1 TI LIU N TP L THUYT N TH L (CB) TNTHPT I H 2011 HN I. DAO N I. DAO N IU HO 1. Phng trnh dao ng:- nh ngha: dh l 1 d c m t bng 1 nh lut dng cos (hoc sin), trong A, e, l nhng hng s - Chu k:T = 1f= 2te= tn (trong n l s dao ng vt thc hin trong thi gian t) + Chu k T: L khong thi gian vt thc hin c 1 d ton phn. n v ca chu k l giy (s). + Tn s f: L s d ton phn thc hin c trong 1 giy. n v l Hc (Hz). - Tn s gc: e = 2tf = 2tT ; - Phng trnh dao ng: x = Acos(et + ) + x : Li d, l khong cch t VTCB n v tr ca vt ti thi im t ang xt(cm) + A: Bin d, l li cc i (cm). c trng cho mnh yu ca dh. Bin cng ln nng lng d cng ln. Nng lng ca vt dh t l vi bnh phng ca bin .+ e: Tn s gc ca d (rad/s). c trng cho s bin thin nhanh chm ca cc trng thi ca dh. Tn s gc ca d cng ln th cc trng thi ca d bin i cng nhanh. + : Pha ban u ca d (rad). xc nh trng thi ban u ca d, l i lng quan trng khi tng hp d. + (et + ) : Pha ca d ti thi im t ang xtLu : Trong qu trnh vt d th li bin thin iu ha theo hm s cos (x thay i theo thi gian t), nhng cc i lng A, e, l nhng hng s. Ring A, e l nhng hng s dng. 2. n tc tc thi: v = x = -eAsin(et + ) = eAcos(et + +t/2)vlun cng chiu vi chiu chuyn ng (vt chuyn ng theo chiu dng th v>0, theo chiu m th v v = -eAsin(et + ) = eAcos(et + + t/2) ==> a = -e2Acos(et + ) = e2Acos(et + + t) 8. hiu di qu o: s = 2A 9. Qung ng i trong 1 chu k lun l 4A; trong 1/2 chu k lun l 2A Qung ng i trong l/4 chu k khi vt i t VTCB n v tr bin hoc ngc li l A. 10.c bc lp phng trnh dao ng dao ng iu ho: x = Acos(et + ) - Tm A : + T VTCB ko vt 1 on x0 ri bung tay cho d th A = x0 + T pt: A2 = x2 + v2e2hoc A2 = x2 + mv2k

+ A = s/2 vi s l chiu di qu o chuyn ng ca vt 2 A-Ax1 x2M2M1M'1M'2OAA+ T ct :vmax = eA ==> A = vmaxe + A = smax-smin2

+ Tm e : e = km; e = gAl;e = 2tf = 2tT... + Tm : Ty theo u bi. Chn t = 0 l lc vt c li x = [ ] , vn tc v = [ ] ==>x = Acos = [ ] v = -eAcos = [ ]==> = [ ? ] Lu : + Vt chuyn ng theo chiu dng th v > 0, ngc li v < 0 + C th x bng cch v ng trn lng gic v k ban u.11. Khong thi gian ngn nht vt i t v tr c li x1 n x2 - S dng mi lin h gia dao ng iu ho v chuynng trn u. - Da vo cng thc ca c trn u: A = e.At==> .Tt2A AA = =e t

- Ch : A l gc qut c ca bk ni vt c trong khong tgian At v do ta phi x ta u x1 tng ng gc 1 v ta cui x2 tng ng gc 2. 12. Qung ng vt i c t thi im t1 n t2.- S ln vt dao ng c trong khong thi gian t: 0tn ...T= ===> t = t2 t1 = nT + At (n e N; 0 At < T)- Qung ng i c trong thi gian nT l S1 = 4nA, trong thi gian At l S2. - Qung ng tng cng l S = S1 + S2 - Lu : + NuAt = T/2 th S2 = 2A + Tnh S2 bng cch nh v tr x1, x2 v chiu chuyn ng ca vt trn trc Ox + Trong mt s trng hp c th gii bi ton bng cch s dng mi lin h gia dao ng iu ho v chuyn ng trn u s n gin hn. + Tc trung bnh ca vt i t thi im t1 n t2: 2 1tbSvt t= vi S l qung ng tnh nh trn. 13. Bi ton tnh qung ng ln nht v nh nht vt i c trong khong thi gian 0 < At < T/2. -VtcvntclnnhtkhiquaVTCB,nhnhtkhiquavtrbinnntrongcngmtkhongthigian qung ng i c cng ln khi vt cng gn VTCB v cng nh khi cng gn v tr bin. - S dng mi lin h gia dao ng iu ho v chuynng trn u.Gc qut A = eAt. - Qung ng ln nht khi vt i t M1 n M2 i xng qua trc sin (hnh 1) max2Asin2A= S - Qung ng nh nht khi vt i t M1 n M2 i xng qua trc cos (hnh 2) 2 (1 os )2A= minS A c - Lu : Trong trng hp At > T/2Tch'2Tt n t A = + Atrong *; 0 '2Tn N t e < A Al (Vi Ox hng xung): - Thi gian l xo nn 1 ln l thi gian ngn nht vt i t v tr x1 = -Al n x2 = -A. - Thi gian l xo gin 1 ln l thi gian ngn nht vt i t v tr x1 = -Al n x2 = A, Lu : Trong mt dao ng (mt chu k) l xo nn 2 ln v gin 2 ln . Lc ko v hay lc hi phc- c im:* L lc gy dao ng iu ha cho vt. * Lun hng v VTCB * Bin thin iu ho cng tn s vi li - Lc lm vt dh l lc hi phc: Fhp = -kx = -me2x===> Fhp max = kA = me2A l lc vt i qua cc v tr bin. Fhp min = 0 lc vt qua VTCB. . Lc n hi l lc a vt v v tr l xo khng bin dng: x A -A Al Nn 0 Gin Hnh v th hin thi gian l xo nn v gin trong 1 chu k Al gin O x A -A nn Al gin O x A -A Hnh a (A < Al) Hnh b (A > Al)4 C ln Fh = kx (x l bin dng ca l xo) * Vi con lc l xo nm ngang th lc ko v v lc n hi l mt (v ti VTCB l xo khng bin dng) * Vi con lc l xo thng ng: + ln lc n hi c biu thc: * Fh = k,Al + x, vi chiu dng hng xung * Fh = k,Al - x,vi chiu dng hng ln + Lc n hi cc i (lc ko): FMax = k(Al + A) = FKmax(lc vt v tr thp nht) + Lc n hi cc tiu: * Nu A < Al FMin = k(Al - A) = FKMin * Nu A Al FMin = 0 (lc vt i qua v tr l xo khng bin dng) ==>Lc y (lc nn) n hi cc i: FNmax = k(A - Al) (lc vt v tr cao nht) 6. Lu :- Trong mt dao ng (mt chu k) l xo nn 2 ln v gin 2 ln - Vt dh i chiu chuyn ng khi lc hi phc t gi tr ln nht. - Th nng ca vt dh bng ng nng ca n khi 2Ax = 7. Mt l xo c cng k, chiu di l c ct thnh cc l xo c cng k1, k2, v chiu di tng ng l l1, l2, th c: kl = k1l1 = k2l2 = 8. Ghp l xo:* Ni tip 1 21 1 1...k k k= + + cng treo mt vt khi lng nh nhau th: T2 = T12 + T22 * Song song: k = k1 + k2 + cng treo mt vt khi lng nh nhau th:2 2 21 21 1 1...T T T= + +9.nlxokvovtkhilngm1cchukT1,vovtkhilngm2cT2,vovtkhilngm1+m2 c chu k T3, vo vt khi lng m1 m2 (m1 > m2) c chu k T4. Th ta c:2 2 23 1 2T T T = +v2 2 24 1 2T T T = 10. o chu k bng phng php trng phng xc nh chu k T ca mt con lc l xo (con lc n) ngi ta so snh vi chu k T0 ( bit) ca mt con lc khc (T ~ T0). Hai con lc gi l trng phng khi chng ng thi i qua mt v tr xc nh theo cng mt chiu. Thi gian gia hai ln trng phng 00TTT Tu = Nu T > T0 u = (n+1)T = nT0.Nu T < T0 u = nT = (n+1)T0. vi n e N* III. ON L N 1. Tn s gc: gle = ; chu k: 22lTgtte= = ; tn s: 1 12 2gfT let t= = =iu kin dao ng iu ho: B qua ma st, lc cn v o0 0 F E || ; cn nu q < 0 F E |+ ) * Lc y csimt: F = DgV( F lung thng ng hng ln) Trong : D l khi lng ring ca cht lng hay cht kh. g l gia tc ri t do. 6 V l th tch ca phn vt chm trong cht lng hay cht kh . - Khi :' P P F = +gi l trng lc hiu dng hay trng lc biu kin (c vai tr nh trng lcP ) 'Fg gm= +gi l gia tc trng trng hiu dng hay gia tc trng trng biu kin. hu k dao ng ca con lc n khi :lT' = 2g' - Cc trng hp c bit: *Fc phng ngang: + Ti VTCB dy treo lch vi phng thng ng mt gc c:tanFPo = + 2 2' ( )Fg gm= +*F c phng thng ng th'Fg gm= + NuFhng xung th'Fg gm= + + NuFhng ln th 'Fg gm= IV. TN HP DAO N 1. Tng hp hai dao ng iu ho cng phng cng tn s x1 = A1cos(et + 1) v x2 = A2cos(et + 2) c mt dao ng iu ho cng phng cng tn s x = Acos(et + ). Vi: - Bin ca d tng hp :A2 = A12 + A22 + 2A1A2cos(2 - 1) - Pha ban u ca d tng hp:tg = A1sin1 + A2sin2 A1cos1 + A2cos2 + Khi 2 d cng pha:A = 2kt ==>A = A1 + A2 + Khi 2 d ngc pha:A = (2k + 1)t==>A =A1 A2 | ,A1 - A2, A A1 + A2 2. Khi bit mt dao ng thnh phn x1 = A1cos(et + 1) v dao ng tng hp x = Acos(et + ) th dao ng thnh phn cn li l x2 = A2cos(et + 2). Trong : 2 2 22 1 1 12 os( ) A A A AAc = + ; 1 121 1sin sintanos osA AAc Ac = 3. Nu mt vt tham gia ng thi nhiu dao ng iu ho cng phng cng tn s x1 = A1cos(et + 1); x2 = A2cos(et + 2) th dao ng tng hp cng l dao ng iu ho cng phng cng tn s x = Acos(et + ).Chiu ln trc Ox v trc Oy Ox . Ta c: 1 1 2 2os os os ...xA Ac Ac A c = = + +1 1 2 2sin sin sin ...yA A A A = = + +2 2x yA A A = +vtanyxAA = vi e[Min;Max] . DAO N TT DN DAO N N B - N HN 1. L thuyt chung: - D tt dn l d c bin gim dn theo thi gian. Nguyn nhn l do ma st, do lc cn ca mi trng. - D cng bc l d chu tc dng ca 1 lc cng bc tun hon. in ca d cng bc h thuc vo A vca lc cng bc. - D duy tr l d c duy tr bng cch gi cho bin khng i m khng lm thay i chu k d ring. - D ring l d vi bin v tn s ring (f0) khng i, ch ph thuc vo cc c tnh ca h d. -Hintngcnghnglhintngbincad cngbctngngitrccikhitns(f)calc cngbcbngtnsdring(f0)cahd.Hin tng cng hng cng r nt khi lc cn, lc ma st ca mi trng cng nh. ==>Hintngcnghngxyrakhi:f=f0haye= e0 hay T = T0 Vif,e,Tvf0,e0,T0ltns,tnsgc,chuk ca lc cng bc v ca h dao ng.2.MtconlcdaongttdnvibinAhs ma st .T AA x t 0 7 ngnn x: - GiS l qung - ng i - c k t lc chuyn ng cho n khi dng hn. C nng ban u bng tng cng ca lc ma st trn ton b qung - ng , tc l: 21 kA2kA = F .SS =ms2 2Fms.- Qung ng vt i c n lc dng li l: 2 2 2 22 2 2= = =mskA kA ASF mg ge - gim bin sau mi chu k l: 24 4 mg gAk eA = = - S dao ng thc hin c: 24 4A Ak ANA mg ge = = =A - Thi gian vt dao ng n lc dng li:.4 2AkT At N Tmg gte A = = = (Nu coi dao ng tt dn c tnh tun hon vi chu k 2Tte= ) ngnnn: + Suy ra, gim bin di sau mt chu k: 4FmsS =2m + S dao ng thc hin - c: SSNA=0 + Thi gian k t lc chuyn ng cho n khi dng hn: l = N.T = N.2g + GiS l qung - ng i - c k t lc chuyn ng cho n khi dng hn. C nng ban u bng tng cng ca lc ma st trn ton b qung - ng , tc l:12 2m S = F .S S =?ms02 HN II. SN V SNG M I. SN H 1. c khi nim: - Sng c l s lan truyn d trong 1 mi trng vt cht (khng truyn c trong chn khng). Khi sng c truyn i ch c pha d c truyn i cn cc phn t vt cht ch d xung quanh VTCB c nh. - Sng dc l sng c c phng dao ng song song hoc trng vi phng truyn sng. Sng dc truyn c trong cht kh, lng, rn.- Sng ngang l sng c c phng d vung gc vi phng truyn sng. Sng ngang truyn c trn b mt cht rn v trn mt nc. 2. Phng trnh sng: - Ti im O:u0 = acos(et + ) - Ti im M1 : uM1 = acos[e(t -d1v ) + ]= acos[2t1d tT | | |\ .+ ] = acos(et + - t12d) - Ti im M2 : uM2 =acos(et + + t22d) vi u : l li ca sng;a: l bin sng ; e : l tn s gc vi: d1 lk/c t ngun pht sng n im M1;d1vl thi gian sng truyn t 0 n M - Bc sng :v = T==> = vT = vf Vi v l vn tc truyn sng (m/s): v ph thuc vo b/c ca mi trng truyn sng. l bc sng (m); T l chu k dao ng ca sng (s) ;f l tn s d ca sng (Hz). - Gi k/c gia 2 im M v N trn phng truyn sng l d, v k/c t 2 im n ngun sng ln lt l d1, d2. Ta c:d =d1 d2 |- Gi lch pha gia 2 im M v N trn phng truyn sng l A, th lch pha l : A = 2td

