Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang1 CHNG VT L HT NHN A: TM TT L THUYT 1. CU TO CA HT NHN NGUYN T- HT KHII.CU TO CA HT NHN NGUYN T 1.Cu ht nhn nguyn t :Ht nhn c cu to bi hai loi ht s cp gi l nucln gm: Ht s cp (nuclon) Ki hiuKhi lng theo kgKhi lng theo u 1u =1,66055.10 -27 kg in tch Prtn: H p11= mp = 2710 . 67262 , 1kgmp =1,00728u+eNtrn: 10n n = mn =2710 . 67493 , 1 kg mn =1,00866ukhng mang in tch 1.1. K hiu ht nhn:AZX -A= s nuctrn : s khi -Z = s prtn = in tch ht nhn (nguyn t s) -N A Z = : s ntrn1.2. Bn knh ht nhn nguyn t:11531, 2.10 R A= (m) V d: + Bn knh ht nhnH11H: R = 1,2.10-15m + Bn knh ht nhnAl2713Al: R = 3,6.10-15m 2.ng vl nhng nguyn t c cng s prtn ( Z ), nhng khc s ntrn (N) hay khc s nucln (A). V d: Hidr c ba ng v:1 2 2 3 31 1 1 1 1; ( ) ; ( ) H H D H T + ng v bn : trong thin nhin c khong 300 ng v . + ng v phng x ( khng bn): c khong vi nghn ng v phng x t nhin v nhn to . 3.n v khi lng nguyn t -u : c gi tr bng 1/ 12khi lng ng v cacbon 126C- 27 2231 12 1 121 . . 1, 66055.10 931, 5 /12 12 6, 0221.10= = =Au g g kg MeV cN;131 1, 6.10= MeV J 4. Khi lng v nng lng: H thc Anhxtanh gia nng lng v khi lng: E = mc2 => m = 2cE

=> khi lng c th o bng n v nng lng chia cho c2:eV/c2 hay MeV/c2.-TheoAnhxtanh,mtvtckhilngm0khitrngthinghthkhichuynngvitcv,khi lng s tng ln thnh m vi: m = 2201cvm trong m0 gi l khi lng ngh v m gi l khi lng ng. 5.Mt s cc ht thng gp: Tn giK hiuCng thcGhi chprtnp 11Hhay 11phir nh teriD 21Hhay 21D hir nng tritiT 31Hhay 31Thir siu nng anpha 42HeHt Nhn Hli bta tr- 01e electron bta cng+ 01e+ Pzitn (phn electron) ntronn 10nkhng mang in ntrin khng mang in, m0 = 0, v c + - Nguyn t Hidr, Ht nhn c 1 nucln l prtn Ht nhnHli c 4 nucln: 2 prtn v 2 ntrn+ + - - Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang2 II. HT KHI NNG LNG LIN KT CA HT NHN 1.Lc ht nhn-Lc ht nhn l lc tng tc gia cc nucln, bn knh tng tc khong 1510 m. -Lc ht nhn khng cng bn cht vi lc hp dn hay lc tnh in; n l lc tng tc mnh.2. ht khim ca ht nhn AZXKhi lng ht nhn hnmlun nh hn tng khi lng cc nucln to thnh ht nhn mt lngm : Khi lng ht nhnKhi lng Z PrtnKhi lng N Ntrn ht khimmhn (mX)Zmp (A Z)mn m = Zmp + (A Z)mn mhn 3.Nng lng lin kt lkW ca ht nhn AZX-Nnglinktlnnglngtarakhitothnhmthtnhn(haynnglngthuvophvmthtnhn thnh cc nucln ring bit). Cng thc :2.lkW m c = Hay :2. . .lk p n hnW Z m Nm m c( = +

4.Nng lng lin kt ring ca ht nhn-Nng lng lin kt ring l nng lng lin kt tnh trn mt nucln = lkWA.-Ht nhn c nng lng lin kt ring cng ln th cng bn vng. - V d: 5628Fe c nng lng lin kt ring ln = lkWA=8,8 (MeV/nucln) 2. PHN NG HT NHNI.PHN NG HT NHN -Phn ng ht nhn l mi qu trnh dn ti s bin i s bin i ca ht nhn.

1 2 3 41 2 3 41 2 3 4A A A AZ Z Z ZX X X X + + hay 1 2 3 41 2 3 4A A A AZ Z Z ZA B C D + + -C hai loi phn ng ht nhn +Phn ng t phn r ca mt ht nhn khng bn thnh cc ht nhn khc (phng x) +Phn ng tng tc gia cc ht nhn vi nhau dn n s bin i thnh cc ht nhn khc. Ch : Cc ht thng gp trong phn ng ht nhn:1 11 1p H = ;10n ; 42= He ; 01e = ; 01e ++=II. CC NH LUT BO TON TRONG PHN NG HT NHN 1.nh lut bo ton s nucln (s khi A) 1 2 3 4A A A A + = +2.nh lut bo ton in tch (nguyn t s Z) 1 2 3 4Z Z Z Z + = +3.nh lut bo ton ng lng: = s P Pt 4.nh lut bo ton nng lng ton phn Ws Wt= Ch :-Nng lng ton phn ca ht nhn: gm nng lng ngh v nng lng thng thng( ng nng):

2 212W mc mv = + - nh lut bo ton nng lng ton phn c th vit: W1 + W2 + m1.c2 + m2.c2 = W3 + W4 + m3.c2 + m4.c2 => (m1 + m2 - m3 - m4) c2 = W3 + W4 - W1 - W2= Q ta /thu - Lin h gia ng lng v ng nng22dP mW = hay 22dPWm= III.NNG LNG TRONG PHN NG HT NHN: + Khi lng trc v sau phn ng:m0 = m1+m2 v m = m3 + m4 + Nng lng W: -Trong trng hp( ) ; ( ) m kg W J :2020) ( ) ( c m m c m m W = = (J) -Trong trng hp( ) ; ( ) m u W MeV : 5 , 931 ) ( 5 , 931 ) (0 0m m m m W = =Nu m0 > m:0 W > : phn ng ta nng lng; Nu m0 < m :0 W< : phn ng thu nng lngChuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang3 3. PHNG XI.PHNG X:Phng x l hin tng ht nhn khng bn vng t phn r, pht ra cc tia phng x v bin i thnh cc ht nhn khc. II. CC TIA PHNG X 1.1 Cc phng trnh phng x: -Phng x 42( ) He : ht nhn con li hai so vi ht nhn m trong bng tun hon:4 42 2A AZ ZX He Y + -Phng x 01( ) e : ht nhn con tin mt so vi ht nhn m trong bng tun hon:01 1A AZ ZX e Y + + -Phng x 01( ) e ++: ht nhn con li mt so vi ht nhn m trong bng tun hon: 01 1A AZ ZX e Y+ + -Phng x : Sng in t c bc sng rt ngn: * 00A AZ ZX X + 1.2. Bn cht v tnh cht ca cc loi tia phng x Loi TiaBn ChtTnh Cht () -LdnghtnhnnguyntHeli(42He ),chuyn ng vi vn tc c 2.107m/s. -Ion ho rt mnh. -m xuyn yu. (-)-L dng ht lectron 01( ) e, vn tcc (+) -Ldnghtlectrondng(cngilpozitron) 01( ) e+, vn tcc . -Ionhoyuhnnhngmxuynmnh hn tia . () -Lbcxintcbcsngrtngn(di10-11 m), l ht phtn c nng lng rt cao -Ion ho yu nht, m xuyn mnh nht. III. CC NH LUT PHNG X 1.Chu k bn r ca cht phng x (T) Chu k bn r l thi gian mt na s ht nhn hin c ca mt lng cht phng x b phn r, bin i thnh ht nhn khc.2.Hng s phng x: ln 2T = (c trng cho tng loi cht phng x) 3.nh lut phng x: Theo s ht (N)Theo khi lng (m) phng x (H)10(1 3, 7.10 ) Ci Bq = Trongqutrnhphnr,sht nhn phng x gim theo thi gian : Trongqutrnhphnr,khi lng ht nhn phng x gim theo thi gian : -ilngctrngchotnhphngx mnh hay yu ca cht phng x.-S phn r trong mt giy. ( ) 0 0.2 .= =ttTtN N N e ( ) 0 0.2 .= =ttTtm m m e ( ) 0 0.2 .= =ttTtH H H e H N =0N : s ht nhn phng x thi im ban u.( ) tN : s ht nhn phng x cn li sau thi giant . 0m : khi lng phng x thi im ban u. ( ) tm : khi lng phng x cn li sau thi giant . 0H : phng x thi im ban u. ( ) tH : phng x cn li sau thi gian t Hay: i lng Cn li sau thi gian tB phn r sau thi gian tN/N0haym/m0(N0 N)/N0 ; (m0 m)/m0

Theo s ht N N(t)= N0 e- t ; N(t) = N0 Tt 2 N0 N = N0(1- e- t ) Tt 2 (1- e- t ) Theo khi lng (m) m = m0 e- t ; m(t) = m0 Tt 2 m0 m = m0(1- e- t ) Tt 2(1- e- t ) Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang4 * phng x: ca mt lng cht phng x c trng cho tnh phng x mnh hay yu ca n, c xc nh bi s ht nhn b phn r trong 1 giy: H = - tN = N = N0Tt 2= N0e- t

hay H = H0Tt 2= H0e- t . n v o phng x l becren (Bq): 1 Bq = 1 phn r/giy. Thc t cn dng n v curi (Ci): 1 Ci = 3,7.1010 Bq, xp x bng phng x ca mt gam rai. IV. NG DNG CA CC NG V PHNG X -Theo di qu trnh vn chuyn cht trong cy bng phng php nguyn t nh du. -Dng phng xtm khuyt tt trong sn phm c, bo qun thc phm, cha bnh ung th -Xc nh tui c vt. 4. PHN NG PHN HCH - PHN NG NHIT HCHI.PHN NG PHN HCH 1.Phnngphnhch:lmthtnhnrtnngnhUrani(23592U )hpthmtntrnchmsvthnhhaiht nhn trung bnh, cng vi mt vi ntrn mi sinh ra. 1 21 2235 1 236 192 0 92 0200A AZ ZU n U X X k n MeV + + + + 2.Phnngphnhchdychuyn:Nusphnhchtipdinthnhmtdychuynthtacphnngphn hchdychuyn,khisphnhchtnglnnhanhtrongmtthigianngnvcnnglngrtlnctara.iukinxyraphn ngdychuyn:xt sntrntrungbnhksinhrasaumiphnngphnhch( k l hs nhn ntrn). - Nu1 k < : th phn ng dy chuyn khng th xy ra. - Nu1 k= : th phn ng dy chuyn s xy ra v iu khin c. - Nu1 k> : th phn ng dy chuyn xy ra khng iu khin c.- Ngoi ra khi lng 23592Uphi t ti gi tr ti thiu gi l khi lng ti hn thm . 3.Nh my in ht nhn (nguyn t)B phn chnh ca nh my in ht nhn l l phn ng ht nhn PWR.(Xem sch GK C BN trang 199 nh XB-GD 2007, hoc SGK NC trang 285-287 Nh XB-GD-2007) II. PHN NG NHIT HCH 1.Phn ng nhit hch Phn ng nhit hch l phn ng kt hp hai ht nhn nh thnh mt ht nhn nng hn.

2 2 3 11 1 2 03, 25 H H H n Mev + + +2.iu kin xy ra phn ng nhit hch -Nhit cao khong t 50 triu ti 100 triu . -Hn hp nhin liu phi giam hm trong mt khong khng gian rt nh. 3.Nng lng nhit hch -Tuy mt phn ng nhit hch ta nng lng thn mt phn ng phnhch nhng nu tnh theo khi lng nhin liu th phn ng nhit hch ta ra nng lng ln hn. -Nhin liu nhit hch l v tn trong thin nhin: l teri, triti rt nhiu trong nc sng v bin. -V mt sinh thi, phn ng nhit hch sch hn so vi phn ng phn hch v khng c bc x hay cn b phng x lm nhim mi trng. Nguyn tc thnh cng: Suy ngh tch cc; Cm nhn am m; Hnh ng kin tr ! Chc cc em hc sinhTHNH CNG trong hc tp!Su tm v chnh l: GV: on Vn Lng Email:doanvluong@yahoo.com ;doanvluong@gmail.com; T: 0915718188 0906848238 Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang5 B. CC HNG S VT L v I N V VT L : 1.Cc hng s vt l : +Vi my tnh cm tay, ngoi cc tin ch nh tnh ton thun li, thc hin cc php tnh nhanh, n gin v chnh xc th phi k ti tin ch tra cu mt s hng s vt l v i mt s n v trong vt l. Cc hng s vt l c ci sn trong b nh ca my tnh vi n v trong h n v SI. +Cc hng s c ci sn trong my tinh cm tay Fx570MS; Fx570ES; 570ES Plus bng cc lnh:[CONST]Number [0 40] ( xem cc m lnh trn np ca my tnh cm tay ).

2.Lu : Khi tnh ton dng my tnh cm tay, ty theo yu cu bi c th nhp trc tip cc hng s t bi cho , hoc nu mun kt qu chnh xc hn th nn nhp cc hng s thng quacc m lnh CONST [0 40] c ci t sn trong my tinh! (Xem thm bng HNG S VT Ldi y)Cc hng s thng dng l: Hng s vt lM sCch nhp my : My 570MS bm: CONST0 40 = My 570ES bm:SHIFT 70 40 = Gi tr hin th Khi lng prton (mp)01Const[01] = 1,67262158.10-27 (kg) Khi lng ntron (mn)02Const[02] = 1,67492716.10-27 (kg) Khi lng lectron (me)03Const[03] = 9,10938188.10-31 (kg) Khi lng 1u (u)17Const[17] = 1,66053873.10-27 (kg) Hng s Faray (F)22Const[22] = 96485,3415 (mol/C) in tch lectron (e)23Const[23] = 1,602176462.10-19 (C) S Avgar (NA)24Const[24] = 6,02214199.1023 (mol-1) Tc nh sng trong chn khng (C0) hay c 28Const[28] = 299792458 (m/s) + i n v ( khng cn thit lm):Vi cc m lnh ta c th tra bng in np ca my tnh. +i n v: 1eV =1,6.10-19J.1MeV=1,6.10-13J.+i n v t uc2 sang MeV:1uc2= 931,5MeV (My 570ES: SHIFT 717 xSHIFT 728x2 :SHIFT 723 :X10X 6= hin th 931,494...) - My 570ESbmShift8Conv[m s] = -V d :T 36 km/h sang ? m/s , bm:36Shift8[Conv]19=Mn hnh hin th : 10m/s My 570MSbm ShiftConstConv [m s] = Nguyn tc thnh cng: Suy ngh tch cc; Cm nhn am m; Hnh ng kin tr ! Chc cc em hc sinhTHNH CNG trong hc tp!Su tm v chnh l: GV: on Vn Lng Email:doanvluong@yahoo.com ;doanvluong@gmail.com; T: 0915718188 0906848238 Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang6 C: CC DNG BI TP I.CU TO HT NHN- HT KHI V NNG LNG LIN KT: Dng 1 : Xc nh cu to ht nhn: a.Phng Php: T k hiu ht nhnAZXZ A, , N = A-Zb.Bi tp Bi 1: Xc nh cu to htnhn 23892U,Na2311,He42( Tm s Z prtn v sN ntron) +23892Uc cu to gm: Z=92 , A = 238 N= A Z = 146.p n: 23892U : 92 prtn ;146 ntron + Na2311 gm : Z= 11 , A = 23 N= A Z = 12p n:Na2311: 11 prtn; 12 ntron + He42gm : Z= 2 , A = 4 N= A Z = 2 p n:Na2311: 2 prtn; 2 ntron c.Trc nghim: Cu 1. Pht biu no sau y l ng? A. Ht nhn nguyn tXAZ c cu to gm Z ntron v A prton. B. Ht nhn nguyn tXAZ c cu to gm Z prton v A ntron. C. Ht nhn nguyn tXAZ c cu to gm Z prton v (A Z) ntron. D. Ht nhn nguyn tXAZ c cu to gm Z ntron v (A + Z) prton. Cu 2.Ht nhnCo6027 c cu to gm: A. 33 prton v 27 ntron B. 27 prtonv 60 ntronC. 27 prton v 33 ntron D. 33 prton v 27 ntronCu 3: Xc nh s ht proton v notron ca ht nhnN147 A.07 proton v 14 notron B.07 proton v 07 notron C.14 proton v 07 notron D.21 proton v 07 notron Cu 4: Trong nguyn t ng v phng x U23592 c:A. 92 electron v tng s proton v electron l 235 B.92 proton v tng s proton v electron l 235C. 92 proton v tng s proton v ntron l 235D.92 proton v tng s ntron l 235 Cu 5: Nhn Uranium c 92 proton v 143 notron k hiu nhn lA.U32792 B. U23592 C.U92235 D.U14392 Cu 6: Tm pht biu sai v ht nhn nguyn tAl A. S prtn l 13.B. Ht nhn Al c 13 nucln.C. S nucln l 27. D. S ntrn l 14. Cu 7: Trong vt l ht nhn, bt ng thc no l ng khi so snh khi lng prtn (mP), ntrn (mn) v n v khi lng nguyn t u. A. mP > u > mnB. mn < mP < uC. mn > mP > u D. mn = mP > u Cu 8. Cho ht nhn 115X. Hy tm pht biu sai. A. Ht nhn c 6 ntrn.B. Ht nhn c 11 nucln. C. in tch ht nhn l 6e.D. Khi lng ht nhn xp x bng 11u. Cu 9(H2007): Pht biu no l sai? A. Cc ng v phng x u khng bn. B. Cc nguyn t m ht nhn c cng s prtn nhng c s ntrn (ntron) khc nhau gi l ng v. C. Cc ng v ca cng mt nguyn t c s ntrn khc nhau nn tnh cht ha hc khc nhau. D. Cc ng v ca cng mt nguyn t c cng v tr trong bng h thng tun hon.Cu 10.(HC-2010 ) So vi ht nhn 2914Si , ht nhn 4020Cac nhiu hn A. 11 ntrn v 6 prtn.B. 5 ntrn v 6 prtn. C. 6 ntrn v 5 prtn.D. 5 ntrn v 12 prtn. Cu 11: (C-2011) Ht nhn 3517Clc: A. 35 ntronB. 35 nuclnC. 17 ntron D. 18 proton. Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang7 Dng 2 : Xc nh ht khi, nng lng lin kt ht nhn, nng lng lin kt ring: a.Phng Php: +S dng cng thc ht khi: 0m m m = ;m = Zmp+ Nmn

