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7/31/2019 Chuyen de VATLY Hat NhanDH
1/58
Chuyn : VT L HT NHN N THI H-C GV: on Vn Lng
Email: [email protected]; [email protected] : 0915718188 - 0906848238 Trang
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 scp gi l nucln gm:Ht scp(nuclon)
Ki hiu Khi lng theo kg Khi lng theo u1u =1,66055.10 -27 kg
in tch
Prtn: Hp 11= mp = 2710.67262,1 kg mp =1,00728u +eNtrn: 1
0n n= mn =2710.67493,1 kg mn =1,00866u khng mang in tch
1.1. K hiu ht nhn:A
ZX - A = s nuctrn : s khi- Z= s prtn = in tch ht nhn (nguyn t s)- N A Z= : s ntrn
1.2. Bn knh ht nhn nguyn t:1
15 31,2.10R A= (m)
V d: + Bn knh ht nhn H1
1
H: R = 1,2.10-15m
+ Bn knh ht nhn Al2713 Al: R = 3,6.10-15m
2.ng v l 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 vkhi lng nguyn t- u : c gi tr bng 1/ 12 khi lng ng v cacbon 126C
- 27 2231 12 1 12
1 . . 1,66055 .10 931,5 / 12 12 6,0221.10
= = =A
u 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 =2c
E
=> khi lng c tho bng n v nng lng chia cho c2: eV/c2 hay MeV/c2.-Theo Anhxtanh, mt vt c khi lng m0 khi trng thi ngh th khi chuyn ng vi tc v, kh
lng s tng ln thnh m vi: m =
2
2
0
1c
v
m
trong m0 gi l khi lng ngh v m gi l khi lng ng.
5.Mt scc ht thng gp: Tn gi K hiu Cng thc Ghi ch
prtn p11H hay
11p hir nh
teri D
2
1H hay
2
1D hir nngtriti T
31H hay
31T hir siu nng
anpha 42He Ht Nhn Hli
bta tr -01e electron
bta cng +01e+ Pzitn (phn electron
ntron n10 n khng mang in
ntrin khng mang in, m0 = 0, v c
+-
Nguyn t Hidr, Ht nhnc 1 nucln l prtn
Ht nhn Hli c 4 nucl2 prtn v 2 ntrn
+ +- -
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Chuyn : VT L HT NHN N THI H-C GV: on Vn Lng
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II. HT KHI NNG LNG LIN KT CA HT NHN1. 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 khi m ca ht nhn AZX
Khi lng ht nhnhn
m lun nh hn tng khi lng cc nucln to thnh ht nhn mt lng m :
Khi lng ht nhn Khi lng Z Prtn Khi lng N Ntrn ht khi m
mhn (mX) Zmp (A Z)mn m = Zmp + (A Z)mn mhn
3. Nng lng lin kt lkW ca ht nhnA
ZX
- Nng lin kt l nng lng ta ra khi to thnh mt ht nhn (hay nng lng thu vo ph vmt ht nh
thnh cc nucln ring bit). Cng thc :2.
lkW m c= Hay :
2. . .lk p n hnW Z m N m 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 = lkW
A.
- Ht nhn c nng lng lin kt ring cng ln th cng bn vng.
- V d:5628 Fe c nng lng lin kt ring ln = lk
W
A =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 4
1 2 3 41 2 3 4
A A A A
Z Z Z ZX X X X+ + hay1 2 3 4
1 2 3 4
A A A A
Z 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= ;10 n ;
42 =He ;
01e
= ;
01e
++=
II. CC NH LUT BO TON TRONG PHN NG HT NHN1. nh lut bo ton snucln (skhi A) 1 2 3 4A A A A+ = +
2. nh lut bo ton in tch (nguyn tsZ) 1 2 3 4Z Z Z Z+ = +
3. nh lut bo ton ng lng: = sPPt
4. nh lut bo ton nng lng ton phn WsW t =
Ch :-Nng lng ton phn ca ht nhn: gm nng lng ngh v nng lng thng thng( ng nng):
2 21
2W mc mv= +
- nh lut bo ton nng lng ton phn c th vit: W1 + W2 + m1.c2 + m2.c
2 = W3 + W4 + m3.c2 + m4.c
2
=> (m1
+ m2 - m3 - m4) c
2
= W3 + W4 - W1 - W2 = Q ta /thu- Lin h gia ng lng v ng nng 2 2 dP mW= hay
2
2dP
Wm
=
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 : 20
20 )()( cmmcmmW == (J)
-Trong trng hp ( ) ; ( )m u W MeV : 5,931)(5,931)( 00 mmmmW ==
Nu m0 > m: 0W> : phn ng ta nng lng;Nu m0 < m : 0W< : phn ng thu nng lng
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Chuyn : VT L HT NHN N THI H-C GV: on Vn Lng
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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 hnhn 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 2
A A
Z 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 1
A A
Z ZX e Y+ +
- Phng x: Sng in t c bc sng rt ngn:* 0
0A A
Z ZX X + 1.2. Bn cht v tnh cht ca cc loi tia phng x
Loi Tia Bn Cht Tnh Cht
()-L dng ht nhn nguyn t Heli ( 42He ), chuyn
ng vi vn tc c2.107m/s.
-Ion ho rt mnh.-m xuyn yu.
(-) -L dng ht lectron 01( )e , vn tc c
(+) -L dng ht lectron dng (cn gi l pozitron)01( )e+ , vn tc c .
-Ion ho yu hn nhng m xuyn mnhhn tia .
()-L bc xin t c bc sng rt ngn (di 10-11
m), l ht phtn c nng lng rt cao-Ion ho yu nht, m xuyn mnh nht.
