Suggested Problem Solutions Associated with Homework #5

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1 Suggesed Problem Soluions Associaed wih Homework #5 431 (a) 8 Si has proons and neurons (b) 85 3 Rb has 3 proons and 48 neurons (c) 5 Tl 81 has 81 proons and neurons 43 IDENTIFY and SET UP: The ex calculaes ha he binding energy of he deueron is 4 MeV A phoon ha breaks he deueron up ino a proon and a neuron mus have a leas his much energy hc hc E = so λ = λ E 15 8 (413 1 ev s)(998 1 m/s) 13 λ = = m = 555 pm 4 1 ev EVALUATE: This phoon has gamma-ray wavelengh 43 (a) IDENTIFY: Find he energy equivalen of he mass defec SET UP: A B 5 aom has 5 proons, 5 = neurons, and 5 elecrons The mass defec herefore is M = 5m + m + 5 m M( B) p equivalen is n e 5 M = 5(15 u) + (1849 u) + 5( u) 935 u = 8181 u The energy E B = (8181 u)(9315 MeV/u) = 1 MeV IDENTIFY In each case deermine how he decay changes A and Z of he nucleus The β and β paricles have charge bu heir nucleon number is A = (a) SET UP: 4 α-decay: Z increases by, A = N + Z decreases by 4 (an α paricle is a He nucleus) Pu He + U (b) SET UP: β decay: Z increases by 1, A = N + Z remains he same (a β paricle is an elecron, ) Na e + Mg (c) SET UP β decay: Z decreases by 1, A = N + Z remains he same (a β + paricle is a posiron, e) O + 1e+ N EVALUATE: In each case he oal charge and oal number of nucleons for he decay producs equals he charge and number of nucleons for he paren nucleus; hese wo quaniies are conserved in he decay 1 e 431 If β decay of C is possible, hen we are considering he decay C N+ β m= M( C) M( N) m e m = (34 u (549 u)) (34 u (549 u)) 5491 u m=+ E = = = u So (18 1 u)(9315 MeV u ) 15 MeV 15 kev 433 IDENTIFY and SET UP: As discussed in Secion 434, he aciviy A = dn / d obeys he same decay equaion as Eq λ (431): A= Ae For C, T = 53 y and λ = ln/ T so A= Ae Calculae A a each ; A = 18 1/ 1/ decays/min (ln ) / T1/ ;

2 (a) = 1 y, A = 159 decays/min (b) = 5, y, A = 43 decays/min EVALUATE: 1 The ime in par (b) is 83 half-lives, so he decay rae has decreased by a facor or () (a) 3 3 H e+ He (b) 1 1, 1 and λ (ln ) λ N = Ne N= N = T 1 (ln ) 1 1 = e T ; (ln ) 1 T = ln(1); ln(1) T 1 = = ln 49 y λ (ln )/ T 1/ 43 A= Ae = Ae T 1 (ln ) = ln( A A ) T1 (ln ) (ln )(4 days) = = = 8 days ln( A A ) ln( ) dn 4331 (a) = = d 5 1 Bq 5 1 decays s 1 dn 5 1 decays s N = = = 4 1 λ d 35 1 s 15 1 nuclei λ = = = 35 1 s T (38 min)( s min) 1 (b) The number of nuclei lef afer one half-life is dn d = 38 1 decays s N = 1 15 nuclei, and he aciviy is half: (c) Afer hree half lives (94 minues) here is an eighh of he original amoun, so dn 1 of he aciviy: = decays s d N = 53 1 nuclei, and an eighh 4333 IDENTIFY and SET UP: Find λ from he half-life and he number N of nuclei from he mass of one nucleus and he mass of he sample Then use Eq(431) o calculae dn / d, he number of decays per second (a) dn / d = λn λ = = = s 9 T (18 1 y)(315 1 s/1 y) 1 1 1/ 4 The mass of K aom is approximaely 4 u, so he number of 4 K nuclei in he sample is kg 13 1 kg 1 N = = = Then dn d 4 u 4(154 1 kg) λ / = N = (115 1 s )(454 1 ) = 41 decays/s 1 (b) dn / d = (41 decays/s)(1 Ci/(3 1 decays/s)) = 4 1 Ci EVALUATE: The very small sample sill conains a very large number of nuclei Bu he half life is very large, so he decay rae is small

