Answer Key, Problem Set 10

Transcription

1 Chemisry 22 Mines, Spring 28 Answer Key, Problem Se. NT. Wrie an equaion describing he radioacive decay of each of he following nuclides. (The paricle produced is shown in parenheses, excep for elecron capure, where an elecron is a reacan.) (a) 68 Ga (elecron capure); (b) 62 Cu (posiron); (c) 22 Fr (); (d) 29 Sb () Answers: (a) (c) Ga -e 3 Zn (b) Cu 62 Ni e 22 Fr 28 A (d) Sb 29 Te 5 52 e Sraegy and Applicaion of Sraegy: In all cases, firs wrie ou he reacan nuclide on he lef and he produc paricle on he righ (excep for he elecron capure case, where i is a second reacan. Then figure ou wha he produc nuclide mus be using he wo conservaion rules (for mass number and charge). (a) 68 3Ga -e 68 3 Zn In elecron capure, he elecron is a reacan. Sum of lower lef subscrips on lef is 3 + (-) = +3, so i mus be so on he righ side as well (conservaion of charge). Since here is only one produc, 3 mus be he number of proons in he produc nuclide Zn. Mass number is 68 + on he lef, so i mus be 68 on he righ as well. In shor: Conservaion of Mass Number: 68 + = x x = 68 Conservaion of Charge: 3 + (-) = y y = +3 (b) 62 Cu e Cu 62 Ni e In posiron emission, he posiron (posiive elecron) is a produc. Conservaion of Mass Number: 62 = x + x = 62 Conservaion of Charge: 29 = y + (+) y = 28 Ni (c) (d) Fr Fr 28 A 22 4 or 4 2 2He is a produc. 87 Conservaion of Mass Number: 22 = x + 4 x = 28 Conservaion of Charge: +87 = y + (+2) y = +85 A Sb 29 5 e Sb 29 Te e is a produc, and is an elecron Conservaion of Mass Number: 29 = x + x = 29 Conservaion of Charge: 5 = y + (-) y = 52 Te 2. NT2. In each of he following nuclear reacions, supply he missing paricle. (a) 73 Ga 73 Ge + (b) 92 P 88 Os + (c) 25 Bi 25 Pb + (d) 24 Cm + 24 Am (e) 37 Ba 37 Ba + Answers: (a) e (b) 4 2 (c) e (d) - e (e) PS-

2 Answer Key, Problem Se Sraegy and Applicaion: In each case, I used he elemen symbols o find he elemens on he periodic able o ge he aomic numbers (#p s; subscrip). Then I used he conservaion rules as noed above (Problem 2). (a) 73 Ga 73 Ge + 73 Ga Ge e Conservaion of Mass Number: 73 = 73 + x x = Conservaion of Charge: 3 = 32 + y y = - Check: -emission is associaed wih a n p conversion. Mass number says he same, bu proon number increases by one. Tha fis wha happened here. (b) 92 P 88 Os + P Os Conservaion of Mass Number: 92 = 88 + x x = 4 Conservaion of Charge: 78 = 76 + y y = 2 Check: -emission is associaed wih a loss of 2 p s and 2 n s. Mass number decreases by 4, and proon number decreases by wo. Tha fis wha happened here. (c) 25 Bi 25 Pb + Bi Pb e 25 Conservaion of Mass Number: 25 = 25 + x x = Conservaion of Charge: 83 = 82 + y y = + Check: posiron emission is associaed wih a p n conversion. Mass number says he same, bu proon number decreases by one. Tha fis wha happened here. (d) 24 Cm + 24 Am 24 Cm Am Conservaion of Mass Number: 24 + x = 24 x = Conservaion of Charge: 96 + y = 95 y = - Check: Elecron capure is associaed wih a p n conversion. Mass number says he same, bu proon number decreases by one. Tha fis wha happened here. (e) 37 Ba 37 Ba + 95 e - Gamma emission does no change he ideniy of he nuclide. 