III. CC NH LUT PHNG X1. 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 thnht nhn khc.
2. Hng sphng x:ln 2
T = (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=
Trong qu trnh phn r, s htnhn phng x gim theo thi gian:
Trong qu trnh phn r, khilng ht nhn phng x gim theothi gian :
- i lng c trng cho tnh phng xmnh hay yu ca cht phng x.
- S phn r trong mt giy.
( ) 0 0.2 .
= =t
tTt
N N N e ( ) 0 0.2 .
= =t
tTtm m m e
( ) 0 0.2 .
= =
t
tTt
H H H e
H N=
0N : s ht nhn phng xthi
im ban u.
( )tN : s ht nhn phng x cnli sau thi gian t.
0m : khi lng phng xthi
im ban u.
( )tm : khi lng phng x cn lisau thi gian t.
0H : phng xthi im ban u.
( )tH : phng x cn li sau thi gian t
Hay:i lng Cn li sau thi gian t B phn r sau thi gian t N/N0 hay m/m0 (N0 N)/N0 ;
(m0 m)/m0
Theo s ht N
N(t)= N0 e-t; N(t) = N0 T
t
2 N0 N = N0(1- e
-t )T
t
2 (1- e-t )
Theo khi lng(m) m= m0 e
-t ; m(t) = m0 Tt
2 m0 m = m0(1- e
-t )T
t
2 (1- e-t )
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Chuyn : VT L HT NHN N THI H-C GV: on Vn Lng
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* phng x: ca mt lng cht phng xc trng cho tnh phng x mnh hay yu ca n, c xc n
bi s ht nhn b phn r trong 1 giy: H = -t
N
= N = N0 Tt
2 = N0e-t
hay H = H0 Tt
2 = H0e-t .
n vo 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 xbng 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 tnh du.- Dng phng x tm 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. Phn ng phn hch: l mt ht nhn rt nng nh Urani ( 23592U) hp th mt ntrn chm s vthnh hai h
nhn trung bnh, cng vi mt vi ntrn mi sinh ra.1 2
1 2
235 1 236 192 0 92 0 200
A A
Z ZU n U X X k n MeV + + + +
2. Phn ng phn hch dy chuyn: Nu s phn hch tip din thnh mt dy chuyn th ta c phn ng phhch dy chuyn, khi s phn hch tng ln nhanh trong mt thi gian ngn v c nng lng rt ln c ta riu kin xy ra phn ng dy chuyn: xt s ntrn trung bnh k sinh ra sau mi phn ng phn hch ( k l hnhn ntrn).
- Nu 1k< : th phn ng dy chuyn khng th xy ra.- Nu 1k= : th phn ng dy chuyn s xy ra v iu khin c.- Nu 1k> : th phn ng dy chuyn xy ra khng iu khin c.- Ngoi ra khi lng 23592U phi 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 HCH1. 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 0 3,25H 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 t hn mt phn ng phn hch nhng nu tnh theo kh
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 nghtch cc; Cm nhn am m; Hnh ng kin tr !Chc cc em hc sinh THNH CNG trong hc tp!Su tm v chnh l: GV:on Vn LngEmail: [email protected] ; [email protected];T: 0915718188 0906848238
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B. CC HNG S VT L v I N V VT L :1.Cc hng svt 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 xth phi k ti tin ch tra cu mt s hng s vt lv i mt sn v trong vt l. Cc hng s vt l c ci strong b nhca my tnh vi n v trong hn v SI.
+Cc hng sc ci sn trong my tinh cm tay Fx570MS; Fx570ES; 570ES Plus bng cc lnh: [CONSTNumber [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 tbi cho , hoc nu mun kt qu chnh xc hn th nn nhp cc hng s thng qua cc m lnh CONS[0 40] c ci t sn trong my tinh! (Xem thm bng HNG S VT L di y)Cc hng s thng dng l:
Hng s vt l M s Cch nhp my :
My 570MS bm: CONST 0 40 =
My 570ES bm: SHIFT 7 0 40 =
Gi tr hin th
Khi lng prton (mp) 01 Const [01] = 1,67262158.10-27
(kg)Khi lng ntron (mn) 02 Const [02] = 1,67492716.10
-27 (kg)
Khi lng lectron (me) 03 Const [03] = 9,10938188.10-31 (kg)
Khi lng 1u (u) 17 Const [17] = 1,66053873.10-27 (kg)
Hng s Faray (F) 22 Const [22] = 96485,3415 (mol/C)
in tch lectron (e) 23 Const [23] = 1,602176462.10-19 (C)
S Avgar (NA) 24 Const [24] = 6,02214199.1023 (mol-1)
Tc nh sng trong chn
khng (C0) hay c
28 Const [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 tuc2 sang MeV: 1uc2 = 931,5MeV(My 570ES: SHIFT 7 17 x SHIFT 7 28 x2 : SHIFT 7 23 : X10X 6 = hin th 931,494...)
- My 570ES bm Shift 8 Conv [m s] =
-V d : T 36 km/h sang ? m/s , bm: 36 Shift 8 [Conv] 19 = Mn hnh hin th : 10m/s
My 570MS bm Shift Const Conv [m s] =
Nguyn tc thnh cng: Suy nghtch cc; Cm nhn am m; Hnh ng kin tr !Chc cc em hc sinh THNH CNG trong hc tp!Su tm v chnh l: GV:on Vn LngEmail: [email protected] ; [email protected];T: 0915718188 0906848238
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C: CC DNG BI TPI.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 nhnA
ZX ZA, , N = A-Z
b.Bi tpBi 1: Xc nh cu to ht nhn
23892U , Na
2311 , He
42 ( Tm s Z prtn v s N ntron)
+ 23892U c 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 = 12 p n: Na2311 : 11 prtn ; 12 ntron
+ He42 gm : 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 t XAZ c cu to gm Z ntron v A prton.