+Nng lng lin kt: 2 2. . . .lk p n hnW Z m N m m c m c( = + = +Nng lng lin kt ring: = AWlkMeV/nuclon. HayAmcAE2== +Chuyn i n v t uc2 sangMeV:1uc2 = 931,5MeVCh :+ So snh : Ht nhn c nng lng lin kt ring cng ln th cng bn vng . + Ht nhn c s khi t 50 70 trong bng HTTH thng bn hn cc nguyn t ca cc ht nhn cn li . b.Bi tp Bi 1 : Khi lng ca ht 104Bel mBe = 10,01134u, khi lng ca ntron l mN = 1,0087u, khi lng ca proton l mP = 1,0073u. Tnh ht khi ca ht nhn 104Bel bao nhiu? HD gii-Xc nh cu to ht nhn104Bec Z = 4proton, N= A-Z = 10-4=6 notron - ht khi:. ( ).p N hnm Z m A Z m m( = + = 4.1,0073u + 6.1,0087u 10,01134u m = 0,07u. p n:m = 0,07u Bi2: Tnh nng lng lin kt ht nhn triD21? Cho mp = 1,0073u, mn = 1,0087u, mD = 2,0136u; 1u = 931 MeV/c2. A.2,431 MeV.B. 1,122 MeV. C. 1,243 MeV.D. 2,234MeV. HD Gii : ht khi ca ht nhn D :m = mp + mn mD = 1.mp +1.mn mD = 0,0024 u Nng lng lin kt ca ht nhn D :Wlk = m.c2 = 0,0024.uc2 = 2,234 MeV . Chn D. Bi 3. Xc nh s Ntrn N ca ht nhn:He42. Tnh nng lng lin kt ring. Bitmn = 1,00866u; mp = 1,00728u;mHe = 4,0015u HD gii : T =HeZ A N422 2 4 = = N .Ta c03038 , 0 0015 , 4 ) ( 2 = + = n pm m m u MeV MeV uc E 29 , 28 5 , 931 . 03038 , 0 03038 , 02= = = MeV 07 , 7429 , 28= = Bi 4. ChoFe5626. Tnh nng lng lin kt ring. Bit mn = 1,00866u; mp = 1,00728u; mFe = 55,9349u HD gii: + Ta cu m m mn p50866 , 0 9349 , 55 30 26 = + = MeV MeV uc E 8 , 473 5 , 931 . 50866 , 0 50866 , 02= = = MeV 46 , 8568 , 473= = Bi5:HtnhnBe104ckhilng10,0135u.Khilngcantrn(ntron)mn=1,0087u,khilngcaprtn (prton) mP = 1,0073u, 1u = 931 MeV/c2. Nng lng lin kt ring ca ht nhn lBe104 A. 0,632 MeV. B. 63,215MeV. C. 6,325 MeV. D. 632,153 MeV.HD Gii :-Nng lng lin kt ca ht nhnBe104: Wlk = m.c2 = (4.mP +6.mn mBe).c2 = 0,0679.c2 = 63,249 MeV. -Suy ra nng lng lin kt ring ca ht nhn Be104: 63,1256, 32510lkWA= = MeV/nucln.Chn: C. Bi 6. Ht nhn heli c khi lng 4,0015 u. Tnh nng lng lin kt ring ca ht nhn hli. Tnh nng lng ta ra khi to thnh 1 gam hli. Cho bit khi lng ca prton v ntron lmp = 1,007276 u v mn = 1,008665 u; 1 u = 931,5 MeV/c2; s avgar l NA = 6,022.1023 mol-1. HDGii:He= AWlk= Ac m m Z A m ZHe n p2). ) ( . ( += 45 , 931 ). 0015 , 4 ) 008685 , 1 007276 , 1 .( 2 ( +=7,0752MeV;W = Mm.NA.Wlk = 0015 , 41.6,022.1023.7,0752.4 = 46,38332.1023 MeV = 7,42133.1011 J. Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang8 Bi 7. Tnh nng lng lin kt ring ca hai ht nhnNa2311 v Fe5626. Ht nhn no bn vng hn? Cho: mNa = 22,983734u; mFe = 55,9207u; mn = 1,008665 u; mp = 1,007276 u; 1u = 931,5 MeV/c2. HD Gii. Na = AWlk=Ac m m Z A m ZHe n p2). ) ( . ( +=235 , 931 ). 983734 , 22 008685 , 1 . 12 007276 , 1 . 11 ( += 8,1114 MeV; Fe = 565 , 931 ). 9207 , 55 008685 , 1 . 30 007276 , 1 . 26 ( += 8,7898 MeV; Fe > Na nn ht nhn Fe bn vng hn ht nhn Na. Bi8.Tmnnglngtorakhimthtnhnurani 234Uphngxtiatothnhngvthori 230Th.Choccnng lng lin kt ring ca ht l 7,10 MeV; ca 234U l 7,63 MeV; ca 230Th l 7,70 MeV. HD Gii . Ta c: W = 230.Th + 4.He - 234.U = 13,98 MeV. Bi 9. Khi lng nguyn t ca rai Ra226 l m = 226,0254 u . a/Hy ch ra thnh phn cu to ht nhn Rai ? b/ Tnh ra kg ca 1 mol nguyn t Rai , khi lng 1 ht nhn , 1 mol ht nhn Rai? c/ Tm khi lng ring ca ht nhn nguyn t cho bit bn knh ht nhn c tnh theo cng thc :r = r0.A1/3 . vi r0 = 1,4.1015m , A l s khi . d/ Tnh nng lng lin kt ca ht nhn , nng lng lin kt ring , bit mp = 1,007276u , mn = 1.008665u ; me = 0,00549u ; 1u = 931MeV/c2 . HD Gii : a/ Rai ht nhn c 88 prton , N = A- Z = 226 88 = 138 ntronb/ Khi lng 1 nguyn t: m = 226,0254u.1,66055.1027 = 375,7.1027 kg Khi lng mt mol :mmol = mNA = 375,7.1027.6,022.1023 = 226,17.103 kg = 226,17g Khi lng mt ht nhn : mhn = m Zme = 259,977u = 3,7524.1025kg Khi lng 1mol ht nhn : mmolhn = mnh.NA = 0,22589kg c/ Th tch ht nhn : V = 4r3/3 = 4r03A/ 3 .Khi lng ring ca ht nhn : D = 317303010 . 45 , 1433 / 4 mkgrrmA rrAmVmp p = = d/ Tnh nng lng lin kt ca ht nhn : E = mc2 = {Zmp + (A Z)mn m}c2 = 1,8197u E = 1,8107.931 = 1685 MeV Nng lng lin kt ring : = E/A = 7,4557 MeV. Bi 10: Bit khi lng ca cc ht nhnu m u m u m u mn p C0087 , 1 ; 0073 , 1 ; 0015 , 4 ; 000 , 12 = = = v 2/ 931 1 c Mev u= . Nng lng cn thit ti thiu chia ht nhnC126 thnh ba ht theo n v Jun l A. 6,7.10-13 JB. 6,7.10-15 JC. 6,7.10-17 JD. 6,7.10-19 J HD Gii: C123 He Nng lng ph v mt ht C12 thnh 3 ht He: W = ( mri - mhn )c2 = (3.4,0015 12). 931= 4.1895MeV Theo n v Jun l: W = 4,1895. 1,6.10-13 = 6,7032.10 -13J; ChnA Bi 11 :Cho bit m = 4,0015u;999 , 15 =Om u;u mp007276 , 1 = ,u mn008667 , 1 = . Hy sp xp cc ht nhnHe42, C126,O168 theo th t tng dn bn vng . Cu tr li ng l: A.C126, ,42He O168. B.C126,O168,,42He C.,42He C126,O168.D.,42He O168, C126. HD Gii: bi khng cho khi lng ca 12C nhng ch y dng n v u, theo nh ngha on v u bng 1/12 khi lng ng v 12C do c th ly khi lng 12C l 12 u.-Suy ra nng lng lin kt ring ca tng ht nhn l : He :Wlk = (2.mp + 2.mn m )c2 = 28,289366 MeV Wlk ring = 7,0723 MeV / nuclon. C :Wlk = (6.mp + 6.mn mC )c2 = 89,057598 MeVWlkring= 7,4215 MeV/ nuclon.O :Wlk = (8.mp + 8.mn mO )c2 = 119,674464 meV Wlk ring = 7,4797 MeV/ nuclon.-Ht nhn c nng lnglin kt ring cng ln th cng bn vng. Vy chiu bn vng ht nhn tng dn l : He < C < O. ChnC.Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang9 c.Trc nghim: Cu1:HtnhnCo6027ckhilngl59,919u.Bitkhilngcaprtonl1,0073uvkhilngcantronl 1,0087u. ht khi ca ht nhnCo6027 l A. 0,565u B. 0,536uC. 3,154uD. 3,637u Cu 2:ng v phng x cban 6027Copht ra tia - v tia . Bit Co nm 55, 940u; m 1, 008665u; = =pm 1, 007276u = . Nng lng lin kt ca ht nhn cban l bao nhiu? A. 10E 6, 766.10 J = B. 10E 3, 766.10 J = C. 10E 5, 766.10 J =D. 10E 7, 766.10 J =Cu3:BitkhilngcahtnhnU238l238,00028u,khilngcaprtnvntronlmP=1.007276U;mn= 1,008665u; 1u = 931 MeV/ c2. Nng lng lin kt ca Urani 23892Ul bao nhiu? A. 1400,47 MeVB. 1740,04 MeVC.1800,74 MeV D. 1874 MeV Cu 4:Bit khi lng ca prtn mp=1,0073u, khi lng ntron mn=1,0087u, khi lng ca ht nhn teri mD=2,0136u v 1u=931MeV/c2. Nng lng lin kt ring ca ht nhn nguyn t teriD21 l A. 1,12MeV B. 2,24MeV C. 3,36MeV D. 1,24MeVCu 5:Khi lng ca ht nhn 104Bel 10,0113u; khi lng ca prtn mp = 1,0072u, ca ntron mn = 1,0086; 1u = 931 MeV/c2. Nng lng lin kt ring ca ht nhn ny l bao nhiu? A. 6,43 MeVB. 6,43 MeVC. 0,643 MeVD. Mt gi tr khc Cu 6:Ht nhn 2010Ne c khi lng Nem 19, 986950u = . Cho bit p nm 1, 00726u; m 1, 008665u; = =21u 931, 5MeV/ c = . Nng lng lin kt ring ca 2010Nec gi tr l bao nhiu? A. 5,66625eVB. 6,626245MeVC. 7,66225eVD. 8,02487MeV Cu 7:Tnh nng lng lin kt ring ca ht nhn3717Cl. Cho bit: mp = 1,0087u; mn = 1,00867u; mCl = 36,95655u; 1u = 931MeV/c2 A. 8,16MeVB. 5,82 MeVC. 8,57MeVD. 9,38MeV Cu 8.Ht nhn hli (42He) c nng lng lin kt l 28,4MeV; ht nhn liti (73Li) c nng lng lin kt l 39,2MeV; ht nhn tri (21D) c nng lng lin kt l 2,24MeV. Hy sp theo th t tng dn v tnh bn vng ca chng: A. liti, hli, tri.B. tri, hli, liti. C. hli, liti, tri.D. tri, liti, hli. Cu 9. Ht c khi lng 4,0015u, bit s Avgar NA = 6,02.1023mol-1, 1u = 931MeV/c2. Cc nucln kt hp vi nhau to thnh ht, nng lng ta ra khi to thnh 1mol kh Hli l A. 2,7.1012JB. 3,5. 1012J C. 2,7.1010J D. 3,5. 1010J Cu 10(H2007): Cho: mC = 12,00000 u; mp = 1,00728 u; mn = 1,00867 u; 1u = 1,66058.10-27 kg; 1eV = 1,6.10-19 J ; c = 3.108 m/s. Nng lng ti thiu tch ht nhnC 126thnh cc nucln ring bit bng A. 72,7 MeV. B. 89,4 MeV. C. 44,7 MeV.D. 8,94 MeV. Cu11(C-2008):HtnhnCl1737ckhilngnghbng36,956563u.Bitkhilngcantrn(ntron) l1,008670u,khilngcaprtn(prton)l1,007276uvu=931MeV/c2. Nnglnglinktringcahtnhn Cl1737 bngA. 9,2782 MeV.B. 7,3680 MeV.C. 8,2532 MeV. D. 8,5684 MeV.Cu 12(H 2008): Ht nhn 104Be c khi lng 10,0135u. Khi lng ca ntrn (ntron) mn = 1,0087u, khi lng ca prtn (prton) mP = 1,0073u, 1u = 931 MeV/c2. Nng lng lin kt ring ca ht nhn 104BelA. 0,6321 MeV.B. 63,2152 MeV. C. 6,3215 MeV.D. 632,1531 MeV. Cu 13(C- 2009): Bit khi lng ca prtn; ntron; ht nhn 168O ln lt l 1,0073 u; 1,0087 u; 15,9904 u v 1u = 931,5 MeV/c2. Nng lng lin kt ca ht nhn 168O xp x bng A. 14,25 MeV.B. 18,76 MeV.C. 128,17 MeV.D. 190,81 MeV. Cu 14. (H- C-2010)Cho khi lng ca prtn; ntron; 4018Ar ; 63Li ln lt l: 1,0073u; 1,0087u; 39,9525u; 6,0145 u v 1u = 931,5 MeV/c2. So vi nng lng lin kt ring ca ht nhn 63Li th nng lng lin kt ring ca ht nhn 4018Ar A. ln hn mt lng l 5,20 MeV. B. ln hn mt lng l 3,42 MeV. C. nh hn mt lng l 3,42 MeV. D. nh hn mt lng l 5,20 MeV. Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang10 Dng 3: Tnh s ht nhn nguyn t v s ntron, prtn c trong mlng cht ht nhn. a.PHNG PHP: Cho khi lng m hoc s mol ca ht nhn XAZ . Tm s ht p , n c trong mu ht nhn . Nu c khi lng m suy ra s ht ht nhn X l : N = ANAm. (ht) . S mol :4 , 22VNNAmnA= = =.Hng S Avgar: NA = 6,023.1023 nguyn t/mol Nu c s mol suy ra s ht ht nhn Xl : N = n.NA(ht).+Khi : 1 ht ht nhn X c Z htproton v (A Z ) ht ht notron.=>Trong N ht ht nhn X c : N.Zht protonv(A-Z) Nht notron. b.BI TP Bi 1:Bit s Avgar l 6,02.10 23 mol-1, khi lng mol ca ht nhn uraniU23892 l 238 gam / mol. S ntron trong 119 gam uraniU23892l : A. 2510 . 2 , 2 htB.2510 . 2 , 1ht C 2510 . 8 , 8 ht D. 2510 . 4 , 4 ht HD Gii:S ht nhn c trong 119 gam uraniU23892l : N = ANAm. 23 2310 . 01 . 3 10 . 02 , 6 .238119= =ht Suy ra s ht ntron c trong N ht nhn uraniU23892 l :(A-Z). N= ( 238 92 ).3,01.1023 = 4,4.1025 ht p n : D Bi 2. Cho s Avgar l 6,02.10 23 mol-1. S ht nhn nguyn t c trong 100 g It 13152I l : A. 3,952.1023 htB. 4,595.1023 ht C.4.952.1023 ht D.5,925.1023 ht HD Gii : S ht nhn nguyn t c trong 100 g ht nhn I l :N = 2310 . 02 , 6 .131100. =ANAm ht. ChnB. c.TRC NGHIM: Cu 1 (C- 2009):BitNA = 6,02.1023 mol-1. Trong 59,50g 23892Uc s ntron xp x l A. 2,38.1023.B. 2,20.1025.C. 1,19.1025.D. 9,21.1024. Cu 2(C 2008): Bit s Avgar NA = 6,02.1023 ht/mol v khi lng ca ht nhn bng s khi ca n. S prtn (prton) c trong 0,27 gamAl1327lA. 6,826.1022. B. 8,826.1022. C. 9,826.1022. D. 7,826.1022. *Dng: Cho tng s ht c bnv hiu s ht mang in trong nguyn t ( Ht mang in gm Prtn v Electrn). Gi tng s ht mang in l S, hiu l a, ta d dng c cng thc sau: Z = (S + a) : 4 Cn c vo Z ta s xc nh c nguyn t l thuc nguyn t ha hc no (cng thc rt d chng minh). VD1: Tng s ht c bn ca 1 nguyn t X l 82, trong tng s ht mang in nhiu hn s ht khng mang in l 22. Vy X l Li gii: Ta c: Z = (82 + 22) : 4 = 26 =>St(Fe) VD2: Tng s ht c bn trongnguyn t Y l 52, trong tng s ht mang in nhiu hn s ht khng mang in l 16. Y l Li gii: Ta c: Z = (52 + 16) : 4 = 17 => Y l Clo (Cl) VD3: Tng s ht c bn trongnguyn t Y l 18, trong tng s ht mang in nhiu hn s ht khng mang in l 6. Y l Li gii: Ta c: Z = (18 + 6) : 4 = 6 => Y l Cacbon(C) Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang11 II.NH LUT PHNG X- PHNG X Dng 1: Xc nh lng cht cn li (N hay m), phng x:a.Phng php: Vn dngcng thc: -Khi lng cn li ca X sau thi gian t : .00 0.2 .2ttTtTmm m m e= = =. -S ht nhn X cn li sau thi gian t :. 00 0.2 .2ttTtTNN N N e= = =