3. NT3. (a) Wha is he zone (valley) of sabiliy? I is a graphical way o indicae which combinaions of proons and neurons resul in a nucleus ha is sable (and, by exension, hose ha are reasonably sable ) wih respec o sponaneous nuclear decay. I is ypically illusraed as a plo of #neurons vs #proons (N vs. Z), in which a poin appears on he plo if he combinaion resuls in a sable nucleus. I is called a zone (or valley) because he plo does no yield a line or a smooh curve, bu raher a zone (or valley) conaining many poins ha cluser ogeher in a paern ha resembles a curve wih widh (like a valley). (b) Sable ligh nuclides have abou equal numbers of neurons and proons. Wha happens o he neuron o-proon raio for sable nuclides as he number of proons increases? n/p (Tro wries his as N/Z) for sable nuclides ges larger as p increases [up o Z = 83]. I is ~. for ligh nuclides (Z ~2), bu increases somewha seadily, bu no linearly, o ~.5 for sable heavy nuclides (Z 8). (For example, when Z is ~5, a sable nuclide has an n/p raio of abou.3. (c) Nuclides ha are no already in he zone of sabiliy undergo radioacive processes o [ulimaely] ge o he zone of sabiliy. If a nuclide has oo many neurons, which process(es) can he nuclide undergo o become more sable? PS-2

3 Answer Key, Problem Se -decay (also called -emission) is he preferred mode of decay in his insance, because when a - paricle is produced, a neuron urns ino a proon, hus lowering he n/p raio. (Noe: If a neuron urns ino a proon, he number of neurons goes down by one uni while he number of proons goes up by one uni. Tha means n/p mus decrease (numeraor decreases and denominaor increases means value of fracion decreases.) (d) Answer he same quesion (as in (c)) for a nuclide having oo many proons. This quesion is acually a bi ambiguous. As such, here are wo answers. For nuclides ha really have oo many proons (i.e., hose wih Z > 83), he mos common resul is -decay. I call his being beyond he zone (or valley) of sabiliy, alhough Tro chooses no o sress (or even address) his for some reason. Losing an -paricle causes boh n and p o decrease, which brings he nuclide down and lef on he N vs. Z plo, which is in he general direcion of he zone of sabiliy if one is, in fac, beyond i. Perhaps Tro does no focus on his because a fair number of nuclides in his region also undergo -decay. However, I sill believe i is sill reasonable o predic -decay here. For nuclides ha have less han 83 proons, bu have oo many proons relaive o he number of neurons in a sable nuclide of ha Z value (i.e., an n/p raio ha is smaller han he nuclides in he zone of sabiliy, a siuaion ha I call being below he zone of sabiliy), posiron emission and elecron capure are he mos common opions. This is because when eiher of hese processes occurs, a proon urns ino a neuron, hus raising he n/p raio. (I should be poined ou ha -decay also occurs for a small percenage of nuclides below he zone of sabiliy. Noe ha if n/p >, he loss of 2 p s and 2 n s does raise he n/p raio.) 4. NT4. (a) The only sable isoope of fluorine is F-9. Predic possible modes of decay for F-2, F-8, and F-7. (b) Also predic possible modes of decay for 2 Po and (c) 95 Au. Answers, wih mos brief explanaions: (a) F-2, oo many neurons -decay (acual is F-8 and F-7, oo few neurons posiron emission or elecron capure (b) Po-2, Z > 83 -decay (acual is (acual is P.E. for boh) (c) Au-95, oo few neurons posiron emission or elecron capure (acual is EC Sraegy, General Approach (wih Explanaion): ) Look o see if he number of proons is greaer han 83. If i is, predic alpha decay (even hough his won' always be he acual decay). If p is less han (or equal o) 83, hen: 2) Firs calculae he n/p raio for he nuclide (alhough see Noe in #2 below for he absolue quickes approach, alhough you need o make sure you don forge wha an n/p raio is!). 