B. Ht nhn nguyn t XAZ c cu to gm Z prton v A ntron.
C. Ht nhn nguyn t XAZ c cu to gm Z prton v (A Z) ntron.
D. Ht nhn nguyn t XAZ c cu to gm Z ntron v (A + Z) prton.
Cu 2. Ht nhn Co6027 c cu to gm:
A. 33 prton v 27 ntron B. 27 prton v 60 ntron C. 27 prton v 33 ntron D. 33 prton v 27 ntron
Cu 3: Xc nh s ht proton v notron ca ht nhn N147
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 tng 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 235 D. 92 proton v tng s ntron l 235
Cu 5: Nhn Uranium c 92 proton v 143 notron k hiu nhn l
A. U32792 B. U23592 C. U
92235 D. U
14392
Cu 6: Tm pht biu sai v ht nhn nguyn t AlA. 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 khlng nguyn t u.
A. mP > u > mn B. mn < mP < u C. mn > mP > u D. mn = mP > u
Cu 8. Cho ht nhn 115 X . 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 xu 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 2914 Si , ht nhn4020Ca c 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 35
17 Cl c:
A. 35 ntron B. 35 nucln C. 17 ntron D. 18 proton.
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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: 0mmm = ; 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: =A
Wlk MeV/nuclon. HayA
mc
A
E 2=
=
+Chuyn i n v tuc2 sang MeV: 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 tpBi 1 :Khi lng ca ht 104Be l mBe = 10,01134u, khi lng ca ntron l mN = 1,0087u, khi lng ca proton
mP = 1,0073u. Tnh ht khi ca ht nhn104Be l bao nhiu?
HD gii-Xc nh cu to ht nhn 104Be c 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
Bi 2: Tnh nng lng lin kt ht nhn tri D21 ? Cho mp = 1,0073u, mn = 1,0087u, mD = 2,0136u; 1u = 931 MeV/c
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 + mnmD = 1.mp +1.mn mD = 0,0024 uNng lng lin kt ca ht nhn D : Wlk = m.c
2 = 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. Bit mn = 1,00866u; mp = 1,00728u;mHe = 4,0015u
HD gii : T =
He
ZAN
42
224 ==N . Ta c 03038,00015,4)(2 =+= np mmm u
MeVMeVucE 29,285,931.03038,003038,0 2 === MeV07,74
29,28==
Bi 4. Cho Fe5626 . Tnh nng lng lin kt ring. Bit mn = 1,00866u; mp = 1,00728u; mFe = 55,9349uHD gii: + Ta c ummm np 50866,09349,553026 =+=
MeVMeVucE 8,4735,931.50866,050866,0 2 === MeV46,856
8,473==
Bi 5: Ht nhn Be104 c khi lng 10,0135u. Khi lng ca ntrn (ntron) mn = 1,0087u, khi lng ca prt
(prton) mP = 1,0073u, 1u = 931 MeV/c2. Nng lng lin kt ring ca ht nhn l Be104
A. 0,632 MeV. B. 63,215MeV. C. 6,325 MeV. D. 632,153 MeV.HD Gii :-Nng lng lin kt ca ht nhn Be104 : Wlk = m.c
2 = (4.mP +6.mn mBe).c2
= 0,0679.c2
= 63,249 MeV.
-Suy ra nng lng lin kt ring ca ht nhn Be104 :
63,1256,32510
lkW
A = = 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 khto thnh 1 gam hli. Cho bit khi lng ca prton v ntron l mp = 1,007276 u v mn = 1,008665 u; 1 u = 931,5MeV/c2; s avgar l NA = 6,022.10
23 mol-1.
HD Gii: He =A
Wlk =A
cmmZAmZ Henp2).)(.( +
=4
5,931).0015,4)008685,1007276,1.(2( += 7,0752 Me
W =M
m.NA.Wlk =
0015,4
1.6,022.1023.7,0752.4 = 46,38332.1023 MeV = 7,42133.1011 J.
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Bi 7. Tnh nng lng lin kt ring ca hai ht nhn Na2311 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 =A
Wlk =A
cmmZAmZ Henp2).)(.( +
=23
5,931).983734,22008685,1.12007276,1.11( += 8,1114 MeV;
Fe =56
5,931).9207,55008685,1.30007276,1.26( += 8,7898 MeV;
Fe > Na nn ht nhn Fe bn vng hn ht nhn Na.
Bi 8. Tm nng lng to ra khi mt ht nhn urani 234U phng x tia to thnh ng v thori 230Th. Cho cc nnlng 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.A
1/3 .vi r0 = 1,4.10
15m , 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/c
2 .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 = 0,22589kgc/ Th tch ht nhn : V = 4r3/3 = 4r0
3A/3 .
Khi lng ring ca ht nhn : D =3
173
03
0
10.45,14
3
3/4 m
kg
rr
m
Arr
Am
V
m pp==
d/ Tnh nng lng lin kt ca ht nhn : E = mc2 = {Zmp + (A Z)mn m}c2 = 1,8197u
E = 1,8107.931 = 1685 MeVNng lng lin kt ring : = E/A = 7,4557 MeV.