- phng x: tNHtb = ; TtTtHHH= = 2 .200 hay tte HeHH= = .00Vi :ln 2T =-Cng thc tm s mol : AmNNnA= =

-Ch : + t v T phi a v cng n v . + m vm0cng n v v khng cn i n vCc trng hp c bit, hc sinh cn nh gii nhanh cc cu hi trc nghim: t Cn liN= N02tT T s N/N0 hay (%) B phn r N0 N(%) T s(N0- N)/N0 T s(N0- N)/N t =T N = N012 = N0/2 1/2 hay ( 50%)N0/2 hay ( 50%)1/21 t =2T N = N022 = N0/4 1/4 hay (25%)3N0/4 hay (75%)3/43 t =3T N = N032 = N0/8 1/8hay (12,5%)7N0/8 hay (87,5%)7/87 t =4T N = N042 = N0/16 1/16 hay (6,25%)15N0/16 hay (93,75%)15/1615 t =5T N = N052 = N0/32 1/32 hay (3,125%)31N0/32 hay (96,875%)31/3231 t =6T N = N062 = N0/64 1/64 hay (1,5625%)63N0/64 hay (98,4375%)63/6463 t =7T N = N072 = N0/128 1/128 hay (0,78125%)127N0/128 hay (99,21875%)127/128127 t =8T N = N082 = N0/256 1/256 hay (0,390625%)255N0/256 hay (99,609375%)255/256255 t =9T.................----------------------------------- Hay: Thi gian tT2T3T4T5T6T7T Cn li: N/N0 hay m/m01/21/221/231/241/251/261/27 r: (N0 N)/N0 1/23/47/815/1631/3263/64127/128 T l % r 50%75%87,5%93,75%96,875%98,4375%99,21875% T l ( t s) ht r vcn li 137153163127 T l ( t s)ht cn li v b phn r11/31/71/151/311/631/127 b. Bi tp: Bi 1: Cht It phng x 13153I dng trong y t c chu k bn r 8 ngy m. Nu nhn c 100g cht ny th sau 8 tun l cn bao nhiu? A. O,87g B.0,78g C.7,8g D. 8,7g HD Gi HD Gi HD Gi HD Gii :t = 8 tun = 56 ngy = 7.T .Suy ra sau thi gian t th khi lng cht phng x 13153I cn li l : 702 . 100 2 .= =Ttm m = 0,78 gam . Chn p n B. Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang12 Bi 2: Mt lng cht phng x c khi lng ban u l 0m . Sau 5 chu k bn r khi lng cht phng x cn li l bao nhiu? A.m= m0/5B.m =m0/8C. m = m0/32D. m = m0/10 HD Gii :t = 5T. Saut = 5T thkhi lng cht phng x cn li l: 322 . 2 .0 50 0mm m mTt= = = p n: C : 0m /32 Bi 3 : Mt cht phng x c chu k bn r l 3,8 ngy. Sau thi gian 11,4 ngy th phng x (hot phng x) ca lng cht phng x cn li bng bao nhiu phn trm so vi phng x ca lng cht phng x ban u?A. 25%.B. 75%.C. 12,5%. D. 87,5%.HD Gii : T = 3,8 ngy ; t = 11,4 = 3T ngy . Do ta a v hm m gii nhanh nh sau : TtTtmmm m = = 2 2 .0081230= =mm = 12,5% Chn p n : C. Bi 4: 22286Rn l cht phng x c chu k bn r T=3,8 ngy. Ban u c 2g. Hy tnh a) S nguyn t ban u b) S nguyn t cn li sau khong thi gian t= 5,7 ngy c) phng x ca lng Rn ni trn.Hd gii: a) p dng N0 =0.Am NA. D dng tnh c N0=5,42.1021 ht b) : t=5,7=1,5T nn N=2otTN =1,91.1021 ht c) p dng cng thc:0 0.2 .ttTH H H e N= = =m 0, 693T =Nn: H=0, 693T.N D dng tnh c: H= 4,05.1015 (Bq) =1,1.105(Ci) Bi5:Plni lnguyntphngx ,nphngramtht vbinithnhhtnhncon X.Chukbnrca Plni l T = 138 ngy. 1.Xc nh cu to, tn gi ca ht nhn con X. 2.Ban u c 0,01g. Tnh phng x ca mu phng x sau 3chu k bn r. HD Gii: 1. Xc nh ht nhn con X + Ta c phng trnh phn r:X He PoAZ+ 4221084 + Theo cc LBT ta c: ==+ =+ =822062 844 210ZAZAPb X20682: 2.T BqA TN mHAmNHm mNAmNN Hm mkAAkATt11 00010 . 08 , 2.2 . . 693 , 02 ..2 .= = ===== Nu trc nghim cn nh:BqA TN mHkA 11 010 . 08 , 2.2 . . 693 , 0= = Bi 6:Pht pho ( )3215Pphng x - vi chu k bn r T = 14,2 ngy v bin i thnh lu hunh (S). Vit phng trnh ca s phng x v nu cu to ca ht nhn lu hunh. Sau 42,6 ngy k t thi im ban u, khi lng ca mt khi cht phng x 3215Pcn li l 2,5g. Tnh khi lng ban u ca n. HD Gii : Phng trnh ca s pht x: 32 0 3215 1 16P e + SHt nhn lu hunh 3216S gm 16 prtn v 16 ntrnChuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang13 T nh lut phng x ta c: ln2 tttT To o om = m e m e m 2= =Suy ra khi lng ban u: t3Tom m.2 2, 5. 2 20g = = = Bi 7 (H -2009): Mt cht phng x ban u c N0 ht nhn. Sau 1 nm, cn li mt phn ba s ht nhn ban u cha phn r. Sau 1 nm na, s ht nhn cn li cha phn r ca cht phng x lA. N0 /6B.N0 /16. C. N0 /9.D. N0 /4.HD Gii :t1 = 1nmth s ht nhn cha phn r (cn li ) l N1, theo ta c : 312101= =TtNN Sau 1nm na tc l t2 = 2t1nm th s ht nhn cn li cha phn r l N2,ta c :

TtTtNN1 22022121= = 9131212202=||

\|=||||

\|=TtNN.Hoc N2 = 9 3 3020 1N N N= = Chn: C Bi 8:Natri ( )2411Nal cht phng x - vi chu k bn r T = 15 gi. Ban u c 12g Na. Hi sau bao lu ch cn li 3g cht phng x trn? Tnh phng x ca 3g natri ny. Cho s Avgar NA = 6,022 x 1023 mol-1 HD Gii : Ta c ttTo om m e m 2= =t2 0Tm 12 t2 4 2 2 t 2T 2x15x30m 3 T = = = = = = =gi. phng x: Aln 2 mH N . .NT N= = Thay s: 2317 6ln 2 3x6, 022x10H x 9, 66x10 Bq 2, 61x10 Ci15x3600 24= = =Bi 9:Gi t l khong thi gian s ht nhn ca mt lng cht phng x gim i e ln (e l c s ca loga t nhin vi lne = 1). T l chu k bn r ca cht phng x. Chng minh rng Ttln 2= . Hi sau khong thi gian 0,15t cht phng x cn li bao nhiu phn trm lng ban u? Cho bit e-0,51 = 0,6 HD Gii : S ht nhn ca cht phng x N gim vi thi gian t theo cng thc toN N e= , vi l hng s phn x, N0 l s ht nhn ban u ti t = 0Theo iu kin u bi: . t oNe eN = =;Suy rat 1 = , do 1 Ttln 2= =

Lng cht cn li sau thi gian 0,15t t l thun vi s ht: 0,15 t 0,15oNe e 0, 6 60%N = = = = c.Trc nghim: Cu 1: C 100g cht phng x vi chu k bn r l 7 ngy m. Sau 28 ngy m khi lng cht phng x cn li l A. 93,75g.B. 87,5g.C. 12,5g.D. 6,25g. Cu 2: Chu k bn r ca 6027Co bng gn 5 nm. Sau 10 nm, t mt ngun 6027Co c khi lng 1g s cn li A. gn 0,75g.B. hn 0,75g mt lng nh.C. gn 0,25g. D. hn 0,25g mt lng nh. Cu 3: C 100g it phng x 13153I vi chu k bn r l 8 ngy m. Tnh khi lng cht it cn li sau 8 tun l. A. 8,7g. B. 7,8g. C. 0,87g.D. 0,78g. Cu 4: Ban u c 5 gam cht phng x radon 22286Rn vi chu k bn r 3,8 ngy. S nguyn t radon cn li sau 9,5 ngy l A. 23,9.1021.B. 2,39.1021.C. 3,29.1021.D. 32,9.1021. Cu 5: Pht phoP3215 phng x - vi chu k bn r T = 14,2 ngy. Sau 42,6 ngy k t thi im ban u, khi lng ca mt khi cht phng xP3215 cn li l 2,5g. Tnh khi lng ban u ca n. A. 15g. B. 20g.C. 25g.D. 30g. Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang14 Cu 6: Cht phng x 210Po ban u c 200 g; Chu k bn r ca Po l 138 ngy . khi lng Po cn li sau thi gian 690 ngy l: A.6,25gB.62,5gC. 0,625g D. 50g Cu7:Gitlkhongthigianshtnhn camtlngchtphngxgimieln(elcscalgat nhin vi lne = 1), T l chu k bn r ca cht phng x. Hi sau khong thi gian 0,51t cht phng x cn li bao nhiu phn trm lng ban u ? A. 40%.B. 50%. C. 60%. D. 70%. Cu 8:Cng thc no di y khng phi l cng thc tnh phng x? A. ( )( )dtdNHtt = B. ( )( )dtdNHtt=C. ( ) ( ) t tN H = D. ( )TttH H= 20 Cu9:MtlngchtphngxRn22286banuckhilng1mg.Sau15,2ngyphngxgim93,75%. phng x ca lng Rn cn li l A. 3,40.1011BqB. 3,88.1011BqC. 3,58.1011BqD. 5,03.1011Bq Cu10:(C2007):Banumtmuchtphngxnguynchtckhi lngm0,chukbn r ca chtnyl3,8 ngy. Sau 15,2 ngy khi lng ca cht phng x cn li l 2,24 g. Khi lng m0 lA.5,60 g.B. 35,84 g.C. 17,92 g. D. 8,96 g. Cu 11: Mt ngun phng x c chu k bn r T v ti thi im ban u c 32N0 ht nhn. Sau cc khong thi gian T/2, 2T v 3T, s ht nhn cn li ln lt bng bao nhiu? A. 0 0 024N ,12N , 6N B. 0 0 016 2N , 8N , 4NC. 0 0 016N ,8N , 4N D. 0 0 016 2N , 8 2N , 4 2NCu 12: Mt ngun phng x c chu k bn r T v ti thi im ban u c 48No ht nhn. Hi sau khong thi gian 3T, s ht nhn cn li l bao nhiu? A. 4N0B. 6N0 C. 8N0D. 16N0 Cu 13: (H-C-2010). Ban u c N0 ht nhn ca mt mu cht phng x nguyn cht c chu k bn r T. Sau khong thi gian t = 0,5T, k t thi im ban u, s ht nhn cha b phn r ca mu cht phng x ny l A. 20N. B. 20N. C. 40N. D. N02 . Cu 14(C- 2009): Gi l khong thi gian s ht nhn ca mt ng v phng x gim i bn ln. Sau thi gian 2 s ht nhn cn li ca ng v bng bao nhiu phn trm s ht nhn ban u? A. 25,25%. B. 93,75%.C. 6,25%.D. 13,5%. Cu 15(H2008): Pht biu no sao y l sai khi ni v phng x (hot phng x)? A. phng x l i lng c trng cho tnh phng x mnh hay yu ca mt lng cht phng x. B. n v o phng x l becren. C. Vi mi lng cht phng x xc nh th phng x t l vi s nguyn t ca lng cht . D. phng x ca mt lng cht phng x ph thuc nhit ca lng cht . Cu 16(C- 2008): Ban u c 20 gam cht phng x X c chu k bn r T. Khi lng ca cht X cn li sau khong thi gian 3T, k t thi im ban u bng A. 3,2 gam. B. 2,5 gam.C. 4,5 gam.D. 1,5 gam.Cu 17(H 2008): Mt cht phng x c chu k bn r l 3,8 ngy. Sau thi gian 11,4 ngy th phng x (hot phng x) ca lng cht phng x cn li bng bao nhiu phn trm so vi phng x ca lng cht phng x ban u? A.25%.B. 75%.C. 12,5%.D. 87,5%. Cu 18(H 2008) : Ht nhn 11AZX phng x v bin thnh mt ht nhn 22AZY bn. Coi khi lng ca ht nhn X, Y bng s khi ca chng tnh theo n v u. Bit cht phng x 11AZX c chu k bn r l T. Ban u c mt khi lng cht 11AZX, sau 2 chu k bn r th t s gia khi lng ca cht Y v khi lng ca cht X l A. 12A4AB. 21A4AC. 21A3AD. 12A3A Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang15 Dng 2: Xc nh lng cht b phn r : a.Phng php:- Cho khi lng ht nhn ban u m0 ( hoc s ht nhn ban u N0 ) v T . Tm khi lng ht nhn hoc s ht nhn b phn r trong thi gian t ? -Khi lng ht nhn b phn r:m = ) 1 ( ) 2 1 (.0 0 0tTte m m m m = =

-S ht nhn b phn r l : N =) 1 ( ) 2 1 (.0 0 0tTte N N N N = = -> Hay Tm s nguyn t phn r sau thi gian t: . .0 0 0 0 0 0 0 .1 1 1. (1 ) (1 ) (1 )2tt tk t teN N N N N e N e N N Ne e = = = = = = - Nu t m0/m = 2t/T = 24

Suy ra t = 4T = 4.138 = 552 ngy m. Bi 2: Tnh s ht nhn b phn r sau 1s trong 1g Rai 226Ra . Cho bit chu k bn r ca 226Ra l 1580 nm. S Avgar l NA = 6,02.1023 mol-1. A. 3,55.1010 ht. B. 3,40.1010 ht. C. 3,75.1010 ht.D..3,70.1010 ht. HD Gii: -S ht nhn nguyn t c trong 1 gam226Ra l :N0 = 21 2310 . 6646 , 2 10 . 022 , 6 .2261. = =ANAm ht . -Suy ra s ht nhn nguyn t Ra phn r sau 1 s l :