3) Then assess he approximae "sable" n/p raio for he proon number of he nuclide ha you're dealing wih. (I.e., find ou "where you are" on he "zone of sabiliy" plo.) Noe: As I menioned in class (and as Tro menions on p. 873), here is anoher ( shorcu ) way o ge a rough idea of where he approximae sable n/p raio for a nuclide is. You can look a he average aomic mass of he elemen on he periodic able, round i o he neares ineger, and assume ha he nuclide wih ha mass number is likely o be sable! For example, I old you ha near Z = 83, he sable n/p raio is abou.5. If you look a Au, and round is average aomic mass of up o 97 and ake ha as he mass number of a sable nuclide of Au (which has 79 p s in every nucleus), he number of neurons in ha nuclide would be = 8, and hus he n/p raio would be.49, which is very close o.5! This works mos of he ime because he average aomic mass is he weighed average of he naurally occurring isoopes on Earh, which means hose isoopes ha are reasonably sable and hus lie in or near he zone of sabiliy. 3) Compare he acual n/p raio of each nuclide wih he "sable" n/p raio o see wheher or no i has "oo many neurons" (acual n/p larger han "sable" n/p) or "oo few neurons" (acual n/p smaller han "sable" n/p). Then you can make your conclusion: PS-3

4 Answer Key, Problem Se a) If here are "oo many neurons" o be sable, he nuclide is expeced o urn a neuron ino a proon, which is associaed wih bea decay. b) If here are "oo few neurons" o be sable, he nuclide is expeced o urn a proon ino a neuron, which is associaed wih eiher posiron emission or elecron capure. NOTE: Alpha decay is also a possibiliy since i will also ge you somewha closer o he zone of sabiliy, bu i is much rarer, and you need never predic i. Applicaion of Sraegy: (a) For he F example, clearly Z < 83 since Z = 9. So no -decays are prediced. There is no need o calculae an n/p raio since hey ell you ha he only sable isoope is F-9. Tha being said, F-2 mus have oo many neurons o be sable, and so -decay (n p) is prediced. And similarly, F-7 and F-8 mus have oo few neurons, and so posiron emission or elecron capure are prediced. (b) For 2 Po, he aomic number is 84, so -decay is prediced. (c) For 95 Au, he aomic number is 79 which is < 83, so no -decay is prediced. Using he shorcu mehod, he assumed sable isoope of Au is Au-97 since he average aomic mass is close o 97 amu. So Au-95, which has fewer neurons han in Au-97, is prediced o have oo few neurons and is prediced o decay by he process ha urns a p ino an n posiron emission or elecron capure. If asked o consider he n/p raio, you should be able o do so. Here, n/p = (95 79)/79 = is very close o he end of he line which is a Z = 83 and a which n/p is close o.5. Alhough you migh say his one is oo close o call, if pushed o guess, you d have o say ha n/p was oo small i clearly isn oo large! So again, you d predic he decay ha would urn a proon ino a neuron. 5. NT5. Complee he following nuclear reacions. [All have been used o synhesize elemens.] (a) 2He 97Bk n (b) 92U 6C 6 n (c) Cf Db 4 n (d) Cf B Lr 2 Answers: (a) Am (b) 98 Cf (c) 5 7 N (d) n Work / Reasoning (using conservaion rules as in Problem 2): (a) He Bk n Answer: Am Conservaion of Mass Number: x + 4 = x = = 24 Conservaion of Charge: y + 2 = 97 + y = 97 2 = 95 Am (b) U 6C 6 n Answer: Cf Conservaion of Mass Number: = x + 6() x = 25 6 = 244 Conservaion of Charge: = y + 6() y = 98 Cf (c) Cf 5Db 4 n Answer: N Conservaion of Mass Number: x = () y = = 5 Conservaion of Charge: 98 + y = 5 + 4() y = 5 98 = 7 (d) Cf 5B 3Lr 2 Answer: n 249 Conservaion of Mass Number: = x 2x = = 2 x = 2 Conservaion of Charge: = 3 + 2y 2y = 3-3 = y = PS-4