Bi 10:Bit khi lng ca cc ht nhn umumumum npC 0087,1;0073,1;0015,4;000,12 === v2/9311 cMevu = . Nng lng cn thit ti thiu chia ht nhn C126 thnh ba ht theo n v Jun l
A. 6,7.10-13 J B. 6,7.10-15 J C. 6,7.10-17 J D. 6,7.10-19 JHD Gii: C12 3 HeNng lng ph vmt ht C12 thnh 3 ht He: W = ( mri - mhn )c
2 = (3.4,0015 12). 931= 4.1895MeVTheo n v Jun l: W = 4,1895. 1,6.10-13 = 6,7032.10 -13J; Chn A
Bi 11 : Cho bit m= 4,0015u; 999,15=Om u; ump 007276,1= , umn 008667,1= . Hy sp xp cc ht nhn H42
C12
6
, O16
8
theo th t tng dn bn vng . Cu tr li ng l:
A. C126 , ,42He O
168 . B. C
126 , O
168 , ,
42He C. ,
42He C
126 , O
168 . D. ,
42He O
168 , C
126 .
HDGii: bi khng cho khi lng ca 12Cnhng ch y dng n vu, theo nh ngha on v u bng 1/12khi lng ng v12C do c th ly khi lng 12Cl 12 u.-Suy ra nng lng lin kt ring ca tng ht nhn l :He : Wlk = (2.mp + 2.mn m )c
2 = 28,289366 MeV Wlk ring = 7,0723 MeV / nuclon.C : Wlk = (6.mp + 6.mn mC )c
2 = 89,057598 MeV Wlkring = 7,4215 MeV/ nuclon.O : Wlk = (8.mp + 8.mn mO )c
2 = 119,674464 meV Wlk ring = 7,4797 MeV/ nuclon.-Ht nhn c nng lng lin kt ring cng ln th cng bn vng. Vy chiu bn vng ht nhn tng dn l :He < C < O. Chn C.
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c.Trc nghim:Cu 1: Ht nhn Co6027 c khi lng l 59,919u. Bit khi lng ca prton l 1,0073u v khi lng ca ntron
1,0087u. ht khi ca ht nhn Co6027 l
A. 0,565u B. 0,536u C. 3,154u D. 3,637u
Cu 2: ng v phng x cban 6027 Co pht 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 =
Cu 3: Bit khi lng ca ht nhn U238 l 238,00028u, khi lng ca prtn v ntron l mP=1.007276U; mn1,008665u; 1u = 931 MeV/ c2. Nng lng lin kt ca Urani 238
92 U l bao nhiu?
A. 1400,47 MeV B. 1740,04 MeV C.1800,74 MeV D. 1874 MeVCu4: Bit khi lng ca prtn mp=1,0073u, khi lng ntron mn=1,0087u, khi lng ca ht nhn temD=2,0136u v 1u=931MeV/c
2. Nng lng lin kt ring ca ht nhn nguyn tteri D21 lA. 1,12MeV B. 2,24MeV C. 3,36MeV D. 1,24MeV
Cu5: Khi lng ca ht nhn 104 Be l 10,0113u; khi lng ca prtn m p = 1,0072u, ca ntron m n = 1,0086; 1u
931 MeV/c 2 . Nng lng lin kt ring ca ht nhn ny l bao nhiu?A. 6,43 MeV B. 6,43 MeV C. 0,643 MeV D. Mt gi tr khc
Cu6: Ht nhn 2010
Nec khi lng
Ne
m 19,986950u= . Cho bitp n
m 1,00726u;m 1,008665u;= = 21u 931, 5MeV / c= . Nng lng lin kt ring ca 2010 Ne c gi tr l bao nhiu?
A. 5,66625eV B. 6,626245MeV C. 7,66225eV D. 8,02487MeVCu7: Tnh nng lng lin kt ring ca ht nhn 37
17Cl . Cho bit: mp = 1,0087u; mn = 1,00867u; mCl = 36,95655u
1u = 931MeV/c2A. 8,16MeV B. 5,82 MeV C. 8,57MeV D. 9,38MeV
Cu 8. Ht nhn hli ( 42 He) c nng lng lin kt l 28,4MeV; ht nhn liti (73 Li) c nng lng lin kt l 39,2MeV;
ht nhn tri ( 21 D) 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 = 6,02.10
23mol-1, 1u = 931MeV/c2. Cc nucln kt hp v
nhau to thnh ht , nng lng ta ra khi to thnh 1mol kh Hli lA. 2,7.1012J B. 3,5. 1012J C. 2,7.1010J D. 3,5. 1010JCu 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 = 3.108 m/s. Nng lng ti thiu tch ht nhn C 126 thnh cc nucln ring bit bng
A. 72,7 MeV. B. 89,4 MeV. C. 44,7 MeV. D. 8,94 MeV.Cu 11(C-2008): Ht nhn Cl17
37 c khi lng ngh bng 36,956563u. Bit khi lng ca ntrn (ntrol1,008670u, khi lng ca prtn (prton) l 1,007276u v u = 931 MeV/c2. Nng lng lin kt ring ca ht nhCl1737 bng
A. 9,2782 MeV. B. 7,3680 MeV. C. 8,2532 MeV. D. 8,5684 MeV.
Cu 12(H 2008): Ht nhn 104 Be c khi lng 10,0135u. Khi lng ca ntrn (ntron) mn = 1,0087u, khi l
ca prtn (prton) mP = 1,0073u, 1u = 931 MeV/c2. Nng lng lin kt ring ca ht nhn 104 Be l
A. 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 168 O 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 168 O xp x bngA. 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; 4018 Ar ;63 Li 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 63 Li th nng lng lin kt ring ca ht nhn4018 Ar
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.
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Dng 3: Tnh sht nhn nguyn tv sntron, prtn c trong m lng cht ht nhn.a.PHNG PHP:Cho khi lng m hoc smol ca ht nhn XAZ . Tm sht p , n c trong mu ht nhn .
Nu c khi lng m suy ra sht ht nhn X l : N = ANA
m. (ht) .
