Pearson Physics Level 30 Unit VIII Atomic Physics: Chapter 16 Solutions

Transcription

1 Pearson Physics Level 30 Uni VIII Aomic Physics: Chaper 6 Soluions Suden Book page 79 Example 6. Pracice Problems. Given Z = A = 4 neuron number (N) Since A Z N, N AZ 4 There are neurons in a nucleus of 4 Mg.. Given Z = 9 N = 46 aomic mass number (A) Since A Z N, A The aomic mass number of he uranium aom is 38. Concep Check All hree nuclei have he same aomic number, Z = 6, bu differen aomic masses and numbers of neurons. Example 6. Pracice Problems. Given 5 r = 5 fm = 5 0 m graviaional force (F g ) Suden Book page 79

2 Gmm Use he equaion Fg, where m and m represen he mass of each proon, r kg. N m kg kg Fg 5 50 m N 36 The graviaional force beween wo proons ha are 5 fm apar is 7 0 N.. Given 5 r = 5 fm = 5 0 m elecrosaic force (F e ) kqq Use he equaion Fe, where q and q represen he charge on each proon, r C. 9 N m C C Fe 5 50 m 9 N The elecrosaic force ha wo proons exer on each oher when hey are 5 fm apar is 9 N. Suden Book page 794 Example 6.3 Pracice Problems. Given m = 0. u energy equivalen in ev Simply muliply 0. u by he appropriae equivalence facor: 93.5 MeV 0. u 06 MeV u A mass of 0. u is equivalen o 06 MeV.. Given E = 50 MeV mass equivalen (u)

3 Use he conversion facor u = 93.5 MeV u 50 MeV 0.68 u 93.5 MeV 50 MeV is equivalen o a mass of 0.68 u. Suden Book page 795 Concep Check Sar wih he expression mzm Nm H neuron maom. Noe ha a hydrogen aom consiss of one proon and one elecron. So, Zm = Z(mass of proon + mass of elecron). H Also, m aom = (mass of nucleus + mass of Z elecrons), because here are Z elecrons in a neural aom. Puing hese equaions ogeher gives: m = Z(mass of proon + mass of elecron) + Nm neuron (mass of nucleus + mass of Z elecrons). The elecron masses subrac, leaving mzmproon Nmneuron mnucleus. Example 6.4 Pracice Problems. Given m Na = u mass defec ( m ) From 3 Na, A = 3 and Z =. Use he equaion N A Z o find he number of neurons, hen use he equaion m Zm Nmneuron maom, where m u and m H H neuron u. Soluion N AZ 3 mzm Nm m H neuron aom ( u) ( u) u u The mass defec for he sodium nucleus is u.. Given he sodium aom, 3 Na binding energy (E b ) 3

4 Use Δm from quesion and he conversion facors: 0 u.490 J u 93.5 MeV J 93.5 MeV Eb u u u or u.9880 J 86.6 MeV The binding energy of he sodium-3 nucleus is J or 86.6 MeV. Suden Book page Check and Reflec Knowledge. To find he number of neurons, N, and he number of proons, Z, use he equaion A = Z + N. (a) 90 38Sr A = 90 and Z = 38, so N A Z Sr has 38 proons and 5 neurons. (b) 3 6C A = 3 and Z = 6, so N AZ C has 6 proons and 7 neurons. (c) 56 6Fe A = 56 and Z = 6, so N AZ Fe has 6 proons and 30 neurons. (d) H A = and Z =, so N AZ 0 H has proon and no neurons. 4

5 0.6 0 J ev =.0 GeV 9 J.60 0 ev 3. u = 93.5 MeV so 93.5 MeV 0.5 u 0.5 u u 33 MeV 8.30 ev 4. Given 9 E = 5.00 GJ J mass (m) Use he equaion E = mc. E m c J 8 (3.000 m/s) kg 8 The quaniy of mass ha is convered o 5.00 GJ of energy is kg. 5. Isoopes are aoms ha have he same aomic number bu differen neuron numbers. They are chemically very similar bu do no have he same aomic mass. 6. A sable nucleus is bound ogeher by he srong nuclear force. Binding energy is he energy required o separae all he proons and neurons in a nucleus and move hem infiniely far apar. The mass equivalence of his binding energy, calculaed using Einsein s equaion, E = mc, accouns for he slighly lower mass of a sable nucleus han ha given by Zm proon + Nm neuron. Applicaions 7. Given 0 Ne, m aom = u binding energy (E b ) Firs calculae he number of neurons using A = N + Z. Then use he equaion Δm = Zm H + Nm neuron m aom o calculae he mass difference. Use your answer for he mass difference and u = 93.5 MeV o calculae he binding energy. Soluion From 0 Ne, Z 0, A N AZ 0 5

6 m 0( u) ( u) u u 93.5 MeV u 77.8 MeV u The binding energy for neon- is 77.8 MeV. 8. (a) Given 40 9 K, m aom = u mass defec ( m ) Use he equaion A = N + Z o calculae he number of neurons. From 40 9K, A = 40 and Z = 9, so N AZ 40 9 Use mzmh Nmneuron maom o find he mass defec. m 9( u) ( u) u u Poassium-40 has a mass defec of u. (b) Given m u (from par (a)) Eb binding energy per nucleon A Conver he mass defec o binding energy using: 93.5 MeV Eb u u MeV Divide by aomic mass number o obain he binding energy per nucleon, From 40 9 K, A = 40, so Eb MeV A MeV/nucleon Poassium-40 has a binding energy per nucleon of MeV. 9. To esimae he binding energy, read values from he graph. Noe ha he binding energy is given in energy per nucleon, so muliply by he aomic mass number o obain he binding energy per nucleus. (a) From he graph in Figure 6.4, 3 6 C has ~7.7 MeV/nucleon, so Eb A. 6