5 Answer Key, Problem Se 6. NT6. Radioacive copper-64 decays wih a half-life of 2.8 days. (a) Wha is he value of k in s -? (b) A sample conains 28. mg 64 Cu. How many decay evens will be produced in he firs second? (i.e., Calculae he iniial rae in decays per second!) Assume he aomic mass of 64 Cu is 64. (amu). (c) A chemis obains a fresh sample of 64 Cu and measures is radioaciviy. She hen deermines ha o do an experimen, he radioaciviy canno fall below 25% of he iniial measured value. How long does she have o do he experimen? Answers: (a) 6.27 x -7 s - (b).65 x 2 dps (c) 25.6 days Work/ Reasoning: (a) Wha is he value of k in s -? k ln.5 /2.693 / h 6 min 2.8 d x x d h 6 s x min x Remember: All nuclear decays are firs order, so he above relaionship applies. -7 s x (b) A sample conains 28. mg 64 Cu. How many decay evens will be produced in he firs second? Assume he aomic mass of 64 Cu is 64. (amu). Firs order R = kn. or A = kn Since we know k (par (a)), o ge R a any given ime, we jus need N a ha ime. In he firs second, N is he iniial value i.e., he number of nuclei in he 28.-mg sample of 64 Cu:.g 28. mg x mg (c) x mol 64. g x x mol aoms aoms Cu nucleus x aom A = x s x Cu nuclei.65x 4 dps (or cps) x Cu nuclei (a = ) A chemis obains a fresh sample of 64 Cu and measures is radioaciviy. She hen deermines ha o do an experimen, he radioaciviy canno fall below 25% of he iniial measured value. How long does she have o do he experimen? Explanaion, Quick Way (for his problem, because of a special case) For firs order nuclear processes, he half-life is a consan (because i is firs order). Thus, he ime i akes for he number of radioacive nuclei o become half of wha i was originally is one half-life, and he ime i akes for he number o become one half of ha, or ¼ he original, is wo half-lives. Since decay rae is proporional o number of nuclei, i also akes wo half-lives for he sample s decay rae (radioaciviy) o drop o ¼ (=25%) of is original value. So in his problem, he chemis has wo half-lives worh of ime o do her experimen. Since he half-life is given as 2.8 days, she has 2 x 2.8 = 25.6 days. Explanaion, More General: A more general soluion, paricularly useful when he ime is no an inegral number of half-lives, is as follows. Whenever you have a quesion ha relaes ime and amoun (or fracion) remaining, you should recognize ha he appropriae equaion o use is he inegraed rae law, and you should UNDERSTAND THE MEANING OF THIS EQUATION, as well as keep rack of unis (as always!): Fracion of radioacive nuclei remaining a ime N N o e NOTE: In his key, A is ofen used for rae because (radio)aciviy is ofen used as a synonym for rae of decay ). k Remember ha here are oher equivalen versions of his equaion. For example, Tro uses N ln N o k -7 s - Equaion [9.3] ALSO remember ha since he aciviy (decay rae) is proporional o he amoun of nuclide presen (A = kn), he raio of aciviies is equal o he raio of number of nuclei. Tha is: PS-5

6 Answer Key, Problem Se A kn N k A e or in oher words, e A kn N A o o o o This means ha he fracion of he aciviy remaining can be used in he inegraed rae law jus like fracion of nuclei remaining is. In his problem, he % remaining is 25%, which equals.25 in erms of a fracion (or decimal). Thus:.25 ln(.25) 6.27 x s ( ) e 6.27 x -7 s - ln(.25) = -(6.27 x -7 s - ) = x min s x 6 s hr day x x 6 min 24 hr k 2.2 x 6 s 25.6 days (which is he same answer as above [2 x /2]) 7. NT7. Fresh rainwaer or surface waer conains enough riium ( 3 H ) o show 5.5 decay evens per minue per. g of waer. Triium has a half-life of 2.3 years. Preend ha he year is 29 (raher han he curren year). You are asked o check a vinage wine ha is claimed o have been produced in 946. How many decay evens per minue should you expec o observe in. g of ha wine if he claim is rue? (Again, assume i is 29 righ now.) Answer:.6 dpm Reasoning/Work: One needs o look a his is as a varian of a Wha is? kind of problem. Tha is, ypically in his kind of problem we wan o dae somehing. So he quesion really becomes How long has i been decaying? The answer, which is a ime, ends up being an age. And ha ime is he in. 693 A k he inegraed rae law:. 