S mol :4,22
V
N
N
A
mn
A
=== . Hng S Avgar: NA = 6,023.1023 nguyn t/mol
Nu c smol suy ra sht ht nhn X l : N = n.NA (ht).+Khi : 1 ht ht nhn X c Z ht proton v (A Z ) ht ht notron.=>Trong N ht ht nhn X c :N.Z ht proton v (A-Z) N ht notron.
b.BI TPBi 1: Bit s Avgar l 6,02.10 23 mol-1, khi lng mol ca ht nhn urani U23892 l 238 gam / mol.
S ntron trong 119 gam urani U23892 l :
A. 2510.2,2 ht B.2510.2,1 ht C 2510.8,8 ht D. 2510.4,4 ht
HD Gii: S ht nhn c trong 119 gam urani U23892 l : N = ANA
m. 2323 10.01.310.02,6.
238
119== ht
Suy ra s ht ntron c trong N ht nhn urani U23892 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 It13152 I l :
A. 3,952.1023 ht B. 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 = 2310.02,6.131
100. =AN
A
mht. Chn B.
c.TRC NGHIM:Cu 1 (C- 2009): Bit NA = 6,02.10
23 mol-1. Trong 59,50g 23892 U c 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 prt
(prton) c trong 0,27 gam Al1327 l
A. 6,826.1022. B. 8,826.1022. C. 9,826.1022. D. 7,826.1022.
*Dng: Cho tng s ht cbn v hiu s ht mang in trong nguyn t( Ht mangin 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) : 4Cn 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 cbn ca 1 nguyn t X l 82, trong tng s ht mang in nhiu hn s ht khng
mang in l 22. Vy X lLi gii: Ta c: Z = (82 + 22) : 4 = 26 => St (Fe)VD2: Tng s ht cbn trong nguyn t Y l 52, trong tng s ht mang in nhiu hn s ht khng
mang in l 16. Y lLi gii: Ta c: Z = (52 + 16) : 4 = 17 => Y l Clo (Cl)VD3: Tng s ht cbn trong nguyn t Y l 18, trong tng s ht mang in nhiu hn s ht khng
mang in l 6. Y lLi gii: Ta c: Z = (18 + 6) : 4 = 6 => Y l Cacbon (C)
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II.NH LUT PHNG X- PHNG XDng 1: Xc nh lng cht cn li (N hay m), phng x:a.Phng php:Vn dng cng thc:
-Khi lng cn li ca X sau thi gian t :.0
0 0.2 .
2
t
tTt
T
mm m m e
= = = .
-Sht nhn X cn li sau thi gian t :.0
0 0.2 .2
t
tTt
T
NN N N e = = =
- phng x:t
NHtb
= ;
T
t
T
tH
HH
== 2.
20
0hay
t
teH
e
HH
== .00
Vi :ln 2
T =
-Cng thc tm smol :A
m
N
Nn
A
==
-Ch : + t v T phi a v cng n v.+ m v m0 cng n vv khng cn i n v
Cc trng hp c bit, hc sinh cn nhgii nhanh cc cu hi trc nghim:t
Cn li N= N0 2t
T
T s N/N0 hay (%)B phn r N0 N (%) T s
(N0- N)/N0T s(N0- N)/N
t =T N = N0 12 = N0/2 1/2 hay ( 50%) N0/2 hay ( 50%) 1/2 1t =2T N = N0 22 = N0/4 1/4 hay (25%) 3N0/4 hay (75%) 3/4 3t =3T N = N0 32 = N0/8 1/8 hay (12,5%) 7N0/8 hay (87,5%) 7/8 7t =4T N = N0 42 = N0/16 1/16 hay (6,25%) 15N0/16 hay (93,75%) 15/16 15t =5T N = N0 52 = N0/32 1/32 hay (3,125%) 31N0/32 hay (96,875%) 31/32 31t =6T N = N0 62 = N0/64 1/64 hay (1,5625%) 63N0/64 hay (98,4375%) 63/64 63t =7T N = N0 72 = N0/128 1/128 hay (0,78125%) 127N0/128 hay (99,21875%) 127/128 127
t =8T N = N0 82 = N0/256 1/256 hay (0,390625%) 255N0/256 hay (99,609375%) 255/256 255t =9T ................. ----------- ---------- ------- -------
Hay:Thi gian t T 2T 3T 4T 5T 6T 7T
Cn li: N/N0 hay m/m0 1/2 1/22 1/23 1/24 1/25 1/26 1/27
r: (N0 N)/N0 1/2 3/4 7/8 15/16 31/32 63/64 127/128T l % r 50% 75% 87,5% 93,75% 96,875% 98,4375% 99,21875%T l ( t s) ht r v cn li 1 3 7 15 31 63 127T l ( t s) ht cn li v bphn r
1 1/3 1/7 1/15 1/31 1/63 1/127
b. Bi tp:Bi 1: Cht It phng x13153 I dng trong y t c chu k bn r 8 ngy m. Nu nhn c 100g cht ny th sau 8 tunl cn bao nhiu?
A. O,87g B. 0,78g C. 7,8g D. 8,7g
HD GiHD GiHD GiHD Gii : t = 8 tun = 56 ngy = 7.T .Suy ra sau thi gian t th khi lng cht phng x13153 I cn li l :
70 2.1002.
== Tt
mm = 0,78 gam . Chn p n B.
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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
bao nhiu?A.m= m0/5 B.m = m0/8 C. m = m0/32 D. m = m0/10
HD Gii : t = 5T. Sau t = 5T th khi lng cht phng x cn li l:
322.2. 0500
mmmm T
t
===
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) clng 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 :
T
t
T
t
m
mmm
== 22.
00
8
12 3
0
==
m
m= 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 ub) S nguyn t cn li sau khong thi gian t= 5,7 ngyc) phng x ca lng Rn ni trn.