7 E b = 3 nucleons 7.7 MeV/nucleon = 00 MeV (b) 56 6 Fe has 8.5 MeV/nucleon, so E b = 56 nucleons 8.5 MeV/nucleon = 476 MeV (c) 38 9 U has 7.3 MeV/nucleon, so E b = 38 nucleons 7.3 MeV/nucleon = 737 MeV 0. MeV is an energy uni, so i is equivalen o joules. The unis of MeV/c are J N m m s m s Ns kg m m s s m kg Exensions. (a) The srong nuclear force is he sronger of he wo forces, by a facor of abou 00 bu, unlike he elecrosaic force, i acs only over very shor disances a few femomeres a mos and acs on boh proons and neurons. The range of he elecromagneic force is infinie, bu i acs only on charged paricles, such as proons and elecrons. (b) The proons repel each oher by he elecromagneic force and are araced o each oher and neurons by he srong nuclear force. If he srong force aced over a larger disance, hen much larger nuclei would exis. If he elecromagneic force were sronger, hen i would be difficul o form large nuclei, especially hose wih many proons.. The nucleus is in a consan sruggle beween he srong nuclear force ha binds i ogeher and he elecrosaic force ha causes like charges o repel. If he elecrosaic force were sronger, hen nuclei would be less sable, and very large nuclei would probably no exis because he srong nuclear force would no be able o overcome he repulsion of he proons in a large nucleus. 3. Given r r 3 0A, where r0.0 fm radius of he nucleus of 90 38Sr aom (r) disance beween adjacen nucleons in 90 38Sr nucleus (d) From 90 38Sr, A 90. Subsiue his value ino he equaion for nuclear radius. The radius of he nucleus of he sronium-90 aom is: 7

8 3 r (.0 fm) fm The disance d is wice he radius of he space occupied by one nucleon, V nucleon. To esimae V nucleon, firs deermine he volume of he nucleus and divide i by 90, he number of nucleons in he nucleus. This gives he volume of he space occupied by one nucleon, V nucleon. The volume of he nucleus is: 4 3 V r 3 4 [(.0 fm)90 3 ] (.0 fm) V Vnucleon A 4 (.0 fm) (.0 fm) 3 3 Bu, 4 3 Vnucleon rnucleon rnucleon (.0 fm) 3 3 r.0 fm nucleon d rnucleon.0 fm ~ fm The radius of he 90 38Sr aom is abou 5.38 fm. The cenres of he nucleons are, on average, abou fm apar. This number ses an upper limi on he possible size of he individual nucleons. The acual size of he proons and neurons mus be less han his value. Concep Check Suden Book page 797 Apply he righ-hand rule for a posiive charge. The humb poins in he direcion of charge moion, o he righ. The fingers poin ou of he page, in he direcion of he magneic field. The palm faces down, oward he boom of he page, and represens he direcion of he magneic force. Since he alpha paricle moves oward he boom of he 8

9 page in Figure 6.5, in he same direcion as he magneic force, i mus have a posiive charge. The bea paricle is defleced in he opposie direcion, oward he op of he page. I mus herefore have a negaive charge, according o he lef-hand rule. The alpha paricle is defleced much less han he bea paricle, so i mus be considerably more massive han he bea paricle. The gamma ray appears o be neural because i is no defleced. Example 6.5 Pracice Problems Suden Book page 799 Po Rn Charge: Aomic mass number: This equaion violaes he laws of conservaion of charge and of aomic mass number. Therefore, his decay process is impossible Pa U Charge: 9 = 9 Aomic mass number: 33 = 33 This equaion is correc boh mass number and charge are conserved. Therefore, his decay process is possible C N H Charge: Aomic mass number: This equaion violaes he laws of conservaion of charge and of aomic mass number. Therefore, his decay process is impossible. Concep Check The conservaion law applies o nucleons, which make up mos of he aom s mass. Elecrons are no nucleons. Example 6.6 Pracice Problems Suden Book page 800 A A4 4 All alpha decay processes fi he paern ZX ZY For horium-30, he daugher elemen is 90 Y 88 Y. On he periodic able, he elemen wih Z = 88 is radium Th 88Ra For uranium-38, he daugher elemen is 9 Y 90 Y. On he periodic able, he elemen wih Z = 90 is horium U 90Th 3. For polonium-4, he daugher elemen is Y 8 Y. On he periodic able, he elemen wih Z = 8 is lead Po Pb

10 Suden Book page 80 Concep Check Alpha decay will only occur if he nucleus can achieve a lower energy by emiing an alpha paricle. The alpha paricle removes energy from he nucleus, so ΔE mus be posiive. Use he mass-energy equivalence o undersand his concep. The mass-energy of he paren nucleus equals he mass-energy of he daugher nucleus and he alpha paricle, plus he kineic energy of hese wo paricles, according o he equaion mparenc mdaugherc m c E where ΔE is he kineic energy of he daugher nucleus and of he alpha paricle. Example 6.7 Pracice Problems. Given Th energy released during -decay (ΔE) A A From ZX ZY, he α-decay process for horium is 90Th 88Ra (see Example 6.6 Pracice Problem ). The energy released is equivalen o he difference beween he mass of he paren aom and he oal mass of he producs. mmparen mproducs m30 ( m6 m4 ) 90Th 88 Ra u u u u J 93.5 MeV E u u u or u J MeV 3 A horium nucleus releases J or MeV of energy when i undergoes α-decay.. Given 38 9 U energy released during -decay (ΔE) Deermine he decay process of uranium-38 o find he daugher nucleus. From Example 6.6 Pracice Problem, he decay equaion for uranium-38 is U 90Th, so he daugher nucleus is horium-34. The energy released is equivalen o he difference beween he mass of he paren aom and he oal mass of he producs. 0

11 mmparen mproducs m38 ( m34 m4 ) 9 U 90Th u u u u J 93.5 MeV E u u u or u J 4.70 MeV 3 Uranium releases J or 4.70 MeV of energy when i undergoes α-decay. 3. Given 4 84 Po energy released during -decay (ΔE) Deermine he decay process of polonium-4 o find he daugher nucleus. From Example 6.6 Pracice Problem 3, he decay equaion for polonium-4 is Po 8Pb, so he daugher nucleus is lead-0. The energy released is equivalen o he difference beween he mass of he paren aom and he oal mass of he producs. mmparen mproducs m4 ( m0 m4 ) 84 Po 8 Pb u u u u J 93.5 MeV E u u u or u.550 J MeV When polonium undergoes α-decay, i releases.55 0 J or MeV of energy. Suden Book page 80 Concep Check Free neurons are unsable and decay ino a proon, elecron, and ani-neurino. In addiion 3 o creaing hese hree paricles, abou.0 J of energy is released (as kineic energy). By using mass-energy equivalence, his iny bi of energy explains why he neuron has a slighly higher res mass. The ypical lifeime of a free neuron is abou 900 s. Example 6.8 Pracice Problems Suden Book page 803 A A 0. Use he paern ZX ZX. The aomic number mus increase by one wihou changing he aomic mass number.

