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Solution Manual for Introduction to Nuclear Engineering, 4th Edition

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    Chapter 2

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    Downloaded from StudyXY.com ® + StudyXY Sd Ye. o> \ | iF ’ pr E \ 3 S Stu dy Anything This ContentHas been Posted On StudyXY.com as supplementary learning material. StudyXY does not endrose any university, college or publisher. Allmaterials posted are under the liability of the contributors. wv 8) www.studyxy.com

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    B 1. How many neutrons and protons are there in the nuclei of the following atoms: (a) "Li, fo oe (d) Bi, (e) Rn? CWE Nh) BR, N=-1350) 2esH N=81; (&) 2=03, NM=/34, (e) Z=54, N= 136. ng S 2/18 Sule aud LoxToime Ceri) x 2.6022 x10 = o. S22%10°% hou. -_— Tm eae te He msloili ii. gurkin aie WH WY on WY and 4%. =z 20) and pz) HM Jackin J tantly secuisiiq Kydigpue Hor Gil fr a Lage Volos of Lyin, : Frachsi Hat Ae a) << Ax 91) = (0.99985) %= 0.9997= 99.97%. Ho frocks. wird wept 4 oi A)xora) ~ 225x852, 250%. adgpdr 3 Can Ze Jrmat <i Fr wage. Me Jackin 73 <i M2) o7R) + 272)97) = RXV) = AX 0, 999F5x 0, 0004S = Zpaxrn¥= 6.03 %. Nase Mak fe frock Mat Frnt Dprbevcton wpb of elo 2, 3,004 ui YE) 30) #282) = (or) 272)% =.

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    7. A beaker contains 50 g of ordinary (i.e., naturally occurring) water. (a) How many moles of water are present? . (b) How many hydrogen atoms? (c) How many deuterium atoms? @) Numdew Fd Inploas = 53/08. 005 = 2, 7s #) Ab. 7 Welw = 2.7] 4p = 167) x106%F Hor me Row 9 H der yusliosls , 2s 7. 2 theme of H = 2X 16701003 3.3¢2 x 02 © "of 3 aMown = fxn Fx 3.3¢ax07F = 57013 x107° —_— a 9. Compute the mass of a proton in amu. 7 “Mass of fret = / 67 265 x5 -a7 Cc amu = 7+ Llox7 x/0 As. Maina. af frit Be amu —2 (ama) = [.L72L5k/0 Ks . 11CL 08 x 10" 2 fs fang, = [0073 ame 11. Show that 1 amu is numerically equal to the reciprocal of Avogadro's number. : Ans goster mandi, 6022 pp?! / _ 2622570 tomas JTS — = Jllosy Yk

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    13. Using Eq. (2.3), estimate the density of nuclear matter in g/om®; in Kg/m?. Take the mass of each nucleon to be approximately 1.5 x 10-2 g. _Yrp3. ¢ -13)3 V=4rR3< £7 (las x0 VA. 2. 3x00 ep fo} Vr (25)3 aut He sco dently 4 3x0 ay. ¥, 3 = 2XL A 2"%"= 1.94 x10 [2738 rye x saw el J = LGex10% Ay fy ® 15. The complete combustion of 1 kg of bituminous coal releases about 3 x 107 J in heat energy. The conversion of 1 g of mass into energy is equivalent to the burning of how ‘much coal? ! gadom pot rm Vo Ix (3x07)% = 9x er = Go’ jrutes . pom Tak. T..8, . ) T= 9 428x007 Btu. fue 8 ~ 9x3 9.478 Xx007% = $530 x 10/0 Blu. Mi 2 Mo eat se - Lael 8. 530,00"Y/13000 x3000= 3380 Lower of Coat. 17. Compute the ncutron—proton mass difference in MeV. Co Usuiy palise firms 7th Zl, 8M = Mp Mp =o C0$665 ~ 1.00777] = 0.00/38 Crus. sence (amar = 93149) MeV dM = 0.001388 x 931.48/ = 1. 293 Mal. mometee. SC © eR

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    19. Derive Eq. (2.18). [Hint: Square both sides of Eq. (2.5) and solve for mu.) h (rome %. (a.s) . Yo mE —— \ 1-v%et » z_ Ho. 1- Yc? me = met met 2 z » Bot mes Gp, pe’ Epp aut MVE, He, JY SE z _ 2 7: & y Shp" Cuat : 21. Using the result derived in Problem 2.20, calculate the speed of a 1-MeV electron, one with a kinetic energy of 1 MeV. Ph bw ick, Et = 0-SI/Y, Bud Sf bpatic butspy <a 11064] Mew Gp 457) 10 Y, ’ Brows fad. 2.44, : ve fi-e=) = 0.9¢/¢ = 2.83x10eassfonc. xe N ¢ =AEIRK IC 23. Show that the wavelength of a relativistic particle is given by Adel JE2y— Ely where Ac = h/m,c = 2.426 x 10~'° cm is called the Compton wavelength. © Pom G- (2.19), yeXe A met Ner-ev c 7 2 CT ER <p = “ \&-c= Et = Gut Study