- Vy 2 im M v N trn phng truyn sng s: + dao ng cng pha khi:d = k vi k = 0, 1, 2 ... O x M1 d2 M2 d1 8 d1 0NN d d2 M + dao ng ngc pha khi:d = (2k + 1)2 + dao ng vung pha khi: d = (2k + 1)4

Lu : n v ca x, x1, x2, v v hi tng ng vi nhau Trong hin tng truyn sng trn si dy, dy c kch thch dao ng bi nam chm in vi tn s dng in lth tn s dao ng ca dy l 2. II. SN DN 1. Mt s ch - Sng dng l s giao thoa ca sng ti v sng phn x, khi sng ti v sng phn x truyn theo cng mt phng. Khi sng ti v sng phn x l sng kt hp v giao thoa to sng dng. - u c nh hoc u dao ng nh l nt sng. - u t do l bng sng - Hai im i xng vi nhau qua nt sng lun dao ng ngc pha. - Hai im i xng vi nhau qua bng sng lun dao ng cng pha. - Cc im trn dy u dao ng vi bin khng i nng lng khng truyn i - Khong thi gian gia hai ln si dy cng ngang (cc phn t i qua VTCB) l na chu k. -Khongcchgiahaibngsnglinkl/2.Khongcchgiahaintsnglinkl/2.Khongcch gia mt bng sng v mt nt sng lin k l /4. - B rng ca bng sng = 2.A = 2.2a = 4.a 2. iu kin c sng dng trn si dy di l: - Hai u l nt sng: *( )2l k k N= e S bng sng = s b sng (mi) = k ; S nt sng = k + 1 - Mt u l nt sng cn mt u l bng sng:(2 1) ( )4l k k N= + eS b (mi) sng nguyn = k = s bng sng tr 1; S bng sng = s nt sng = k + 1 2. Phng trnh sng dng: - Pt sng ti im M trn dy c 2 u c nh, d l k/c t M n u c nh, l l k/c t ngun (d vi bin nh, coi l nt) n im c nh: M2d 2l u = 2aCos( - )Cos(t - + ) 2 2 - Pt sng ti M trn dy c 1 u c nh 1 u t do, d l k/c t M n u t do, l l k/c t ngun (d vi bin nh, coi l nt) n u t do: M2d 2lu = 2aCos( )Cos(t - ) III. GIAO THOA SNG - Hin tng giao thoa sng l s tng hp ca 2 hay nhiu sng kt hp trong khng gian, trong c nhng chbinsngctngcng(ccigiaothoa)hoctrittiu(cctiugiaothoa),tuthucvohiu ng i ca chng. - iu kin xy ra hin tng giao thoa l hai sng phi l hai sng kt hp. - Hai sng kt hp l hai sng c gy ra bi hai ngun c cng tn s, cng pha hoc lch pha nhau mt gc khng i. - V tr cc im dao ng vi bin cc i : d2 d1 = k V tr cc im dao ng vi bin cc tiu:d2 d1 = 2k + 1/2 - Giao thoa ca hai sng pht ra t hai ngun sng kt hp S1, S2 cch nhau mt khong l: + Xt im M cch hai ngun ln lt d1, d2 + Phng trnh sng ti 2 ngun 1 1Acos(2 ) u ft t = +; 2 2Acos(2 ) u ft t = ++ Phng trnh sng ti M (cch 2 ngun ln lt l d1 v d2) do hai sng t hai ngun truyn ti: 11 1Acos(2 2 ) = +Mdu ft t t v22 2Acos(2 2 ) = +Mdu ft t t + Phng trnh giao thoa sng ti M: uM = u1M + u2M ==> 2 1 1 2 1 22 os os 22 + +((= + (( Md d d du Ac c ft t t t 9 + Bin dao ng ti M: 2 12 osMd dA A c t | |= |\ .

- Ch :* S cc i, tnh c 2 ngun:+(k Z)2 2A A + s s el lk t t * S cc tiu, tnh c 2 ngun:s s el 1 l 1 - - + k - +(k Z) 2 2 2 2 1. Hai ngun dao ng cng pha 1 20 A= = ): - im dao ng cc i:d2 d1 = k (keZ) S ng hoc s im (tnh c hai ngun): s sl lk - im dao ng cc tiu (khng dao ng): d2 d1 = (2k+1)2 (keZ) S ng hoc s im (tnh c hai ngun):1 12 2 s s l lk 2. Hai ngun dao ng ngc pha:1 2 t A= = ) - im dao ng cc i:d2 d1 = (2k+1)2 (keZ) S ng hoc s im (tnh c hai ngun):1 12 2 s s l lk - im dao ng cc tiu (khng dao ng): d2 d1 = k (keZ) S ng hoc s im (tnh c hai ngun): s sl lk 3. Ch : Vi bi ton tm s ng dao ng cc i v khng dao ng (cc tiu) gia hai im M, N cch hai ngun ln lt l d1M, d2M, d1N, d2N. t AdM = d1M - d2M ; AdN = d1N - d2N v gi s AdM < AdN. + Hai ngun dao ng cng pha: -Cc i: AdM < k < AdN -Cc tiu: AdM < (k+0,5) < AdN + Hai ngun dao ng ngc pha: -Cc i:AdM < (k+0,5) < AdN -Cc tiu: AdM < k < AdN ==> S gi tr nguyn ca k tho mn cc biu thc trn l s ng cn tm. IV. SNG M - Sng m l nhng sng c truyn trong cc mi trng rn lng kh. Ngun m l cc vt dao ng. - Sng m thanh (gy ra cm gic m trong tai con ngi) l sng c hc c tn s trong khong t 16 Hz n 20000 Hz. < 16 Hz sng h m, > 20000 Hz sng siu m. Sng m truyn c trong cc mi trng rn lng v kh, khng truyn c trong chn khng. - Vn tc truyn m ph thuc vo tnh n hi, mt v nhit ca mi trng. vrn > vlng > vkh. - Khi sng m truyn t mi trng ny sang mi trng khc th vn tc v bc sng thay i. Nhng tn s v do chu k ca sng khng i. -Ngng nghe: l gi tr cc tiu ca cng m gy cm gic m trong tai con ngi. Ngng nghe thay i theo tn s m. - Ngng au: l gi tr cc i ca cng m m tai con ngi cn chu ng c (thng thng ngng au l ng vi mc cng m l 1db) - Cm gic m to hay nh khng nhng ph thuc vo cng m m cn ph thuc vo tn s m. - Tnh h v m n s m, ngm h m ngm v hngm. + ng m: W PI= =tS S(W/m2) Vi W (J), P (W) l nng lng, cng sut pht m ca ngun S (m2) l din tch mt vung gc vi phng truyn m (vi sng cu-ngun m l ngun m im- th S l din tch mt cu, viS=4R2) P=W/t=I.S==>Cngsutmcangun=lngnnglngmmtruynquadintchmtcu trong 1 n v thi gian: P0 = W0 = I.S = I.4R2. 10 Nu ngun m im pht m qua 2 im A v B, th:

2A B A AA B A B 2 2A B B BP P I RI ; I do P P4 R 4 R I R| |= = = = |t t\ . + Mc cng m:0( ) lgIL BI=Hoc 0( ) 10.lgIL dBI= Vi I0 = 10-12 W/m2 = 1Hz: cng m chun.Khi gii thng p dng t/c ca lgart: loga (M.N) = logaM + logaN: loga (M/N) = logaM logaN. - Tnh h sinh m (gn in vi n s), (gn in vi m ngm) v m s (gn in vi hngm) - Tn s do n pht ra (hai u dy c nh hai u l nt sng):( k N*)2vf kl= eng vi k = 1 m pht ra m c bn c tn s 12vfl=k = 2,3,4 c cc ho m bc 2 (tn s 2f1), bc 3 (tn s 3f1) - Tn s do ng so pht ra (mt u bt kn, mt u h mt u l nt sng, mt u l bng sng): (2 1) ( k N)4vf kl= + eng vi k = 0 m pht ra m c bn c tn s 14vfl=k = 1,2,3 c cc ho m bc 3 (tn s 3f1), bc 5 (tn s 5f1) HN III. DN IN XOAY HIU 1. ch to ra dxc: Cho khung dy dn din tch S, c N vng dy, quay u vi tn s gc e trong t trng uB( B trc quay) . Th trong mch c d bin thin iu ha vi tn s gc e gi l dxc. Lu : Khi khung dy quay mt vng mt chu k th dng in chy trong khung i chiu 2 ln. a, T thng qua khung:u = NBScos(et + ) Hin tng cm ng in t: L hin tng khi c s bin thin ca t thng qua mt khung dy kn th trong khung xut hin mt sut in ng cm ng sinh ra mt d cm ng: e = -ut = eNBSsin(et + ) = eNBScos(et + - t/2) = E0 cos(et + - t/2). b, Biu thc in p tc thi v dng in tc thi: u = U0cos(et + u) vi = I0cos(et + i) Trong : i l gi tr cng d ti thi im t; I0 > 0 l gi tr cc i ca i; e > 0 l tn s gc; (et + i) l pha ca i ti thi im t;i l pha ban u ca d. u l gi tr in p ti thi im t; U0 > 0 l gi tr cc i ca u; e > 0 l tn s gc; (et + u) l pha ca u ti thi im t;u l pha ban u ca in p.Vi = u il lch pha ca u so vi i, c 2 2t t s sc c gi tr hiu dng: - Cng hiu dng ca dxc l i lng c gi tr bng cng ca mt d khng i, sao cho khi i qua cng mt in tr R, trongcng mt khong thi gian th cng sut tiuth ca R bi d khngi y bng cng sut tiu th trung bnh ca R bi dxc ni trn. - in p hiu dng cng c nh ngha tng t. - Gi tr hiu dng bng gi tr cc i ca i lng chia cho2 . 0 0 0; ;2 2 2U I EU I E = = =2. Mt s ch :- Dng in xoay chiu i = I0cos(2tft + i) Mi giy dng in i chiu 2f ln * Nu pha ban u i = 2thoc i = 2t th ch giy u tin i chiu 2f-1 ln. - ng thc tnh thi gian n hunh quang sng trong mt chu k: Khi t in p u = U0cos(et + u) vo hai u bng n, bit n ch sng ln khi u U1.