12 (a) (b) Ra Ac Pb Bi Example 6.9 Pracice Problems. (a) Given cobal-60 nucleus producs of decay Idenify he correc srucure for cobal-60 and hen apply he decay paern. A A 0 ZX ZX Co 8 Ni + The decay of cobal-60 produces 60 8 Ni. (b) Given cobal-60 nucleus Co 8 Ni + (from par (a)) energy released by decay ( E ) The energy released is equivalen o he difference beween he mass of he paren aom and he oal mass of he decay producs. Use he equaion mmparen mproducs, hen conver his value o energy using Einsein s mass-energy equivalence principle. mm m E paren producs m60 m60 + m 0 7 Co 8 Ni m60 m60 7 Co 8 Ni u u u 93.5 MeV u u.83 MeV The decay of a cobal-60 nucleus releases.83 MeV of energy. Concep Check Suden Book page 804 The discrepancy in he bea decay ha led o he discovery of he neurino was a small amoun of missing energy (hence mass hrough he mass-energy equivalence). Because

13 charge is conserved in his decay, he neurino does no change he charge. Therefore, i mus be neural. Example 6.0 Pracice Problems Suden Book page 805. (a) During decay, he aomic number decreases by. The aomic number, Z, of hallium-0 is 8. When he aomic number of hallium-0 decreases by, i becomes Z = 8 = 80, which is he aomic number for mercury-0. A A 0 (b) Use he general paern ZX ZY for decay: 8Tl 80Hg (c) Given decrease in mass of he hallium nucleus = u energy released in he decay ( E ) The energy released is equivalen o he difference beween he mass of he paren aom and he oal mass of he producs. mmparen mproducs m ( m m ) Tl 80 Hg m ( m m m ) Tl 80 Hg m ( m m ) Tl 80 Hg You are given he decrease in mass of he hallium-0 nucleus, so he equaion can be wrien as m ( m0 m ) 0 m 0 8Tl 80 Hg u ( u) u Now use he equaion for mass-energy equivalence MeV E u u MeV The energy released by he decay of hallium-0 is MeV. Concep Check Suden Book page 807 Gamma decays canno be shown as pahs on a decay series graph because gamma decay does no change eiher he aomic number (Z) or he aomic mass number (A) of an aom. Gamma decay is he resul of nucleons moving o lower-energy saes and emiing gamma-ray phoons. 3

14 Suden Book page Check and Reflec Knowledge. The hree basic radioacive decay processes are: alpha decay, where he nucleus emis an alpha paricle and is aomic number changes from Z o Z bea decay: (a) decay, where he nucleus emis an elecron and an anineurino, and is aomic number increases by (b) decay, where he nucleus emis a posiron and a neurino, and is aomic number decreases by gamma decay, where he nucleus drops from an excied sae o a lower energy sae and emis a gamma-ray phoon. The heavies sable isoopes end o have a slighly higher number of neurons han proons, so he raio is roughly 3: (see Figure 6.0 on p. 806 of he suden ex). A A (a) Use he paern ZX ZY o ge 9Pa 89Ac. (b) The paren nucleus is proacinium-34 and he daugher nucleus is acinium (a) Since Z increases from 6 o 7, his ype of bea decay mus be decay (b) 6C 7N 5. (a) Since Z decreases by, he decay mus be decay. 0 (b) Na 0 Ne 6. The emission of he bea paricle leaves he daugher nucleus in an excied sae. The gamma-ray phoon is emied as he daugher nucleus drops ino a lower energy sae. 7. Of he hree main radioacive decay processes, gamma rays usually have he greaes peneraing power. Applicaions 8. Given Na decays o produce 0 Ne. m Na = u m Ne = u energy released by he decay process ( E ) The aomic number decreases by, so Na undergoes decay o produce 0 Ne 0 according o he equaion Na 0 Ne. Find he mass defec, Δm, using he equaion mmparen mproducs, and conver i o is energy equivalen by using he conversion facor u = 93.5 MeV. 4

15 mm m E paren producs m m m 0 Na 0 Ne m m m m 0 0 Na 0 Ne m m m 0 Na 0 Ne u u ( u) u 93.5 MeV u u.89 MeV The decay of sodium ino neon liberaes.89 MeV of energy. 9. Yes, aomic number can increase during nuclear decay. A decay causes aomic number o increase by one. For example, carbon-4 can decay o nirogen-4, according o he decay process 6C 7N. 0. You receive abou 300 μsv of radiaion from naural sources each year and abou 75 μsv from medical and denal sources. Hence, you receive on average abou four imes more radiaion from naural background sources han from medical procedures.. Decay process Decay ype Paren elemen Daugher elemen (a) 90Th 88Ra* alpha decay horium radium 0 (b) Na 0 Ne decay sodium neon 8 8 (c) 88Ra* 88Ra + gamma decay radium radium (d) 88Ra 89Ac decay radium acinium (e) 89Ac 90Th decay acinium horium (f) 90Th 88Ra alpha decay horium radium (g) 0 p 0n decay proon neuron Exensions. (a) The absorpion of an elecron causes a decrease in he aomic number. The elecron combines wih a proon in he nucleus o form a neuron. A 0 A (b) ZX ZY (c) Elecron capure is similar o bea decay because boh processes change an aom from one elemen o anoher. However, during elecron capure, an elecron is absorbed, whereas during bea decay, an elecron (or a posiron) is emied. 3. One possible experimen would be o bombard a nucleus wih neurons in an aemp o excie he nucleus and cause i o emi gamma-ray phoons when he nucleus reurns o he ground sae. If he nucleus behaves in a manner analogous o he aom, you should see disinc energies given off in he form of gamma-ray phoons of specific wavelenghs, similar o he brigh line specrum seen in elecron ransiions. 4. Radon is a naural radioacive decay produc of radium. Since radium occurs naurally, so does radon. Concree conains race amouns of radium and hence can 5

16 also produce race amouns of radon. Since radon is a noble gas and canno be held by chemical bonds, i can migrae hrough he cemen (or soil). Furhermore, because i is heavier han air, radon can accumulae in a basemen. Such accumulaion is a healh concern because radon is iself radioacive i undergoes alpha decay. If you inhale radon, you run an increased risk of absorbing an alpha emier ino your lungs or bloodsream, which poses an increased risk of cancer. Suden Book page 8 Example 6. Pracice Problems. Given = s N =.0 0 aoms aciviy (A) Use he equaion A = λn and subsiue he given values. 9 A (4.0 s )(.00 ) s Bq The negaive sign means ha he number of cobal-60 nuclei is decreasing. The sample has a decay rae of Bq.. Given N = aoms A =.50 0 Bq he decay consan ( ) Use he equaion A N (he negaive sign signifies decay). A N.500 Bq aoms s The decay consan for his sample is s. Example 6. Pracice Problems. Given =.6 s Suden Book page 83 6

