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EEE460_Hmwk02_Solution

EEE460_Hmwk02_Solution - 4 4,92 “me.“ v flunmwmmm...

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Unformatted text preview: 4% 4,92 “me...“ v. flunmwmmm “New” _\ MW \_-.‘M_M\_«.._m MW N‘Mmm New», MW“. A .c- c. “WWW 1. An accelerator increases the total energy of electrons uniformly to 10 GeV over a 3000 In path. That means that at 30 m, 300511, and 3000 m, the kinetic energy ; is 108, 109, and 1010 eV, respectively. At each of these distances, compute the "‘17“ velocity, relative to light (v/c), and the mass in atomic mass units. '. \ == 3-. --_ '- 2 , 02 __ . 2 EKMa-l-r c C‘Ao‘FaI L/‘em‘ M C M0 C Elfin ENC ._ m ‘ c1 ya (0 #rom Talc/e [,5 3 me 2 5,4858K10_4 :4;me rat/so raCaH +5ia+ 93/u5—M5V/amu I ‘ gé/OBeV// WW I . La) “9.69 v/Met/ (a3; S‘Wev + 0' 0005568 58 3.4 zxdflqjfiffl): = of??? @237 [07 8V mb‘ ' ' Eléb) m =6? b-WNeV 93/.5‘MeV) + (LOCO 54838 ' l 07% aqu ' /- (0°,‘i‘é‘;i‘fgs———-—--—§) flame? (7’97 237 2 %> + 0.000 S‘ei-BSB /‘ 0 720%:356 9‘ +0.“??? 799 9987 ,L/_ c H . Error ,1: 3. In fission reactors one deals with neutrons having kinetic energies as high as 10 MeV. How much'error is incurred in computing the speed of 10—MeV neu— trons by using the classical expression rather than the relativistic expression for kinetic energy? mfl 2 747+ 9272x/o"" k5 = “r39, 555:3? ’75“ L2 EKIWC+IC "’ (0 Me V>([/‘Z:T/V >(l éOQQ/(IO “,9 if) .= Aéozzx/o ”V /< , v Z 25" = (2 El 6022x103/‘QJ—l afl) _ I 67'4— 9272x10"27/<3 V = ‘7‘ 373017x/07 m/5 £6, [a +I\/(J‘+IC‘ “We 57—7“? J I (V :1 V \ ._. -— __ ._ LT 52 / a ([+FK—mdcl [63 MeV @3956 533’ C1 =(Q,99egIOSg—> ‘/~-'— 03*] 'v: 4,,339'5X'IO7m/3 Vc/Qs-S _ erla+ (Viz-)0 ‘70) Vrefcc'} : w5(mo 9%) a 0,795 (3/0 error Cal 01 91:22 :at— 9_ A 1 MeV photon is Compton scattered at an angle of 55 degrees. Calculate (a) the energy of the scattered photon, (b) the change in wavelength, and (c) the recoil energy of the electron. Gt) "EL .- 'L ,— L': +“j—’_ 2 mac [email protected]$>=ffl%a+aéfim/(Il—cos(§5§) {25’ = 0545 M-e\/ E’ = / Mav-— @9595 Mel/ Z: O.‘+SS Mel/2 . _ '— Chap—3 2. Consider an electron in the first Bohr orbit of a hydrogen atom. (a) What is the radius (in meters) of this orbit? (b) What is the total energy (in eV) of the _ electron in this orbit? (c) How much energy is required to ionize a hydrogen atom when the electron is in the ground state? I hqé .1, ‘ . 1 ___,_____£___._. (Kslnj E?‘ (394) In: fl‘mg Zea 8854/88x/0"/‘°_E. C): {‘_' 'J'I’I/J Value ijcred‘ bur-FL I +Ao+ oneaeven cm: Pad” 53 . _' 2 (9 use 5—21 (35> En= mag :1?“ (3 1402:7é5x/6""c ‘2 '- g e%o7</O“3”*J—s f ‘ ~t8___ f V - {2.