4teAA =Vi 10osUcU A = ,(0 < A < t/2)(At: thi gian n sng trong 1 chu k) 11 - C// = C1 + C2; Cnt = (C1C2) : (C1 + C2); L// = (L1L2) : (L1 + L2);Lnt = L1 + L2. 3. Dng in xoay chiu trong on mch RL - on mch ch c in tr thun R: uR cng pha vi i, = u i = 0, UIR=v 00UIR=Lu : in tr R cho dng in khng i i qua v c UIR= - on mch ch c cun thun cm L: uL nhanh pha hn i l t/2, = u i = t/2 LUIZ=v 00LUIZ=vi ZL = eL l cm khng Lu : Cun thun cm L cho dng in khng i i qua (khng cn tr). - on mch ch c t in C: uC chm pha hn i l t/2, = u i = -t/2

CUIZ=v 00CUIZ=vi 1CZC e=l dung khng Lu : T in C khng cho dng in khng i i qua (cn tr hon ton). - on mch RL khng phn nhnh 2 2 2 2 2 20 0 0 0; ( ) ( ) ( )L C R L C R L CUI Z R Z Z U U U U U U U UZ= = + = + = + tan ; sin ; osL C L CZ Z Z Z RcR Z Z = = = vi 2 2t t s s+ Khi ZL > ZC hay 1LCe > > 0 th u nhanh pha hn i , mch c tnh cm khng. + Khi ZL < ZC hay 1LCe 0,85. - Ch :+ vi mch LC th cos = 0 , mch khng tiu th in!P = 0 + in nng tiu th: A = P.t vi A tnh bng J, P tnh bng W, t tnh bng s. + K c cng hng in: 21 1L CZ Z LC LCe ee= = =+ Khi c cng hng in th:. d t cc i Imax = UR v cng sut tiu th t cc i Pmax = U2R . u cng pha vi i: = 0, u = i; U = UR ;UL = UC; cos = RZ= 1==> R = Z. 5. My pht in xoay chiu mt pha: - Hot ng da trn hin tng cm ng in t, bin c nng thnh in nng.- Cu to gm 3 b phn :+ B phn to ra t trng gi l phn cm : L cc nam chm + B phn to ra dng in gi l phn ng: L khung dy 12 + B phn a d ra ngoi gi l b gp: Gm 2 vnh khuyn v 2 chi qut - Trong cc my pht in: Rto l phn cm ; Stato l phn ng. - Trong my pht in cng sut nh khng trnh by trong chng trnh ph thng: Rto (b phn chuyn ng) l phn ng ; Stato (b phn ng yn) l phn cm. - Tn s dng in do my pht pht ra :f = np60. Vi p l s cp cc, n l s vng quay ca rto/pht. = np. Vi p l s cp cc, n l s vng quay ca rto/giy. - T thng gi qua khung dy ca my pht in u = NBScos(et +) = u0cos(et + )Vi u0 = NBS l t thng cc i, N l s vng dy, B l cm ng t ca t trng, S l din tch ca vng dy, e = 2tf- Sut in ng trong khung dy: e = - | = eNBSsin(et +)= eNSBcos(et + - 2t) = E0cos(et + - 2t) Vi E0 = eNSB l sut in ng cc i. 6. My pht in xoay chiu ba pha: - My pht in xc ba pha l my to ra ba s xc hnh sin cng tn s, cng bin v lch nhau mt gc 2t3

- Cu to: Phn ng l ba cun dy ging nhau gn c nh trn mt ng trn tm 0 ti ba v tr i xng, t lch nhau 1 gc 1200. Phn cm l mt nc c th quay quanh trc 0 vi tc gc e khng i. - Hot ng da trn hin tng cm ng in t, bin c nng thnh in nng. Khi nam chm quay t thng qua mi cun dy l ba hm s sin ca thi gian, cng tn s gce, cng bin v lch nhau 1200. Kt qu trong ba cun dy xut hin ba s xc cm ng cng bin , cng tn s v lch pha nhau gc 1200. (Lu : khi dng in1 trongcun dy t cc i I0 th dng in trong 2 cun cn li = ,I0) - Dng in xoay chiu ba pha l h thng ba dng in xoay chiu, gy bi ba sut in ng xoay chiu cng tn s, cng bin nhng lch pha tng i mt l 23t 1 02 03 0os( )2os( )32os( )3e E c te E c te E c tetete== = +trong trng hp ti i xng th1 02 03 0os( )2os( )32os( )3i I c ti I c ti I c tetete== = + - My pht mc hnh sao: Ud =3 Up - My pht mc hnh tam gic: Ud = Up - Ti tiu th mc hnh sao: Id = Ip - Ti tiu th mc hnh tam gic: Id =3 Ip 7. My bin p: - Hot ng: Da trn hin tng cm ng in t. - Cu to: + Li bin p: L cc l st non pha silic ghp li. Tc dng dn t. + Hai cun dy qun: -Cun dy s cp D1 c hai u ni vi ngun in c N1 vng. -Cun dy th cp D2 c hai u ni vi ti tiu th c N2 vng. -Tc dng ca hai cun dy l dn in. - Tc dng ca MBA: bin i in p ca dxc m vn gi nguyn tn s. MBA khng c tc dng bin i nng lng (cng). - Cng thc my bin p: 1 1 2 12 2 1 2U E I NkU E I N= = = =-Nu k > 1: N1 > N2 U1 > U2 : MBA h p. -Nu k < 1: N1 < N2 U1 < U2 : MBA tng p. - Ch : MBA tng in p bao nhiu ln th lm gim d i by nhiu ln v ngc li. - Hiu sut MBA:H = P2P1 = U2I2cos2 U1I1cos1

- ng dng ca MBA: Trong truyn ti v s dng in nng.V d: Ch cn tng in p u ng dy ti in ln 10 ln th c th gim hao ph i 102 = 100 ln. 13 8. ng sut hao ph trong qu trnh truyn ti in nng: 22 idy dy 2iPP R I R(U cos )A = = - Trong : P: cng sut truyn i ni cung cp;U: in p ni cung cp; cos: h s cng sut ca dy ti in (thng thng cos = 1);dlRS= l in tr tng cng ca dy ti in(u : dn in bng 2 dy)- gim in p trn ng dy ti in: AU = RdI - Hiu sut ti in: n ii iP P PHP PA= =9. ng c khng ng b ba pha: - Hot ng : Da trn hin tng cm ng in t v t trng quay. - Cu to: Gm hai b phn chnh l: -Rto (phn cm): L khung dy c th quay di tc dng ca t trng quay. -Stato (phn ng): Gn 3 cun dy ging ht nhau t ti 3 v tr nm trn 1 vng trn sao cho 3 trc ca 3 cun dy y ng qui ti tm 0 ca vng trn v hp nhau nhng gc 1200. - Khi cho dxc 3 pha vo 3 cun dy y th t trng tng hp do 3 cun dy to ra ti tm 0 l t trng quay.B = 1,5B0 vi B l t trng tng hp ti tm 0, B0 l t trng do 1 cun dy to ra. T trng quay ny s tc dng vo khung dy l khung quay vi tc nh hn tc quay ca t trng. Chuyn ng quay ca rto (khung dy) c s dng lm quay cc my khc. (Lu : khi dng in1 trongcun dy t cc i I0 th dng in trong 2 cun cn li = ,I0) - u im: + Cu to n gin, d ch to.+ S dng tin li, khng cn vnh khuyn chi qut. + C th thay i chiu quay d dng. 10. on mch c LR e thay i:a on mch RL c R thay i - Khi thay i R Pmax,t pt: P = RI2 = RU2Z2 = U2R + (ZL - ZC)2R(1) Pmax khi

((R + (ZL - ZC)2R min

((R + (ZL - ZC)2R min theo bt ng thc cosi ==> R = (ZL - ZC)2R R =ZL ZC | Pmax khi R =ZL ZC |(2) Lc ny t (1) v (2) ta c : Pmax = U22R =L C2U2 - ;cos =RR 2=22

- Khi R = R1 hoc R = R2 th P c cng 1 gi tr ta c R1 R2 tha mn pt bc 2: PR2 - U2R + P(ZL-ZC)2 = 0 ==> R1 + R2 = U2/P ;R1R2 = (ZL ZC)2. b, on mch RL c L thay i: * Khi ZL = ZC hay 21LC e=th IMax URmax; PMax cn ULCMin * Khi 2 2CLCR ZZZ+=th 2 2axCLMU R ZUR+=* Vi L = L1 hoc L = L2 th UL c cng gi tr th ULmax khi 1 21 21 22 1 1 1 1( )2L L LL LLZ Z Z L L= + =+ * Khi 2 242C CLZ R ZZ+ +=th ax2 22 R4RLMC CUUR Z Z=+ c. on mch RL cthay i: * Khi ZL = ZChay 21CL e=th IMax URmax; PMax cn ULCMin * Khi 2 2LCLR ZZZ+=th 2 2axLCMU R ZUR+=* Khi C = C1 hoc C = C2 th UC c cng gi tr th UCmax khi 1 21 21 1 1 1( )2 2C C CC CCZ Z Z+= + =14 * Khi C = C1 hoc C = C2 th cng sut P c cng gi tr th: 1 2C C LZ Z 2.Z + = * Khi 2 242L LCZ R ZZ+ +=th ax2 22 R4RCML LUUR Z Z=+

d. on mch RL c e thay i: * Khi ZL = ZChay1LCe =th IMax URmax; PMax cn ULCMin * Khi 21 12CL RCe = th ax2 22 .4LMU LUR LC R C= * Khi 212L RL Ce = th ax2 22 .4CMU LUR LC R C= * Vi e = e1 hoc e = e2 th I hoc Phoc UR c cng mt gi tr th IMax hoc PMax hoc URMax khi

1 2e ee = tn s 1 2f f f =11. Hai on mch AM gm R1L1C1 ni tip v on mch MB gm R2L2C2 ni tip mc ni tip vi nhau c UAB = UAM + UMB uAB; uAM v uMB cng pha tanuAB = tanuAM = tanuMB 12. Hai on mch R1L1C1 v R2L2C2 cng u hoc cng i c pha lch nhau A Vi 1 111tanL CZ ZR=v 2 222tanL CZ ZR=(gi s 1 > 2) C 1 2 = A 1 21 2tan tantan1 tan tan = A+ Trng hp c bit A = t/2 (vung pha nhau) th tan1tan2 = -1. HN I. DAO N SN IN T 1. Kin thc chung: - Mch dao ng l 1 mch in gm 1 cun cm c t cm L mc ni tip vi 1 t in c in dung C thnh 1 mch in kn. - Nu in tr ca mch rt nh, coi nh bng khng, th mch l 1 mch ao ng l tng. - T in c nhim v tch in cho mch, sau n phng in qua li trong mch nhiu ln to ra mt dxc trong mch. - Khi trong mch c 1 dao ng in t vi cc tnh cht : + Nng lng ca mch d gm c nng lng in trng tp trung t in v nng lng t trng tp trung cun cm. + Nng lng in trng v nng lng t trng cng bin thin tun hon theo 1 tn s chung. + Ti mi thi im, tng ca nng lng in trng v nng lng t trng l khng i, ni cch khc nng lng ca mch dao ng c bo ton. - Dao ng in t t do: S bin thin iu hotheo thi gian ca in tch q v cng dng in i (hoc cng in trng E v cm ng t B) trong mch dao ng c gi l dao ng in t t do. - Khi 1 t trng bin thin theo thi gian th n sinh ra 1 in trng xoy (l 1 in trng m cc ng sc bao quanh cc ng cm ng t). Ngc li khi mt in trng bin thin theo thi gian n sinh ra 1 t trng xoy (l 1 t trng m cc ng cm ng t bao quanh cc ng sc ca in trng) - Dng in qua cun dy l d dn, d qua t in l d dch (l s bin thin ca in trng gia 2 bn t) - in trng v t trng l 2 mt th hin khc nhau ca 1 loi trng duy nht l in t trng. - Sng in t l s lan truyn trong khng gian ca in t trng bin thin tun hon theo thi gian. Sng in t l 1 sng ngang do n c 2 thnh phn l thnh phn in Ev thnh phn t Bvung gc vi nhau v vung gc vi phng truyn sng. Cc vect E, B,v lp thnh 1 tam din thun (xoay inh c vect Etrng vect B th chiu tin ca inh c trng vi chiu ca vect v)- Sng in t c mi t/c nh sng c hc (phn x, giao thoa, to sng dng...), ngoi ra n cn truyn c trong chn khng. - pht sng in t ngi ta mc phi hp 1 my pht dao ng iu ho vi 1 ngten (l 1 mch d h) - thu sng in t ngi ta mc phi hp 1 ngten vi 1 mch dao ng c tn s ring iu chnh c ( xy ra cng hng vi tn s ca sng cn thu). 15 - Nng lng ca sng t l vi bnh phng ca bin , vi lu tha bc 4 ca tn s. Nn sng cng ngn (tn s cng cao, do =cf ) th nng lng sng cng ln.+ Sng di : dng thng tin di nc. + Sng trung: dng thng tin mt t, vo ban m thng tin tt hn ban ngy.+ Sng ngn: dng thng tin mt t, k c ngy hay m. Do t b khng kh hp th, mt khc sng ngn phn x tt trn mt t v trn tng in li, nn c th truyn i xa.+ Sng cc ngn: dng thng tin v tr. - Sng di: bc sng 103 m; tn s 3.105 Hz. Sng trung: bc sng 102 m; tn s 3.106Hz. Sng ngn: bc sng 101 m; tn s 3.107 Hz. Sng cc ngn: bc sng vi mt; tn s 3.108 Hz. 2. Dao ng in t - in tch tc thi:q = q0cos(et + ) - Hiu in th (in ) tc thi:00os( ) os( )q qu c t U c tC Ce e = = + = +- Dng in tc thi: i = q = -eq0sin(et + ) = I0cos(et + +2t) ==> u, q dao ng cng pha; i sm pha hn u, q 1 gc t/2. - Cm ng t:0os( )2B B c tte = + + Trong : 1LCe =l tn s gc ring2 T LC t =l chu k ring