17 ime required for 99% of asaine in a sample o decay () Use he equaion N N 0. Re-wrie as N N 0 and le : N N 0 Since 99% of he sample has undergone decay, 0.0 N 0.0 N. Choose values for α and plug hem ino your calculaion unil α = 6.64 Solve for : (6.64) (6.64)(.6 s) 0.6 s I akes only abou s for 99% of a sample of asaine-8 o undergo radioacive decay.. Given = 600 years = 8000 years N he percenage of a sample of radium-6 remaining afer 8000 years N0 Use he equaion 0 N N N N 0. 7

18 N N % Afer 8000 years, only 3.5% of he original radium-6 sample will remain. Example 6.3 Pracice Problems. Given = 9. years = 00 years Suden Book page 84 proporion of he sample remaining afer 00 years Use he equaion N N N N 0. N N = % Afer 00 years, 9.4% of he original amoun of sronium-90 remains.. Given =.3 years m = 00 mg = 5.0 years N proporion of riium remaining afer 5.0 years N0 8

19 Use he equaion N N N N % 75% of 00 mg is 75 mg. Afer 5.0 years, 75%, or 75 mg, of riium will remain.. Suden Book page 86 Concep Check Mos maerials weakly absorb bea radiaion. The amoun of absorpion depends on he hickness of he absorber. Therefore, by measuring he amoun of radiaion emied by a known source of bea radiaion, he hickness of he absorber can be deermined. Gamma radiaion can easily penerae mos maerials, and i akes very large hickness differences o produce an appreciable difference in he absorpion of gamma rays. Because of is grea peneraing qualiy, gamma radiaion can be used o provide a deailed image of he inerior of such objecs as meal pipes and panels, o help reveal flaws or cracks in srucures. Suden Book page Check and Reflec Knowledge. For each half-life, one-half of a sample undergoes decay while he oher half remains. So, afer four half-lives, here remains of he original amoun of 6 radioacive maerial.. Given = s N = aoms aciviy of he sample (A) Subsiue he given values ino he equaion A N. 9

20 AN s s The radioacive sample decays a a rae of aoms/s. 3. Since boh samples have he same mass, assume ha hey have roughly he same number of aoms. The firs sample has a half-life ha is much shorer han ha of he second sample. Therefore, he firs sample mus undergo more radioacive decays per second, and hence have a greaer aciviy, han he second sample. Applicaions 4. Given N N 6 0 = years he age of he rock sample () Use he equaion N N N N (3.0 0 years) 6. 0 years The rock sample is approximaely. million years old. 5. Given =.6 h = 4 h he proporion of he racer remaining in a paien afer 4 h N N0 0

21 Use he equaion N N N N Afer 4 h, only.7 0. ( 0.7%) of he original quaniy of he racer remains. 6. Given N N = 5% = 0 4 he age of he arrow () Assume ha he arrow has one-quarer of is original carbon-4 conen. Use he equaion N N 0. Look up he half-life of carbon-4 in a source. I is 5730 years. N N 0 4 (5730 years) 460 years 4.0 years The arrow appears o be wo half-lives old. 4 The arrow is approximaely. 0 years old. 7. Given A =.5 MBq = h = 0.5 d

22 8. he aciviy of he sample one week (7 d) laer (A) The aciviy depends on he number of radioacive nuclei presen (A = λn), bu N depends on ime, according o he equaion Le he original aciviy be A 0 = λn 0. A A N N 0. Therefore, AN0 6 (.50 Bq) 53 Bq.50 Bq Afer one week, he aciviy will have dropped from Bq o jus.5 0 Bq. Time (h) Aciviy (decays/min) (a) The aciviy drops from 3000 decays/min o 500 decays/min in abou 4 h, so he half-life is abou 4 h. (b) Given A = 307 a = h = 4 h aciviy a = 0 (A 0 )

23 From quesion 7, A A 0. A A A = 0, he aciviy of he sample is 3600 decays/min. Exensions 9. Carbon-4 is useless for daing exremely old arefacs because i has a half-life of only 5730 years. A 65-million-year-old objec would be abou 84 carbon-4 halflives old! If you sared wih one mole of carbon-4, hen afer 84 half-lives, here would be no carbon-4 aoms lef! You should be suspicious because carbon-4 could no have been used o dae his bone. 0. For Irradiaion kills baceria, hereby prolonging he shelf-life of foods and allowing hem o be ranspored longer disances wihou spoiling. Agains Irradiaion may cause changes o he geneic srucure of food. The major argumen agains irradiaion is ha is long-erm effecs are unknown.. (a) Depleed uranium is uranium lef over from spen fuel rods or from decommissioned nuclear weapons. (b) Uranium is a very heavy, dense meal. Is high densiy makes i very effecive a piercing armour. I is also useful as ballas because i akes up less room while sill providing much needed weigh (due o is high densiy). (c) There are a leas hree major concerns wih he use of depleed uranium. (i) Depleed uranium is a heavy meal and herefore presens he same kinds of healh concerns ha oher, including non-radioacive, heavy meals pose for healh, including nervous sysem disorders. (ii) Depleed uranium is sill radioacive. Exposure o i poses he same healh risks as exposure o any radioacive subsance does. (iii) Depleed uranium can be physically changed by impac ino a dus, especially when used in weaponry. The dus, when airborne and inhaled, poses an even greaer healh risk. Concep Check Suden Book page 88 If he binding energy per nucleon increases, hen he nucleons are more ighly bound o each oher. In order for nucleons o bind more ighly, hey mus each lose energy. 3

24 Consequenly, a nuclear reacion ha increases he binding energy per nucleon releases energy. As an analogy, imagine ha you dropped your backpack while hiking and i rolled down he side of an embankmen, ending up 00 m below you. Your backpack is now more ighly bound o Earh because i is closer o i, and i will ake even more energy o bring i back o your elevaion. Binding energy represens he energy needed o separae nucleons from each oher. The nucleus wih he highes binding energy per nucleon is iron, which makes i he mos sable of nuclei. Suden Book page 89 Example 6.4 Pracice Problems. Given Iniial mass: 35 9U plus one neuron Final mass: Zr, 5Te, and hree neurons energy released ( E ) Use he aomic mass daa on p. 88 of he suden ex o calculae he ne change in mass resuling from he reacion. m m m i 35 9 U n u u u mf 40Zr 5Te 3 0n u u + 3( u) u mi mf u u u Use mass-energy equivalence o calculae he energy released. u is equivalen o 93.5 MeV/u, so 93.5 MeV E u u 7.9 MeV The nuclear fission of U-35 ino ellurium and zirconium releases 7.9 MeV of energy U 0n 35Br 57La? 0n To deermine he number of neurons released, balance he aomic mass numbers on boh sides of he equaion: 35 + = ?? = 3 Therefore, he balanced equaion for he reacion is U n Br La 3 n