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    25. An electron moves with a kinetic energy equal to its rest-mass energy. Calculate the electron’s (a) total energy in units of m.c%; (b) mass in units of m,; (c) speed in units of ¢; (d) wavelength in units of the Compton wavelength. T (@) Epp = ee? (B) me Gop = Qc’, aul M= Ine. (&) oy 214, a= cy 1-2 uy = o.86be. MC? (2) Fao jndb. 2.19, 2 A= De = Mee. 0.577dc Z 2 Name) (mec) 27. The first three excited states of the nucleus of Hg are at 0.158 MeV, 0.208 MeV, and 0.403 MeV above the ground state. If all transitions between these states and ground occurred, what energy y-rays would be observed? # yy © 7887, Ey, £ G.20FH UL I 0. ISENY &, [Tay I= °o LS N= 0.1sYrV J 2 = 00S He ¥y,= 0.1580 Vv XY, = 0.208 a Lb. 295) YT oto3ny Study

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    29. Tritium CH) decays by negative beta decay with a half-life of 12.26 years. The atomi weight of °H is 3.016. (2) To what nucleus does 3H decay? . (b) What is the mass in grams of 1 mCi of tritium? (=) Tht 3H pre a peilon. ad Lon mewdirmn, Boda ey erly C TAsnnfirnnl ion Ff oo 9 uls «Jeon plan ton elesron 5s Protest Lae R putin dl [sbi ofl Hon H — He +5~ 5) Abb, od, oo An. Fn Aude). 1- bd 2, PR AZ FT oe— 12.2b ou» 368 Her 2Vhen 005ec 2 “oy “hne = 1793x007 se B “3. 70 » K= L07CK 30D Lose 3.7%" digg pa r/R EE _ Fx 107s fee, /,293% 76" Tse! = 2.0 X00" ons Mara = My x an Tad: af

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    2 2.069) sc P85 pb (6022500 boy Ft = [03x04 F 31. Carbon tetrachloride labeled with C is sold commercially with an activity of 10 millicuries per millimole (10 mCi/mM). What fraction of the carbon atoms is 14C? Corben Close dtercd, ¢, Cely Tretorn adr cod fre adele. bo. Pit feo ors lonrt of Hour ou 6022570 dom font, x 107 aot, f pg = 2 ; Gore xo elon Cf, py Jf Fhe Said ondary 10, C lide, yy ne 3. 510" fre, r08Le Lar ® 372x007 dey, dat’, A Wor oL=QN. 2.453, dw i Tas s3g 5 Ei ZTE patient, A=3 p35 —z,. -) - a 10" “fee Nz > 3 sd = 4.40, 2.82 x 40 42 w/e 7 ade te) oud irs Putt ial, Tex" (or25p0¥ = Lexrtoy [67

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    33. After the initial cleanup effort at Three Mile Island, approximately. 400,000 gallons of radioactive water remained in the basement of the containment building of the Three Mile Island Unit 2 nuclear plant. The principal sources of this radioactivity were ¥7Cs at 156 uCi/cm? and Cs at 26 uCi/em?, How many atoms per cm® of these radionuclides were in the water at that time? 137, , / (Ps) = 15 eiffn3 ad hd 30. 2 GF ¢ LR 4.2 - Nisy® 3 —_— = 0-0230¢4 = 30. 2ynn 134 1324 ALG) 26pm fas £ = 2.06 yw, ‘ . La. ~ Nise Dot 2d 0.33454.” BY or ; Le PYJA7S A, 39 = i377 %,y, Nyy “137 = (1st xn0bc).. 3)(3, 7x00" 34, -&Y) ee ge ee in (0023040 =) Xx (% 15X70 ee ) Pp y (s = 7,90 % 167 aims fn 2 _ Ny V3 = i = (2¢ x10 Cha), Txp'y ) A124 — T= Ou 33GT yt 7 /4 5 xe’) —_ /3 = 9.00 x70 atoms),

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    35. Polonium-210 decays to the ground state of Pb by the emission of a 5.305-MeV a-particle with a half-life of 138 days. What mass of 2/%Po is required to produce 1 MW of thermal energy from its radioactive decay? 206 2p —= PL + 24 (°F )= tof disret [sec s = Leg nliancdd [ebnind =S, 20S Hel/ Bors =A E =P ol= Doss ”, = ila lio = ©" Xan€ 1, GR lad (Ida je ) 210 €'° 36th} TEOOIe 2 —5 — = 81 x/0 see”! pe lew — Ly i = Gr poe) (5.308 FV ) v vs c V4 10fW 13/5 vv Me Naan (Fr Gar Tae” 2/0 Mss "p= "Maso _ (2.03 70 ons ) (2705 /mole ) Stille isha NR (¢.0 2 x 76 Tatras! mole ) = 9.05% 07527 0F Ks.