12fLC t=l tn s ring 00 0qI = q =LC

0 00 0 0q I LU LI IC C Cee= = = =- Nng lng in trng:22C1 1 qW = Cu = qu =2 2 2C22 0os ( )2qc tCe = +- Nng lng t trng:22 0q= sin (t +)2C2L1W = Li2 - Nng lng in t: W=W Wt+ 22 2 00 0 0 0q 1 1 1W= CU = q U = = LI2 2 2C 2 s h : - Mch dao ng c tn s gc e, tn s f v chu k T th W v Wt bin thin vi tn s gc 2e, tn s 2f v chu k T/2- Mch dao ng c in tr thun R = 0 th dao ng s tt dn. duy tr dao ng cn cung cp cho mch mt nng lng c cng sut: 2 2 2 22 0 0 C U U RCP = I R = R =2 2L - Khi t phng in th q v u gim v ngc li Quy c: q > 0 ng vi bn t ta xt tch in dng th i > 0 ng vi dng in chy n bn t m ta xt. - Mi lin h gia cc gi tr u i U0 v I0: 2 2 202 2 20Lu + i = UCCu +i = IL - c quay ca t xoay: + Cng thc x in dung ca t in phng: 9.SC4 .9.10 .dc=t + Khi t quay t omin n o ( in dung t Cmin n C) th gc xoay ca t l: 16 minmin max minmax minC C.( )C CAo = oo = o o + Khi t quay t v tr omax v v tr o ( in dung t C n Cmax) th gc xoay ca t l: maxmax max minmax minC C.( )C CAo = o o = o o - ch cp nng lng ban u cho mch dao ng: + Cp nng lng ban u cho t: 2C1W C.E2= ; E l sut in ng ca ngun, C l in dung t + Cp nng lng ban u cho cun dy: 2 2L 01 1 EW LI L( )2 2 r= = ; r l in tr trong ca ngun - Cho mch dao ng vi L c nh. Mc L vi C1 c tn s dao ng l f1, mc L vi C2 c tn s l f2. + Khi mc ni tip C1 vi C2 ri mc vi L ta c tn s f tha :22212f f f + =+ Khi mc song song C1 vi C2 ri mc vi L ta c tn s f tha :222121 1 1f f f+ =3. S tng t gia dao ng in v dao ng c i lng ci lng inDao ng cDao ng in xqx + e 2x = 0q + e 2q = 0 vi kme =1LCe =mLx = Acos(et + )q = q0cos(et + ) k 1Cv = x = -eAsin(et + )i = q = -eq0sin(et + ) Fu 2 2 2( )vA xe= +2 2 20( )iq qe= +RF = -kx = -me2x2qu L qCe = =WWt (WC)W =12mv2 Wt = 12Li2 WtW (WL)Wt = 12kx2 W =22qC 4. Sng in t - Vn tc lan truyn trong khng gian v = c = 3.108m/s - My pht hoc my thu sng in t s dng mch dao ng LC th tn s sng in t pht hoc thuc bng tn s ring ca mch. - Bc sng ca sng in t cc2 LCf = = tVi: c: vn tc as trong chn khng;C: in dung ca t in (F);L: t cm ca cun dy (H). - Lu : Mch dao ng c L bin i t LMin LMax v C bin i t CMin CMax th bc sng ca sng in t pht (hoc thu) Min tng ng vi LMin v CMinMax tng ng vi LMax v CMax . 5. S khi ca my pht v thu thanh v tuyn n gin: - S khi ca my pht thanh v tuyn n gin: Micr (1) to ra dao ng in c tn s m; Mch pht sng in t cao tn (2) pht ra sng in t c tn s cao (c MHz) ; Mch bin iu (3) trn dao ng in t cao tn vi dao ng in t m tn ; Mch khuch i (4) khuch i dao ng in t cao tn bin iu ; anten (5) to ra in t trng cao tn lan truyn trong khng gian. - S khi ca my thu thanh v tuyn n gin: Anten (1) thu sng in t cao tn bin iu ; Mch khuch i dao ng in t cao tn (2) khuch i dao ng in t cao tn t anten gi ti ; Mch tch sng (3) tch dao ng in t m tn ra khi dao ng in t cao tn ; Mch khuch i (4) khuch i dao ng in t m tn t mch tch sng gi n ; Loa (5) bin dao 1234 5 3 2 1 45 17 ng in thnh dao ng m. - ng dng ca sng in t: Sng v tuyn in c s dng trong thng tin lin lc. i ht thanh, dao ng m tn c dng bin iu (bin hc tn s) dao ng cao tn. Dao ng cao tn c bin iu s c pht x t ng ten di dng sng in t. mt thu thanh, nh c ng ten thu, s thu c dao ng cao tn c bin iu, v sau dao ng m tn li c tch ra khi dao ng cao tn bin iu nh qu trnh tch sng, ri a ra loa. - Nguyn tc chung ca thng tin lin lc bng sng v tuyn:-Phi dng cc sng in t cao tn lm sng mang -Phi bin iu sng mang. - nn thu phi tch sng m tn ra khi sng cao tn (sng mang). -Khi tn hiu thu nh phi khuych i chng bng mch khuych i. - Lu un ng: ng mnginng in sng m n,n s ng n ssng n HN . SNG NH SNG 1. Hin tng tn sc nh sng. - /n: L hin tng nh sng b tch thnh nhiu mu khc nhau khi i qua mt phn cch ca hai mi trng trong sut. - i vi as trng sau khi i qua lng knh th b tn sc thnh mt di mu nh cu vng, tia lch t nht tia tm b lch nhiu nht. - Lu :+ Hin tng tn sc nh sng s xy ra khi nh sng trng i qua lng knh, thu knh, git nc ma, lng cht phng, bn mt song song ... (cc mi trng trong sut) + Hin tng cu vng l do hin tng tn sc nh sng. + nh sng phn x trn cc vng du, m hoc bong bng x phng (c mu sc s) l do hin tng giao thoa nh sng khi dng nh sng trng. Lu : + Nu tia ti l as trng i song song vi y lng knh, m tia l l chm tia sng cng song song vi y ca lng knh. Th tia tm trn tia di. + Nu tia ti l as trng sau khi qua lng knh c 1 tia i lch l l mt bn ca lng knh, th cc tia cn li c bc sng di hn. VD: Sau khi qua LK tia vng i l l mt bn th cc tia cn li l , da cam. - nh sng n sc l nh sng khng b tn sc + nh sng n sc c tn s xc nh, ch c mt mu.+ Bc sng ca nh sng n sc vf, truyn trong chn khng 0cf

0 0cv n vi c cnv .f= = l trit sut ca mi trng. t tt ttcvnv n1 v vc v nvn= = > >= y trong cng 1 mt as truyn nhanh hn as tm hit sut ca mi trng ph thuc vo bc sng v tn s as. Thng th chit sut gim khi tng. - Chit sut ca mi trng trong sut ph thuc vo mu sc nh sng. i vi nh sng mu chit sut ca mi trng l nh nht, mu tm l ln nht. - nh sng trng l tp hp ca v s nh sng n sc c mu bin thin lin tc t n tm. Bc sng ca nh sng trng: 0,38 m s s 0,76 m. - ng thc lng knh: + Tng qut: sini1 = nsinr1 ; sini2 = nsinr2 ; A = r1 + r2 ; D = (i1 + i2) A. + Gc trit quang nh:i1 = n.r1 ; i2 = n.r2 ; A = r1 + r2 ; D = (n-1).A + Gc lch cc tiu: i1 = i2 , r1 = r2 = A/2 ,Dmin =2.i A; minD A Asin n.sin2 2+=2. Hin tng nhiu x nh sng. Hin tng nh sng b lch phng truyn khi nh sng truyn qua l nh, hoc gn mp nhng vt trong sut 18 hoc khng trong sut gi l hin tng nhiu x nh sng. 3. Hin tng giao thoa nh sng (ch xt giao thoa nh sng trong th nghim Ing). - /n: L s tng hp ca hai hay nhiu sng nh sng kt hp trong khng gian trong xut hin nhng vch sng v nhng vch ti xen k nhau. Cc vch sng (vn sng) v cc vch ti (vn ti) gi l vn giao thoa. - H thng vn giao thoa i vi as n sc: L 1 h thng cc vch mu n sc v cc vch ti nm xen k. i vi as trng: Chnh gia l vn sng trung tm, 2 bn l nhng di mu tm trong ngoi. - Hiu ng i ca nh sng (hiu quang trnh): 2 1axd d dD Trong : a = S1S2 l khong cch gia hai khe sng D = OI l khong cch t hai khe sng S1, S2 n mn quan st S1M = d1; S2M = d2

x = OMl (ta ) khong cch t vn trung tm n im M ta xt - tr to vn sng: Ad = k ; k ZDx = k = k.i a k = 0:Vn sng trung tm; k = 1: Vn sng bc (th) 1; k = 2: Vn sng bc (th) 2;k > 0 khi d2 > d1, k < 0 khi d2 < d1. - tr to vn ti: Ad = (k + 0,5) ; k ZDx = (k +0, 5) = (k +0, 5).i a Vi cc vn ti khng c khi nim bc giao thoa. (Vn ti th 3 ng vi k = 2, th 5 ng vi k = 4 ...) - Khong vn i: L khong cch gia hai vn sng hoc hai vn ti lin tip: Di =a - Nu th nghim c tin hnh trong mi trng trong sut c chit sut n th bc sng v khong vn i vi mi trng l: nn nD iin a n - tm s vn sng v s vn ti trn b rng trng giao thoa c chiu di L i xng qua vn trung tm): + S khong vn trn na trng giao thoa:n : Pha n nguye np: Pha n tha p pha nL= n, p2.i + S vn sng trn c trng giao thoa: (2n + 1) + S vn ti trn c trng giao thoa:(2n) nup < 0,5 2(n + 1)nup > 0,5 + V dL/2i = 4,5 ==> n = 4; p = 0,5 ==> s vn sng l 9, s vn ti l 10. L/2i = 5,45 ==> n = 5; p = 0,45 ==> s vn sng l 11, s vn ti l 11. L/2i = 3,72 ==> n = 3; p = 0,72 ==> s vn sng l 7, s vn ti l 8. - Bit khong vn i bit v tr ca im M (xM) th: + Ti M l vn sng khi: xMi = n (n e N); + Ti M l vn ti khi: xMi= n + 12

- Xc nh s vn sng vn ti gia hai im M N c to x1, x2 gi s x1 < x2) + Vn sng: x1 s ki s x2 ( v ) + Vn ti:x1 s (k+0,5)i s x2 ( v ) S gi tr k e Z l s vn sng (vn ti) cn tm Lu : M v N cng pha vi vn trung tm th x1 v x2 cng du. M v N khc pha vi vn trung tm th x1 v x2 khc du. - Xc nh khong vn i trong khong c b rng L. Bit trong khong L c n vn sng. + Nu 2 u l hai vn sng th: 1Lin + Nu 2 u l hai vn ti th: Lin S1 D S2 d1 d2 I O x M a 19 + Nu mt u l vn sng cn mt u l vn ti th: 0, 5Lin