1086400 . 365 . 1580121010 . 70 , 3 2 1 10 . 6646 , 2 ) 2 1 ( =||||

\| = = TtN N ht . Chn D.Bi 3: Mt cht phng x c chu k bn ra T. Sau thi gian t = 3T k t thi in ban u, t s gia s ht nhn b phn r thnh ht nhn ca nguyn t khc vi s ht nhn ca cht phng x cn liA. 7B. 3C. 1/3D. 1/7HD Gii :Thi gian phn r t = 3T; S ht nhn cn li : 787812030= = = = =NNN N NNN Bi 4: ng v phng x Cban 6027Co pht ra tia v vi chu k bn r T = 71,3 ngy. Trong 365 ngy, phn trm cht Cban ny b phn r bng A. 97,12%B. 80,09%C. 31,17%D. 65,94%HD Gii: % lng cht 60Co b phn r sau 365 ngy : Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang16 m =) 1 (.0 0te m m m = % 12 , 97 103 , 712 ln . 365= =emm . Hoc m = ) 2 1 (0 0Ttm m m = ==TtTtmm22 1097,12% Chn A.Bi 5: Mt cht phng xc chu k bn r l 20 pht. Ban u mt mu cht c khi lng l 2g. Sau 1h40pht, lng cht phn r c gi tr no? A: 1,9375 g B: 0,0625g C: 1,25 g D: mt p n khc HD Gii: S lng cht phn r) 2 1 .(Tt0 = m m =1,9375 g Chn A.Bi 6: Ht nhn 21084Po phng x anpha thnh ht nhn ch bn. Ban u trong muPo cha mt lngmo (g). B qua nng lng ht ca photon gama. Khi lng ht nhn con to thnh tnh theo m0 sau bn chu k bn r l? A.0,92m0B.0,06m0 C.0,98m0D.0,12m0 HD Gii: Pb Po2068221084+ p dng nh lut phng xN =N0 /24 .s ht nhn ch to thnh ng bng s ht nhn Po bi phn r =16152 /0 40NN N N = = ( N0 = ANm.2100) .Suy ra mPb = 206 .ANN = 206 * .210 * . 16150m = 0,9196m0. c. TRC NGHIM: Cu 1: ng v Co6027 l cht phng x vi chu k bn r T = 5,33 nm, ban u mt lng Co c khi lng m0. Sau mt nm lng Co trn b phn r bao nhiu phn trm? A. 12,2%B. 27,8%C. 30,2% D. 42,7% Cu 2: Cban 6027Co l cht phng x vi chu k bn r 316nm. Nu lc u c 1kg cht phng x ny th sau 16 nm khi lng 6027Co b phn r l A. 875g.B. 125g.C. 500g.D. 250g. Cu 3: Chu k bn r 21084Pol 318 ngy m. Khi phng x tia , plni bin thnh ch. C bao nhiu nguyn t plni b phn r sau 276 ngy trong 100mg 21084Po ? A. 200, 215.10 B. 202,15.10C. 200, 215.10D. 201, 25.10Cu 4. Chu k bn r ca U 238 l 4,5.109 nm. S nguyn tb phn r sau 106 nm t 1 gam U 238 ban u l bao nhiu? Bit s Avgadr NA = 6,02.1023 ht/mol. A. 2,529.1021B. 2,529.1018 C. 3,896.1014 D. 3,896.1017 Cu 5:Chu k bn r ca cht phng x 9038Sr l 20 nm. Sau 80 nm c bao nhiu phn trm cht phng x phn r thnh cht khc ? A. 6,25%.B. 12,5%.C. 87,5%.D. 93,75%. Cu 6: ng v phng x 6629Cu c chu k bn r 4,3 pht. Sau khong thi gian t = 12,9 pht, phng x ca ng v ny gim xung bao nhiu : A.85 %B. 87,5 % C. 82, 5 %D.80 %Cu 7:Gi l khong thi gian s ht nhn ca mt ng v phng x gim i bn ln. Sau thi gian 2 s ht nhn cn li ca ng v bng bao nhiu phn trm s ht nhn ban u? A. 25,25%. B. 93,75%. C. 6,25%.D. 13,5%. Cu8: Chtphngx 2411Nacchukbnr15gi.Sovikhi lngNabanu,khilngchtnybphn r trong vng 5h u tin bng A. 70,7%.B. 29,3%.C. 79,4%.D. 20,6% Cu 9:Gi t l khong thi gian s ht nhn ca mt lng cht phng x gim i e ln (e l c s ca lga t nhin vi lne = 1), T l chu k bn r ca cht phng x. Hi sau khong thi gian 0,51t cht phng x cn li bao nhiu phn trm lng ban u ? A. 40%.B. 50%.C. 60%.D. 70%. Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang17 Dng3 : Xc nh khi lng ca ht nhn con : a.Phng php:- Cho phn r : Y XBZAZ ' + tia phng x . Bit m0 , T ca ht nhn m. Ta c : 1 ht nhn m phn rth s c 1 ht nhn con tao thnh. Do :NX (phng x) = NY (to thnh) -S mol cht b phn r bng s mol cht to thnh YXXnAmn ==-Khi lngcht to thnh l AB mmXY. =. Tng qut :mcon = conmemeAAm. -Hay Khi lng cht mi c to thnh sau thi gian t

1 0 11 1 0(1 ) (1 )t tA AA N A Nm A e m eN N A = = = Trong : A, A1 l s khi ca cht phng x ban u v ca cht mi c to thnh NA = 6,022.10-23 mol-1 l s Avgar. -Lu : Ttrong phn r : khi lng ht nhn con hnh thnh bng khi lng ht nhn m b phn r (Trng hp phng x +, - th A = A1 m1 = m ) b. Bi tp: Bi 1: ng v2411 Na l cht phng x - to thnh ht nhn magi 2412Mg. Ban u c 12gam Na v chu k bn r l 15 gi. Sau 45 h th khi lng Mg to thnh l : A. 10,5gB. 5,16 gC. 51,6g D. 0,516g HD Gii: Nhn xt : t = 3.T nn ta dng hm m 2 gii cho nhanh bi ton : - Khi lng Na b phn r sau 45 = 3T gi:m ) 2 1 ( 12 ) 2 1 (310 = =Ttm m = 10,5 g . -Suy ra khi lng ca mg to thnh :mcon = 5 , 10 24 .245 , 10 .= =mecon meAA mgam. Chn p nABi 2 : Cht phng x PoloniPo21084c chu k bn r T = 138 ngy phng ra tia v bin thnh ng v chPb20682,ban u c 0,168g poloni . Hi sau 414 ngy m c : a.Bao nhiu nguyn t poloni b phn r? b.Tim khi lng ch hnh thnh trong thi gian HD Gii :t = 414 ngy = 3T a.S nguyn t b phn r sau 3 chu ki: 030 0 0872 N N N N N N = = = haykhi lng cht b phn r m = 087m = 0,147g 20 23 010 . 214 , 4 10 . 023 , 6 .210 . 8168 , 0 . 787= = = ANAmNnguyn t b.Khi lng ch hnh thnh trong 414 ngy m:mcon = conmemeAAm.=g 144 , 0 206 .210147 , 0= ] Bi 3 :Ht nhn 22688Ra c chu k bn r 1570 nm phn r thnh 1 ht v bin i thnh ht nhn X. Tnh s ht nhn X c to thnh trong nm th 786. Bit lc u c 2,26 gam radi. Coi khi lng ca ht nhn tnh theo u xp x bng s khi ca chng v NA = 6,02.1023 mol-1. HD Gii 3. Phng trnh phn ng: 22688Ra 42He + 22286Rn. Trong nm th 786: khi lng 22688Ra b phn r l: mRa = m0(15707852-15707862) = 7.10-4g; khi lng 22286Rn c to thnh: mRn = mRa.RaRnAA= 6,93g;s ht nhn 22286Rn c to thnh l: NRn = RnRnAm.NA = 1,88.1018 ht. Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang18 Bi 4 : Plni 21084Po l mt cht phng x c chu k bn r 140 ngy m. Ht nhn plni phng x s bin thnh ht nhn ch (Pb) v km theo mt ht . Ban u c 42 mg cht phng x plni. Tnh khi lng ch sinh ra sau 280 ngy m. HD Gii 4. Ta c: mPb = m0.PoPbAA(1 - Tt2 ) = 31,1 mg. Bi 5 :ng v 23592U phn r thnh ht nhn AZTh . 1) Vit y phng trnh phn r trn. Nu r cu to ca ht nhn c to thnh. 2) Chui phng x trn cn tip tc cho n ht nhn con l ng v bn 20782Pb . Hi c bao nhiu ht nhn Hli v ht nhn in t c to thnh trong qu trnh phn r . HD Gii 5. 1) Phng trnh phn r 235 4 A92 2 ZU Th +T nh lut bo ton s khi: 235 = 4 + A => A = 231. T nh lut bo ton in tch: 92 = 2 + Z => Z = 90.Vy phng trnh phn ng: 235 4 23192 2 90U Th +Cu to ht nhn 23190Thgm 231 ht nuclen vi 90 ht prtn v 231 90 = 141 ht ntrn. 2) Gi x l s phn r , y l s phn r . T nh lut bo ton s khi:235 = 207 + 4x + 0y -> x = 7T nh lut bo ton in tch:90 = 82 + 2x y -> y = 4 Mi h phn r s to ra mt ht nhn Hli, mi phn r s to ra mt ht in t. Vy c 7 ht nhn Hli v 4 ht in t c to thnh. Bi 6 :Cho chm ntron bn ph ng v bn 5525Mnta thu c ng v phng x 5625Mn . ng v phng x 56Mnc chu tr bn r T = 2,5h v pht x ra tia-. Sau qu trnh bn ph 55Mn bng ntron kt thc ngi ta thy trong mu trn t s gia s nguyn t 56Mnv s lng nguyn t 55Mn = 10-10. Sau 10 gi tip th t s gia nguyn t ca hai loi ht trn l: A. 1,25.10-11B. 3,125.10-12C. 6,25.10-12D. 2,5.10-11 Gii:Sauqutrnhbnph 55Mnbngntronktthcthsnguyntca 5625Mn gim,csnguynt 5525Mn khng i, Sau 10 gi = 4 chu k s nguyn t ca 5625Mngim 24 = 16 ln. Do th t s gia nguyn t ca hai loi ht trn l: 5556MnMnNN= 161010 = 6,25.10-12 Chn p n C c.TRC NGHIM: Cu 1: Urani (23892U ) c chu kbnr l 4,5.109nm. Khi phng x , urani bin thnh thri (23490Th ). Khi lng thri to thnh trong 23,8 g urani sau 9.109 nm l bao nhiu? A. 17,55gB. 18,66gC. 19,77gD. Phng n khc Cu 2: Chu k bn r 21184Po l 138 ngy. Ban u c 1mmg 21184Po . Sau 276 ngy, khi lng 21184Pob phn r l: A. 0,25mmg B. 0,50mmg C. 0,75mmgD. p n khc * Cht phng x 21084Po c chu k bn r 140 ngy,bin thnh ht nhn ch(Pb). Ban u c 42mg.Tr li cc cu 3,4,5 Cu 3 : S prtn v ntron ca Pb nhn gi tr no sau y. A. 80notron v 130 protonB. 84 notron v 126 proton C. 84notron v 124 protonD. 82 notron v 124 proton Cu 4 : phng x ban u ca 21084Po nhn gi tr no ? A. 6,9.1016 BqB. 6,9.1012 BqC. 9,6.1012 BqD. 9,6.1016 Bq Cu 5 : Sau 280 ngy m phng x, khi lng ch trong mu l ? A. 10,5mgB. 21mgC. 30,9mgD. 28mg Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang19 Dng 4: Xc nh chu k bn r T. a.Phng php:1) Cho m & m0 ( hoc N & N0) hay H&H0 : - Bit sau thi gian t th mu vt c t l m/m0 ( hay N/N0 ) . Tm chu k bn r T ca mu vt ? a) T s s nguyn t ban uv s nguyn t cn li sau thi gian phng x t N= N0 te. => T=NNt0ln2 ln .Hocm=m0 te. => T=t lnmlnm02 NuxNN=02=> x = tT Hoc:xmm=02=> x = tT Nu 0mm= 0NN= n21(vi n N * ) ntT nTt= = . Nu t s : 0mm= 0NN khng p th:m Ttm= 2 .0|||

\|= =020log 2mmTtmmTt T=. Tng t chos nguyn tv phng x: N TtN= 2 .0|||

\|= =020log 2NNTtNNTt T=. H TtH= 2 .0|||

\|= =020log 2HHTtHHTt T=. b)T s s nguyn t ban u v s nguyn t b phn r sau thi gian phng x t N= N0(1-te. ) =>0NN =1-te. =>T= -) 1 ln(2 ln .0NNt 2)Tm chu k bn r khi bit s ht nhn(hay khi lng) cc thi im t1 v t2 -Theo s ht nhn:N1= N0 1.te ; N2=N0 2.te ;21NN=) .(1 2t te =2 1ln2.( ) t tTe=>T = 211 2ln2 ln ) (NNt t -Theo s khi lng: m1= m0 1.te ; m2= m0 2.te => 12mm=) .(1 2t te =2 1ln2.( ) t tTe=>T = 2 112( ) ln 2lnt tmm 3)Tm chu k bn khi bit s ht nhn b phn r trong hai thi gian khc nhau 1N l s ht nhn b phn r trong thi gian t1 Sau t (s) :2N l s ht nhn b phn r trong thi gian t2-t1 -Ban u : H0=11tN ;-Sau t(s) H=22tN m H=H0te. => T=21ln2 ln .NNt Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang20 b. S dng lnhSOLVE trong my tnh Fx-570ES tm nhanh mt i lng cha bit : -My Fx570ES Ch dng trong COMP:MODE1 ) SHIFT MODE 1Mn hnh: Math Cc bc Chn ch Nt lnh ngha- Kt qu Dng COMPBm: MODE1COMP l tnh ton chung Ch nh dng nhp / xut ton MathBm: SHIFT MODE 1Mn hnh xut hin Math Nhpbin X (i lng cn tm)Bm: ALPHA )Mn hnh xut hin X. Nhp du =Bm: ALPHA CALC Mn hnh xut hin du= Chc nng SOLVE: Bm: SHIFTCALC = hin th ktqu X= ..... V d: Mt muNa2411 ti t=0 c khi lng 48g. Sau thi gian t=30 gi, muNa2411 cn li 12g. BitNa2411 l cht phng x- to thnh ht nhn con l Mg2412.Chu k bn r caNa2411 l A: 15hB: 15ngy C: 15phtD: 15giy Ta dng biu thc00.2 :2tTtTmm m Hay m= =Vii lng cha bit l:T ( Tl bin X)Nhp my :3012 48.2X=Bm: SHIFTCALC=(ch khong thi gian 6s) Hin th: X= 15 .Chn A T v d ny ta c th suy lun cch dng cc cng thc khc!!! b. Bi tp: Bi 1 :Mt lng cht phng x sau 12 nm th cn li 1/16 khi lng ban u ca n. Chu k bn r ca cht l A. 3 nmB. 4,5 nmC. 9 nmD. 48 nm HD Gii :Ta c 0mm=n21=421161=ntT nTt= = . =412= 3 nm .Chon p n A. 3 nm Bi 2: Sau thi gian t, phng x ca mt cht phng x - gim 128 ln. Chu k bn r ca cht phng x lA. 128t.B. 128t.C. 7t. D.128 t. HD Gii: Ta c=0HHn21=7211281= 77tTTt= =p n C Bi 3: Sau khong thi gian 1 ngy m 87,5% khi lng ban u ca mt cht phng x b phn r thnh cht khc. Chu k bn r ca cht phng x l A. 12 gi.B. 8 gi. C. 6 gi. D. 4 gi. Tm ttGii : ?24% 5 , 870===Th tmmTa c :30 00218 87871005 , 87= = = = = mmmmmm Hay . htTTt832433 = = = = . ChnBBi 4. (C-2011) : Trong khong thi gian 4h c 75% s ht nhn ban u ca mt ng v phng x b phn r. Chu k bn r ca ng v l: A. 1h B. 3hC. 4hD. 2h HD:htTTtkNNk k222412175 . 02110= = = = = = = Bi5.PhngtrnhphngxcaPlnicdng:21084PoAZPb + .ChochukbnrcaPlni T=138ngy.Khi lng ban u m0=1g. Hi sau bao lu khi lng Plni ch cn 0,707g? Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang21 A: 69 ngyB: 138 ngyC: 97,57 ngyD: 195,19 ngyHd gii: Tnh t: 0mm=te. => t=2 lnln .0mmT=2 ln707 , 01ln . 138=69 ngy (Chn A) Bi 6. Vo u nm 1985 phng th nghim nhn mu qung cha cht phng xCs17355 khi phng x l : H0 = 1,8.105Bq . a/ Tnh khi lng Cs trong qung bit chu k bn d ca Cs l 30 nm . b/ Tm phng x vo u nm 1985. c/ Vo thi gian no phng x cn 3,6.104Bq . HDGii : a/ Ta bit H0 = N0 , vi N0 = AmNA => m = A0A0N . 693 , 0AT HN .A H=Thay s m = 5,6.108g b/ Sau 10 nm : H = H0te ; t =231 , 03010 . 693 , 0==> H = 1,4.105 Bq . c/ H = 3,6.104Bq=> HH0 = 5=> t = ln5 = Tt . 693 , 0 => t = 693 , 05 ln T = 69 nm . Bi 7. ng v Cacbon 146C phng xv bin thnh nito (N). Vit phng trnh ca s phng x . Nu cu to ca htnhnnito.Muchtbanuc2x10-3gCacban 146C.Saukhongthigian11200nm.KhilngcaCacbon 146C trong mu cn li 0.5 x 10-3 g . Tnh chu kbn r ca cacbon 146C. HD Gii: Phng trnh ca s phng x : 14 o 146 1 7C e N +-Ht nhn nit 147N gm Z = 7 prtn V N = A Z= 14 7 = 7 ntrn - Ta c: t toT Tomm m 2 2m= = (1) Theo bi: 32 o3m 2 104 2m0.5 10= = =(2) T (1) v (2) t t 112002 T 5600T 2 2 = = = = nm Bi 8.Ht nhnC146 l cht phng x - c chu k bn r l 5730 nm. Sau bao lu lng cht phng x ca mt mu ch cn bng 81 lng cht phng x ban u ca mu . HD gii . Ta c: N = N0Tt2 0NN= Tt2 ln0NN= - Ttln2t = 2 lnln .0NNT= 17190 nm. Bi 9: Tnh chu k bn r ca Thri, bit rng sau 100 ngy phng x ca n gim i 1,07 ln. Bi gii: phng x ti thi im t.: H = H0.e - t=> e t= HH0 => t= ln(HH0) =) ln(10HHtm==T2 ln) ln(10HHt T = 07 , 1 ln. 2 ln t= 067658 , 0693 , 0.100ngy1023 ngy. Bi 10.Bit ng v phng x 146C c chu k bn r 5730 nm. Gi s mt mu g c c phng x 200 phn r/pht v mt mu g khc cng loi, cng khi lng vi mu g c , ly t cy mi cht, c phng x 1600 phn r/pht. Tnh tui ca mu g c. Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang22 HD gii . Ta c: H = H0.Tt2 = TtH20 Tt2 = HH0= 8 = 23