693 / 2 e e, because k for firs order processes. In his problem, A o /2 however, we are no asked for ime, bu effecively are given he ime and asked o predic a value of aciviy. The same inegraed rae law applies. The only horny issues here are ha you have o make some assumpions o solve he problem. One is ha you have o assume ha. g of wine is essenially % waer, which is obviously no rue, bu is probably no oo far off (our bodies are mosly [bu no %!] waer as well!). Also, you mus assume ha he waer ulimaely go ino he wine via rainwaer or surface waer (because i came from grapes, which go heir waer from naure as he vines grew). The idea is ha once he wine is boled, no new waer ges added, so he riium in he waer jus decays as he wine sis. The aciviy will go away a a rae dicaed by is k (and hus half-life), and hus now (i.e., in 29) he rae will be lower han i was originally since here aren as many radioacive nuclei lef (since hey ve been decaying since 946 abou 63 years!). The aciviy now will be relaed o he original aciviy (which is assumed o be equal o he curren aciviy of fresh rainwaer) by he inegraed rae law, so you can plug ino ha equaion as follows o find he curren (29) aciviy (prediced):. 693 now / 2 e A A o A now e 5.5 dpm y 2 3 y. A 5.5 dpm now e y 2. 3 y.58.6 dpm 8. NT8 Assume a consan 4 C/ 2 C raio of 3.6 couns per minue per gram of living maer. A sample of perified ree was found o give.2 couns per minue per gram. How old was he ree? ( /2 of 4 C is 573 years.) Answer: 2 y Work/Reasoning: Use he inegraed rae law again, as in he prior wo problems. The raio of aciviies is equal o e -k : PS-6

7 Answer Key, Problem Se. 693 now / 2 e A A o.2 cpm 3.6 cpm e y.8824 e. 29 y - ln(.8824) = -(.29 y - ) = y - 28 y 2 y 9. NT9. A small aomic bomb releases energy equivalen o he deonaion of 2, ons of TNT; a on of TNT releases 4 x 9 J of energy when exploded. Using 2 x 3 J/mol as he energy released by fission of 235 U, (a) approximaely wha mass of 235 U undergoes fission in his aomic bomb? (b) Does he bomb have more mass or less mass afer he explosion? By how much (in mg)? x J mol U 235 g 235 (a) 2 on TNT x x x 94 g U on TNT 2 x J mol U (b) Answers: Less mass; ~9 mg less Reasoning: Because of he concep of mass / energy inerconversion (Einsein), when energy leaves a sysem, mass is los as well! We inerpre Einsein s E = mc 2 equaion o mean ha when energy leaves a sysem, i is said o have come from he conversion of a lile bi of is mass. Conversely, when energy is added o a sysem, mass is added. Any exohermic process mus be associaed wih a mass loss in he sysem; any endohermic process resuls in a main gain. The magniude of he mass loss/gain depends on he change in energy according o: E = mc 2. In his case, E = -2 on x (4 x 9 J/on) = -8 x 3 J (sign is negaive because process is exohermic), so: E c 8 x x J m/s 3 m kg g mg -.89 kg x x 89 9 mg 9 mg was los kg g The negaive sign indicaes ha 9 mg of mass were los (convered ino energy) when he bomb exploded (reaced). Tha s nearly.% of is mass! Tha is no insignifican!. NT. (a) Wha is mean by he erms binding energy and mass defec? Binding energy (Eb [my abbreviaion]) is he energy required o break apar a nucleus compleely ino free proons and neurons (or he energy released when a nucleus is formed from free proons and neurons). Mass defec is he mass increase ha occurs when a nucleus is compleely broken apar ino free proons and neurons (or he mass loss ha accompanies he formaion of a nucleus from free proons and neurons). NOTE: Tro s definiion is consisen wih mine, alhough he phrases i differenly, as