25 3. Given Iniial mass: 35 9 U plus one neuron Final mass: Br, 57La, and hree neurons energy released ( E ) Use he aomic mass daa on p. 88 of he suden ex o calculae he ne change in mass resuling from he reacion. 35 mi 9U mn u u u mf 35Br 57La 3 0n u u + 3( u) u mi mf u u u Use mass-energy equivalence o calculae he energy released. u is equivalen o 93.5 MeV/u, so E u 93.5 MeV u 67.8 MeV This reacion releases 67.8 MeV of energy. Suden Book page 80 Example 6.5 Pracice Problems. (a) Given E = 600 MJ d = 500 km he mass of gasoline (m) The chemical energy in gasoline is J/kg (from Example 6.5). Divide he given energy (600 MJ) by his value J 36 kg 7 J kg I akes 36 kg of gasoline o provide 600 MJ of energy. (b) Given E = 600 MJ d = 500 km 5

26 mass of U-35 required o deliver he same amoun of energy (m) Do he same calculaion as in par (a) using he energy conen for U-35: J/kg (from Example 6.5) J 5.30 kg 3 mg 3 J 7.00 kg To ravel 500 km in an average family car would require only 3 mg of uranium- 35! Suden Book page 8 Concep Check The small nuclear dimensions mean ha he de Broglie wavelenghs of nucleons are much smaller han he de Broglie wavelenghs of elecrons in he aom. Since momenum depends inversely on he wavelengh, nucleons have much higher momena, and hence higher energy, han do he elecrons in he aom. The paricle-in-a-box example from Chaper 4 (Example 4.) could be used here as an analogy. Nucleons are confined o much smaller boxes, and hence mus have greaer energies. Anoher reason why nuclear reacions are more energeic has o do wih he srong nuclear force, which operaes only in aomic nuclei. I akes a grea deal of energy o overcome he srong force. Also, he elecric poenial energy of proons in a nucleus is enormous because he proons are packed so ighly ogeher. Changing nuclear srucure involves manipulaing hese forces, which can lead o enormous energy changes. Suden Book page 8 Example 6.6 Pracice Problems. (a) Given P = W number of helium nuclei produced per second Use he resuls of Example 6.6. Every fusion of four hydrogen aoms o a helium nucleus releases 6.7 MeV. Conver his value o joules and hen divide power (J/s) by energy o deermine he number of helium nuclei formed each second ev J/eV = J 5.60 J/s Rae J s The sar produces new helium nuclei each second by nuclear fusion. 6

27 (b) Given = 4 billion years mass of helium produced (m) Muliply he resul in (a) by he ime (4 billion years) o ge he oal number of he nuclei formed. 40 years days year 4 h day 3600 s s h The number of helium nuclei produced = ( s )( s) = Muliply by he mass of a helium nucleus, kg, o obain he oal mass of helium produced. The mass of helium produced = ( nuclei)( kg/nucleus) = kg Afer 4 billion years, his sar will have produced kg of helium. Suden Book page 83 Concep Check For fusion o occur, proons mus be close enough so ha he srong nuclear force can cause hem o fuse. In order o come sufficienly close, he proons mus firs overcome he repulsive elecrosaic force beween hem, called he Coulomb barrier. To penerae he Coulomb barrier, hey mus have very high kineic energy, ha is, hey mus be moving very fas, which is he same as saing ha hey mus be a a very high emperaure. For his reason, fusion reacions can occur only a exremely high emperaures. Suden Book page 84 Concep Check Triium is a radioacive isoope of hydrogen. If inhaled, i poses a serious healh concern because riium undergoes bea decay. If absorbed ino he body, riium becomes a longerm, low-level radiaion source ha can cause cellular damage, including geneic muaion. Because riium is a form of hydrogen, i can form chemical bonds wih oher elemens ha can be incorporaed ino organic molecules in our issues. 6.4 Check and Reflec Knowledge. (a) The oal number of nucleons mus add up o 35. There are 40 nucleons in he Xe nucleus, plus wo more neurons, so A = = 93 Also, charge mus be conserved, so 7

28 9 = 54 + Z Z = 38 From he periodic able, he missing elemen is sronium-93, Sr. (b) This reacion is a sponaneous fission reacion because a larger nucleus splis ino wo smaller fragmens.. In order for energy o be released during a nuclear reacion, he binding energy per nucleon mus increase. The release of energy means ha he nucleons in he producs of a nuclear reacion are more ighly bound per nucleon han he nucleons in he paren nucleus. 3. The combined binding energy of he iron and silicon nuclei is 49 MeV + 37 MeV = 79 MeV Because his value is greaer han he binding energy of he final nucleus, 78 MeV, his reacion requires energy, so energy will no be released. 4. (a) Heavy elemens (A > 0) will release energy in fission reacions because hey can spli ino nuclei ha each have binding energies per nucleon ha are higher han he binding energies of he paren nucleus. (b) Ligh elemens (A < 50) ha can combine o form a nucleus wih a higher binding energy per nucleon han he original nuclei are mos likely o undergo fusion. A A (a) Use he alpha decay paern, ZX ZY. Aluminium-7 is 7 3Al, so he absorpion equaion is 7 4 A 3Al ZY 0n = A + A = = Z + 0 Z = 5 From he periodic able, his elemen is phosphorus (b) 3Al 5P 0n Applicaions (a) He 8O 0Ne This fusion reacion produces neon-0. Since he neon-0 nucleus will be in an excied sae, a gamma ray will be emied. (b) Given helium-4 oxygen-6 energy released (ΔE) Use he nuclear process equaion and he aomic mass daa on page 88 of he suden ex o calculae he ne change in mass resuling from he reacion He 8O 0Ne m i = m He + m O = u u m f = m Ne = u 8