Tt= 3t = 3T = 17190 (nm). Bi 11.Silic 3114Sil cht phng x, pht ra ht v bin thnh ht nhn X. Mt mu phng x 3114Siban u trong thi gian 5 pht c 190 nguyn t b phn r, nhng sau 3 gi cng trong thi gian 5 pht ch c 85 nguyn t b phn r. Hy xc nh chu k bn r ca cht phng x. HD Gii: -Ban u: Trong thi gian 5 pht c 190 nguyn t b phn r H0=190phn r/5pht -Sau t=3 gi:Trong thi gian 5 pht c 85 nguyn t b phn r. H=85phn r /5pht H=H0te. =>T=HHt0ln2 ln .=85190ln2 ln . 3= 2,585 gi Bi 12. Mt mu phng xSi3114 ban u trong 5 pht c 196 nguyn t b phn r, nhng sau 5,2 gi (k t lct = 0) cng trong 5 pht ch c 49 nguyn t b phn r. Tnh chu k bn r caSi3114. HD gii . Ta c: H = H0Tt2 = TtH20

Tt2 =HH0= 4 = 22 Tt= 2T = 2t = 2,6 gi. Bi 13. Ht nhn Plni l cht phng x,sau khi phng x n tr thnh ht nhn ch bn. Dng mt mu Po no ,sau 30 ngy ,ngi ta thy t s khi lng ca ch v Po trong mu bng 0,1595.Tnh chu k bn r ca Po HD Gii:Tnh chu k bn r ca Po: PoPbmm=mm' =tAte m NA e N.0.. 0' ) 1 ( =AA'(1-te. ) T=-)' ..1 ln(2 ln .A mA mtPoPb= )206210 . 1595 , 01 ln(2 ln . 30= 138 ngy Bi 14.Ban u (t = 0) c mt mu cht phng x X nguyn cht. thi im t1 mu cht phng x X cn li 20% ht nhn cha b phn r. n thi imt2 = t1 + 100 (s) s ht nhn X cha b phn r ch cn 5% so vi s ht nhn ban u. Tnh chu k bn r ca cht phng x . HD gii .. Ta c: N = N0Tt2Tt2 = 0NN. Theo bi ra: Tt12= 01NN= 20% = 0,2 (1); Tt22= 02NN= 5% = 0,05 (2). T (1) v (2) suy ra: TtTt2122= Tt t1 22= 05 , 02 , 0= 4 = 22

Tt t1 2 = 2T = 210021 1 1 2t t t t +== 50 s. Bi 15. o chu k ca mt cht phng x ngi ta cho my m xung bt u m t thi im t0=0. n thi im t1=2 gi, my m c n1 xung, n thi im t2=3t1, my m c n2 xung, vi n2=2,3n1. Xc nh chu k bn r ca cht phng x ny. HD Gii: -S xung m c chnh l s ht nhn b phn r: N=N0(1-te. ) -Ti thi im t1:N1= N0(1-1.te )=n1 -Ti thi im t2 :N2= N0(1-2.te )=n2=2,3n1 1-2.te =2,3(1-1.te )1-1. 3 te =2,3(1-1.te )1 +1.te +1. 2 te =2,3 1. 2 te +1.te -1,3=0 => 1.te =x>0 X2 +x-1,3= 0 => T= 4,71 h Bi 16.ng v coban 6027Co l cht phng x -; sinh ra ht nhn con l niken (Ni). phng x ca 0.2g6027Co l H = 225 Ci.a.Hy vit phng trnh ca phng x v nu r thnh phn cu to ca ht nhn con.b.Tm chu k bn r ca 6027Cov tm thi gian c 75% 6027Cob phn r. Bit sNA = 6.032 x 1023mol-1. HD gii . a) Phng trnh ca s phng x: +60 - 6027 28Co NiChuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang23 Thnh phn ca ht nhn 6028Nil: 28 proton v 60 28 = 32 neutron. b) phng x: = t t0 0H = H .e H .e ; vi = = =2321 0 A0m N 0.2x6.022x10N 2.0073x10M 60 ht. T (1) suy ra:= =0 0 0ln2H l.N N .T => = = = =218 0100N l n2 2.0073x10 l n2T 1.67x10 s 5.3H225x3.7x10 nm. Theo nh lut phng x:= =lt 00mm m .et2Tsuy ra= = = =t2 0 0T0m m2 4 2m (1 0.75)m

t = 2T = 2 x 5.3=10.6 nm. Bi 17. Cban( )6027Cophng x - vi chu k bn r T = 5,27 nm v bin i thnh niken (Ni).a.Vit phng trnh phn r v nu cu to ca ht nhn con.b.Hi sau thi gian bao lu th 75% khi lng ca mt khi to cht phng x( )6027Cophn r ht? HD Cch 1:a.Phng trnh phn r: 60270 60Co e Ni1 28 + . Ht nhn Ni c 28 prtn v 32 ntrn b.Lng cht phng x cn li so vi ban u: 100% - 75% = 25% =1/4Hay00m m 14m 4 m= =nh lut phng x: ln2 t.ttT T0 0 0m m .e m e m 2 = = =Hay t0Tm2 4 t 2T 10, 54m= = = =nm HD Cch 2 . Ta c: m = m0 - m = m0Tt2t =2 ln'ln .00mm mT= 10,54 nm. Bi 18 : C 0,2(mg) RadiRa22688phng ra 4,35.108 ht trong 1 pht. Tm chu k bn r ca Ra ( cho T >> t). Cho x BqA TN mHkA 11 010 . 578 , 3.2 . . 693 , 0= = c.TRC NGHIM: Cu 1: Mt ht bi 22688Ra c khi lng 1,8.108 (g) nm cch mn hunh quang 1cm. Mn c din tch 0,03cm2. Hi sau 1 pht c bao nhiu chm sng trn mn, bit chu k bn r ca Ra l 1590 nm: A.50. B.95.C.100.D.150. Dng5: Xc nh thi gian phng x t, tui th vt cht. a.Phng php:Tng t nh dng 4 :Lu : cc i lng m & m0 , N & N0 , H &H0phi cng n v ..Tui ca vt c: 0 0ln lnln 2 ln 2N m T TtN m= = hay 0 01 1ln lnN mtN m = = . b. Bi tp: Bi 1: Mt ng v phng x c chu k bn r T. C sau mt khong thi gian bng bao nhiu th s ht nhn b phn r trong khong thi gian bng ba ln s ht nhn cn li ca ng v y?A. 2T. B. 3T.C. 0,5T. D. T. Tm tt Giim=3m Theo , ta c :32 .) 2 1 (00==TtTtmmmm

t = ?T 4 2 3 1 2 = = TtTt t = 2T. Chn p n : A Bi2: Mt cht phng x c chu k bn r l 360 gi. Sau bao lu th khi lng ca n ch cn 1/32 khi lng ban u : A. 75 ngyB. 11,25 giC. 11,25 ngyD. 480 ngy Gii T = 360h ; 3210=mm. t? Ta c3210=mm521=5 =Tt t = 5Tt = 1800 gi = 75 ngy. ChnA.Bi3:Lc u mt mu Plni21084Po nguyn cht, c khi lng 2g, cht phng x ny pht ra ht v bin thnh ht nhn X. a) Vit phng trnh phn ng. Nu cu to ht nhn X. b) Ti thi im kho st, ngi ta bit c t s gia khi lng X v khi lng Plni cn li trong mu vt l 0,6. Tnh tui ca mu vt. Cho bit chu k bn r ca Plni l T = 138 ngy, NA = 6,023 x 1023 ht/mol. Gii a) Vit phng trnh : 210 1 A84 2 ZPo He X +Ap dng nh lut bo ton s khi : 210 = 4 + A A = 206 Ap dng nh lut bo ton in tch: 84 = 2 + Z Z = 82 Vy 210 1 20684 2 82Po He Pb + .Ht nhn21084Po c cu to t 82 prtn v 124 ntrn b) Ta c :- S ht Plni ban u : o Aom NNA= ; - S Plni cn li : toN N .e= -S ht Plni b phn r : oN N N = ;toN N (1 e ) = ;- S ht ch sinh ra : tPb oN N N (1 e )= = Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang28 - Khi lng ch to thnh : Pb PbPbAN .AmN= (1); - Khi Plni cn li :( )tom m e 2= ( )( )( ) ( )( )t tPbPb Pb Pbt t tA otA 1 e 1 e1 m N .A 2060, 62 m N .m e A e 210 ee 0, 62 t 95,19 = = = = nga y Bi 4: phng x ca mt tng g bng 0,8 ln phng x ca mu g cng loi cng khi lng va mi cht. Bit chu k ca 14C l 5600 nm. Tui ca tng g l : A. 1900 nmB. 2016 nmC. 1802 nm D. 1890 nm Tm tt Gii m nh nhau Theo ta c :32 , 0 8 , 0 log 8 , 0 220 = = = =TtHHTt. H= 0,8H0t = 0,32T = 0,32.5600 = 1802 nm Chn p nC t = ? Bi 5:Plni 21084Pol cht phng xto thnh ht nhn AZX bn theo phn ng: +210 4 A84 2 ZPo He X. 1) Xc nh tn gi v cu to ht nhn AZX. Ban u c 1gPlni, hi sau bao lu th khi lng Plni ch cn li 0,125g? Cho chu k bn r ca Plni T = 138 ngy. 2) Sau thi gian t bng bao nhiu th t l khi lng gia AZX v Plni l 0,406? Ly= 2 1, 4138. Gii :1) Vit phng trnh phn ng: 210 4 A84 2 ZPo He X +Ap dng nh lut bo ton in tch v s khi ta c:210 4 A A 20684 2 Z Z 82= + = = + =

A 206Z 82X Pb = .Vy X l Pb. 20682Pbc 82 ht prtn v 206 82 = 124 ht ntrnTheo nh lut phng x ta c:ot moTmtTm 1m 2 80,1252= = =hay t3T2 2 = t = 3T = 3 x 138 = 414 ngy2) Gi No l s ht ban u, N l s ht Plni thi im t, ta c N = No - N l s ht Plni b phn r bng s ht ch to ra Theo bi:oPb o APoAN N.206m N N N 206. 0, 406Nm N 210.210N= = =o oN N N 85, 561N N 206 = =oN1 0, 4138 1, 4138 2N = + = = Vy 1 1oT 2N T 1382 2 t 69N 2 2= = = = =ngy Bi 6:Cht phng x urani 238 sau mt lot phng x v th bin thnh ch 206. Chu k bn r ca s bin i tng hp ny l 4,6 x 109 nm. Gi s ban u mt loi ch cha urani khng cha ch. Nu hin nay t l cc khi lng ca urani v ch trong l=u(Pb)m37m th tui ca l bao nhiu? Gii :S ht U 238 b phn r hin nay bng s ht ch pb 206 c to thnh: = = to oN N N N (1 e )Khi lng Pb 206: = (Pb) t(Pb) oAAm N (1 e )N; Khi lng U 238: = =t(U)o(U) (U)A AAN em .N A .N N Gi thit=(U)pbm37m = =tte 37 20632, 0252381 e

=t t(1 e )32, 025.e 1 = =t33, 025e 1, 03132, 025 Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang29 = t ln1, 031 0.03 = 9 80.03t 4, 6 10 2 10 na m0.693 Bi7: Tnh tui ca mt ci tng c bng g, bit rng phng x ca C14 trong tng g bng 0.707 ln phng x trong khc g c cng khi lng va mi cht. Bit chu k bn r C14 l 5600 nm. Gii : Khi lng ca g (mi cht) bng khi lng ca tng g nn phng x ca C14 trong khc g mi cht hin nay l Ho.Do ta c = = = = =t(t ) tToH1 Te 2 0, 707 t 2800H 2 2 nm. Bi8: C hai mu cht phng x A v B thuc cng mt cht c chu kbnr T = 138,2 ngy v c khi lng ban u nh nhau . Ti thi im quan st , t s s ht nhn hai mu cht2, 72BANN= .Tuica mu A nhiu hn mu B l A. 199,8 ngyB. 199,5 ngy C. 190,4 ngyD. 189,8 ngy Gii :NA = N0 1te ; NB = N0 2te . 2 1( )1 2ln 22, 72 ( ) ln 2, 72t t BANe t tN T = = = => t1 t2 = ln 2, 72199, 506 199, 5ln 2T= = ngy. p nB :199,5 ngy Bi9:Mt pho tng c bng g bit rng phng x ca n bng 0,42 ln phng x ca mt mu g ti cng loi va mi cht c khi lng bng 2 ln khi lng ca pho tng c ny. Bit chu k bn r ca ng v phng xC146 l 5730 nm. Tui ca pho tng c ny gn bng A. 4141,3 nm.B. 1414,3 nm.C. 144,3 nm. D. 1441,3 nm. Gii:Theo bita c: H = 0,42.2 H0 = 0,84 H0. Theo L phng x: H = H0 e-t. => e-t = 0,84 -t = ln0,84=> t = - ln0,84.T/ln2 = 1441,3 nm Bi10:Trong qung urani t nhin hin nay gm hai ng v U238 v U235. U235 chim t l 7,143000. Gi s lc u trI t mihnhthnh t l 2 ng v ny l 1:1. Xc nh tui ca tri t. Chu k bn r ca U238 l T1= 4,5.109 nm.Chu k bn r ca U235 l T2= 0,713.109 nm A: 6,04 t nmB: 6,04 triu nmC: 604 t nmD: 60,4 t nm Gii S ht U235 v U238 khi tri t mi hnh thnh l N0 . S ht U238by gi1Tt2 .0 1= N N S ht U235by gi 2Tt2 .0 2= N N => 92110 . 04 , 61000143 , 7= = tNN(nm)= 6,04 t nm Bi 11. Plni l nguyn t phng x vi chu k bn r l T = 138ngy. 1.Vit phng trnh phng x v khi lng ban u ca polni. Bit H0 = 1,67.1011Bq. 2.Sau thi gian bao lu phng x ca n gim i 16ln. 3.Tm nng lng ta ra khi cht phng x trn phn r ht. HD: 1.X He PoAZ+ 4221084==82206ZA mg gNTA HmA TN mAN mHAA A1 10 . 1. 693 , 0 .. . 693 , 03 000 00= = = = = 2. T ngay T tHHHHTtTt552 4 2 222 16404 0= = = == = 3. Nng lng ta ra do mt phn r l: q = (209,9828-4,0026-205,9744)uc2 = 5,8.10-3.931,5 = 5,4MeV Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang30 Trong m0 = 1mg c N0 = 183 2310 . 867 , 221010 . 10 . 022 , 6= Nng lng ta ra khi phn r N0 ht l: Q = N0.q = 2,867.1018.5,4.1,6.10-13 = 2,47.106J = 2,47MJ Bi 12. Pnli l cht phng x (210Po84) phng ra tia bin thnh 206Pb84, chu k bn r l 138 ngy. Sau bao lu th t s s ht gia Pb v Po l 3 ? A. 276 ngyB. 138 ngyC. 179 ngyD. 384 ngy Gii cch 1:Ti thi im t, t s gia s ht nhn ch v s ht nhn plnitrong mu l 3.Suy ra 3 phn b phn r ,cn li 1 phn(trong 4 phn)Hay cn 1/4 => t1 = 2T=2.138=276 ngy . Gii cch 2:Ta c phng trnh: 210 4 20684 2 82Phong XaPo Pb + Sau thi gian t = ? th3 3PbPb PoPoNN NN= = (1) S ht nhn ch sinh ra l NPb:.206PbPb AmN N =S ht nhn Po cn li NPo:0 o o. .2 .2 210.2oP oPt A A t t tT T TmeN m mN N NA= = = Thay vo ( 1) ta c: o oo. 3. . 210. 3 .206 210. .2 3.206 (2)206210.2 2tPb oP oPTA A Pb Pb oP t tT Tm m mN N m m m = = = vi . .206210Po Pb PoPbPom A mmA = = M: oPooPo1.(1 ).20612.(1 )2102tTPo PbtTmm m m = = Thay vo (2)ta c: oPooPo oPo oPo21.(1 ).20612210. .2 3.206. .(1 ).2 3.21021(1 ).2 3 2 1 3 2 4 2 2 2 2.138 276 y2tt tTT TtTt t tT T TtTmm m mtt T ngT= = = = = = = = = = Bi 13:PlniPo21084l cht phng xv bin thnh chPb20682.Chu k bn r l 138 ngy m. Ban u c 0,168g Po. Hy tnh.a, S nguyn t Po b phn r sau 414 ngy m. b, xc nh lng ch c to thnh trong khong thi gian ni trn. Gii:a, S nguyn t Po cn li sau 414 ngy m: N = T tN/02 = 302Nvi N0 = AN mt 0= 232310 . 004816 , 021010 . 02 , 6 . 168 , 0= ngt. N= 1932310 . 02 , 6210 . 004816 , 0=S nguyn t b phn r: N = N0 N = 48,16.1019 6,02.1019 = 42,14.1019ngt b, S nguyn t Pb c to thnh bng s nguyn t Po b phn r bngN. Khi lng Ch c to thnh:mPb = N .NA N =g 1442 , 010 . 02 , 6206 . 10 . 14 , 422319=Bi 14: xc nh hng s phng x caCo55. Bit rng s nguyn t ca ng v y c mi gi gim i 3,8%. Gii:p dng nh lut phng x:N = N0 e - t