Hd gii: a) p dng N0 = 0. Am N
A
. D dng tnh c N0=5,42.1021 ht
b) : t=5,7=1,5T nn N=
2
o
t
T
N=1,91.1021 ht
c) p dng cng thc: 0 0.2 .t
tTH H H e N
= = = m0,693
T= Nn: H=
0,693
T.N
D dng tnh c: H= 4,05.1015 (Bq) =1,1.105(Ci)Bi 5: Plni l nguyn t phng x , n phng ra mt ht v bin i thnh ht nhn con X. Chu k bn r cPlni 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: XHePo AZ+42
21084
+ Theo cc LBT ta c:
=
=
+=
+=
82
206
284
4210
Z
A
Z
APbX 20682:
2.T BqAT
NmH
A
mNH
mm
NA
mN
NH
mmk
A
A
k
A
T
t
11000
10.08,2.
2..693,02.
.
2.
==
=
=
=
=
=
Nu trc nghim cn nh: BqAT
NmH
kA 110 10.08,2
.2..693,0 ==
Bi 6: Pht pho ( )3215 P phng x- vi chu k bn r T = 14,2 ngy v bin i thnh lu hunh (S). Vit phng trnhca 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 3215
P cn li l 2,5g. Tnh khi lng ban u ca n.
HD Gii :Phng trnh ca s pht x: 32 0 3215 1 16
P e + S
Ht nhn lu hunh 3216
S gm 16 prtn v 16 ntrn
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Tnh lut phng x ta c:
ln 2 ttt T T
o o om = m e m e m 2
= =
Suy ra khi lng ban u:
t3T
om m.2 2,5.2 20g= = = Bi 7(H -2009): Mt cht phng x ban u c N
0ht nhn. Sau 1 nm, cn li mt phn ba s ht nhn ban u ch
phn r. Sau 1 nm na, s ht nhn cn li cha phn r ca cht phng x lA. N0 /6 B. N0/16. C. N0 /9. D. N0/4.
HD Gii : t1 = 1nm th s ht nhn cha phn r (cn li ) l N1, theo ta c :31
21
0
1 ==T
tNN
Sau 1nm na tc l t2 = 2t1 nm th s ht nhn cn li cha phn r l N2, ta c :
T
t
T
tN
N12 2
0
2
2
1
2
1==
9
1
3
1
2
12
2
0
2 =
=
=
T
tN
N. Hoc N2 =
9330
201 NNN == Chn: C
Bi 8: Natri ( )2411Na l cht phng x- vi chu k bn r T = 15 gi. Ban u c 12g Na. Hi sau bao lu ch cn li 3cht phng x trn? Tnh phng x ca 3g natri ny. Cho s Avgar NA = 6,022 x 10
23 mol-1
HD Gii :Ta ctt T
o om m e m 2= =
t 20T m 12 t2 4 2 2 t 2T 2x15x30m 3 T
= = = = = = = gi.
phng x: Aln 2 m
H N . .NT N
= = Thay s:23
17 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 cs ca loga t nhi
vi lne = 1). T l chu k bn r ca cht phng x. Chng minh rngT
tln 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: . to
Ne e
N = = ; Suy ra t 1 = , do
1 Tt
ln 2 = =
Lng cht cn li sau thi gian 0,15t t l thun vi s ht: 0,15 t 0,15
o
Ne e 0,6 60%
N = = = =
c.Trc nghim:Cu1: 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 A. 93,75g. B. 87,5g. C. 12,5g. D. 6,25g.
Cu2: Chu k bn r ca 6027 Co bng gn 5 nm. Sau 10 nm, t mt ngun6027 Co 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.
Cu3: C 100g it phng x13153 I 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.
Cu4: Ban u c 5 gam cht phng x radon 22286 Rn vi chu k bn r 3,8 ngy. S nguyn t radon cn li sau 9
ngy lA. 23,9.1021. B. 2,39.1021. C. 3,29.1021. D. 32,9.1021.
Cu5: Pht pho P3215 phng x- vi chu k bn r T = 14,2 ngy. Sau 42,6 ngy k t thi im ban u, khi l
ca mt khi cht phng x P3215 cn li l 2,5g. Tnh khi lng ban u ca n.
A. 15g. B. 20g. C. 25g. D. 30g.
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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 gian690 ngy l:A. 6,25g B. 62,5g C. 0,625g D. 50g
Cu7: Gi t l khong thi gian s ht nhn ca mt lng cht phng x gim i e ln (e l cs ca lga 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 nhiphn trm lng ban u ?
A. 40%. B. 50%. C. 60%. D. 70%.Cu8: Cng thc no di y khng phi l cng thc tnh phng x?
A. ( ) ( )dt
dNH tt = B. ( ) ( )dtdNHt
t = C. ( ) ( )tt NH = D. ( ) Tt
t HH
= 20
Cu9: Mt lng cht phng x Rn22286 ban u c khi lng 1mg. Sau 15,2 ngy phng x gim 93,75%.
phng x ca lng Rn cn li lA. 3,40.1011Bq B. 3,88.1011Bq C. 3,58.1011Bq D. 5,03.1011BqCu10:(C 2007): Ban u mt mu cht phng x nguyn cht c khi lng m0 , chu k bn r ca cht ny l 3ngy. Sau 15,2 ngy khi lng ca cht phng x cn li l 2,24 g. Khi lng m0 l
A.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 giT/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 , 4N
C. 0 0 016N ,8N , 4N D. 0 0 016 2N ,8 2N ,4 2N Cu 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 3s ht nhn cn li l bao nhiu?
A. 4N0 B. 6N0 C. 8N0 D. 16N0Cu13: (H-C-2010). Ban u c N0 ht nhn ca mt mu cht phng x nguyn cht c chu k bn r T. Sau khothi 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.2
0N . B.20N . C.