29 mi mf = u u u u Use mass-energy equivalence o calculae he energy released. u is equivalen o 93.5 ev, so 93.5 MeV E u u MeV The fusion reacion of helium-4 and oxygen-6 releases MeV of energy. 7. (a) Given Deuerium H and riium 3 H fuse o form helium 4 He. paricle emied Use he laws of conservaion of aomic mass number and of charge. 3 4 H H He? Since here is no change in charge bu a change in aomic mass number, a neuron has been emied. The fusion equaion is herefore 3 4 H H He 0n This reacion emis a neuron. (b) Given Deuerium H and riium 3 H fuse o form helium 4 He. energy released (ΔE) Use he aomic mass daa on p. 88 of he suden ex o calculae he ne change in mass resuling from he reacion. From 3 4 H H He 0n, 3 mi H H.04 0 u u 4 mf He 0n u u mi mf.04 0 u u ( u u) u Use mass-energy equivalence o calculae he energy released. u is equivalen o 93.5 MeV, so 93.5 MeV E u u 7.59 MeV The reacion produces 7.59 MeV of energy. 9

30 8. (a) Given P = 700 MW efficiency = 7% average energy release per fission = 00 MeV number of fission reacions per second Firs deermine how much nuclear power mus be generaed o produce 700 MW of elecrical power. 6 P W 9 P.60 W 9.60 J/s Divide he energy needed each second by 00 MeV per fission o ge he number of U-35 nuclei needed each second. 9 J.60 s s 6 9 J 000 ev.60 0 ev The number of nuclei ha undergo fission each second when he reacor is running 9 a full power is8. 0. (b) Given A nuclear reacor produces 700 MW of elecrical power. The average energy release per fission is 00 MeV and CANDU is 7% efficien. mass of U-35 used per year (m) Muliply he answer in (a) by he oal ime (one year) and by he mass per U-35 nucleus. In one year, he number of fission reacions ha occur is 9 d s = a s a d The mass of U-35 used is: 7 7 kg m u u kg The CANDU reacor will consume one onne of U-35 in a year o produce an elecrical power oupu of 700 MW. The assumpion made is ha only pure U-35 is used. In fac, he U-35 conen is much smaller, so a considerably greaer mass of nuclear fuel mus be used. Exensions (a) 6Fe 6Fe 5Te The elemen formed is ellurium-. 30

31 (b) Given m u 5 Te Show ha his reacion absorbs energy. Use he aomic mass daa on p. 88 of he suden ex o calculae he ne change in mass resuling from he reacion. m i = m 56 + m 6 Fe 56 6 Fe = u u = u m f = m 5 Te = u mi mf u u u Use mass-energy equivalence o calculae he energy released. u is equivalen o 93.5 MeV, so 93.5 MeV E u u 43.9 MeV A negaive answer means ha energy is absorbed. The fusion of wo iron-56 nuclei resuls in he formaion of ellurium-, which is heavier han he iniial wo nuclei. Therefore, his reacion absorbs 43.9 MeV of energy. (c) Sars similar o our Sun do no have sufficien mass o reach he core emperaures and densiies needed o cause he producion of iron, le alone he fusion of iron. The producion of elemens ha are heavier han iron occurs during supernova explosions. The Sun is no massive enough o undergo a supernova explosion. 0. (a) The major radioacive wase isoopes produced by nuclear reacors, and heir halflives, are lised in he able below. Wase Isoope Half-life cesium years sronium years uranium million years pluonium years (b) A curren shor-erm mehod of soring nuclear wase is he use of on-sie, sealed conainers. Highly radioacive maerials are sored in waer-filled sorage anks. A long-erm sorage mehod is virificaion, where radioacive maerial is sored in a glass-like marix as logs ha can be deposied in deep wells bored in a sable landmass such as he Canadian Shield. One proposed long-erm disposal mehod, called subducive wase disposal, would place radioacive wase in subducion zones along acive undersea fauls. The radioacive maerial would be subduced or absorbed in he magma layers of Earh s manle. 3

32 . Coal power (risk/benefis) Benefi: Coal provides a relaively low-cos energy source ha is available in many areas of he world and is easy o mine. Producion of power from coal is an old and proven echnology. Risk: Burning coal is a major source of greenhouse gases, which pose an environmenal risk. Risk: Coal combusion releases polluans ino he amosphere, including naurally occurring radioacive isoopes. Wihou cosly scrubbing of exhaus gas from a coal generaion plan, more radioacive maerial is inroduced ino he environmen han direc emissions from a nuclear power plan. Benefi: Coal is a safe energy source in poliically unsable regions of he world. Risk: Coal is a non-renewable resource ha requires large-scale pi or sub-surface mining. Boh mehods presen environmenal hazards. Nuclear power (risk/benefis) Risk: Nuclear power is a complex and cosly echnology o implemen. Incidens such as Three Mile Island in he U.S. and Chernobyl in he Ukraine remind us ha his mehod of power generaion is poenially dangerous. Nuclear accidens can pollue large regions of a counry wih wases ha have half-lives of ens o housands of years. Benefi: Nuclear power is a clean energy source wih few direc polluans inroduced ino he environmen (no including wase disposal, uranium mining, or a poenial nuclear acciden). Risk: Disposal of spen fuel rods and radioacive wase is a major environmenal concern for which no fool-proof long-erm mehod currenly exiss. Risk: Nuclear power generaion poses a number of poenial risks in poliically unsable regions of he world. Acs of errorism could lead o saboage of nuclear faciliies. As well, nuclear reacors can be modified o become breeder reacors for producing weapons-grade fissionable maerial. Benefi: Alhough uranium is also a nonrenewable resource, i is many imes more efficien as an energy source han coal combusion. Chaper 6 Review Suden Book pages Knowledge. No all nuclei conain more neurons han proons. For example, a hydrogen- nucleus conains no neurons, hydrogen- conains one proon and one neuron, and helium-4 conains wo proons and wo neurons. 3

33 . In mos cases, he aomic mass number for an aom is greaer han he aomic number. The excepion is hydrogen-, where A = Z. 3. Elemens wih he same aomic number bu differen neuron numbers are isoopes. 4. These nuclei are isoopes of uranium. Each aom conains he same number of elecrons and proons bu differen numbers of neurons. 5. For 5 55Cs, A = Z + N. N = 5 55 = 60 neurons Z = 55 proons 6. ev = J so 6 50 MeV 50 0 ev.60 0 ev 9 J J 7. Use he equaion E mc. 8 E (0.00 kg)( m/s) 3 90 J 3 One gram of maer is equivalen o 9 0 J of energy. 8. u = kg The energy equivalen of his mass is E mc kg m/s J 0.3 u J J 0 The energy equivalen of.3 u is J. 9. Use he conversion facor u = 93.5 MeV. u 300 MeV 0.3 u 93.5 MeV 0. Use binding energy as he energy equivalen of he mass defec. Since u = 93.5 MeV, 93.5 MeV 0.0 u 0 MeV u. Gamma decay processes do no change he aomic number of a nucleus.. (a) Bea paricles can be negaive (elecrons) or posiive (posirons). (b) An alpha paricle is a helium nucleus, so is charge is +. (c) A gamma ray is a high-energy phoon, so i has no charge. 3. (a) Alpha decay emis a helium nucleus, which removes wo neurons and wo proons from he nucleus. (b) decay appears as he emission of an elecron and he appearance of a new proon in he nucleus. decay appears as he emission of a posiron and he creaion of a new neuron in he nucleus. (c) Gamma decay does no change he number of nucleons. I represens he change in inernal energy of he nucleus. 33