3 Suggesed Problem Soluions Associaed wih Homework # rad = 1 Gy, so 1 Gy = 1 rad and he dose was 5 rad rem = (rad)(rbe) = (5 rad)(4) = rem 1 Gy = 1 J kg, so 5 J kg 433 IDENTIFY and SET UP: For x rays RBE = 1 and he equivalen dose equals he absorbed dose (a) 15 krad = 15 krem = 15 kgy = 15 ksv 3 (15 1 J/kg)(15 kg) = 1 J (b) 15 krad = 15 kgy ; (15)(15 krad) = krem = ksv The energy deposied would be 1 J, he same as in (a) EVALUATE: The energy required o raise he emperaure of 15 kg of waer 1 C is 8 J, and 1 J is less han his The energy deposied corresponds o a very small amoun of heaing 4339 (a) We need o know how many decays per second occur λ = = = T (13 y) (315 1 s y) s 1 1 dn (35 Ci) (3 1 Bq Ci) The number of riium aoms is N = = = λ d 19 1 s The number of remaining nuclei afer one week is N = N e = e = λ 18 (19 1 s ) () (4) (3s) 18 (5 1 ) 4 1 nuclei nuclei 15 N = N N = 8 1 decays So he energy absorbed is Eoal = N E γ = (8 1 ) (5 ev) (1 1 J ev) =4 J The absorbed dose is (4 J) = 15 J kg = 15 rad Since RBE = 1, hen he equivalen dose is 15 rem (5 kg) (b) In he decay, aninerinos are also emied These are no absorbed by he body, and so some of he energy of he decay is los (abou 1 kev ) 4341 (a) IDENTIFY and SET UP: Deermine X by balancing he charge and nucleon number on he wo sides of he reacion equaion X mus have A= + 1 = and Z = 1+ 5 = 3 Thus X is 3 and he reacion is 1 1H+ N 3Li+ 5B (b) IDENTIFY and SET UP: Calculae he mass decrease and find is energy equivalen The neural aoms on each side of he reacion equaion have a oal of 8 elecrons, so he elecron masses cancel when neural aom masses are used The neural aom masses are found in Table 43 mass of H + N is 1 u + 34 u = 111 u 1 mass of 1 3Li + 5B is 15 u + 93 u = 1858 u The mass increases, so energy is absorbed by he reacion The Q value is (111 u 1858 u)(9315 MeV/u) = 4 MeV (c) IDENTIFY and SET UP: The available energy in he collision, he kineic energy Kcm in he cener of mass reference frame, is relaed o he kineic energy K of he bombarding paricle by Eq (434) The kineic energy ha mus be available o cause he reacion is 4 MeV Thus K = 4 MeV The mass M of he saionary arge ( N) is M = u The mass m of he colliding paricle ( H) 1 is u Then by Eq (434) he minimum kineic energy K ha he H 1 mus have is cm