40N . D. N0 2 .
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 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 vo 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 khonthi 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 (hotphng x) ca lng cht phng x cn li bng bao nhiu phn trm so vi phng x ca lng cht phng x bu?
A. 25%. B. 75%. C. 12,5%. D. 87,5%.
Cu 18(H 2008) : Ht nhn 1
1
AZ
X phng x v bin thnh mt ht nhn 2
2
AZ
Y bn. Coi khi lng ca ht nhn X,
bng s khi ca chng tnh theo n v u. Bit cht phng x 1
1
A
ZX c chu k bn r l T. Ban u c mt khi l
cht 1
1
A
ZX, sau 2 chu k bn r th t s gia khi lng ca cht Y v khi lng ca cht X l
A. 1
2
A4
AB. 2
1
A4
A C. 2
1
A3
AD. 1
2
A3
A
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Dng 2: Xc nh lng cht b phn r :a.Phng php:- Cho khi lng ht nhn ban u m0 ( hoc sht nhn ban u N0 ) v T . Tm khi lng ht nhn hoc shtnhn bphn r trong thi gian t ?
-Khi lng ht nhn bphn r: m = )1()21(.
000tT
t
emmmm
==
-Sht nhn bphn r l : N = )1()21( .000 tT
t
eNNNN
==
-> Hay Tm s nguyn tphn r sau thi gian t:
. .0 0 0 0 0 0 0.
1 1 1. (1 ) (1 ) (1 )
2
tt t
k t t
eN N N N N e N e N N N
e e
= = = = = =
-
Nu t B + C . Th: mAmB + mC
b. Bi tp:Bi 1. Cht phng x Po21084 phng ra tia thnh ch Pb
20682 .
a/ Trong 0,168g Plni c bao nhiu nguyn t b phn d trong 414 ngy m, xc nh lng ch to thnh trong thgian trn ?
b/ Bao nhiu lu lng Plni cn 10,5mg ? Cho chu k bn d ca Plni l 138 ngy m .HDGii :
a/ S nguyn t Plni lc u : N0 = m0NA/A , vi m0 = 0,168g , A = 210 , NA = 6,022.1023
Ta thy t/T = 414/138 = 3 nn p dng cng thc : N = N02t/T = N02
3 = N0/8 .S nguyn t b phn d l : N = N0 N = N0(1 2
t/T) = 7N0/8 = 4,214.1020 nguyn t .
S nguyn t ch to thnh bng s nguyn t Plni phn r trong cng thi gian trn .V vy thi gian trn khi lng ch l : m2 = N.A2/NA , vi A2 = 206 . Thay s m2 = 0,144g .
b/ Ta c : m0/m = 0,168/0,0105 = 16 = 24
. T cng thc m = m02t/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 226 Ra . Cho bit chu k bn r ca 226 Ra l 1580 nm. SAvgar l NA = 6,02.10
23 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 gam 226Ra l : N0 =2123 10.6646,210.022,6.
226
1. ==AN
A
mht .
Suy ra s ht nhn nguyn t Ra phn r sau 1 s l :
1086400.365.1580
121
0 10.70,32110.6646,2)21( =
==
T
t
NN 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 phr thnh ht nhn ca nguyn t khc vi s ht nhn ca cht phng x cn li
A. 7 B. 3 C. 1/3 D. 1/7
HD Gii :Thi gian phn r t = 3T; S ht nhn cn li : 78
7
8
1
2 030 =
====
N
NNNN
NN
Bi 4:ng v phng x Cban 6027 Co pht ra tiav vi chu k bn r T = 71,3 ngy. Trong 365 ngy, phn trm ch
Cban ny b phn r bngA. 97,12% B. 80,09% C. 31,17% D. 65,94%
HD Gii: % lng cht 60Co b phn r sau 365 ngy :
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m = )1( .00temmm = %12,971
03,71
2ln.365
==
em
m.
Hoc m = )21(00 Tt
mmm
= =
=
T
t
T
t
m
m
2
21
0
97,12% Chn A.
Bi 5: Mt cht phng x c 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 )21.( Tt
0
= mm =1,9375 g Chn A.
Bi 6: Ht nhn210
84Po phng x anpha thnh ht nhn ch bn. Ban u trong mu Po cha mt lng mo (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.0,92m0 B.0,06m0 C.0,98m0 D.0,12m0
HD Gii: PbPo 2068221084 +
p dng nh lut phng x N = N0/24 .s ht nhn ch to thnh ng bng s ht nhn Po bi phn r
=16
152/ 040
NNNN == ( N0 = AN
m.
2100 ) .Suy ra mPb = 206.
AN
N= 206*.
210*.16
15 0m = 0,9196m0.
c. TRC NGHIM:Cu1: ng v Co6027 l cht phng x
vi chu k bn r T = 5,33 nm, ban u mt lng Co c khi lng m
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%
Cu2: Cban 6027 Co l cht phng x vi chu k bn r 3
16nm. Nu lc u c 1kg cht phng x ny th sau 16 n
khi lng 6027 Co b phn r l
A. 875g. B. 125g. C. 500g. D. 250g.
Cu3: Chu k bn r 21084 Po l 318 ngy m. Khi phng x tia , plni bin thnh ch. C bao nhiu nguyn t pl
b phn r sau 276 ngy trong 100mg
210
84 Po ?A. 200,215.10 B. 202,15.10 C. 200,215.10 D. 201,25.10 Cu 4. Chu k bn r ca U 238 l 4,5.109 nm. S nguyn t b phn r sau 106 nm t 1 gam U 238 ban u l bao nhiBit s Avgadr NA = 6,02.10
23 ht/mol.A. 2,529.1021 B. 2,529.1018 C. 3,896.1014 D. 3,896.1017
Cu5: Chu k bn r ca cht phng x 9038 Sr l 20 nm. Sau 80 nm c bao nhiu phn trm cht phng x phn
thnh cht khc ?A. 6,25%. B. 12,5%. C. 87,5%. D. 93,75%.