34 K 0Ca This equaion is an example of decay due o he emission of an elecron and an anineurino. The paren nucleus is poassium and he daugher elemen is calcium. 5. Gamma radiaion is he mos peneraing. Alpha radiaion is he leas peneraing. 6. The mos imporan evidence involves he energy of radioacive decay producs. Typical radioacive decays are measured in hundreds of housands o millions of elecron vols. Aomic processes are many orders of magniude less energeic. Also, one elemen changes ino anoher elemen during radioacive decay. This is evidence ha radioaciviy originaes in he nucleus, where he proons are locaed. If ionizaion and redox reacions involved nucleons, chemical reacions would creae new elemens. 7. Radon exposure averages approximaely 00 Sv/year, compared o 73 Sv/year for denal X rays. Of hese wo sources, radon exposure is much more significan because radon is a gas ha, when inhaled, can be absorbed ino body issue. Radon decays via alpha paricle emission and remains in he body for long periods of ime, in close proximiy o vial organs. 8. Given N = =. 0 s Use he definiion for aciviy, N A N = (. 0 s )( ) = s = decays/s 9. Given N 0 = =.5 h = 6 h he number of nuclei remaining afer 6 h (N) The number of radioacive nuclei has decreased by exacly one-half in.5 h: So, he half-life,, of he sample is.5 h. Since 6 h = 4.5 h, he sample will decay by an addiional 4 half-lives. The number of radioacive nuclei remaining afer 6 h is:

35 Alernaively, use he equaion 6.5 N N 0 0 N The sample has a half-life of.5 h so, afer 6 h, i will have decayed from o 0 9 nuclei. 0. The half-life of carbon-4 is 5730 years. For his reason, carbon-4 is usually useful for daing iems ha are no older han years. Ages of mos rock samples are measured in millions of years. Also, carbon is no commonly found in mos rocks, excep for carbonaes.. Fission produces energy when a heavy nucleus splis ino wo smaller nuclei ha have greaer binding energy per nucleon han he original nucleus. This process is mos effecive for he mos massive nuclei, wih A > 00.. The fusion of hydrogen nuclei ino helium is he mos imporan energy source in sars. 3. The seps in he proon-proon chain are: i) hydrogen + hydrogen produces deuerium ii) hydrogen + deuerium produces helium-3 iii) helium-3 + helium-3 produces helium-4 plus proons Applicaions 4. (a) Given 4 He binding energy per nucleon (E b ) Use Δm = Zm H + Nm neuron m aom o find he mass defec, where Z = aomic number, N = neuron number, m H = mass of he hydrogen aom, m neuron = mass of neuron, and m aom = mass of he aom. From 4 He, A = 4, Z = N = A Z = 4 = m = u 4 He Δm = Zm H + Nm neuron m aom = ( u) + ( u) u = u Use he mass-energy equivalence o calculae binding energy from he mass defec. u = 93.5 MeV. 35

36 93.5 MeV E b = u u = 8.96 MeV Divide his number by he aomic number o find he binding energy per nucleon MeV MeV 4 The binding energy per nucleon in a helium aom is MeV. (b) Given 8 4 Si binding energy per nucleon (E b ) Use Δm = Zm H + Nm neuron m aom o find he mass defec, where Z = aomic number, N = neuron number, m H = mass of he hydrogen aom, m neuron = mass of neuron, and m aom = mass of he aom. Use mass daa from p. 88 of he suden ex. From 8 4Si, A = 8, Z = 4 N = A Z = 8 4 = 4 m = u 8 4 Si Δm = Zm H + Nm neuron m aom = 4( u) + 4( u) u = u Use he mass-energy equivalence o calculae binding energy from he mass defec. u = 93.5 MeV 93.5 MeV E b = u u = MeV Divide his number by he aomic number o find he binding energy per nucleon MeV MeV 8 The binding energy per nucleon in a silicon aom is MeV. (c) Given 58 6 Fe binding energy per nucleon (E b ) Use Δm = Zm H + Nm neuron m aom o find he mass defec, where Z = aomic number, N = neuron number, m H = mass of he hydrogen aom, m neuron = mass of neuron, and m aom = mass of he aom. Use mass daa from p. 88 of he suden 36

37 ex. From 58 6 N = A Z = 58 6 = Fe Fe, A = 58, Z = 6 m = u Δm = Zm H + Nm neuron m aom = 6( u) + 3( u) u = u Use he mass-energy equivalence o calculae binding energy from he mass defec. u = 93.5 MeV 93.5 MeV E b = u u = MeV Divide his number by he aomic number o find he binding energy per nucleon MeV 8.79 MeV 58 The binding energy per nucleon in an iron aom is 8.79 MeV. (d) Given 35 9 U binding energy per nucleon (E b ) Use Δm = Zm H + Nm neuron m aom o find he mass defec, where Z = aomic number, N = neuron number, m H = mass of he hydrogen aom, m neuron = mass of neuron, and m aom = mass of he aom. Use mass daa from p. 88 of he suden ex. From 35 9 N = A Z = 35 9 = U U, A = 35, Z = 9 m = u Δm = Zm H + Nm neuron m aom = 9( u) + 43( u) u = u Use he mass-energy equivalence o calculae binding energy from he mass defec. u = 93.5 MeV 93.5 MeV E b = u u = MeV Divide his number by he aomic number o find he binding energy per nucleon. 37