- S trng nhau ca cc bc x 1, 2 ... khong vn tng ng l i1, i2 ...) + Trng nhau ca vn sng: xs = k1i1 = k2i2 = ...k11 = k22 = ... + Trng nhau ca vn ti: xt = (k1 + 0,5)i1 = (k2 + 0,5)i2 = ...(k1 + 0,5)1 = (k2 + 0,5)2 = ... - Lu : V tr c mu cng mu vi vn sng trung tm l v tr trng nhau ca tt c cc vn sng ca cc bc x. - Trong hin tng giao thoa nh sng trng 038m s s 0,76m) + B rng quang ph bc k:( )A = = k t tDx k ( ) ki ia + Xc nh s vn sng, s vn ti v cc bc x tng ng ti mt v tr xc nh ( bit x): + Vn sng:76 , 0138 , 0 s = sDaxk cc gi tr ca k + Vn ti: 76 , 05 . 0138 , 0 s+= sDaxk cc gi tr ca k 4. S x dch ca h vn giao thoa: a, X dch do s x dch ca ngun S: = IOOO' .SS'IS ++Va n trung ta md / c ng c chie u d / c cu a nguo nS' IO' tha ng ha ng b, X dch do bn mt song song: 1 = (n )eDOO'a;Vn trung tm dch v pha bn. 5. ch to ra ngun kt hp: a) Khe Yng ( hc). b Lng lng knh rexnen:S1S2 = a = 2.d1.A(n - 1);i = D/a==> ) 1 n ( A . d . 2) d d (i12 1+ =Chiu rng min giao thoa: MN = 2.d2.A(n -1) c) Lng thu knh Bi: Gm mt thu knh c ca i qua quang tm ri: + C1: Ht i mi na mt phn nh l e ri ghp st vo nhau.Hai nh hi l o th to ra giao thoa.Khong cch hai nh l : /1 11 21d - da = S S = 2e.d;B rng min giao thoa l: ///d . dd ) d d ( edda MN1 12 1 1122 = =; khong vn a) d d (i2/1 + = ; Cch 1Cch 2 O O S1 S2 I S S S S1 S2 O O e, n E S S1 S2 O2 O1 M N O d1 d2 D d1/ S S1 S2 O1 O2 M N O d1 d2 D d1/ E S S1 S2 O M N Io d1 d2 D 20 + Hoc 2 m mt ming ba mng 2 na thu knh cch nhau 1 khong l b. Hai nh hi l tht s cho giao thoa, khong cch hai nh l: /1 11b.(d +d )a =d;Min giao thoa l: 12 1d) d d .( bMN+= ; Khong vn: aDi= .d Lng gng phng rexnen: gm hai gng phng t lch nha mt gc o nh. S1S2 = a = 2.d1.tgo = 2.d1.o. Chiu rng min giao thoa: MN = 2.d2.o.Khong vn o+ =. d . 2) d d (i12 1.6. c loi quang ph: a, Quang ph pht x: L quang ph ca nh sng do cc cht rn lng kh khi c nung nng nhit cao pht ra. Quang ph pht x ca cc cht chia lm hai loi: quang ph lin tc v quang ph vch. * Quang ph lin tc:- L 1 di sng c mu bin i lin tc t n tm, ging nh quang ph ca nh sng mt tri. - Tt c cc vt rn, lng, kh c t khi ln khi b nung nng u pht ra quang ph lin tc- c im : quang ph lin tc khng ph thuc bn cht ca ngun sng m ch ph thuc vo nhit ca vt pht sng. Khi nhit ca vt cng cao th min quang ph cng m rng v as c bc sng ngn- ng dng: cho php xc nh nhit ca ngun sng * Quang ph vch: - L 1 h thng cc vch mu ring r ngn cch nhau bi nhng khong ti. - Khi kch thch khi kh hay hi p sut thp chng pht sng th chng pht ra quang ph vch pht x. - c im: Cc nguyn t khc nhau th pht ra cc qp vch px khc nhau: = v s lng vch, sng, v tr, mu sc ca cc vch v sng t i ca cc vch. - ng dng: Dng phn tch thnh phn mu vt. b, Quang ph hp th: - L 1 h thng cc vch ti ring r nm trn 1 nn quang ph lin tc. - Cn 1 ngun sng trng pht ra QPLT,gia ngun sng v my qp l m kh hay hi c t chy pht ra qp vch hp th. (Qpm i mhu n i p hp h B m Ti phung ph in ) - c im: Nhit ca ngun pht ra qp vch hp th phi nh hn nhit ca ngun pht ra qp lin tc. - ng dng: Trong php phn tch quang ph. * Hin tng o sc nh sng: L hin tng khi ngun pht ra qplt t nhin mt i th nn qplt mt i, cc vch ti ca qp vch hp th tr thnh cc vch mu ca qp vch pht x. Lc ngun pht ra qp vch hp th tr thnh ngun pht ra qp vch pht x. hng t m hi c kh nng pht ra nhng as n sc no th cng c kh nng hp th as . Tia hng ngoitia t ngoi v tia X: a Tia hng ngoi:- nh ngha : L nhng bc x khng nhn thy c c bc sng ln hn bc sng ca nh sng :> 0,76 m - Bn cht : l sng in t . S S1 S2 I O M N d2 d1 G2 G1 o o 21 - Ngun pht sinh : Tt c cc vt nung nng u pht ra tia hng ngoi (mt tri, c th ngi, bng n . . .) C50% nng lng Mt Tri thuc v vng hng ngoi. - c im : Tc dng nhit, td ln knh nh hng ngoi, td ha hc, c th bin iu nh sng in t cao tn. - ng dng : Dng si m, sy kh, chp nh hng ngoi, trong ci iu khin t xa: tivi, t. b Tia t ngoi: - nh ngha : L nhng bc x khng nhn thy c, c bc sng nh hn bc sng ca nh sng tm :< 0,38 m - Bn cht : l sng in t . - Ngun pht sinh :-Vt b nung nng trn 20000C pht ra tia t ngoi Ngun pht ra tia t ngoi : mt tri, h quang in. . . C9% nng lng Mt Tri thuc v vng t ngoi. - c im :-Tc dng mnh ln knh nh, lm pht quang mt s cht, lm ion ha khng kh, gy ra nhng phn ng quang ha, quang hp.-B thy tinh v nc hp th mnh.-C mt s tc dng sinh hc- ng dng : -Dng kh trng, cha bnh ci xng. (ng dng ca td sinh hc: hy dit t bo) -Pht hin vt nt, vt xc trn b mt sn phm. (ng dng ca td lm pht quang mt s cht) c, Tia Rnghen: - Pht hin tia X: Mi khi mt chm tia catt tc l chm tia eelectron c nng lng ln p vo mt vt rn th vt pht ra tia X. - Bn cht : l sng in t c bc sng rt ngn c 10-11 m 10-8m- Tnh cht : -C kh nng m xuyn ln, c th truyn qua giy, g . . . nhng truyn qua kim loi th kh hn. Kim loi c khi lng ring cng ln th ngn cn tia Rnghen cng tt (ch . . )-Tc dng mnh ln phim nh.-Lm pht quang mt s cht-Lm ion h cht kh-C tc dng sinh l, hy hoi t bo, dit vi khun- Cng dng : -Trong y hc : dng chiu in, chp in, cha mt s bnh ung th.-Trong cng nghip : dng d khuyt tt bn trong sn phm, ch to my o liu lng tia rnghen.6. Thang sng in t: - Sng v tuyn: Bc sng t vi chc km n vi mm. - Tia hng ngoi: Bc sng t vi mini mt n 0,76m. - nh sng kh kin: Bc sng t 0,76m n 0,38m. - Tia t ngoi: Bc sng t 3,8.10-7m n 10-9m. - Tia X: Bc sng t 10-8m n 10-11m. - Tia gamma: Bc sng t 10-12 m n 10-15 m. Sng v tuyn, tia hng ngoi, nh sng kh kin, tia t ngoi, tia X v tia gamma u c bn cht l sng in tnhngcbcsngkhcnhaunnctnhcht,tcdngkhcnhauvngunpht,cchthuchngcng khc nhau. HN I. LN T NH SN 1. Hin tng quang in: - Hin tng quang in ngoi: Hin tng nh sng lm bt cc lectron ra khi mt kim loi gi l hin tng quang in. - Hin tng quang in trong (quang dn): Hin tng nh sng gii phng cc lectron lin kt thnh cc lectron dn v cc l trng cng tham gia vo qu trnh dn in, gi l hin tng quang in trong. - nh lut v gii hn quang in: i vi mi kim loi, nh sng kch thch phi c bc sng ngn hn hoc bng gii hn quang in 0 ca kim loi , mi gy ra c hin tng quang in. ==> Cc hin tng quang in v cc nh lut quang in chng t nh sng c tnh cht ht. Vng : 0, 640 0, 760 m m Vng cam : 0, 590 0, 650 m m Vng vng : 0, 570 0, 600 m m Vng lc : 0, 500 0, 575 m m Vng lam : 0, 450 0, 510 m m Vng chm : 0, 440 0, 460 m m Vng tm : 0, 38 0, 440 m m 22 - ng dng ca cc hin tng quang in trong cc t bo quang in, trong cc dng c bin i cc tn hiu nh sng thnh tn hiu in, trong cc quang in tr, pin quang in. 2. Thuyt lng t nh sng. - Gi thuyt ca Plng: Lng nng lng m mi ln mt nguyn t hay phn t hp th hay pht x c gi tr hon ton xc nh v bng hf; trong f l tn s ca nh sng b hp th hay pht x, h l mt hng s. (h = 6,625.10-34Js). - Nng lng mt lng t nh sng (ht phtn) hchfTrong h = 6,625.10-34 Js l hng s Plng;c = 3.108m/sl vn tc nh sng trong chn khng; f, l tn s, bc sng ca nh sng (ca bc x).Khi as truyn i cc lng t as khng b thay i khngph thuc k/c ti ngun sng - Mi phtn ca as n sc c nng lng:c = hf = hc/ = mc2

==> Khi lng tng i tnh ca phtn:m = c/c2 = h/(c) ==> ng lng ca phtn: p = mc = h/ - Lu : hng c htn ng yn, htn ch tn ti khi n chuyn ng khi ng yn khi lng ca n bng khng. - Thuyt lng t nh sng: + AS c to thnh bi cc ht gi l phtn. + Vi mi as n sc c tn s f, cc phtn u ging nhau, mi phtn mang nng lng bng hf. + Phtn bay i vi vn tc c = 3.108 m/s dc theo cc tia sng. + Mi ln 1 nguyn t hay phn t pht x hay hp th as th chng pht ra hay hp th 1 phtn. 3. Hin tng quang in *Cng thc Anhxtanh v hin tng quang in: 20Maxmv hchf A2 Trong 0hcA = l cng thot ca kim loi dng lm catt; 0 l gii hn quang in ca kim loi dng lm catt v0Max l vn tc ban u ca electron quang in khi thot khi catt f, l tn s, bc sng ca nh sng kch thch* dng quang in trit tiu th UAK s Uh (Uh < 0), Uh gi l hiu in th hm 20MaxhmveU =2 Lu : Trong mt s bi ton ngi ta ly Uh > 0 th l ln. i vi tia Rnghen X: - Cng dng in trong ng Rnghen:i = NeVi N l s electron ti p v i catt trong 1 giy. - nh l ng nng: E Eo = eUAK Vi E = mv2/2 l ng nng ca electron ngay trc khi p vo i catt v Eo = mvo2/2 l ng nng ca electron ngay sau khi bt ra khi catt, thng th Eo = 0. ==> E = eUAK - nh lut bo ton nng lng: E = c + Q = hf + Q+ ng nng ca electron bin thnh nng lng tia X v lm nng i catt. + Vi c l nng lng tia X v Q l nhit lng lm nng i catt. - Bc sng nh nht ca bc x do ng Rnghen pht ra ng vi trng hp ton b ng nng ca electron E (ngay trc khi p vo i catt) bin thnh nng lng c ca tia X: T E = c + Q = hf + Q==> E > hf = hc/ ==> > hc/ E==>min = hc/ E Vi: h = 6,625.10-34 Js l hng s Plng, c = 3.108m/s l vn tc as trong chn khng. 4. 1 s cng thc lin quan: * Xt vt c lp v in, c in th cc i VMaxv khong cch cc i dMax m electron chuyn ng trong in trng cn c cng E c tnh theo cng thc:

2Max 0Max Max1e V = mv = e Ed2 * Vi U l hiu in th gia ant v catt, vA l tc cc i ca electron khi p vo ant, vK = v0Max l tc ban u cc i ca electron khi ri catt th:2 2AK A K1 1e U = mv - mv2 2= EA EK = EA (c - A) 23 * Cng sut chiu sng:P = Nc =N.hc/ Trong N l s phtn ti b mt KL hoc c pht bi ngun trong 1 giy. * Cng dng quang in bo ha:Ibh = n.e Trong n l s electrn quang in n ant trong mi giy, e = 1,6.10-19C * Hiu sut lng t (hiu sut quang in): nH =N Vi n v N l s electron quang in bt khi catt v s phtn p vo catt trong 1 giy.* Bn knh qu o ca electron khi chuyn ng vi vn tc v trong t trng u B :o sin B emvR = ( ( ) B v, = o ) Lu : Hin tng quang in xy ra khi c chiu ng thi nhiu bc x th khi tnh cc i lng: Tc ban u cc iv0Max, hiu in th hm Uh, in th cc i VMax, u c tnh ng vi bc x cMin (hoc fMax) 5. Quang tr v pin quang in: - Quang in tr l 1 in tr lm bng cht quang dn. in tr ca n c th thay i t vi mgam khi khng c chiu sng xung n vi chc m khi c chiu sng. - Pin quang in (cn gi l pin mt tri) l 1 ngun in chy bng nng lng as. N bin i trc tip quang nng thnh in nng. Pin hot ng da vo hin tng quang in trong xy ra bn cnh 1 lp chn. 6. S pht quang: - S pht quang l mt s cht c kh nng hp th as c bc sng ny pht ra as c bc sng khc. - c im ca s pht quang: l n cn ko di 1 thi gian sau khi tt as kch thch. - Hunh quang: L s pht quang ca cc cht lng v cht kh, c c im l as pht quang tt rt nhanh sau khi tt as kch thch. nh sng hunh quang c bc sng di hn bc sng ca as kch thch: hq > kt. - Ln quang: L s pht quang ca cc cht rn, c c im l as pht quang c th ko di 1 khong thi gian no sau khi tt as kch thch. ng dng: ch to cc loi sn trn cc bin bo giao thng, tng pht sng... 7. Tin Bo - Quang ph nguyn t Hir - Tin v trng thi dng: Nguyn t ch tn ti trong nhng trng thi c nng lng xc nh, gi l cc trng thi dng. Trong trng thi dng nguyn t khng bc x. Trong cc trng thi dng ca nguyn t, lectrn ch chuyn ng quanh ht nhn trn cc qu o c bn knh hon ton xc nh gi l cc qu o dng. - Tin v s bc x v haapf th nng lng ca nguyn t: + hi nguyn t chuyn t trng thi dng c nng lng Ecao sang trng thi dng c mc nng lng Eth (vi Ecao > Eth) th nguyn t ht ra 1 htn c nng lng ng bng hiuEcao - Eth:c = hf = hc = Ecao - Ehp + Ngc li, nu 1 nguyn t angtrng thi dng c nng lng thEth m h thu c 1 htn c nng lng h ng bng hiu Ecao - Eth th n chuyn ln trng thi dng c nng lng Ecao ln hn. ==> Nguyn t lun c xu hng chuyn t mc nng lng cao v mc nng lng thp hn. * Bn knh qu o dng th n ca electron trong nguyn t hir:rn = n2r0 Vi r0 =5,3.10-11m l bn knh Bo ( qu o K); n = 1, 2, 3, 4, 5, 6... * Nng lng electron trong nguyn t hir: n 213, 6E = - (eV)n Vi n e N*. * S mc nng lng - Dy Laiman: Nm trong vng t ngoi ng vi e chuyn t qu o bn ngoi v qu o K Lu : Vch di nht LK khi e chuyn t L K Vch ngn nht K khi e chuyn t K. - Dy Banme: Mt phn nm trong vng t ngoi, mt phn nm trong vng nh sng nhn thy ng vi e chuyn t qu o bn ngoi v qu o L nhn phtn pht phtn Ecao Ethp Laiman K M N O L P Banme Pasen Ho H| H Ho n=1 n=2 n=3 n=4 n=5 n=6 24 Vng nh sng nhn thy c 4 vch: Vch Ho ng vi e: M L Vch lam H| ng vi e: N L Vch chm H ng vi e: O L Vch tm Hong vi e: P LLu : Vch di nht ML (Vch Ho ) Vch ngn nht L khi e chuyn t L. - Dy Pasen: Nm trong vng hng ngoi ng vi e chuyn t qu o bn ngoi v qu o M Lu : Vch di nht NM khi e chuyn t N M. Vch ngn nht M khi e chuyn t M. - Mi lin h gia cc bc sng v tn s ca cc vch quang ph ca nguyn t hir: 13 12 231 1 1 = +v f13 = f12 +f23 (nh cng vct) 8. S lc v laze: - Laze l phin m ca LASER, ngha l my khuychi as bng s ht x cm ng. - Laze l 1 ngun sng pht ra 1 chm sng c cng ln da trn ng dng ca hn tng pht x cm ng - c im ca tia laze c tnh n sc, tnh nh hng, tnh kt hp rt cao v cng ln. - Ty vo vt liu pht x ngi ta ch to ra laze kh, laze rn v laze bn dn. i vi laze rn, laze rubi (hng ngc) l Al2O3 c pha Cr2O3 mu ca tia laze l do as ca hng ngc do ion crm pht ra khi chuyn t trng thi kch thch v trng thi c bn 9. Lng tnh sng ht ca nh sng: - nh sng va c t/c sng, va c t/c ht vy as c lng tnh sng ht. - Khi bc sng ca as cng ngn (th nng lng ca phtn cng ln), th t/c ht th hin cng m nt: Tnh m xuyn, td quang in, td in ha, td pht quang.Ngc li khi bc sng ca as cng di (th nng lng ca phtn cng nh), th t/c sng th hin cng m nt: d quan st thy hin tng giao thoa, hin tng tn sc ca cc as . HN VII. T L HT NHN 1. u to ht nhn nguyn t n v khi lng nguyn t: a u to ht nhn nguyn t: - Cu to: + Ht nhn nguyn t c cu to t cc prtn (mang in tch nguyn t dng), v cc ntron (trung ho in), gi chung l nucln. + Ht nhn ca cc nguyn t c nguyn t s Z th cha Z prton v N ntron; A = Z + N c gi l s khi. + Cc nucln lin kt vi nhau bi lc ht nhn. Lc ht nhn khng c cng bn cht vi lc tnh in hay lc hp dn; n l loi lc mi truyn tng tc gia cc nucln trong ht nhn (lc tng tc mnh). Lc ht nhn ch pht huy tc dng trong phm vi kch thc ht nhn (c 10-15m). - Bn knh hn nhn tng chm theo s khi A:r = r0.A1/3 .Vir0 = 1,2 Fecmi;1 Fecmi = 10-15m. - ng v: Cc nguyn t m ht nhn c cng s prton Z nhng khc s ntron N gi l cc ng v. b 1 s n v hay dng trong LHN: - n v khi lng nguyn t: n v u c gi tr bng 112khi lng nguyn t ca ng v 126C , c th: 1u = 1,66055.10-27kg;1u = 931,5 Mevc2 ==> 1uc2 = 931,5MeV - u xp x bng khi lng ca mt nucln nn ht nhn c s khi A th c khi lng xp x bng Au. - n v nng lng:1 eV = 1,6.10-19J ==> 1 MeV = 106.1,6.10-19J = 1,6.10-13J - 1 s n v n/t thng gp:mP = 1,67262.10-27 kg = 1,007276 u ;mn = 1.67493.10-27 kg = 1,008665 u ;me = 9,1.10-31 kg = 0,0005486 u; - Cc c v bi :G 109; M 106; k 103 ; m 10-3 ; 10-6 ; n 10-9 ; p 10-12

2. H thc Anhxtanh ht khi nng lng lin kt: - Ht nhn c khi lng ngh m0, chuyn ng vi vn tc v, c nng lng tnh theo cng thc:E = m0c2 + W 1223131 2 3 25 Trong W = m0v2/2 = (1 1-v2c2- 1)m0c2l ng nng ca ht nhn. - Mt vt c khi lng m0 trng thi ngh, khi chuyn ng vi vn tc v, khi lng ca vt s tng ln thnh m vim = m0 1-v2c2

- Ta c th vit h thc Anhxtanh:E = mc2. ==> W = E E0 ; Vi E0 = m0c2 l nng lng ngh ca vt. - ht khi:: Am = [Z.mp + (A Z).mn] mx hi lng ca mt ht nhn lun nh hn tng khi lng ca cc nucln to thnh ht nhn - Nng ng in t:E = mc2

S to thnh ht nhn to nng lng tng ngE, gi l nng lng lin kt ca ht nhn (v mun tch ht nhn thnh cc nucln th cn tn mt nng lng bng E).- Nng lng lin kt ring :c = E/A(l nng lng lin kt tnh cho 1 nucln). Nng lng lin kt ring cng ln th ht nhn cng bn vng. 3. Phn ng ht nhn a, nh ngha:- Phn ng ht nhn l qu trnh bin i ca cc ht nhn. - Phn ng ht nhn c chia lm hai loi: + Phn ng ht nhn t pht: l qu trnh t phn r ca mt ht nhn khng bn vng thnh cc ht nhn khc.A C + DTrong A: ht nhn m;C: ht nhn con;D: tia phng x (o, |, ...) + Phn ng ht nhn kch thch: l qu trnh cc ht nhn tng tc vi nhau thnh cc ht nhn khc. A + B C + D - Phng trnh phn ng: 3 1 2 41 2 3 41 2 3 4A A A AZ Z Z ZX X X X Trong s cc ht ny c th l ht s cp nh nucln, electrn, phtn ... - Trng hp c bit l s phng x: X1 X2 + X3; X1 l ht nhn m, X2 l ht nhn con, X3 l ht o hoc | b, c nh lut bo ton trong phn ng ht nhn: + 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 ==> 3 4 1 2X X X XE = K +K - (K +K) Trong : AE l nng lng phn ng ht nhn 212X x xK m vl 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: 2X X Xp = 2m K- Nng lng phn ng ht nhn:AE = (M0 - M)c2

Trong : 1 20 X XM m m l tng khi lng cc ht nhn trc phn ng.

3 4X XM m ml tng khi lng cc ht nhn sau phn ng. Lu : + Nu M0 > M th phn ng to nng lng AE di dng ng nng ca cc ht X3, X4 hoc phtn . Cc ht sinh ra c ht khi ln hn nn bn vng hn. + Nu M0 < M th phn ng thu nng lng ,AE, di dng ng nng ca cc ht X1, X2 hoc phtn . Cc ht sinh ra c ht khi nh hn nn km bn vng. - Trong phn ng ht nhn 3 1 2 41 2 3 41 2 3 4A A A AZ Z Z ZX X X X Cc ht nhn X1, X2, X3, X4 c: Nng lng lin kt ring tng ng l c1, c2, c3, c4. Nng lng lin kt tng ng l AE1, AE2, AE3, AE4 ; ht khi tng ng l Am1, Am2, Am3, Am4