he difference in mass beween a nucleus and he sum of he separaed paricles. (b) Describe clearly in words how you could calculae he binding energy for a nuclide using he precise masses of a nuclide as well as he mass of a (free) proon and a (free) neuron. Hin: wrie ou he nuclear equaion corresponding o he binding energy. Binding energy and mass defec are associaed wih he same process (described in par (a)). For a nuclide wih Z proons, and A Z neurons (A is he mass number), he equaion ha represens he process associaed wih boh BE and mass defec is: A Z X anzp (A - Z) n ; E b, de m mass defec Since Eb and m are associaed wih he same process, hey mus be relaed by E mc 2 for a given nuclide (In fac, his is how Tro defines (nuclear) binding energy). Thus, o calculae binding energy (if he mass of he nuclide is known), wrie ou he equaion corresponding o he break up of he PS-7

8 Answer Key, Problem Se nucleus of ineres ino is consiuen nucleons. Find he difference in mass beween he (sum of he masses of he) free nucleons and he mass of he nucleus. Then calculae binding energy using (E b ) E = mc 2 (being careful wih unis! ). NT. (a) Wha is he difference beween fusion and fission? (define each) Fusion refers o he process in which wo (or more) nuclei come ogeher o form one (larger) nucleus. Fission refers o he process in which one nucleus splis apar o form wo (or more) smaller nuclei. (b) If he binding energy of Ar-4 is MeV, wha is he binding energy per nucleon? E b nucleon binding energy number of nucleons MeV 4 nucleons MeV/nucleon NOTE: The number of nucleons equals he mass number (A), since A n + p (by definiion) (c) Wha is he difference beween binding energy and binding energy per nucleon? Which one helps you predic wheher or no a nuclide will undergo fission or fusion o lower is energy? Binding energy is defined in NT; he binding energy per nucleon is he binding energy for a paricular nuclide divided by he number of nucleons in he nuclide (see par (b) above). I is he binding energy per nucleon which indicaes how hermodynamically sable a nuclide is (compared o oher nuclides). If he binding energy per nucleon for a nuclide is smaller han he maximum in Figure 9.2, hen he nuclide is no as sable as i could be by eiher breaking up (if he nuclide has a mass number larger han ~56) or fusing ogeher wih anoher (if he nuclide has a mass number smaller han ~56). (d) Do larger nuclides always have larger binding energies? Do larger nuclides always have larger binding energies per nucleon? Answers: Yes, No Explanaion / Reasoning: Larger nuclides always have larger binding energies han smaller nuclides (because wih each exra nucleon added / bound, energy mus be lowered [because nucleons arac one anoher!]), bu hey do NOT always have he larges binding energy per nucleon. Afer a cerain poin, i becomes less sable (per nucleon) o coninue o ge bigger. This is wha is shown in Figure 9.2. There may seem o be a discrepancy here, bu here isn. Comparing he binding energy of one nucleus of U-238 o one nucleus of Fe-56 is like comparing apples o oranges since he number of nucleons is no he same. So even hough U-238 has a larger binding energy han Fe-56, i is no more sable because if he nucleons were rearranged ino four Fe-56 nuclei (plus some exra nucleons), energy would be lowered. How can you know his wihou acually calculaing he E for his hypoheical reacion? You jus look a he binding energy per nucleon and ha ells you his mus be rue! If he same number of nucleons is involved, i is clear ha he nucleus having he larger binding energy per nucleon will be he lower-energy (overall) arrangemen of nucleons. (because E/nucleon x # nucleons Eoverall). Bu even if he number of nucleons is no he same, he Eb per nucleon will reflec inrinsic sabiliy of a nucleus. LOOK AT THE POWERPOINT SLIDES IN WHICH I ASK WHAT IS THE MOST STABLE WAY TO ARRANGE 3 (OR 24) NUCLEONS TO HELP MAKE THESE IDEAS CLEAR TO YOU. PS-8