38 MeV 7.59 MeV 35 The binding energy per nucleon in a uranium aom is 7.59 MeV. I is useful o noe ha iron has he greaes binding energy of he four nuclei. 5. (a) Given 5 6 Fe β + decay process Use he basic form for β + decay: A A 0 ZX ZY Fe 5Y The elemen wih A = 5 and Z = 5 is manganese. Therefore, Fe 5Mn The β + decay of 5 6Fe convers iron ino manganese. (b) To demonsrae charge conservaion, noe ha 6 = = 5 + = 6. The aomic mass number remains a 5. β + decay urns a proon ino a neuron and yields a posiron in he process. Charge and aomic mass number are herefore conserved. 6. (a) Given α-decay o produce lead-08 he paren nucleus Use he sandard form for α-decay. A A 4 4 ZX ZY Le A 4 = 08 and Z = 8. A =, Z = 84 Therefore, he paren nucleus is polonium-. Polonium- decays ino lead-08 by emiing an alpha paricle. (b) Given α-decay o produce lead-08 kineic energy of alpha paricle (E k ) The energy released is equivalen o he difference beween he mass of he paren aom and he oal mass of he producs. mmparen mproducs Use he aomic masses from p. 88 of he suden ex. 38

39 mm ( m m ) Po Pb He u u u u u is equivalen o 93.5 MeV, so 93.5 MeV E u u MeV The approximae kineic energy of he alpha paricle is MeV. 7. (a) Given phosphorus is convered ino silicon decay equaion Wrie boh phosphorus and silicon in aomic form and inspec o deermine wha decay process has occurred. 30 phosphorus P silicon Si Phosphorus decays ino a silicon isoope. Since he aomic number decreases by, his process mus be β + decay P Si The complee decay process for he ransmuaion of P Si P ino 4Si is (b) Given phosphorus is convered ino silicon energy released (ΔE) The energy released is equivalen o he difference beween he mass of he paren aom and he oal mass of he producs P 4Si To deermine energy released, compare masses of paren and daugher nuclei. mmparen mproducs m ( m m ) P 4Si m ( m m m ) P 4Si Since elecrons and posirons have he same mass, mm30 ( m30 m 0 ) 5 P 4Si u [ u ( u)] u u is equivalen o 93.5 MeV, so 39

40 E 93.5 MeV u u 3.0 MeV The energy released during his decay is 3.0 MeV. 8. Given Arifacs conain he original amoun of carbon-4. 4 approximae age of arifacs The half-life,, of carbon-4 is 5730 years. Use he equaion N N0 4 N N 0. (5730 years) 460 years 4.0 years 4 The oldes archaeological sies in Albera appear o be approximaely. 0 years old. 9. (a) Given λ =.98 0 s A = 0.0 MBq number of radium aoms (N) Use he equaion A = λn. A N Bq.980 s (The negaive sign indicaes ha he number of aoms is decreasing.) The clock dial conains aoms of radium-6. (b) Given aoms of radium-6 (from par (a)) 40

41 mass of radium (m) The mass of a radium aom is 6 u, where u = kg. Muliply he mass per aom by he number of aoms kg 5 m 6 u 5. 0 u kg The clock dial conains kg of radium. (c) Given = 600 years = 5000 years λ =.98 0 s N 0 = aoms of radium-6 (from par (a)) aciviy afer 5000 years (A) Use he equaion A = λn, where A N N0 5 N N s (5. 0 ) 4.0 Bq Afer 5000 years, he aciviy of he dials will have dropped by abou a facor of 4 0 o abou. 0 Bq.. 4

42 30. (a) By inspecion, when = 5 h, he aciviy of he sample is 400 Bq. (b) To esimae he half-life, noe how long i akes he aciviy o drop from 400 Bq o 00 Bq. A = 5 h, he aciviy is 400 Bq. A = 9 h, he aciviy is 00 Bq. I has aken approximaely 4 h for he aciviy o drop by a facor of one-half, so he halflife is approximaely 4 h. 3. Given Three helium-4 nuclei combine o form a carbon- nucleus. energy released (ΔE) Use he aomic mass daa on p. 88 of he suden ex o calculae he ne change in mass resuling from he reacion. mi 3mHe 3( u) u mf mc u mi mf u u u Use mass-energy equivalence o calculae he energy released. u is equivalen o 93.5 MeV, so E 93.5 MeV u u 7.74 MeV The riple-α process ha convers helium ino carbon releases 7.74 MeV of energy. 3. (a) The key advanage of polonium over a chemical fuel is ha you will obain a much higher energy yield per uni mass by using nuclear decay raher han energy conversion hrough chemical means. 4

43 (b) Given P = 0 W = 4.5 years efficiency = 5% fuel = polonium-08 polonium-08 decays ino lead-04 = s =.9 years mass of polonium-08 (m) Since Po decays ino 8Pb, he paern is ha of an alpha decay he aomic mass numbers and he aomic numbers of he wo elemens differ by a helium nucleus. The energy released is equivalen o he difference beween he mass of he paren aom and he oal mass of he producs. mmparen mproducs Look up he masses of he elemens on p. 88 of he suden ex. m Po = u m Pb = u m = u mmpo ( mpb m ) u u u u Deermine he energy released during he alpha decay of polonium ino lead using he mass-energy equivalence, where u = 93.5 MeV MeV E u u 5.5 MeV Conver his value o joules. 6 9 J E 5.50 ev.600 ev J Calculae he oal number of decays needed o produce 0 W of elecrical power. Remember ha he energy conversion is only 5% efficien! 0 J/s Since you need 0 J/s of elecrical energy, you will need = 33 J/s of 0.5 hermal energy produced by he decay of polonium. Since each decay releases J, you require J 33 s decays per second J This value is he same as he aciviy of he polonium fuel afer 4.5 years. 43

44 The aciviy is A = λn, where λ is he decay consan, s, and N is he oal number of polonium aoms presen. Solve for N: A N s s.00 Since you require.0 0 aoms of polonium o be presen afer 4.5 years o ensure an elecrical power oupu of 0 W, you mus sar wih more polonium. To deermine he original number of polonium aoms required, N 0, use he equaion N N 0. N N Divide by Avogadro s number o find he number of moles of polonium required: aoms. mol 3 aoms 6.00 mol Since one mole of polonium has a mass of 08 g, he mass of polonium required g is. mol g 0.3 kg mol Powering he space probe for 4.5 years requires only 0.3 kg of polonium. This mass is much less han ha required if chemical fuels were used. Exensions 33. Sar wih he aomic noaion for all he nuclei involved. Make sure ha charge and aomic mass number are conserved. Remember ha mos elemens have isoopes. To deermine which reacion is more likely, look a he mass defecs (differences beween iniial and final masses). The smalles mass defec will also represen he reacion ha requires he leas energy. This reacion will be he more probable one. Soluion 4 4 7N He A A 8O H A and A need o be adjused so ha aomic mass number is conserved. A simple soluion is o leave A as 6 ( normal oxygen). Then, A =. The reacion is N He 8O H, which means ha deuerium is produced. 44