Nng lng ca phn ng ht nhn :AE = A3c3 +A4c4 - A1c1 - A2c2

AE = AE3 + AE4 AE1 AE2

AE = (Am3 + Am4 - Am1 - Am2)c2 c, Quy tc dch chuyn ca s phng x + Phng x o (42He ): 4 42 2A AZ ZX He Y-So vi ht nhn m, ht nhn con li 2 trong bng tun hon v c s khi gim 4 n v. 26 -L hn Hli(eH42), mang in tch dng (+2e) nn b lch v bn m khi bay qua t in. -Chuyn ng vi tc c 2.107m/s, qung ng i c trong khng kh c 8cm, trong vt rn c vi mm. ==> kh nng m xuyn km, c kh nng in ha cht kh. + Phng x |- (10e ): 01 1A AZ ZX e Y-So vi ht nhn m, ht nhn con tin 1 trong bng tun hon v c cng s khi. -Thc cht ca phng x |- l mt ht ntrn bin thnh mt ht prtn, mt ht electrn v mt ht ntrin:n p e v-Bn cht (thc cht) ca tia phng x |- l ht electrn ( e01 ), mang in tch m (-1e) nn b lch v pha bn dng ca t. -Ht ntrin (v) khng mang in, khng khi lng (hoc rt nh) chuyn ng vi vn tc ca nh sng v hu nh khng tng tc vi vt cht.-Phng ra vi vn tc gn bng vn tc as. -In ha cht kh yu hn tia o. -Kh nng m xuyn mnh, i c vi mt trong khng kh v vi mm trong kim loi. + Phng x |+ (10e ): 01 1A AZ ZX e Y-So vi ht nhn m, ht nhn con li 1 trong bng tun hon v c cng s khi. -Thc cht ca phng x |+ l mt ht prtn bin thnh mt ht ntrn, mt ht pzitrn v mt ht ntrin:p n e v-Bn cht (thc cht) ca tia phng x |+ l ht pzitrn (e+), mang in tch dng (+e) nn lch v pha bn m ca t in (lch nhiu hn tia o v i xng vi tia |-). -Phng ra vi vn tc gn bng vn tc as. -In ha cht kh yu hn tia o. -Kh nng m xuyn mnh, i c vi mt trong khng kh v vi mm trong kim loi. + Phng x gamma (ht phtn) -C bn cht l sng in t c bc sng rt ngn (< 0,01nm). L chm phtn c nng lng cao. -Ht nhn con sinh ra trng thi kch thch c mc nng lng cao E1 chuyn xung mc nng lng thp E2 ng thi phng ra mt phtn c nng lng: 1 2hchf E E-L bc x in t khng mang in nn khng b lch trong in trng v t trng. -C cc t/c nh tia Rnghen, c kh nng m xuyn ln, i c vi mt trong b tng v vi centimt trong ch v rt nguy him. -Trong phng x khng c s bin i ht nhn phng x thng i km theo phng x o v |. 4. nh lut phng x:- nguyn (h nhn) h phng x n i su hi gin : 00 0 kt-N-tTN = N .2 = N .e =2 - h nguyn phn r bng s ht nhn con c to thnh v bng s ht (o hoc e- hoc e+) c to thnh: -t0 0N= N - N= N (1- e )- Khi ng h phng x n i su hi gin : t--t 0 T0 0 kmm= m .2 = m .e =2 Trong : + Vi NA = 6,0221.1023mol-1 l s Avgar. + A l s khi ca nguyn t. + N0, m0 l s nguyn t (ht nhn), khi lng cht phng x ban u. + T l chu k bn r ln 2T = l khong thi gian mt na s ht nhn phn r. + ln2 0, 693 = =T T l hng s phng x, c trng cho cht phng x ang xt. + v T khng ph thuc vo cc tc ng bn ngoi (nh nhit , p sut ...) m ch ph thuc bn cht bn trong ca cht phng x. + k = tT : s chu k bn r trong thi gian t 27 - Khi ng hphng x su hi gin : -t0 0m= m - m= m (1- e )- Phn m ( gim) h phng x phn :01tmem - Phn m h phng x n i: 02tmtTem - i in h gi hi ng v s h nhn:ANN = m.A - Khi lng cht mi c to thnh sau thi gian t:-t -t 1 0 11 1 0A AA N A Nm = A = (1- e ) = m (1- e )N 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 : Trng hp phng x |+, |- th A = A1 m1 = Am- phng x H: L i lng c trng cho tnh phng x mnh hay yu ca mt lng cht phng x, o bng s phn r trong 1 giy: t--t 0 T0 0 kHH= H .2 = H .e = N =2