Sau t = 1h s nguyn t b mt i:N = N0 N = N0( 1 -e - t ) (1) Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang31 theo : 0NN = 3,8% (2) T ( 1) v ( 2) ta c: 1 -e - t = 3,8% = 0,038 e - t = 0,962 - t = ln(0,962) = -0,04 Hng s phng x caCo55 l: = 0,04 (h-1). Bi 15: U238 phn r thnh Pb 206 vi chu k bn r 4,47.10^9 nam .Mt khi cha 93,94.10^-5Kg v 4,27.10^-5 Kg Pb .Gi s khi lc u hon ton nguyn cht ch c U238.Tui ca khi l: A.5,28.10^6(nm) B.3,64.10^8(nm) C.3,32.10^8(nam) B.6,04.10^9(nm) Gii: Gi N l s ht nhn U238 hin ti , N0 l s ht U238 lc u Khi N0 = N + N = N + NPb; N = 238m NA; NPb = 206Pb Am N; Theo L phng x: N = N0 e-t------> 238m NA = (238m NA+206Pb Am N)e-t => et = 2062381238206 238mmm Nm N m NPbAPb A A+ =+ = 1,0525 =>0525 , 1 ln2 ln= tT=> t = 3,3 .108 nm. Chn p n C Bi 16:Tim vo mu bnh nhn 10cm3 dung dch chaNa2411 c chu k bn r T = 15h vi nng 10-3mol/lt. Sau 6h ly 10cm3 mu tm thy 1,5.10-8 mol Na24. Coi Na24 phn b u. Th tch mu ca ngi c tim khong: A. 5 lt.B. 6 lt.C. 4 lt. D. 8 lt.Gii:S mol Na24 tim vo mu: n0 = 10-3.10-2 =10-5 mol. S mol Na24 cn li sau 6h:n = n0 e- t =10-5.Tte. 2 ln= 10-5156 . 2 lne= 0,7579.10-5 mol. Th tch mu ca bnh nhnV =lit l 5 05 , 55 , 1578 , 710 . 5 , 110 . 10 . 7579 , 082 5 = = Chn p n A c.TRC NGHIM: Cu 1: o phng x ca mt mu tng c bng g khi lng M l 8Bq. o phng x ca mu g khi lng 1,5M mi cht l 15 Bq. Xc nh tui ca bc tng c. Bit chu k bn r ca C14 l T= 5600 nm A 1800 nm B 2600 nm C 5400 nmD 5600 nm Cu 2: ng v phng x c chu k bn r 14,3 ngy c to thnh trong l phn ng ht nhn vi tc khng i q=2,7.109 ht/s.Hi k t lc bt u to thnh P32, sau bao lu th tc to thnh ht nhn ca ht nhn con t gi tr N= 109 ht/s (ht nhn con khng phng x)A: 9,5 ngy B: 5,9 ngyC: 3,9 ngy D: Mt gi tr khc Gii Cu 2: Tc phn r trong thi gian t l: Tt2 .0 1= N N ;Tc to thnh trong thi gian t l N0= q.t Tc to thnh ht nhn trong thi gian t l) 2 1 (Tt0 = N N =109 .Thu c t 0,667.T= 9,5 ngy * Poloni 21084Pophng x bin thnh ht nhn Pb vi chu k bn r 138 ngy. Lc u c 1g Po cho NA= 6,02.1023 ht. Tr li cc cu 3,4, .Cu 3: Tm tui ca mu cht trn bit rng thi im kho st t s gia khi lng Pb v Po l 0,6.A. 95 ngyB. 110 ngyC. 85 ngy D. 105 ngyCu 4 Sau 2 nm th tch kh He c gii phng KTC . A. 95cm3B. 103,94 cm3C. 115 cm3D.112,6 cm3

Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang32 Cu 5( H- C-2010)Bit ng v phng x 146C c chu k bn r 5730 nm. Gi s mt mu g c c phng x 200 phn r/pht v mt mu g khc cng loi, cng khi lng vi mu g c , ly t cy mi cht, c phng x 1600 phn r/pht. Tui ca mu g c cho l A. 1910 nm.B. 2865 nm. C. 11460 nm. D. 17190 nm. Cu 6. Mt cht phng x c chu k bn r T =10s. Lc u c phng x 2.107Bq cho phng x gim xung cn 0,25.107Bq th phi mt mt khong thi gian bao lu: A.30s. B.20s. C.15s.D. .25s. Dng 6: XC NH PHNG X H a.Phng php: p dng cng thc:H = H0te. vi H0 = .N0; H = .N n v phng x l Bq hoc Ci:1 Ci = 3,7.1010 Bq. Do phi tnh theo n v (j-1); thi gian n v l giy. Bi 1: Mt cht phng x lc u c 7,07.1020 nguyn t. Tnh phng x ca mu cht ny sau 1,57 ( T l chu k bn r bng 8 ngy m) theo n v Bq v Ci. Bi gii:S ht nhn ngt sau t = 1,5T: N = N0te. = 2 2 2 205 , 10/0N N NT t= = => N = 202010 . 5 , 22 210 . 07 . 7=ngt. phng x ti thi im t.: H = .N =Ci Bq NT310142010 . 77 , 610 . 7 , 310 . 056 , 2506 , 2 10 . 2 .3600 . 24 . 8693 , 0.2 ln = = =Bi 2: Cht PlniPo210c chu k bn r T = 138 ngy m. a, Tm phng x ca 4g Plni. b, Hi sau bao lu phng x ca n gim i 100 ln. Bi gii: a, phng x ban u ca 4g Po.H0 = .N0(1) vi 3600 . 24 . 138693 , 0 2 ln= =T(j-1) v21010 . 02 , 6 . 42300= =AN mNAthay s vo (1) =>:H = 6,67.1014 Bq. b, Tm thi gian:H = H0te. HHet 0 .= t =916 100 ln .693 , 0ln10= = ||

\| THHngy. Bi 3:2411Na l mt cht phng x . Sau thi gian 105 h th phng x ca n gim i 128 ln. Chu k bn r ca 2411Na lA. 7,5h B. 15h C. 30h D. 3,75h Gii : Theo , ta c : 1281210= =TtHH 7 =Tt. Vy chu k bn r ca23Na: T =157=t h . ChnB.b.Trc nghim: Cu 1(H 2008): Mt cht phng x c chu k bn r l 3,8 ngy. Sau thi gian 11,4 ngy th phng x (hot phng x) ca lng cht phng x cn li bng bao nhiu phntrm so vi phng x ca lng cht phng x ban u? A.25%.B. 75%.C. 12,5%.D. 87,5%. Cu 2 : Mt lng cht phng xRn22286 ban u c khi lng 1mg. Sau 15,2 ngy phng x gim 93,75%. phng x ca lng Rn cn li l A. 3,40.1011Bq B. 3,88.1011Bq C. 3,58.1011BqD. 5,03.1011Bq Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang33 III. PHN NG HT NHN: 1. Phng trnh phn ng: 3 1 2 41 2 3 4A A A AZ Z Z ZA B C D + + Trng hp phng x:3 1 41 3 4A A AZ Z ZA C D +A l ht nhn m, C l ht nhn con, D l ht hoc + Cc nh lut bo ton - Bo ton s nucln (s khi): A1 + A2 = A3 + A4 - Bo ton in tch (nguyn t s): Z1 + Z2 = Z3 + Z4 - Bo ton ng lng: 1 2 3 4 1 1 2 2 4 3 4 4 m m m m p p p p hay v v v v + = + + = + - Bo ton nng lng: 1 2 3 4X X X XK K E K K + + = +; Trong : E l nng lng phn ng ht nhn;212X x xK m v =l ng nng chuyn ng ca ht X Lu : - Khng c nh lut bo ton khi lng. - Mi quan h gia ng lng pX v ng nng KX ca ht X l: 22X X Xp m K =- Khi tnh vn tc v hay ng nng K thng p dng quy tc hnh bnh hnh V d: 1 2p p p = +

bit

1 2, p p =

2 2 21 2 1 22 p p p p p cos = + +hay 2 2 21 1 2 2 1 2 1 2( ) ( ) ( ) 2 mv mv m v mm v v cos = + +hay1 1 2 2 1 2 1 22 mK m K m K mm K K cos = + +Tng t khi bit

1 1 , p p =

hoc

2 2 , p p =

Trng hp c bit:1 2p p 2 2 21 2p p p = +Tng t khi 1p p hoc 2p p v = 0 (p = 0) p1 = p2 1 1 2 22 2 1 1K v m AK v m A= = Tng t v1 = 0 hoc v2 = 0. 2. Nng lng phn ng ht nhn:E = (M0 - M)c2

Trong : 0 A BM m m = + l tng khi lng cc ht nhn trc phn ng. E0 = m0c2

3 4X XM m m = +l tng khi lng cc ht nhn sau phn ng.E = mc2 Lu : - Nu M0 > M th phn ng to nng lng |E| = |E0-E| di dng ng nng ca cc ht C, D hoc phtn . Cc ht sinh ra c ht khi ln hn nn bn vng hn. -NuM0

2 21 3 4( ) ( ) p p p = +=>21p =2 23 3 4 3 4 42 cos( ; ) p p p p p p = + +2.Mun tnh gc gia htX1v X3 : T( 1 )=> 2 21 3 4 1 3 4( ) ( ) p p p p p p = = 2 21 1 3 1 3 32 cos( ; ) p p p p p p +24p =5. Cc hng s v n v thng s dng + S Avgar:NA = 6,022.1023 mol-1 + n v nng lng: 1eV = 1,6.10-19 J; 1MeV = 1,6.10-13 J + n v khi lng nguyn t (n v Cacbon): 1u = 1,66055.10-27kg = 931,5 MeV/c2 + in tch nguyn t: |e| = 1,6.10-19 C + Khi lng prtn: mp = 1,0073u + Khi lng ntrn: mn = 1,0087u + Khi lng electrn: me = 9,1.10-31kg = 0,0005u Dng 1: Xc nh ht nhn cha bit v s ht (tia phng x) trong phn ng ht nhn. a.Phng php:a)Xc nh tn ht nhn cha bit ( AZXcn thiu) : - p dng nh lut bo ton s khi v in tch . Ch : nn hc thuc mt vi cht c s in tch Zthng gp trong phn ng ht nhn (khng cn quan tm n s khi v nguyn t loi no ch ph thuc vo Z : s th t trong bng HTTH-Mt vi loi ht phng x v c trng v in tch, s khi ca chng : Ht 42He , ht ntron 10n , ht proton 11p , tia 01 e , tia + 01 . +e , tia c bn cht l sng int. b) Xc nh s cc ht ( tia ) phng x pht ra ca mt phn ng : - Thng thng th loi bi tp ny thuc phn ng phnr ht nhn . Khi ht nhn m sau nhiu ln phng x to ra x ht v y ht ( ch l cc phn ng ch yu to loi v ngun phng x + l rt him ) . Do khi gii bi tp loi ny c cho l , nu gii h hai n khng c nghim th mi gii vi+ - Vic gii s ht hai loi tia phng x th da trn bi tp dng a) trn. b. Bi tp: Bi 1 : Tm ht nhn X trong phn ng ht nhn sau : 105Bo + AZX + 84Be A. 31T B. 21D C. 10n D.11p Gii: Xc nhht c Z= ? v A= ? . 42He p dng nh lut bo ton s khi v in tch.Khi suy ra : X c in tch Z =2+ 4 5 =1 v s khi A = 4 + 8 10 = 2. Vy X l ht nhn21Dng v phng x caH. Chn p n B. Bi 2. Trong phn ng sau y :n + 23592U 9542Mo + 13957La + 2X + 7; ht X l A. ElectronB. ProtonC. HliD. NtronGii :Ta phi xc nh c in tch v s khi ca cc tia & ht cn li trong phn ng :10n; 01 p dng nh lut bo ton in tch v s khi ta c : 2 ht X c2Z = 0+92 42 57 7.(-1) = 0 2A = 1 + 235 95 139 7.0 = 2 .Vy suy ra X c Z = 0 v A = 1. l ht ntron10n. Chn p n : D Bi 3 . Ht nhn 2411Na phn r v bin thnh ht nhn X . S khi A v nguyn t s Z c gi trA. A. A. A. A = 24 ; Z =10 B. B. B. B. A = 23 ; Z = 12C CC C. A = 24 ; Z =12D DD D. A = 24 ; Z = 11 Gi Gi Gi Gii : -T bi, ta c din bin ca phn ng trn l : 2411Na X + 01 . -p dng nh lut bo ton in tch v s khi , ta c : X c Z = 11 (-1) = 12. Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang36 v s khi A = 24 0 = 24 ( ni thm X chnh l2412Mg ). Chn p n C. Bi4. Urani 238 sau mot loat pho ng xa va bien tha nh ch. Phng trnh cu a phan ng la: 23892U 20682Pb +x 42He + y01 . y co gia tr l :A AA A. y = 4 B BB B. y = 5 C CC C. y = 6 D DD D. y = 8 Gi Gi Gi Gii: -Bi tp ny chnh l loi ton gii phng trnh hai n , nhng ch l ht c s khi A = 0 , do phng trnh bo ton s khi ch c n x ca ht . Sau thay gi tr x tm c vo phng trnh bo ton in tch ta tm c y.-Chi tit bi gii nh sau : === == = += = +6810 2810 82 92 ). 1 ( 232 206 238 . 0 4yxy xxy xy x. gi tr y = 6. Chn : CBi 5. Sau bao nhiu ln phng x v bao nhiu ln phng x th ht nhn 23290Th bin i thnh ht nhn20882Pb ?A. 4 ln phng x ;6 ln phng x B. 6 ln phng x ; 8 ln phng x C. 8 ln phng x;6 ln phng x D. 6 ln phng x ; 4 ln phng x Gi Gi Gi Gii . - Theo ta c qu trnh phn ng : 23290Th20882Pb+ x42He + y01 .- p dng nh lut bo ton in tch v s khi , ta c : === == = += = +468 268 82 90 ). 1 ( 224 208 232 . 0 4yxy xxy xy x. Vy c 6 ht v 4 ht . Chn p n : D. Bi 6. Cho phn ng ht nhn : T + X + n . X l ht nhn . A.ntronB. proton C. TritiD. tri Gii: - Ta phi bit cu to ca cc ht khc trong phn ng :31T,42He, 10n. -p dng nh lut bo ton in tch v s khi , ta c : X cin tch Z = 2 + 0 1 = 1 & s khiA = 4 + 1 3 = 2. Vy X l21D Chn : D c.TRC NGHIM: Cu 1. Cho phn ng ht nhn X O p F + +168199 , ht nhn X l ht no sau y? A. ;B. -;C. +;D. N. Cu 2. Cho phn ng ht nhn + + Na X Mg22112512 , ht nhn X l ht nhn no sau y? A. ;B. T31 ;C. D21 ;D. P. Cu 3. Cho phn ng ht nhn n Ar X Cl + +37183717 , ht nhn X l ht nhn no sau y? A. H11 ; B. D21 ;C. T31 ;D. He42 . Cu 4. Cho phn ng ht nhn n X T + + 31 , ht nhn X l ht nhn no sau y? A. H11 ;B. D21 ;C. T31 ;D. He42 . Cu 5. Trong dy phn r phng xY X2078223592 c bao nhiu ht v c pht ra? A. 3 v 7. B. 4 v 7. C. 4 v 8. D. 7 v 4 Cu 6. ng vU23492 sau mt chui phng x v bin i thnhPb20682. S phng x v trong chui l A. 7 phng x , 4 phng x ; B. 5 phng x , 5 phng x C. 10 phng x , 8 phng x ; D. 16 phng x , 12 phng x Cu 7. Ht nhn 22688Rabin i thnh ht nhn 22286Rndo phng x A. v -.B. -.C. .D. + Cu 8. Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang37 Dng2: Tm nng lng to ra ca phn ng phnhch, nhit hch khi bit khi lng v tnh nng lng cho nh my ht nhn hoc nng lng thay th : a.Phng php:- Lu phn ng nhit hch hay phn ng phn hch l cc phn ng ta nng lng - Cho khi lng ca cc ht nhn trc v sau phn ng : M0 v M . Tm nng lng to ra khi xy 1 phn ng: Nng lng to ra : E = ( M0 M ).c2 MeV.-Suy ra nng lng to ra trong m gam phnhch (hay nhit hch ) : E = Q.N = Q.ANAm.MeVb. Bi tp: Bi 1: 23592U + 10n 9542Mo + 13957La +210n + 7e-l mt phn ng phnhch ca Urani 235. Bit khi lng ht nhn : mU = 234,99 u ; mMo = 94,88 u ; mLa = 138,87 u ; mn = 1,0087 u.Cho nng sut to nhit ca xng l 46.106 J/kg . Khi lng xng cn dng c th to nng lng tng ng vi 1 gam U phn hch ? A.1616 kg B. 1717 kgC.1818 kg D.1919 kgTm ttGii mU = 234,99 u S ht nhn nguyn t 235U trong 1 gam vt cht U l : mMo = 94,88 u N=ANAm.=21 2310 . 5617 , 2 10 . 02 , 6 .2351=ht . mLa = 138,87 uNng lng to ra khi gii phng hon ton 1 ht nhn235U mn = 1,0087 uphnhch l:E = ( M0 M ).c2= ( mU + mn mMo mLa 2mn ).c2 = 215,3403 MeV q = 46.106 J/kgNng lng khi 1 gam U phn ng phn hch : E = E.N = 5,5164.1023 MeV = 5,5164.1023 .1,6.10 3 J = 8,8262J Khi lng xng m? Khi lng xng cn dng c nng lng tng ngQ = E =>

