemical efficiency and radioactive decay. The standard deviation in the original activity in the sample is : ( ) (8.45) = (c) The lower limit of detection for the counting portion of the analysis is : ( √ ) ( √ ) ~4.653[ ( ) = Correcting for yield, efficiency, and radioact ive decay as in pa rt)b), gives an LLD for the determination of the original activity of 4.1 min - 1 (d) The activity level of the sample does not exceed the LLD since A n = 0.33 min - 1 which is less than 0.49 min - 1 2.27 Prove that for radioactivity measurements the value of MDA is given by the equation MDA = k 2 + 2CDL, if k α = k β = k . Hint: when n = MDA, the variance 2 0 2     MDA . Answer: Starting with Equation 2.104: 2 2 2               k CDL MDA k CDL MDA k CDL k CDL MDA D D                2 2 2 2 k CDL MDA k CDL MDA (Equation 2.103 : CDL = k α σ 0 ) Now use: k α = k β :        CDL k MDA k CDL MDA MDA CDL MDA k CDL MDA CDL MDA 2 0 2 2 2 2 2 2 2 2              2.28 A sample was counted for 5 min and gave 2250 counts; the background, also recorded for 5 min, gave 2050 counts. Is this sample radioactive? Assume confidence limits of both 95% and 90%. Answer: (a) 95% confidence limit: k=1.645, G=2250, B=2050, min c B 9 5 2050    Using Equation 2.106: MDA = 1.645 2 + 4.053 σ β = 44.8 ≈ 45 min c n 40 5 2050 5 2250    40<45, not radioactive with 95% confidence level (b) 90% confidence limit k=1.285 3 . 34 63 . 3 95 . 1 9 285 . 1 2 2 285 . 1 2        B MDA  40>34.2, yes it is, with 90% confidence limit 2.29 Determine the dead time of a detector based on the following data obtained with the two - source method: g 2 = 14,000 counts/min g 12 = 26,000 counts/min

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    g 2 = 15,000 counts/min b = 50 counts/min Answer: Use Equation 2.110     0 50 26000 15000 14000 50 26000 15000 14000 2 50 26000 15000 50 26000 14000 50 15000 14000 26000 15000 14000 2                        0 10 95 2 10 174 4 10 43 5 3 8 2 12         . . .   12 3 12 2 8 8 10 43 5 2 10 75 2 10 43 5 4 10 174 4 10 174 4             . . . . . τ 1 = 6.9 × 10 - 5 min or 1.15 × 10 - 6 sec or 1.15 μ sec τ 2 = 7.87 × 10 - 6 min or 0.13 × 10 - 6 sec or 0.13 μ sec ( τ 2 is the correct one since τ 1 g >> 1) Note : One could use the simpler E quation 2.111, since b<<g i which gives the same results. 2.30 If the dead time of a detector is 100 μ s, what is the observed counting rate if the loss of counts due to dead time is equal to 5 % ? Answer: s c . g . g 500 10 100 05 0 05 0 6        Proof: f . n g n    05 0 but 2 1 1 g n g gc g n t      f . n g    05 0 1 95 0 . n g  f n g   1         g f g . f 1 95 0 1 05 0 95 0 1 . . g      f g  2.31 Calculate the true net activity and its standard percent error for a sample that g ave 70,000 counts in 2 min. The dead time of the detector is 200 μ s. The background is known to be 100 ± 1 counts/min. Answer: Use Equation 2.113: min c r 39523 100 10 200 10 2 70000 1 2 70000 6         For σ r use Equation 2.114 : % . . r min c . r r 4 0 10 3 4 39523 170 5 169 1 28743 1 2 70000 10 200 60 2 70000 1 1 3 2 2 4 6                              2.32 Calculate the true net activity and its standard error based on the following data: G = 100,000 counts obtained in 10 min B = 10,000 counts obtained in 100 min The dead time of the detector is 150 μ s.