0tHeH =+ Vi: H0 = N0 l phng x ban u. + n v:Becren (Bq); 1Bq = 1 phn r/giy; hoc Curi (Ci);1 Ci = 3,7.1010 Bq ==> gimphng x (%): -t 00 0 0H - H H H= =1- =1- eH H H Lu : Khi nhphng x H, H0 (B) h hu phng x T phi in v giy(s) Bng quy lut phn r t = T2T3T4T5T6T S ht cn li N0/2N0/4N0/8N0/16N0/32N0/64 S ht phn rN0/23 N0/47 N0/815 N0/1631 N0/3263 N0/64 T l % phn r50%75%87.5%93.75%96.875% T l r v cn li 137153163 - ng dng ca cc ng v phng x: trong phng php nguyn t nh du, trong kho c nh tui c vt da vo lng cacbon 14. . Phn ng phn hch phn ng nhit hch: a Phn ng phn hch: - P. phn hch: mt ht nhn rt nng khi hp th mt ntron s v thnh hai ht nhn nh hn, km theo 1 vi ntrn. Nng lng ta ra trong phn ng c 210 MeV.S phn hch ca 1g 235U gii phng mt nng lng bng 8,5.1010J tng ng vi nng lng ca 8,5 tn than hoc 2 tn du ta ra khi chy ht. - P. dy truyn: Gi k l h s nhn ntrn, l s ntrn cn li sau 1 p. h.n n kch thch cc h.n khc. Khi k > 1 xy ra p. phn hch dy chuyn: + Khi k < 1, p. phn hch dy chuyn tt nhanh. + Khi k = 1, p. phn hch dy chuyn t duy tr v nng lng pht ra khng i theo thi gian. + Khi k > 1, p. phn hch dy chuyn t duy tr v nng lng pht ra tng nhanh v c th gy ra bng n. - Khi lng ti hn: l khi lng ti thiu ca cht phn hch p. phn hch dy chuyn duy tr.Vi 235U khi lng ti hn c 15 kg, vi 239Pu vo c 5 kg. b Phn ng nhit hch p. tng hp h.n: - Hai hay nhiu ht nhn rt nh, c th kt hp vi nhau thnh mt ht nhn nng hn. Phn ng ny ch xy ranhitrtcao,nngilphnngnhithch.Conngimichthchincphnngnydi dng khng kim sot c (bom H). - iu kin p. kt hp h.n xy ra: + Phi a hn hp nhin liu sang trng thi plasma bng cch a nhit ln ti 108 . 28 + Mt h.n trong plasma phi ln + Thi gian duy tr trng thi plasma nhit cao phi ln. HN III. T I M N MI. HT S P: 1.Thgiivimvmcspxptheokchthclndn:Htscp,htnhnnguynt,nguynt, phn t, hnh tinh, h Mt Tri, thin h ... 2. Ht s cp: L ht c kch thc v khi lng nh hn ht nhn nguyn t. - Cc ht s cp c chia lm ba loi: + phtn c m0 = 0 + Cc leptn: C khi lng t 0 n200 me. Bao gm: ntrin v, electron e-, pzitron e+,+ Cc harn: C khi lng trn 200me. c chia thnh ba nhm con: - Mzn t, K: C khi lng trn 200me nhng nh hn khi lng nucln. - Nucln p, n. - Hipron: C khi lng ln hn khi lng cc nucln. hm nun v hipn n gi in - Tt c cc harn u c cu to t cc ht nh hn gi l quac. C 6 loi quac (k hiu l: u, d, s, c, b, t) cng vi 6 phn quac tng ng. Cc quac c mang in phn s: e3 , 2e3 . Mt trong cc thnh cng v gi thuyt v quac l d on v ht mga tr O-. - Phn ln cc ht s cp u to thnh cp gm ht v phn ht. Phn ht c cng khi lng ngh v spin nh ht nhng cc c trng khc c tr s bng v ln v tri du.- Lu :+ Sp xp theo th t tng dn v khi lng ca cc ht s cp bit: Phtn, leptn, mzn v barion. + Theo quan nim hin nay v cc ht thc s l s cp gm: Cc quac, cc leptn v cc ht truyn tng tc l glun, phtn, W, Z0 v gravitn. + Ht prton c cu to bi cc quac nn prton c th b ph v. 3. Bn loi tng tc c bn trong v tr: mnh in t yu hp dn. - Tng tc hp dn: L tng tc gia cc ht (cc vt) c khi lng khc khng. Bn knh ln v cng, lc tng tc nh.Vd: Trng lc, lc ht ca T v mt trng... - Tng tc in t: l tng tc gia cc ht mang in v gia cc vt tip tip xc gy nn ma st. Bn knh ln v hn, lc tng tc mnh hn tng tc hp dn c 3810ln. Tng tc in t l bn cht ca cc lc Culng, lc in t, lc Lo ren, lc ma st, lc lin kt ha hc... - Tng tc yu cc leptn: l tng tc gia cc leptn. Bn knh tc dng rt nh c 1810 m, lc tng tc yu hn tng tc hp dn c 1110ln. V d: cc qu trnh hn r |: p n + e+ + ve ;n p + e- + ~ev-Tng tc mnh: L tng tc gia cc hadrn; khng k cc qu trnh phn r ca chng. Bn knh tc dng rt nh c 1510 m, lc tng tc yu hn tng tc hp dn c 210ln.Mt trng h ring ca tng tc mnh l lc ht nhn. 4. Kch thc ca nguyn t, ht nhn, prton ln lt l: 10-10m, 10-14m, 10-15m. - Theo th t kch thc gim dn: Phn t > nguyn t > ht nhn > nucln > quac. II. MT TRI H MT TRI: 1. H mt tri: Gm Mt Tri v 8 hnh tinh, cc tiu hnh tinh v cc v tinh, cc sao chi v thin thch. -Cchnhtinh:Thytinh,Kimtinh,Trit,Hatinh,Mctinh,Thtinh,ThinVngtinh,HiVng tinh. - o n v gia cc hnh tinh ngi ta dng n v thin vn:=61vtv 150.10 km. - Nm nh sng: l qung ng m as i c trong 1 nm. 121 na m a nh sa ng =9,46.10 Km 29 - Cc hnh tinh u quay quanh mt tri theo chiu thun trong cng mt phng, Mt Tri v cc hnh tinh t quay quanh n v u quay theo chiu thun tr Kim tinh. 2. Mt tri: - L thin th trung tm ca h mt tri. C bn knh > 109 ln bk tri t; khi lng = 333 000 ln kl T. - C khi lng ln, lc hp dn ca Mt Tri c vai tr quyt nh s hnh thnh, pht trin v chuyn ng ca h. - L mt qu cu kh nng sng, khong 75% l hir v 23% l heli. Nhit b mt 6000K, trong lng n hng chc triu . Trong lng mt tri lun xy ra . nht hch l . tng h ht nhn hir thnh hn heli. -Cu trc ca mt tri: Nhn tng qut, Mt tri c cu to gm hai phn l quang cu v kh cu. +Quangcu.NhntTrittathyMttricdngmtasngtrnvbnknhgc16pht.khicu nng sng nhn thy ny c gi l quang cu ( cn gi l quang quyn, c bn knh khong 7.105 km). +Kh quyn Mt tri (kh cu). Bao quanh quang cu c kh quyn Mt tri. Kh quyn Mt tri c cu to ch yu bi hir, heli v c nhit rt cao nn kh quyn c c tnh rt phc tp. Kh quyn c phn ra hai lp c tnh cht vt l khc nhau l sc cu v nht hoa. Sc cu l lp kh nm st mt quang cu c dy trn 10 000 km v c nhit khong 4500k. Phangoiscculnhthoa.Vtchtcutonhthoatrngthiionhomnh(giltrngthi plaxma). Nhit khong 1 triu . Nht hoa c hnh dng thay i theo thi gian. - Cng sut pht xMt Tri l=26P 3,9.10 W. Lu : ng sut bc x ca mt tri P = 3,9.1026W, MP = At = Et ==> E = P.t==> Khi Lng mt tri gim i l : m = E/c2 = Pt/c2 3. Tri t:auto:Tritcdnghnhphngcu,bnknhxchobng6378km,bnknhhaiccbng 6357km, khi lng ring trung bnh 35515kg/m . + Li Tri t: bn knh3000km; ch yu l st, niken; nhit khong 03000 - 4000 C. + V Tri t: dy khong35km; ch yu l granit; khi lng ring 33300kg/m . - 1 vi s liu v T:BK = 6400km,KL= 5,98.1024kg, BK qu o quanh mt tri 150.106km. Chu k quay quanh trc 23h56ph004giy. Chu k quay quanh mt tri 365,2422 ngy. Gc nghing 23027 b Mt Trng- v tinh ca Tri t- Mt trng cch Tri t 384 000 km c bn knh 1738 km, c khi lng 7, 35.1022 kg. Gia tc trng trng caMttrngl1,63m/s2.MttrngchuynngquanhTritvichuk27,32ngy.Trongkhichuyn ng caTri t, Mt Trng cn quay quanh trc ca n vi chu k ng bng chu k chuyn ng quanh Tri t. Hn na, do chiu t quay cng chiu vi chiu quay quanh Tri t, nn Mt Trng lun hng mt na nht nh ca n v pha Tri t. - Do lc hp dn b nn Mt Trng khng gi c kh quyn. Ni cc khc, Mt Trng khng c kh quyn. -BmtMttrngcphmtlpvtchtxp.TrnbmtMtTrngcccdynicao,cccvng bng phng c gi l bin (bin , khng phi l bin nc), c bit l c rt nhiu l trn trn cc nh ni (c th l ming ni la tt, hoc vt tch va chm ca cc thin thch). - Nhit trong mt ngy m trn Mt Trng chnh lch nhau rt ln ; vng xch o ca mt Mt Trng, nhit lc gia tra l trn 1000C nhng lc na m li l-150 0C .- Mt Trng c nhiu nh hng n Tri t, m r rt nht l gy ra hin tng thu triu. Cn lu rng kh quyn Tri t cng b tc dng ca lc triu (triu), dng ln v h xung vi bin ln hn bin ca thu triu rt nhiu ln. 3. Hnh tinh chuyn ng xung quanh Mt Tri theo mt qu o xc nh. -Cchnhtinh:Thytinh,Kimtinh,Trit,Hatinh,Mctinh,Thtinh,ThinVngtinh,HiVng tinh. - Cc hnh tinh c kch thc nh c vi trm km hoc nh hn gi l cc tiu hnh tinh. - V tinh chuyn ng quanh hnh tinh.- Nhng hnh tinh thuc nhm Tri t l: Thu tinh, Kim tinh, Tri t v Ho tinh. l cc hnh tinh nh, rn, c khi lng ring tng i ln. Nhit b mt tng i cao. - Nhng hnh tinh thuc nhm Mc tinh l: Mc tinh, Th tinh, Hi vng tinh v Thin vng tinh. Chng l cc hnh tinh ln, c th l khi kh hoc nhn rn v xung quanh l cht lng. Nhit b mt tng di thp. 30 - Cc c trng c bn ca cc hnh tinh Thin th Khong cch n Mt Tri (vtv) Bn knh (km) Khi lng (so vi Tri t) Khi lng ring (103kg/m3) Chu k t quay Chu k chuyn ng quanh Mt Tri S v tinh bit Thy tinh0,3924400,0555,459 ngy87,9 ngy0 Kim tinh0,7260560,815,3243 ngy224,7 ngy0 Tri t1637515,523g56ph 365,25 ngy (1 nm) 1 Ha tinh1,5233950,113,924g37ph1,88 nm2 Mc tinh5,2714903181,39g50ph11,86 nm63 Th tinh9,5460270950,714g14ph29,46 nm34 Thin Vng tinh 19,1925760151,217g14ph84,00 nm27 Hi Vng tinh 30,0725270171,716g11ph164,80 nm13 . Sao chi v thin thch: - Sao chi: L nhng khi kh ng bng ln vi , c ng knh vi km, chuyn ng quanh Mt Tri theo qu o elp rt dt m mt tri l 1 tiu im. Khi sao chic trn qu o gn mt tri vt cht trong sao b nng sng v bay hi thnh m kh v bi quanh sao. m kh v bi bao quanh sao b p sut do as mt tri gy ra y dt v pha i din vi mt tri to thnh ci ui sao chi. ng trn T ta nhn thy c u v ui sao chi: u s hi gn m i, ui s hi x T hn - Thin thch: L nhng tng chuyn ng quanh mt tri. Trng h thin thch bay v bu kh quyn ca tri t th n b ma st mnh nu nng sng v bc chy, li mt vt di m ta gi l s ng III. CC SAO V THIN H: 1. Sao: -Sao l mt thin th nng sng ging nh Mt Tri. Cc sao rt xa, hin nay bit ngi sao gn nht cch chngtanhngchctkm(trn4nmas);cnngisaoxanhtcchxan14tnmnhsng (1219, 46.10 na m a nh sa ng Km = ). - Xung quanh mt s sao cn c cc hnh tinh chuyn ng, ging nh h Mt Tri. Khi lng ca cc sao c gi tr nm trong khong t 0,1 ln khi lng Mt Tri n vi chc ln (a s khong 5 ln ) khi lng Mt Tri. Bn knhca cc sao cgi tr nm trongmt khong rt rng, tkhong mt phn nghn ln bn knh Mt Tri ( sao cht) n gp hng ngn ln bk mt tri ( sao knh). 2. c loi sao:- a s cc sao tn ti trong trng thi n nh; c kch thc, nhit , khng i trong mt thi gian di.- Ngoi ra; ngi ta pht hin thy c mt s sao c bit nh sao bin quang, sao mi, sao ntron, + Sao bin quang c sng thay i, c hai loi:- Sao bin quang do che khut l mt h sao i (gm sao chnh v sao v tinh), sng tng hp m ta thu c s bin thin c chu k. - Sao bin quang do nn dn c sng thay i thc s theo mt chu k xc nh. + Sao mi c sng tng t ngt ln hng ngn, hng vn ln ri sau t t gim. L thuyt cho rng sao mi l mt pha t bin trong qu tnh bin ha ca mt h sao. 31 + Punxa, sao ntron ngoi s bc x nng lng cn c phn bc x nng lng thnh xung sng v tuyn.- Sao ntron c cu to bi cc ht ntron vi mt cc k ln 14 310 g/cm . - Punxa (pulsar) l li sao ntron vi bn knh10km t quay vi tc gc640 vo ng/sv pht ra sng v tuyn.Bc x thu c trn Tri t c dng tng xung sngging nhng sng ngn hi ng m tu bin nhn c. - Ngoi ra, trong h thng cc thin th trong v tr c cc l en v cc tinh vn. + L en l mt thin th c tin on bi l thuyt, cng c cu tao bi cc ntron, c trng hp dn ln n ni thu ht mi vt th, k c nh sng. V vy, thin th ny ti en khng pht bt k sng in t no. Ngi ta ch pht hin c mt l en nh tia X pht ra, khi l en ht mt thin th gn . + Tinh vn ta cn thy nhng m my sng, gi l. l cc m bi khng l c ri sng bi cc ngi sao gn , hoc l cc m kh b ion ho c phng ra t mt sao mi hay sao siu mi. 3. Khi qut v s tin ho ca cc sao Khi nhin liu trong sao cn kit, sao bin thnh cc thin th khc. L thuyt cho thy cc sao c khi lng c Mt Tri c th sng ti 10 t nm, sau bin thnh sao cht trng (hay sao ln ), l sao c bn knh ch bng mt phn trm hay mt phn nghn bn knh Mt Tri nhng li c nhit b mt ti 50 000 K. Cn cc sao c khi lng ln hn mt tri (t nm ln tr ln) th ch sng c khong 100 triu nm, nhit ca saogimdnvsaotrthnhsaoknh,saulitiptctinhovtrthnhmtsaontron(punxa), hoc mt l en.4. Thin h:- Cc sao tn ti trong v tr thnh nhng h thng tng i c lp i vi nhau.H thng saogm nhiu loi sao v tinh vn gi l thin h. a. c loi thin h:- Thin h xon c c hnh dng dt nh cc a, c nhng cnh tay xon c, cha nhiu kh. - Thin h elip c hnh elip, cha t kh v c khi lng tri ra trn mt di rng. C mt loi thin h elip l ngun pht sng v tuyn in rt mnh. - Thin h khng nh hnhtrng nh nhng m my (thin h Ma gien-lng). - ng knh ca cc thin h vo khong 100 000 nm nh sng . - Ton b cc sao trong mi thin h u quay xung quanh trung tm thin h. b. Thin H ca chng ta. Ngn h:-Thinhcachngtalloithinh xonc,cngknhkhong100nghnnmnhsngvckhi lng bng khong 150 t ln khi lng Mt Tri. N l mt h phng ging nh mt ci a, dy khong 330 nm nh sng, cha vi trm t ngi sao. H Mt Tri nm trong mt cnh tay xon ra thin h, cch trung tm trn 30 nghn nm nh sng v quay quanh tm thin h vi tc khong 250 km/s. Gia cc sao c bi v kh. Phn trung tm thin h c dng mt hnh cu dp, gi l vng li trung tm (dy khong 15 000 nm nh sng ), c to bi cc sao gi kh v bi. Ngay trung tm thin h c mt ngun pht x hng ngoi vcnglngunphtxsngvtuynin;ngunnyphtrannglngtngngvisngca chng 20 triu ngi sao nh mt tri v phng ra mt lung gi mnh. - T Tri t, Chng ta ch nhn c hnh chiu ca Thin H trn vm tri, nh mt di sng tri ra trn bu tri m, c gi l di Ngn H. Mt phng trung tm ca di Ngn H tr nn ti do mt ln bi di. Vo u m ma h, ta thy di Ngn H nm trn nn tri sao theo hng ng Bc- Ty Nam . c. Nhm thin h. Siu nhm thin h: -Vtrchngtrmtthinh,ccthinhthngcchnhaukhongmilnkchthcThinHca chng ta. Cc thin h c xu hng hp li vi nhau thnh tng nhm t vi chc n vi nghn thin h. -ThinHcachngtavccthinhlnlnthucvhmhinhphng,gmkhong20thnh vin, chim mt th tch khng gian c ng knh gn mt triu nm nh sng. Nhm ny b chi phi ch yu bi ba thin h xon c ln: Tinh vn Tin N (thin h Tin N M31 hay NGC224); Thin H ca chng ta; Thin h Tam gic, cc thnh vin cn li l Nhm cc thin h elip v cc thin h khng nh hnh t hon. - khong cch c khong 50 triu nm nh sng l Nhm Trinh Ncha hng nghn thin h tri rng trn bu tri trong chm sao Trinh N. - Cc nhm thin h tp hp li thnh Siu nhm thin h hay i hin h. Siu nhm thin h a phng c tm nm trong Nhm Trinh N v cha tt c cc nhm bao quanh n, trong c nhm thin h a phng ca chng ta. I. THUYT N LN BI BAN 1. c s kin thin vn quan trng a tr dn n: Cc thin h dch chuyn ra xa nhau, l bng chng ca s kin thin vn quan trng : v tr ang dn n. 32 b Bc x v tr Bc x ny c pht ng u t pha trong khng trung v tng ng vi bc x pht ra t vt c nhit khong 3K (chnh xc l 2,735K); bc x ny c gi l bc x 3K. Kt qu thu c chng t bc x l bc x c pht ra t mi pha trong v tr (nay ngui) v c gi l bc x nn v tr. 2. nh lut Hp-bn:- Tc li ra xa ca thin h t l vi khong cch gia thin h v chng ta:v = H.d Vi: v l tc chy xa ca thin h d l k/c t thin h ang xt n thin h ca chng ta =2H 1,7.10 m/s.na m a nh sa ng gi l hs Hp - bn =121 na m a nh sa ng 9,46.10 Km 3. Thuyt v n ln Big Bang: -Theothuytvnln,vtrbtudnntmtimkd.tnhtuivbnknhvtr,tachn im k d lm mc (gi l im zr Big Bang). - Ti thi im ny cc nh lut vt l bit v thuyt tng i rng khng p dng c. Vt l hc hin i da vo vt l ht s cp d on cc hin tng xy ra bt u t thi im tp= 10-43ssau V n ln gi l thi im Planck. - thi im Planck, kch thc v tr l 3510 m, nhit l 3210 Kv mt l 91 310 kg/cm . Cc tr s cc ln cc nh ny gi l tr s Planck. T thi im ny V tr dn n rt nhanh, nhit ca V tr gim dn. Ti thi im Planck, V tr b trn ngp bi cc ht c nng lng caonh electron, notrino vquark, nng lng t nht bng 1510 GeV .-Tithiimt=10-6s,chuynngccquarkvphnquarkchmcclctngtcmnhgom chng li v gn kt chng li to thnh cc prtn v ntrn, nng lng trung bnh ca cc ht trong v tr lc ny ch cn 1GeV . - Ti thi im= t 3 phu t , cc ht nhn Heli c to thnh. Trc , prtn v ntrn kt hp vi nhau to thnh ht nhn teri 21H . Khi , xut hin cc ht nhn teri 21H , triti 31H , heli 42Hebn. Cc ht nhn hir v hli chim98% khi lng cc sao v cc thin h, khi lng cc ht nhn nng hn ch chim 2%. mi thin th, c 14 khi lng l hli v c 34 khi lng l hir. iu chng t, mi thin th, mi thin h c cng chung ngun gc.- Ti thi im= t 300000 na m, cc loi ht nhn khc c to thnh, tng tc ch yu chi phi v tr l tng tc in t. Cc lc in t gn cc electron vi cc ht nhn to thnh cc nguyn t H v He.- Ti thi im=6t 10na m, cc nguyn t c to thnh, tng tc chyu chi phi v tr l tng tc hp dn. Cc lc hp dn thu gom cc nguyn t li, to thnh cc thin h v ngn cn cc thin h tip tc n ra. Trong cc thin h, lc hp dn nn cc m nguyn t li to thnh cc sao. Ch c khong cch gia cc thin h tip tc tng ln. - Ti thi im=9t 14.10na m, v tr trng thi nh hin nay vi nhit trung bnh= T 2, 7K. Lu :- Theo hiu ng p-le vi sng as th nu 1 ngun ng yn pht ra 1 bc x n sc bc sng 0, khi ngun chuyn ng vi tc v i vi my thu th bc sng ca bc x m my thu nhn c l .- dch chuyn bc sng ca bc x lA = - 0 = 0vc + Nu ngun ra xa my thu th v > 0 ==> A = - 0 > 0 ==> > 0 , bc sng ca bc x d/c v pha , bs di hn. + Nu ngun li gnmy thu th v < 0 ==> A = - 0 < 0 ==> < 0, bc sng ca bc x d/c v pha tm, bs ngn hn. --- Ht --- hc cc em hc tt t kt qu cao trong cc k thi sp ti