m191910 . 466 =Q kg. Chn p nD Bi 2 : Cho phn ng ht nhn: X He T D + +423121. Ly ht khi ca ht nhn T, ht nhn D, ht nhn He ln lt l 0,009106 u; 0,002491 u; 0,030382 u v 1u = 931,5 MeV/c2 . Nng lng ta ra ca phn ng xp x bng : A. 15,017 MeV. B. 17,498 MeV.C. 21,076 MeV. D. 200,025 MeV.Tm tt Gii T= 0,009106 uy l phn ng nhit hch to nng lng c tnh theo D= 0,002491 u ht khi ca cc cht. He = 0,030382 u Phixc nh y ht khi cc cht trc v sau phn ng.1u = 931,5 MeV/c2Ht nhn X l n10lntron nn c m = 0.E ? E = ( m sau m trc)c2 = (mHe + mn mH + mT ).c2 = 17,498 MeV Chn p n : B Bi3: Tmnnglngtarakhimthtnhn 23492UphngxtiavtothnhngvThri 23090Th .Choccnng lng lin kt ring ca ht l 7,1 MeV, ca 234U l 7,63 MeV, ca 230Th l 7,7 MeV. A.10,82 MeV.B. 13,98 MeV. C. 11,51 MeV. D. 17,24 MeV. Tm tt Gii Wr = 7,1 MeV y l bi ton tnh nng lng to ra ca mt phn rWrU =7,63 MeVphng x khi bit Wlk ca cc ht nhn trong phnng . WrTh =7,7 MeV.Nn phi xc nh c Wlk t d kinWlk ring ca bi. E ? Wlk U = 7,63.234 =1785,42 MeV ,Wlk Th = 7,7.230 = 1771 MeV , Wlk = 7,1.4= 28,4 MeV E= Wlk sau Wlk trc = Wlk Th+ Wlk Wlk U = 13,98 MeV Chn p n : B Bi 4: Cho phn ng ht nhn sau:MeV n He H H 25 , 310422121+ + +.Bit ht khi caH21 l 2/ 931 1 0024 , 0 c MeV u v u mD= = . Nng lng lin kt ht nhnHe42 l A. 7,7188 MeV B. 77,188 MeVC. 771,88 MeV D. 7,7188 eV Tm tt:Gii Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang38 u mD0024 , 0 = MeV n He H H 25 , 310422121+ + +

2/ 931 1 c MeV u = Nng lng ta ra ca phn ng: lkW E = ( m sau m trc)c2 =Wlksau 2mDc2 Wlk = E +2mDc2 = 7,7188MeVChn p n ABi 5: cho phn ng ht nhn: 31T + 21D 42He + X +17,6MeV . Tnh nng lng to ra t phn ng trn khi tng hp c 2g Hli.A. 52,976.1023 MeV B. 5,2976.1023 MeVC.2,012.1023 MeV D.2,012.1024 MeV Gii: -S nguyn t hli c trong 2g hli: N =AN mA. = 410 . 023 , 6 . 223= 3,01.1023 MeV -Nng lng to ragp N ln nng lng ca mt phn ng nhit hch:E = N.Q = 3,01.1023.17,6 = 52,976.1023 MeV Chn p nA. Bi 6. Cho phn ng ht nhn 31H + 21H 42He + 10n + 17,6 MeV. Tnh nng lng ta ra khi tng hp c1 gam kh heli. Gii 6. Ta c: W = Am.NA. W = 41.6,02.1023.17,6.1,6.10-13 = 4,24.1011 (J). Bi 7. Cho phn ng ht nhn: 31T + 21D 42He + X. Cho ht khi ca ht nhn T, D v He ln lt l 0,009106 u; 0,002491 u; 0,030382 u v 1u = 931,5 MeV/c2. Tnh nng lng ta ra ca phn ng. Gii 7. Phng trnh phn ng: 31T + 21D 42He + 10n. V ht ntron 10n khng c ht khi nn ta c nng lng ta ra l: W = (mHe mT mD)c2 = 17,498 MeV. Bi 8. Cho phn ng ht nhn 3717Cl + X n + 3718Ar. Hy cho bit l phn ng ta nng lng hay thu nng lng. Xc nh nng lng ta ra hoc thu vo. Bit khi lng ca cc ht nhn: mAr = 36,956889 u; mCl = 36,956563 u; mp = 1,007276 u; mn = 1,008665 u; u = 1,6605.10-27 kg; c = 3.108 m/s. Gii 8. Phng trnh phn ng: 3717Cl + 11p 10 + 3718Ar. Ta c: m0 = mCl + mp = 37,963839u; m = mn + mAr = 37,965554u.V m0 < m nn phn ng thu nng lng. Nng lng thu vo:W = (m m0).c2 = (37,965554 37,963839).1,6605.10-27.(3.108)2 = 2,56298.10-13 J = 1,602 MeV. Bi 9. Cho phn ng ht nhn 94Be + 11H 42He + 63Li. Hy cho bit l phn ng ta nng lng hay thu nng lng. Xc nh nng lng ta ra hoc thu vo. Bit mBe = 9,01219 u; mp = 1,00783 u; mLi = 6,01513 u;mX = 4,0026 u; 1u = 931 MeV/c2. Gii 9. Ta c: m0 = mBe + mp = 10,02002u; m = mX + MLi = 10,01773u. V m0 > m nn phn ng ta nng lng; nng lng ta ra: W = (m0 m).c2 = (10,02002 10,01773).931 = 2,132MeV. Bi 10: Cht phng x 21084Po pht ra tia v bin thnh 20682Pb. Bit khi lng ca cc ht l205,9744Pbm u = , 209, 9828Pom u = ,4, 0026 m u= . Tnh nng lng ta rakhi mt ht nhn Po phn r. p n: 5,4 MeV Bi 11:cho phn ng ht nhn:3 2 4 A1 1 2 ZT D He X 17, 6MeV + + + .Hy xc nh tn ht nhn X (s A, s Z v tn) v tnh nng lng to ra khi tng hp c 1 mol He t phn ng trn. Cho sAvgar: NA = 6,02x1023 mol-1 Gii :p dng nh lut bo ton s khi v din tch ta c: + = + = = + = 3 2 4 A A 111 2 Z Z 0 Vy ht X l ht ntron 10n . 23AE N 17.6 105.95x10 MeV = = Bi 12:Cho phn ng ht nhn:+ +37 3717 18Cl X n Ar1) Vit phng trnh phn ng y . Xc nh tn ht nhn X. Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang39 2) Phn ng ta hay thu nng lng. Tnh nng lng ta (hay thu) ra n v MeV. Cho= = =Cl Ar nm 36, 9566u; m 36, 9569u; m 1, 0087u;= =X 2MeVm 1, 0073u;1u 931c Gii 12:1) Phn ng ht nhn:+ +37 A 1 3717 Z 0 18Cl X n Arnh lut bo ton s khi: 37 + A = 1 + 37 => A = 1 nh lut bo ton in tch: 17 + Z = 0 + 18 => Z = 1 Vy=11X H (Hir) + +37 1 1 3717 1 0 18Cl H n Ar2) Nng lng phn ng:Tng khi lng 1M v 2M ca ht trc v sau phn ng l= + =1 Cl HM m m 37, 9639u= + =2 n ArM m m 37, 9656uTa thy phn ng thu nng lng , Nng lng thu vo = 22 1E (M M )cThay s = = 2E 0, 0017uc 0, 0017 931MeV 1, 58MeV c.TRC NGHIM: Cu 1. Cht phng xPo21084 pht ra tia v bin i thnhPb20682. Bit khi lng cc ht l mPb = 205,9744u, mPo = 209,9828u, m = 4,0026u. Nng lng ta ra khi 10g Po phn r ht l A. 2,2.1010J; B. 2,5.1010J; C. 2,7.1010J; D. 2,8.1010J Cu2.ChophnnghtnhnMeV 6 , 17 n H H2131+ + + ,bitsAvgarNA=6,02.1023.Nnglngtorakhi tng hp c 1g kh hli l bao nhiu? A. E = 423,808.103J.B. E = 503,272.103J. C. E = 423,808.109J.D. E = 503,272.109J. Cu3.Chophnnghtnhn n Ar p Cl37183717+ + ,khilngcacchtnhnlm(Ar)=36,956889u,m(Cl)= 36,956563u, m(n) = 1,008670u, m(p) = 1,007276u, 1u = 931MeV/c2. Nng lng m phn ng ny to ra hoc thu vo l bao nhiu? A. To ra 1,60132MeV. B. Thu vo 1,60132MeV. C. To ra 2,562112.10-19J.D. Thu vo 2,562112.10-19J. Cu4.NnglngtithiucnthitchiahtnhnC126thnh3htlbaonhiu?(bitmC=11,9967u,m= 4,0015u). A. E = 7,2618J. B. E = 7,2618MeV.C. E = 1,16189.10-19J.D. E = 1,16189.10-13MeV. Cu 5. Trong phn ng v ht nhn urani U235 nng lng trung bnh to ra khi phn chia mt ht nhn l 200MeV. Khi 1kg U235 phn hch hon ton th to ra nng lng l: A. 8,21.1013J; B. 4,11.1013J; C. 5,25.1013J; D. 6,23.1021J. Cu6.Phnnghtnhn:He He H Li42421173+ + .BitmLi=7,0144u;mH=1,0073u;mHe4=4,0015u,1u= 931,5MeV/c2. Nng lng to ra trong phn ng l: A. 7,26MeV; B. 17,42MeV; C. 12,6MeV; D. 17,25MeV.Cu 7. Phn ng ht nhn:He H T H42113221+ + . Bit mH = 1,0073u; mD = 2,0136u; mT = 3,0149u;mHe4 = 4,0015u, 1u = 931,5MeV/c2. Nng lng to ra trong phn ng l: A. 18,35MeV; B. 17,6MeV; C. 17,25MeV;D. 15,5MeV. Cu8.Phnnghtnhn:He He H Li42422163+ + .BitmLi=6,0135u;mD=2,0136u;mHe4=4,0015u,1u= 931,5MeV/c2.Nng lng to ra trong phn ng l: A. 17,26MeV; B. 12,25MeV; C. 15,25MeV; D. 22,45MeV. Cu 9. Phn ng ht nhn:He He H Li42321163+ + . Bit mLi = 6,0135u; mH= 1,0073u; mHe3 = 3,0096u, mHe4 = 4,0015u,1u = 931,5MeV/c2. Nng lng to ra trong phn ng l: A. 9,04MeV; B. 12,25MeV; C. 15,25MeV;D. 21,2MeV. Cu 10. Ht nhn triti (T) v teri (D) tham gia phn ng nhit hch sinh ra ht v ht ntrn. Cho bit ht khi cahtnhntritilmT=0,0087u,cahtnhnterilmD=0,0024u,cahtnhnXlm=0,0305u;1u= 931MeV/c2. Nng lng to ra t phn ng trn l. A. E = 18,0614MeV.B. E = 38,7296MeV. C. E = 18,0614J. D. E = 38,7296J. Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang40 Cu 11. Trong phn ng v ht nhn urani U235 nng lng trung bnh to ra khi phn chia mt ht nhn l 200MeV. Mt nh my in nguyn t dng nguyn liu u rani, c cng sut 500.000kW, hiu sut l 20%. Lng tiu th hng nm nhin liu urani l: A. 961kg;B. 1121kg; C. 1352,5kg;D. 1421kg. Cu 12. Trong phn ng tng hp hli:He He H Li42421173+ +Bit mLi = 7,0144u; mH = 1,0073u; mHe4 = 4,0015u,1u = 931,5MeV/c2. Nhit dung ring ca nc l c = 4,19kJ/kg.k-1. Nu tng hp hli t 1g liti th nng lng to ra c th un si mt nc 00C l: A. 4,25.105kg; B. 5,7.105kg; C. 7,25. 105kg;D. 9,1.105kg. Dng3: Xc nh phn ng ht nhn ta hoc thu nng lng a.Phng php:-Xt phn ng ht nhn :A + B C + D .-Khi : + M0 = mA + mBl tng khi lng ngh ca cc ht nhn trc phn ng . + M =mC + mDl tng khi lng ngh ca cc ht nhn sau phn ng . - Ta c nng lngca phn ng c xc nh : E = ( M0 M)c2 +nuM0 > M E > 0 : phn ng to nhit . + nuM0 < M E < 0 : phn ng thu nhit . b. Bi tp: Bi 1 :Thc hin phn ng ht nhn sau :2311Na + 21D 42He+ 2010Ne .Bit mNa = 22,9327 u ; mHe = 4,0015 u ; mNe = 19,9870 u ; mD = 1,0073 u.Phn ngtrn to hay thu mt nng lngbng bao nhiu J ? A.thu 2,2375 MeV B. to 2,3275 MeV. C.thu 2,3275 MeV D. to 2,2375 MeVGii -Ta c nng lngca phn ng ht nhn trn l :E= ( M0 M ).c2 = ( mNa + mHe mNe mD )c2 =2,3275 MeV> 0 y l phn ng to nng lng. Chn p n B. Bi2:Chophnnght nhn: n Ar H Cl103718113717+ + phnngtrn tahaythubaonhiunnglng?BitmCl= 36,956563u, mH = 1,007276u, mAr =36,956889u, 1u = 931MeV/c2 Tm ttGii: Xc nh phn ngTnh E ta hay thu nng lngE= ( mCl + mH mAr mn ) 931= -1,6 MeV mCl = 36,956563uPhn ng thu nng lng 1,6MeV mH = 1,007276u, mAr =36,956889u1u = 931MeV/c2. . E ? Bi 3 : ng v Plni 21084Pol cht phng x v to thnh ch (Pb). 1) Vit phng trnh phn r v nu thnh phn cu to ca ht nhn ch to thnh. 2) Nng lng ta ra trong phn ng trn di dng ng nng ca ht v ht nhn ch. Tnh ng nng mi ht. Gi thit ban u ht nhn Plni ng yn. ChomPo = 209,9828u; mHe =4,0015u; mPb = 205,9744u;=2MeV1u 931c. Gii bi 3: 1) Phng trnh: +210 4 A84 2 ZPo He X trong Z = 84 - 2 = 82; A = 210 -4 =20620682X: Pb Phng trnh phn ng: +210 4 206Po He Pb84 2 82 Ht nhn 20682Pbc 82 prtn v 206 -82 = 124 ntrn. 2) Nng lng ta ra trong mi phn ng: = = 2o0E (M M)C 209, 9828 4, 0015 205, 9744)x931MeVM = + + =Pb PbE K K K K 6, 24MeV Ap dng nh lut bo ton ng lng: = + =