4 M m u u K + K + = cm = (4 MeV) = 59 MeV M u EVALUATE: The projecile ( ) H is much ligher han he arge 1 ( ) N so K is no much larger han K cm The K we have calculaed is wha is required o allow he mass increase We would also need o check o see if a his energy he projecile can overcome he Coulomb repulsion o ge sufficienly close o he arge nucleus for he reacion o occur 4343 IDENTIFY and SET UP: Deermine X by balancing he charge and he nucleon number on he wo sides of he reacion equaion X mus have A = = and Z = = 3 Thus X is and he reacion is 1H+ 4Be= 3Li+ He (b) IDENTIFY and SET UP: Calculae he mass decrease and find is energy equivalen If we use he neural aom masses hen here are he same number of elecrons (five) in he reacans as in he producs Their masses cancel, so we ge he same mass defec wheher we use nuclear masses or neural aom masses The neural aoms masses are given in Table 43 9 H + Be has mass 1 u + 98 u = 84 u Li + He has mass 13 u + 43 u = 18 u The mass decrease is 84 u 18 u = 8 u This corresponds o an energy release of 8 u(9315 MeV/1 u) = 15 MeV (c) IDENTIFY and SET UP: Esimae he hreshold energy by calculaing he Coulomb poenial energy when he H and Be nuclei jus ouch Obain he nuclear radii from Eq (431) The radius 9 of he Be nucleus is ( m)(9) 1/3 15 R R = = 5 1 m Be 4 Be 15 1/3 15 H 1 H The radius R of he H nucleus is R = (1 1 m)() = 15 1 m The nuclei ouch when heir cener-o-cener separaion is 15 R= RBe + RH = 4 1 m The Coulomb poenial energy of he wo reacan nuclei a his separaion is 1 qq 1 1 e(4 e) U = = 4πP r 4πP r (1 1 C) U = ( N m / C ) = MeV (4 1 m)(1 1 J/eV) This is an esimae of he hreshold energy for his reacion EVALUATE: The reacion releases energy bu he oal iniial kineic energy of he reacans mus be MeV in order for he reacing nuclei o ge close enough o each oher for he reacion o occur The nuclear force is srong bu is very shorrange 434 The energy liberaed will be 3 4 M( He) + M( He) M( Be) = (319 u + 43 u 199 u)(9315 MeV u) = 158 MeV IDENTIFY and SET UP: Find he energy equivalen of he mass decrease Par of he released energy appears as he emied phoon and he res as kineic energy of he elecron Au Hg + e The mass change is u 1995 u = 3 1 u (The neural aom masses include 9 elecrons before he decay and 8 elecrons afer he decay This one addiional elecron in he producs accouns correcly for he elecron emied by he nucleus) The oal energy released in he decay 3 is (3 1 u)(9315 MeV/u) = 13 MeV This energy is divided beween he energy of he emied phoon and he kineic energy of he β paricle Thus he β paricle has kineic energy equal o 13 MeV 41 MeV = 9 MeV

5 198 EVALUATE: The emied elecron is much ligher han he nucleus, so he elecron has almos all he final kineic 8 Hg energy The final kineic energy of he 198 Hg nucleus is very small 43 IDENTIFY: Use Eq (431) o relae he iniial number of radioacive nuclei, N, o he number, N, lef afer ime SET UP: We have o be careful; afer 8 Rb has undergone radioacive decay i is no longer a rubidium aom Le N 85 be he number of 85 Rb aoms; his number doesn change Le N be he number of aoms on earh when he solar sysem was formed Le N be he presen number of 8 Rb aoms The presen measuremens say ha 83 = N/( N + N85) ( N + N85)(83) = N, so N = 385N85 The percenage we are asked o calculae is N/( N + N ) 85 N and N are relaed by N = N e so N =e N λ + λ λ λ λ N Ne (3855 e ) N85 385e Thus N N = Ne N = λ λ (385 e ) N N = e λ = 4 1 y; λ= = = 59 1 y T1/ 45 1 y 1 9 λ (59 1 y )(4 1 y) 1 e = e = e = 194 N (385)(194) Thus 9% N + N = (385)(194) + 1 = 85 EVALUATE: The half-life for 8 Rb is a facor of 1 larger han he age of he solar sysem, so only a small fracion of he 8 Rb nuclei iniially presen have decayed; he percenage of rubidium aoms ha are radioacive is only a bi less now han i was when he solar sysem was formed 8 Rb 439 IDENTIFY and SET UP: Find he energy emied and he energy absorbed each second Conver he absorbed energy o absorbed dose and o equivalen dose decays/s (a) Firs find he number of decays each second: 1 Ci = 9 1 decays/s 1 Ci The average energy per decay is 15 MeV, and one-half of his energy is deposied in he umor The energy delivered o 1 19 he umor per second hen is (9 1 decays/s)(15 1 ev/decay)(1 1 J/eV) = 9 1 J/s (b) The absorbed dose is he energy absorbed divided by he mass of he issue: 9 1 J/s 4 = (19 1 J/kg s)(1 rad/(1 J/kg)) = 19 1 rad/s 5 kg (c) equivalen dose (REM) = RBE absorbed dose (rad) 4 4 In one second he equivalen dose is (19 1 rad) = 13 1 rem 4 (d) ( rem/13 1 rem/s) = 1/5 1 s(1 h/3 s) = 4 h = 1 days EVALUATE: The aciviy of he source is small so ha absorbed energy per second is small and i akes several days for an equivalen dose of rem o be absorbed by he umor A rem dose equals Sv and his is large enough o damage he issue of he umor