9 Answer Key, Problem Se (e) Of he nuclides in he able shown on he firs page of his PS shee, which one(s) are more hermodynamically sable han Ar-4? [see par (b)] (Noe: You may need o calculae some values omied from ha able.) Answer: Ni-6 is he only one, since i is he only one ha has a Eb/nucleon greaer han MeV/nucleon. Noe: You had o calculae he E b /nucleon values for hree of he nuclides in he able by dividing E b by he mass number ( # of nucleons) I pu hose in bold underlined ialics in he able o he righ. 2 and 3 H. The 2. NT2. (a) Calculae he binding energy per nucleon for H Nuclide Mass # E b (MeV) E b /nucleon He Be N Mg Ni Cd Cm aomic masses (in amu) are 2 H, 2.4, and 3 H, m elecron.549 amu; m p.728 amu; m n.866 amu Answers: 2 H,.78 x -3 J OR. MeV (per nucleon) Sraegy: 3 H, 4.53 x -3 J OR 2.83 MeV (per nucleon) ) Wrie ou he balanced equaion corresponding o breaking up he nuclide ino free nucleons. 2) Find m mnucleons - mnuclide. NOTE: When maom (aomic mass) is given, I find mnuclide by subracing he mass of he elecrons from maom: mnucleus [w/o e s] maom [w/ e s] - melecrons. This is how I did i in class. I find his more sraighforward han wha Tro does. Tro uses he mass of an H aom in place of he mass of a proon, and hen uses he mass of he aom (wihou subracing away he mass of he elecron(s)): m mh aoms + neurons - maom 3) Conver m (which will be in unis of amu [per nucleus]) ino kg [per nucleus] using: amu =.665 x -27 kg (o 5 SF) 4) Subsiue ino E = mc 2 (here, i ends up being Eb m.d. x c 2 ). Recognize ha unis will be J (if c is used in unis of m/s and m.d. in kg), and ha his is sill per nucleus. 5) For binding energy per nucleon, recognize ha you mus divide Eb by he number of nucleons. You can do ha a his poin if you wan he resul in J/nucleon. If you wan he resul in MeV/nucleon, you can conver J o MeV firs, and hen divide by he number of nucleons. (Acually, i doesn maer he order in which you do he conversion from J o MeV. I jus prefer o do my division by he # nucleons las.) Use: MeV =.622 x -3 J Execuion of Sraegy for 2 H: H 2 2 p n ; E Eb( H), and m mass defec(of H) 2 m.d. m m (p + n) - mh-2 nucleus ( ) ( ).2389 amu Conver o kg:.665 x.2389 amu x amu -27 kg x -3 kg (per nucleus) Eb m.d. x c 2 ( x -3 kg)( x 8 m/s) x -3 J (per nucleus) x J Eb per nucleon =.7825 x 2 nucleons OR (convering Eb o MeV firs): x J/nucleon PS-9

10 Answer Key, Problem Se Eb = -3 MeV x J x x J MeV (per nucleus) Eb per nucleon = MeV 2 nucleons MeV/nucleon (Noe: his maches he value in Fig. 9.2) Execuion of Sraegy for 3 H: 3 H 2 p n m.d. m m (p + 2n) - mh-3 nucleus ( (.866)) ( ).999 amu Conver o kg:.665 x.999 amu x amu -27 kg.58..x -29 kg (per nucleus) Eb m.d. x c 2 (.58.. x -29 kg)( x 8 m/s) x -2 J (per nucleus) x J -3-3 Eb per nucleon = x x J/nucleon 3 nucleons OR (convering Eb o MeV firs): -2 MeV Eb = x J x MeV (per nucleus) x J Eb per nucleon = MeV 3 nucleons MeV/nucleon (b) Which nuclide is more hermodynamically sable, 2 H or 3 H? Explain briefly. Answer: 3 H is more sable. Explanaion: A greaer Eb per nucleon means more hermodynamically sable. I represens an average amoun of energy lowering per nucleon in ha paricular nucleus, and lower energy (per nucleon) implies more sable. NOTE: In his case, he E b iself also happens o be greaer for 3 H bu ha is no why 3 H is more sable. One mus always compare E b per nucleon o assess relaive hermodynamic sabiliy of wo nuclei. PS-