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    37. Since the half-life of Bs (7.13 x 108 years) is less than that of 28U (4.51 x 10° years), the isotopic abundance of ZU has been steadily decreasing since the earth 2s formed about 4.5 billion years ago. How long ago was the isotopic abundae, FU equal 10 3.0 a/o, the earichment of the uranium used in many nuclear power plants? . 235 9 4 <7:/3% 70 gan - at] 2 x35 —10 J Ln. /e: = 9720/0; 23¢ ? 4, = IS x fon 2 235. 237 ~/0 ATELY gr IY16 4 23s 38 237 4 Aw, eT 23§, 238 25% 1% ea= 7,2% oF Qt to Gelgic abucdones “Li = 0.632) 237 237 23y a3 Sele 5 ad 21 3 5 n 275 kdl 23 Xr = rr = Ys 7 46:03 3Y., 1 2 2) Ne 42,7, 7% 7, . 23 = 03g BL 03 Ny e 5 5) z3 on 0 77 9= 037, => XT. 03. pp3093.teo rn 6.97 a Toby 0072 cl 5 — my ahi Tue 137 £38 VA 23Y = 0,992Fn 23 0.6012 » 238 7 = 0 F92Y_ [ EELS 2,007% 9,066228 Study

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    7%) 71 23 (17 a5) —_— En o 23% ET ¥ A (t) 2.5 ; 55° 1 Sy fn t / afl i + PE 23 n>, = 9 Pp / £7 0 ap - 15Y tp 9725/0 3 6.03057 Ex [77x00%, ae 39. Consider the chain decay A> B—C—, with no atoms of B present at t = 0. (a) Show that the activity of B rises to a maximum value at the time £,, given by 1 Ag w=nene(): at which time the activities of A and B are equal. (b) Show that, for t < ,, the activity of B is less than that of A, whereas the reverse is the case for t > 1,,. Chum &. 2.33 Ho ach of Ba gen Ty: ~ Tet 3 Any 3 ~t Qt oy < gp © EE [e wt At) Aga Assume ne whied torn of B dpa Ho ft Ferns din cdg, 20 am, Tho ahh of [8 50 6 morcomin

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    (rn mniimiares) Hoe dels ou (Ee durin wort, fu of 4g ail dots _ of), A Ot doe) TE Te 3 PA a 2 ; t= —— As ” PN (4 2) BD) Bn cb, dog, dro Stemot T ¢ £-1, dey _ “ SA pmakln—: Zr AE . a dng 8 a0 Hole fre 45 50 a PIR WR drs Mg # Palin = —c{p +25 Thier oly 3 oy For ley cg rac foe

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    41. shew i rk reid oy can be explained by assuming that this isotope orig- 23¢ 234 UL ans U ane nacdilon i Va Unies Sake, The adie ot fot pokuim EAS srl ong “uy, ri, 27% 7, f EE Po, TR mad TL The Lolly Use of Tu 4.51% 107, vs a hadf- Lore. of, Ho Ah, elope IRVIN Shanks CM thi, Ho diag cd of 7 iy Lisi Ma Bet diay cmkehs of tlw hte pesseds q4 Fase yo dan LI, fprorduh BY pur Uo Seoul Agplrios ok alt dwg kV nly Ly "UY 0 fom Hot . ANE Corr po Me = LEN [2% y = MNisy aay Nosy Grp I “poy Nz T, N,v ke31% s \ _ (9 Se Ty 2 2.47 x10 yy 2.4% x 16% gus TI omy = Sox ' [0 yv} From Mppudic 11 £040. 10.2 Ho order w 0.605F ale cok of TU 90.27% Ua nde Hor 5x07 Qlor YU oo be spied 3 71%;

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    43. According to U.S. Nuclear Regulatory Commission regulations, the maximum per- missible concentration of radon-222 in air in equilibrium with its short-lived daugh- ters is 3 pCifliter for nonoccupational exposure. This corresponds to how many atoms of radon-222 per cm’? An ach ff 30C [Lt ~ 320020 ut, Frome. Ho foi of C; = 3,7%¢0 eles fre es - -f Fh crnmlinBio € Se Cr hb guilds 111x107 gy I asf pty, fl SHO PPTRL achTuhy oy a LL Buk tu achiiby o Ag, Na. boo Ho 0 4 | es ~ AA lk odd ll Js ee wo = i ishe N r,. = | [ ( oA Shee 2. 2 = 205700 bei! Tl pont f omen “ S52 svof Le 45. Complete the following reactions and calculate their Q values. [Note: The atomic weight of C is 14003242.) (@) *He(p, db (b) *Be(e, n) (© “N(n, p) (d) "In(d, p) (e) *Pb(y, n) @) Tge cp) 3 2 Fhe +o —> d fre P Het , Q= Et rp )=Cer my D311 Use mats Ly olirten Ti My &s act fn ec om deteru = [002007 1.007925) — (3<0/(63 + 2014102) 93) =~1¥.FtHely .

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