PbPb Pb PbmO m V m V V Vm (1) Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang41 Hay | |= = = |\ 22 Pb PbPb Pbm m 1 1K m V m V K2 2 m m= 51,5KPb(2) T (1) v (2) => KPb = 0,12MeV,= K 6,12MeV Bi 4 :Cho phn ng ht nhn:+ +37 3717 18Cl X n Ar1) Vit phng trnh phn ng y . Xc nh tn ht nhn X. 2) Phn ng ta hay thu nng lng. Tnh nng lng ta (hay thu) ra n v MeV. Cho= = =Cl Ar nm 36, 9566u; m 36, 9569u; m 1, 0087u; = =X 2MeVm 1, 0073u;1u 931c Gii Bi 4 1) Phn ng ht nhn:+ +37 A 1 3717 Z 0 18Cl X n Arnh lut bo ton s khi: 37 + A = 1 + 37 => A = 1 nh lut bo ton in tch: 17 + Z = 0 + 18 => Z = 1 Vy=11X H (Hir)+ +37 1 1 3717 1 0 18Cl H n Ar2) Nng lng phn ng:Tng khi lng 1M v 2M ca ht trc v sau phn ng l= + =1 Cl HM m m 37, 9639u= + =2 n ArM m m 37, 9656u Ta thy phn ng thu nng lng ; Nng lng thu vo = 22 1E (M M )cThay s = = 2E 0, 0017uc 0, 0017 931MeV 1, 58MeV Bi 5(H-2011) : Gi s trong mt phn ng ht nhn, tng khi lng ca cc ht trc phn ng nh hn tng khi lng ca cc ht sau phn ng l 0,02 u. Phn ng ht nhn ny A. ta nng lng 1,863 MeV.B. ta nng lng 18,63 MeV. C. thu nng lng 1,863 MeV.D. thu nng lng 18,63 MeV. HD:m0 < m : phn ng thu nng lng. Nng lng phn ng thu vo :W = ( m m0 ).c2|= 0,02.931,5 = 18,63MeV c.TRC NGHIM: Cu 1. Cho phn ng ht nhnn P Al30152713+ + , khi lng ca cc ht nhn l m = 4,0015u, mAl = 26,97435u, mP = 29,97005u, mn = 1,008670u, 1u = 931Mev/c2. Nng lng m phn ng ny to ra hoc thu vo l bao nhiu? A. To ra 4,275152MeV.B. Thu vo 2,67197MeV.C. To ra 4,275152.10-13J. D. Thu vo 2,67197.10-13J. Cu 2. Cho ht prtn c ng nng KP = 1,8MeV bn vo ht nhnLi73 ng yn, sinh ra hai ht c cng ln vn tcvkhngsinhratiavnhitnng.Chobit:mP=1,0073u;m=4,0015u;mLi=7,0144u;1u=931MeV/c2= 1,66.1027kg. Phn ng ny thu hay to bao nhiu nng lng? A. To ra 17,4097MeV. B. Thu vo 17,4097MeV. C. To ra 2,7855.10-19J. D. Thu vo 2,7855.10-19J. Dng4. ng nng v vn tc ca cc ht trong phn ng ht nhn . a.Phng php:a) Xt phn ng ht nhn :A + B C + D . Hay:11AZX1 + 22AZX2 33AZX3 + 44AZX4. Bo ton s nucln: A1 + A2 = A3 + A4. Bo ton in tch: Z1 + Z2= Z3 + Z4. Bo ton ng lng: m11v + m22v= m33v+ m44v . Bo ton nng lng: (m1 + m2)c2 + 21m1v21+ 21 m2v22 = (m3 + m4)c2 +21 m3v23+ 21 m4v24. Lin h gia ng lng p = mvv ng nng W = 21mv2: p2 = 2mW. b) Khi bit khi lng y ca cc cht tham gia phn ng . -Ta s p dng nh lut bo ton nng lng : Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang42 M0c2 + KA +KB = Mc2 + KC +KD E = (M0 M )c2 Nn:E + KA + KB = KC + KD -Du ca E cho bit phn ng thu hay ta nng lng -Khi nng lng ca vt (nng lng ton phn) lE = mc2 = 22201cvc m -Nng lng E0 = m0c2 c gi l nng lng ngh v hiu sE E0 = (m - m0)c2 chnh l ng nng ca vt. c) Khi bit khi lng khng y v mt vi iu kin v ng nng v vn tc ca ht nhn . -Ta s p dng nh lut bo ton ng lng : D C B AP P P P + = +-Lu : mPK mK P2222= = (K l ng nng ca cc ht ) d) Dng bi tp tnh gc gia cc ht to thnh. Cho ht X1 bn ph ht X2 (ng yn p2 = 0) sinh ra ht X3 v X4 theo phng trnh: X1+ X2 =X3+X4

Theo nh lut bo ton ng lng ta c: 1 3 4(1) p p p = +Mun tnh gc gia hai ht no th ta quy v vect ng lng ca ht ri p dng cng thc:

2 2 2( ) 2 cos( ; ) a b a ab a b b = +1.Mun tnh gc gia htX3v X4 ta bnh phng hai v (1) =>2 21 3 4( ) ( ) p p p = +=>21p =2 23 3 4 3 4 42 cos( ; ) p p p p p p = + +2.Mun tnh gc gia htX1v X3 : T( 1 )=> 2 21 3 4 1 3 4( ) ( ) p p p p p p = = 2 21 1 3 1 3 32 cos( ; ) p p p p p p +24p = b. Bi tp: Bi 1: Ht bn vo ht nhn Alng yn gy ra phn ng : + 2713Al 3015P + n. phn ng ny thu nng lng Q= 2,7 MeV. Bit hai ht sinh ra c cng vn tc, tnh ng nng ca ht . ( coikhi lnght nhnbng s khi ca chng). A. 1,3 MeVB. 13 MeV C. 3,1 MeVD. 31 MeV Gii :Ta c nPnpmmKK==30 Kp = 30 KnM Q = K ( Kp + Kn )(1) p dngnh lut bo ton ng lng: m .v = ( mp + mn)v n Pm mv mv+= M tng ng nng ca h hai ht :Kp + Kn = ) 2 () ( 2) ( 12) (21222n P n Pn Pn Pn Pm mK mm mv mm mv m m mv m m+=+=|||

\|++= + Th(2) vo (1) ta c K = 3,1MeV Chn p nC. Bi 2: ngi ta dng ht prtn c ng nng Wp= 2,69 MeVbn vo ht nhnLiti ng yn thu c 2 ht c cng ng nng . cho mp = 1,,0073u; mLi = 7,0144u; m =4,0015u ; 1u = 931 MeV/c2 . tnh ng nng v vn tc ca miht to thnh? A. 9,755 MeV ; 3,2.107m/sB.10,05 MeV ; 2,2.107 m/s C. 10,55 MeV ; 3,2.107 m/s D. 9,755.107 ; 2,2.107 m/s.Gii: Nng lng ca phn nght nhnl :Q= ( M0 M ).c2= 0,0187uc2 = 17,4097 MeV. Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang43 p dng nh lut bo tonnng lng ta c Q +Wp= 2W W = MeVW Qp05 , 102=+ Vn tc ca mi ht l: v = 0015 , 4 . 9312Wc =2,2.107m/s. Chn p nB. Bi 3: Mt ntron c ng nng Wn = 1,1 MeV bn vo ht nhn Liti ng yn gy ra phn ng: 10 n + 63Li X+ 42He . Bit ht nhn He bay ra vung gc vi ht nhnX.ng nng ca ht nhn X v He ln lt l :?Chomn = 1,00866 u;mx = 3,01600u ; mHe = 4,0016u; mLi = 6,00808u. A.0,12MeV & 0,18MeVB. 0,1MeV & 0,2MeV C.0,18MeV & 0,12 MeVD. 0,2MeV & 0,1MeVGii: Ta c nng lng ca phn ng:Q= ( mn+ mLi m x m He).c2= - 0,8 MeV(y l phn ngthu nng lng ) -p dng nh lut bo tonng lng: + =X Hep p pn 2 2 2X He nP P P + = 2mnWn= 2mHe .W He + 2mx Wx (1) -p dng nh lut bo ton nng lng:Q =Wx +W He Wn = -0,8 (2) T (1),(2) ta c h phngtrnh:=== += +1 , 02 , 03 , 01 , 1 3 4XHeX HeX e HWWW WW WMeV Chn p nB. Bi 4. Cho phn ng ht nhn 23090Th 22688Ra + 42He + 4,91 MeV. Tnh ng nng ca ht nhn Ra. Bit ht nhn Th ng yn. Ly khi lng gn ng ca cc ht nhn tnh bng n v u c gi tr bng s khi ca chng. 1. Theo nh lut bo ton ng lng ta c: +He Rap p = 0pRa = pHe = p. V W =22mv= mp22, do : W = WRa + WHe = He Rampmp2 22 2+ = 5 , 56222 2RaRampmp+ = 57,5Ramp22= 57,5WRaWRa = 56 , 57W= 0,0853MeV. Bi 5. Dng ht prtn c ng nng 1,6 MeV bn vo ht nhn liti (73Li ) ng yn. Gi s sau phn ng thu c hai htgingnhauccngngnngvkhngkmtheotia.Bitnnglngtaracaphnngl17,4MeV.Vit phng trnh phn ng v tnh ng nng ca mi ht sinh ra. Gii . Phng trnh phn ng: 11p + 73Li 242He.Theo nh lut bo ton nng lng ta c: Wp + W = 2WHeWHe = 2W Wp += 9,5 MeV. Bi 6. Bn ht c ng nng 4 MeV vo ht nhn 147N ng yn th thu c mt prton v ht nhn 108O. Gi s hai ht sinh ra c cng tc , tnh ng nng v tc ca prton. Cho: m = 4,0015 u; mO = 16,9947 u; mN = 13,9992 u; mp = 1,0073 u; 1u = 931 MeV/c2; c = 3.108 m/s. Gii. Theo LBT ng lng ta c: mv = (mp + mX)vv2 = 22 2) (X pm mv m+ = 2) (2X pdm mW m+ ;Wp = 21mpv2 = 2) (X pd pm mW m m+ = 12437,7.10-6W = 0,05MeV = 796.10-17 J;v = pdpmW 2= 271710 . 66055 , 1 . 0073 , 110 . 796 . 2= 30,85.105 m/s. Bi 7. Dng mt prtn c ng nng 5,45 MeV bn vo ht nhn 94 Be ang ng yn. Phn ng to ra ht nhn X v ht . Ht bay ra theo phng vung gc vi phng ti ca prtn v c ng nng 4 MeV. Tnh ng nng ca ht nhn X v Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang44 nnglngtaratrongphnngny.Lykhilngcchttnhtheonvkhilngnguyntbngskhica chng. Gii . Theo nh lut bo ton ng lng ta c: X pp p p + =. Vp vv pp pp2X= p2p+ p2 2mX21mXv2X=2mp21mpv2X+2m21mv2Xhay2mXWX=2mpWp+2mWWX= 64 pW W +=3,575MeV. Theo nh lut bo ton nng lng ta c: (mp + mBe)c2 + Wp = (m + mX)c2 + W + WX Nng lng ta ra: W = (mp + mBe - m - mX)c2 = W + WX - Wp = 2,125 MeV. Bi 8. Ht nhn 23492U ng yn phng x pht ra ht v ht nhn con 23090Th (khng km theo tia ). Tnh ng nng ca ht . Cho mU = 233,9904 u; mTh = 229,9737 u; m = 4,0015 u v 1 u = 931,5 MeV/c2. Gii . Theo nh lut bo ton ng lng:p+ Thp= 0p = mv = pTh = mThvTh2mW = 2mThWTh WTh = ThmmW. Nng lng ta ra trong phn ng l: W = WTh + W = ThThm mm+W = (mU mTh-m)c2 W = ( )Th U ThThm m m mm m +c2 = 0,01494 uc2 = 13,92 MeV. Bi 9. Ht nhn 22688Ra ng yn phn r thnh ht v ht nhn X (khng km theo tia ). Bit nng lng m phn ng ta ra l 3,6 MeV v khi lng ca cc ht gn bng s khi ca chng tnh ra n v u. Tnh ng nng ca ht v ht nhn X. Gii . Phng trnh phn ng: 22688Ra 42 + 22286Rn. Theo nh lut bo ton ng lng:p+ Xp= 0p = mv = pX = mXvX2mW = 2mXWX

WX = XmmW. Nng lng ta ra trong phn ng l: W = WX + W = XXm mm+W W = WXXmm m+= 3,536 MeV; WX = XmmW = 0,064 MeV. Bi 10. Ngi ta dng mt ht c ng nng 9,1 MeV bn ph ht nhn nguyn t N14 ng yn. Phn ng sinh ra ht phtn p v ht nhn nguyn t xy O17

1)Hi phn ng thu hay ta bao nhiu nng lng (Tnh theo MeV)? 2)Gi s ln vn tc ca ht prtn ln gp 3 ln vn tc ca ht nhn xy. Tnh ng nng ca ht ? Cho bitmN = 13,9992u;m 4, 0015u;=mp = 110073u; O17m 16, 9947u; = 1u = 931MeV/C2 Gii . 1.Phng trnh phng x: 4 14 1 172 7 1 8He N H O + +M = M0 M = mHe + mN mH - mO M = 4,0015u + 13,9992u 1,0073u 16,9947y = -13 X 10-3u M < 0: phn ng thu nng lng. Nng lng thu vo l: 2 3E M c 1, 3 10 931MeV = = Hay E = 1,21MeV. 2.Tng ng nng ca prtn v ht nhn xy l:Tp + To = 9,1 1,21 = 7,89 MeV M 2 2P p 0 0 O 01 1T m (3V ) v T m v2 2= =2p0200p p 00p0T1.9v 9T 17vT T TT 7, 899 17 9 17 267, 89T 9 2, 73MeV267, 89T 17 5,16 MeV26 = =+ = = =+ = = = = Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang45 Bi 11. Bn htvo ht nhn 147N th ht nhn xy v ht prtn sau phn ng. Vit phng trnh ca phn ng v cho bit phn ng l phn ng ta hay thu nng lng? Tnh nng lng ta ra (hay thu vo) v hy cho bit nu l nng lng ta ra th di dng no, nu l nng lng thu th ly t u? Khi lng ca cc ht nhn: Nm 4, 0015u; m 13, 9992u;= =2O Pm 16, 9947u; m 1, 0073u;1u 931MeV/ c = = = . Gii . Phng trnh phn ng:+ +4 14 17 12 7 8 1He N O H = + He N O Hm m m m mm = (4,0015 + 13,9992 16,9947 1,0073)u = -1,3 x 310u< 0m < 0 phn ng thu nng lng.Nng lng thu vo l: = = =2 3E m c 1,3x10 x9,31 1, 2103MeV .Nng lng thu vo ly t ng nng ca ht n . Bi 12: Ht nhn Plni 21084Pong yn, phng x chuyn thnh ht nhn AZX. Chu k bn r ca Plni l T = 138 ngy. Mt mu Plni nguyn cht c khi lng ban u=om 2g . a) Vit phng trnh phng x. Tnh th tch kh Heli sinh ra iu kin tiu chun sau thi gian 276 ngy. b) Tnh nng lng ta ra khi lng cht phng x trn tan r ht. c) Tnh ng nng ca ht .Cho bit=Pom 209, 9828u , = m 4, 0015u ,=Xm 205, 9744u , =21u 931MeV/ c , =23 1AN 6, 02x10 mol . Gii . a) Phng trnh ca s phng x: +210 4 A84 2 ZPo He X Ta c = + = = + = 210 4 A A 20684 2 z Z 82Vy ht nhn AZX l 20682PbVy phng trnh phng x l: +210 4 20684 2 82Po He PbS ht Plni ban u:=oo APomN .Nm S ht Plni cn li thi im t: = = =t oPo o2N NN N e42 S ht Hli sinh ra thi im t bng sht Plni b phn r = = = = =20 o oHe o o o APoN m 3 3N N N N N .N 43x104 4 4 m ht.Lng kh Hei sinh ra iu kin tiu chun: =HeANV x22, 4N th sV = 0,16 lt b) Nnglng ta ra khi mt ht Po phn r:[ ] = = 2 2Po PbE mc m m m c[ ] = = E 209, 9828 205, 9744 4, 0045 x931 6, 424MeV. Nng lng ta ra khi 2g Po phn r ht:= =22oE N E 3, 683x10 MeV MeV c) Tnh ng nng ca ht .Theo nh lut bo ton nng lng v ng lng: = + =XE K K 6, 424 (1) + = X P P 0 =2 2XP P =X X2m .K 2m .K =XXmK Km(2) Thay (2) vo (1) ta c: = +XmE K Km

=+XXm . EKm mThay s K= 6,3 MeV. Chuyn : VT L HT NHN N THI H-C GV:on Vn Lng Email: doanvluong@gmail.com; doanvluong@yahoo.comT : 0915718188 - 0906848238Trang46 Bi 13. Ngi ta dng prtn c ng nng WP = 5,58MeV bn ph ht nhn 2311Nang yn, to ra phn ng: 23 A11 Np Na Ne + + 1) Neu cac nh luat bao toan trong phan ng hat nhan va cau tao cua hat nhan Ne. 2) Biet ong nang cu a hat a la Wa = 6,6 MeV, tnh ong nang cua hatnha n Ne. Cho mp = 1,0073u; mNa = 22,985u; mNe = 19,9869u; m = 4,9915; lu = 931MeV / c2. Gii . 1) Trong pha n ng hat nhan so nucleon c bao toan Trong phan ng hat nha n ie n tch c bao toan Trong phan ng hat nhan ong lng va na ng lng c ba o toa n Ta co: 1 + 23 = A + 4 A = 20 1 + 11 = Z + 2 Z = 10 Hat nha n Neo n (Ne) co 10 proton va 10 ntron 1)Ta co ( mp + mNa )c2 + Wp = ( mNe + m)c2 + WNe + W WNe = (mp + mNa mNe - m)c2 + Wp W. The so: WNe = 39 x 10-4 x 93