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    Answer: Use Equation 2.113: min c r 10156 100 10256 100 10000 10 150 60 10 100000 1 10 100000 6                         4 6 10 5 2 10 150 60 100 . b For r  , use Equation 2.114: % . . r min c g r r 3 0 10 2 3 10156 33 33 1 1106 100 10000 10 100000 1 1 3 2 2 4                      

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    PROBLEMS 3.1 What 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: 2 1            * M M 99882 0 2 511 10 511 0 1 . . .           Total Mass = T + mc 2 = 10.511 MeV, 519 20 511 0 511 10 . . .  times larger than its rest mass. 3.2 What is the speed of a proton with a total mass equal to 2 Mc 2 ? (M is the proton rest mass). Answer: Same as above: 866 0 2 1 1 2 .           s m C 8 10 598 . 2      3.3 What is the kinetic energy of a neutron that will result in 1% error difference between relativistic and classical calculation of its speed? Answer: Let that kinetic energy be T. The classical equation for speed is: C M T cl    2 . The relativistic equation is: C M T M rel            2 1 Solve for T the equation: c M T M c M T M c M T . rel rel cl                                2 2 1 1 2 01 0 M takes into account the speed of light, c 2 Set:   2 2 1 1 01 . 0 1 1 1 2 :            M M T Or:      1 1 01 1 1 01 1 1 2 2            . .   1 02 1 2 1 01 1 2 2 2          . . 0133 1 .   MeV . . . M ) ( T 5 12 55 939 0133 0 1       (Or T = 1.33% of its rest mass) 3.4 What is the mass of an astronaut traveling with speed υ = 0.8 c ? Mass at rest is 70 kg. s m . C 8 10 9964 2     

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    Answer: kg . . M M * M 66 116 8 0 1 1 2 2       3.5 What is the kinetic energy of an alpha particle with a total mass 10 % greater than its rest mass? Answer: M M M    10 . 1 * ,   MeV . MeV . . M T 59 372 478 931 4 1 0 1        3.6 What would the density of graphite be if the atomic radius were 10 −13 m? [Atomic radius (now) 10 −10 m; density of graphite (now) 1600 kg/m 3 .] Answer: V M   Let 3 3 10 6 1 m kg . N    and ρ C = density of compact graphite 3 3 2 3 4 3 4               C N N atom C atom N C R R R M R M Where: m R N 10 10   € R C = 10 − 13 m 9 3 13 10 10 10 10               N C 3 12 10 6 1 m kg . C    3.7 Calculate the binding energy of the deuteron. [ M ( 1 H) = 1.007825 u; M ( 2 H) = 2.01410 u.] Answer: Use E quation 3.24 :         MeV . MeV . u . u . . . , M M M , B n H 224 2 481 931 10 388 2 014102 2 008665 1 007825 1 1 2 1 2 3            3. 8 One of the most stable nuclei is 55 Mn. Its nuclidic mass is 54.938 amu. Determine its total binding energy and average binding energy per nucleon. Answer: : Mn 55 25 25 protons + 25 electrons 25(1.00783 u = 25.19575 u 30 neutrons 30(1.00867 u) = 30.26010 u Total mass = 55.45585 u Mn 55 25 mass = 54.938 u Difference = 0.518 u BE = (0.158 u) (931.5 MeV / u ) = 483 MeV and : nucleon Mev . nucleons MeV nucleon BE 78 8 55 483   3.9 Calculate the separation energy of the last neutron of 241 Pu [ M ( 240 Pu) = 240.053809 u; M ( 241 Pu) = 241.056847 u ] . Answer: Using E quation 3.25 :       MeV MeV u u M M M B n n 241 . 5 481 . 931 005627 . 0 ) 056847 . 241 008665 . 1 053809 . 240 ( 94 , 241 94 , 240          3.10 Assume that the average binding energy per nucleon (in some new galaxy) changes with A as shown in the following figure: (a) Would fission or fusion or both release energy in such a world? (b) How much energy would be released if a tritium ( 3 H) nucleus and a hel ium ( 4 He) nucleus combined to form a lithium nucleus? [ M ( 3 H) = 3.016050 u; M ( 4 He) = 4.002603 u; M ( 7 Li) = 7.016004 u.] Answer:

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