‘1778x/O J "(flé2217é‘iléxll)"l J‘> ‘ I: .5— }3; 6.66 3V 4.- :/_L,75 vo/ue Cfgf‘é’eJ‘ cyH-Q “Fidel on 4415 op one pea? <54 3% 6 égzéo‘fx /0 3:5 (a/0938i9x10"3‘1<% (1)8 /,¢;o;2176+é><lo"9 '9 I0938qu10 3’1» Q1) 7:: {Oaize 4—146 elec‘;r‘o,q —p,-,~OM m: / 710 0(1) =/c/ 3. What photon energy (eV) is required to excite the hydrogen electron in the innermost (ground state) Bohr orbit to the first excited orbit? _.‘ ‘ __ “Me (263)“ We. use. "<7, (3‘5) En , W Z=i q- _. _ ~— -: “‘ Mae 5/ _ .1. Eexcfl‘e - EQ El 8 602/4 a 1:421 ”I :2 éf‘o unol- 54*u4'e INS i”), :1 Fl'rs+ end/(«Feel s+a+e I": I”).2 3': 2 A; _. = -_L CPI/Lg ('01:. . I V = 62 é3‘7‘9x lo ed :>([_____,__r________._‘ goglgés‘x/O‘WJD 6. From the difference in mass of a hydrogen atom (Appendix B) to the mass of a proton and an electron (Table 1.5), estimate the binding energy of the electron in the hydrogen atom. Compare this to the ienization‘energy of the ground state electron as calculated by the Bohr model. What fraction of the total mass is lost as the electron binds to the proton? fir (guy) = f» 0078230332} mp: [00727éefé3‘f m = 5u4es799x/o’4' =(m. +m¢>—' MALI—l) (L007: 76463??- 5148 579% 0"") - (I 007 82:03.21) = I 47K/0u8 amu :26 *7X/0 gomoflls” 5"“ aria 11/3. (593 evl From Problem «2 ancl +43 40,0 070 F073 54‘, +ke Ionl‘za—I-lon energy is [3 éOé eV _. Am ./.,47x/0“3 ””7625: Fracvt;0n Z—o.r7‘ m 1. Complete the following nuclear reactions based on the conservation of nucleons: (a) 23§U + on —* 0) (b) 1$N+on _4 (?) +lH ) ) (c 2gSRa——>(?)+ 4He (d (?) 233Th+§He 238 I r 2301 (i + ——> O mu m /‘m u (A) ’:N+ ’m —> ’H+/’:C ‘ , l 2. Determine the binding energy (in MeV) per nucleon for the nuclides: (a) 120, (b) £0, (c) SgFe, and (d) ZggU. Usfnj Eg. (41.4) 2 BEéiX> :[iMflI—I>[email protected]_Z> mn—M/ZAXYCQ M[{—/-l) = I 007825032 3/71,, : lUUUbe‘I|b (a) $=E8>04O0782503éb+(/ééigflaOOBééq-‘i/é)‘ Isi‘rqtifilsté 35545“ = W 7,974 MeV ' E§Cb>4g€=fie>fl4oo781303§+fl7_e>fl,ooeé44914>~1499985][7‘33”‘5 Cc) E=[(24)[ 007325035§+[s4 24)[ 009444914)— «=35 93%ng 343/5 5'6 :[84 790 MeV; G5 BE: :(‘mY 00782503455 (235.94% 008669-916) 22350439423 3;? ‘ 7 562.1 MW (e 5. A nuclear scientist attempts to perform experiments on the stable nuclide ggFe. Determine the energy (in MeV) the scientist will need to 1. remove a single neutron. 2. remove a single proton. 3. completely dismantle the nucleus into its individual nucleons. 4. fission it symmetrically into two identical lighter nuclides ngl. '56 r" 55.. 25":9 gon‘kggl—e (1th Eat (it 13) 1, +4; safgqrdxm 546,77}, ”gee/9C! ,7: $1") iZ—M :er +mVI- M ASE-Fey‘s}? = [5% 7382‘?8+ Looeéééfim— 55: 43549151] 93/5 ' S (3 5‘5. 1. -—? 1+ 26F}: ()0 .S' _ . :15 Mm . Usr‘nj EC} (Ll? I5>/ Me selvam‘l—foh ewe/<97 neeclecf 13‘ ‘ ‘I ' ‘ Sé ._ S, {—{M (EMOWC’QWQ Fe) 8 = [St-faweeré + /..oo78:zso3~ 55: ‘234-942 1 93/45 3" 11.0.18 Marv "FAG err-Hf: Ernd’fi energy oi 4478 HUG/fur can 4", 6:94am“! «prom Eye (*4 51) .BEi—Z M(}"H>+ (A'2>mn" M if Fail“ :1 [(éé){i,€>o782303)+(Sé~£é)fi,008654+91) . . , ~ 5 5.93474g/j 93.105 _=[- 4-92.26 Mgvl GI =[M (fife — a? M. ffAchZ- _ = -55257349c+.'2/—(£2)(972-z<?8i9102)]973L'S" ' . 6. Write formulas for the Q-values of the reactions shown in Section 4.4. With these formulas, evaluate the Q-values. :f/V>+ Maj ‘17” ((2:0>+ MP] €51 ;307‘F+q_e-02603j [/4 999/3i5+1007%i‘93/253 (d VIMC —-/,/‘1Me\/ @ g2“ {“0222} [m + ”(M/l]; ~j[_?2 0222102] “[0086654-i OC78Q57i(93i s>=[ -2 2:2 Nev V7 20> ‘30- (NW * _§E/oléoo<++roo7&gj~5oos3os§aal§> {Wow/find H 20200 00>: - 02 222:» >022] 22000224223 3 :55; 990mg] [OOBééS-l- 2500:2203 + Ixjiéms) =[—// 3/ mx/ 00 2200 20>000J [00,0 xx 'vaJic :?[S‘??tf9(§+/ ooeééfl—Eoofixsa— r4, ooé;o:j§[c73/ 53 ‘B I: E E .2 g“ egg 7. What is the Q-value (in MeV) for each of the following possible nuclear reac- tions? Which are exothermic and which are_endother.mic? Nani 4v mic! “QB + A, elécwlror/ §B+6n 1.?ch m 5126.7 .. e 4, hr} ZBe __) 2139+ :P 444 i; (m C ’ 4Be+ 1H H v :1 ~ _ ZBe + §H 01415qu 311i + 3H6 ‘@ Q = LMremam” flame-:7 c1 M547 H>+M/ 5?) Now? _L(zw752503I—90/3/8;{/_ MflmJszfig /0 0.26007/ llllll $53 = 6,5, 8 é 2X (3‘7E’AErn/H‘c/ . "4V - ’- /o 851’21740‘11'L/c2r‘I74/(C (a) Sigma, r‘eac c1479; aria/[Fro clue-1L: / 5‘0 Z 632:9 N L—- ; \ ; fi Cg) (Q =L/0C0220007i— MK? 6’s) ‘" ”7(1)” HDJCQ ‘3 =Eoiozooo7i—8m’205‘303i ~ 2,014 /0/ 81(623/15‘) @=> (’31 = L70,02<§oc>7/ ~ M (471%) _ M (3 HI] [793A 5-) =L76v0200071— :10 I é: 9.29:2 — Bio/50493:] (93/5) 2 — [52,083 Me C efla‘c¢germ,‘c (4”) (i =fiomozwo7/ _ M ($3 14')— M {Z 7%)] (93/5) =[j/0C02000‘7/ — éao IS i223 ,_ 4,002éo31i7 [93/45) £20 ['22 3 Me V _, e:<o%l78;'yVI/E’ H g; M _ c A ' .. . ’ Gt) (>4 s [fwiozmow —~ M 3»; 3>jczri7ca,o;zaoo~7/._ rag/:2 93:] (9345/11ch (15} Q ‘50 02(3007/ ~i‘4/j B)wm ‘ "3-[0 ogoowi-crc/ijgea-x 00866 gflflsj C: A74 I *_"L_ 9. What is the net energy released (in MeV) for each of the following fusion . reactions? (a) 11’H + %H —> SHe +(1,n and (b) ifH + §H ——» §He + 5n [Ora d Ga) 62 =. [:2 M(y—2)~(w(ye-3)+ mm] c9 2 [:2 (3‘0’4101é>bé001602931+LOOBééSQj 93/0 5 242:, : / 3g96‘7 Mev / a» 61 = [M(~52>+ ms) {mm mg] a = 2.014101% amaow ~(Laooaéo-324 1,008é63)j?315 531/ amv = r/‘Z‘scr Mev [email protected] Q zkm—zm C2 ve flea/“frf/ro fl; [“440 >5 r470 :7 {‘X a x Io‘f‘Dj $1215 go ...
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