Cu6: ng v phng x 6629 Cu c chu k bn r 4,3 pht. Sau khong thi gian t = 12,9 pht, phng x ca ng
ny gim xung bao nhiu :A. 85 % B. 87,5 % C. 82, 5 % D. 80 %
Cu7: Gi l khong thi gian s ht nhn ca mt ng v phng x gim i bn ln. Sau thi gian 2 s ht nhcn 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: Cht phng x 2411 Na c chu k bn r 15 gi. So vi khi lng Na ban u, khi lng cht ny b phn trong vng 5h u tin bng
A. 70,7%. B. 29,3%. C. 79,4%. D. 20,6%Cu9: Gi t l khong thi gian s ht nhn ca mt lng cht phng x gim i e ln (e l cs ca lga 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 nhiphn trm lng ban u ?
A. 40%. B. 50%. C. 60%. D. 70%.
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Dng 3 : Xc nh khi lng ca ht nhn con :a.Phng php:- Cho phn r : YX BZ
AZ ' + tia phng x . Bit m0 , T ca ht nhn m.
Ta c : 1 ht nhn m phn r th sc 1 ht nhn con tao thnh.Do : NX (phng x) = NY (to thnh)
-Smol cht bphn r bng smol cht to thnh YX
X nA
mn =
=
-Khi lng cht to thnh lA
Bmm XY .= . Tng qut : mcon = conme
me AA
m .
-Hay Khi lng cht mi c to thnh sau thi gian t
1 0 11 1 0(1 ) (1 )
t t
A A
A N ANm A e m e
N N A
= = =
Trong : A, A1 l s khi ca cht phng x ban u v ca cht mi c to thnhNA = 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 bphn r
(Trng hp phng x+, - th A = A1 m1 = m )
b. Bi tp:
Bi 1:ng v 2411 Na l cht phng x- to thnh ht nhn magi 2412 Mg. Ban u c 12gam Na v chu k bn r l 15gi. Sau 45 h th khi lng Mg to thnh l :
A. 10,5g B. 5,16 g C. 51,6g D. 0,516gHD 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 )21(12)21( 31
0
== T
t
m m = 10,5 g .
-Suy ra khi lng ca mg to thnh : mcon = 5,1024.24
5,10.==
me
conme
A
Amgam. Chn p n A
Bi 2 : Cht phng x Poloni Po21084 c chu k bn r T = 138 ngy phng ra tia v bin thnh ng v ch Pb20682 ,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 = 3Ta.S nguyn t b phn r sau 3 chu ki:
03
000 8
72 NNNNNN === hay khi lng cht b phn r m = 08
7m = 0,147g
20230 10.214,410.023,6.210.8
168,0.7
8
7=== AN
A
mN nguyn t
b.Khi lng ch hnh thnh trong 414 ngy m: mcon = conme
me AA
m.
= g144,0206.
210
147,0= ]
Bi 3 : Ht nhn 22688 Ra c chu k bn r 1570 nm phn r thnh 1 ht v bin i thnh ht nhn X. Tnh s ht nhX 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 xbs khi ca chng v NA = 6,02.10
23 mol-1.
HD Gii 3. Phng trnh phn ng: 22688 Ra 42 He +
22286 Rn. Trong nm th 786: khi lng
22688 Ra b phn r l:
mRa = m0( 1570785
2
- 1570786
2
) = 7.10-4g; khi lng 22286 Rn c to thnh: mRn = mRa.Ra
Rn
A
A= 6,93g;
s ht nhn 22286 Rn c to thnh l: NRn =Rn
Rn
A
m.NA = 1,88.10
18 ht.
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Bi 4 : Plni 21084 Po l mt cht phng x c chu k bn r 140 ngy m. Ht nhn plni phng x s bin thnh h
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 ngm.
HD Gii 4. Ta c: mPb = m0.Po
Pb
A
A(1 - T
t
2 ) = 31,1 mg.
Bi 5 : ng v 23592U phn r thnh ht nhnAZTh .
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 htnhn in tc to thnh trong qu trnh phn r .
HD Gii 5. 1) Phng trnh phn r 235 4 A92 2 ZU Th + Tnh lut bo ton s khi: 235 = 4 + A => A = 231.Tnh 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 23190Th gm 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 .Tnh lut bo ton s khi: 235 = 207 + 4x + 0y -> x = 7
Tnh lut bo ton in tch: 90 = 82 + 2x y -> y = 4Mi 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 tc to thnh.
Bi 6 : Cho chm ntron bn ph ng v bn 5525 Mn ta thu c ng v phng x5625Mn . ng v phng x
56 Mn c chu tr bn r T = 2,5h v pht x ra tia -. Sau qu trnh bn ph 55Mn bng ntron kt thc ng
ta thy trong mu trn t s gia s nguyn t 56 Mn v s lng nguyn t 55Mn = 10-10. Sau 10 gitip th t s gia nguyn t ca hai loi ht trn l:
A. 1,25.10-11 B. 3,125.10-12 C. 6,25.10-12 D. 2,5.10-11
Gii: Sau qu trnh bn ph 55Mn bng ntron kt thc th s nguyn t ca 5625 Mn gim, c s nguyn
5525 Mn khng i, Sau 10 gi= 4 chu k s nguyn t ca 5625 Mn gim 24 = 16 ln. Do th t s gia nguy
t ca hai loi ht trn l:55
56
Mn
Mn
N
N=
16
10 10= 6,25.10-12 Chn p n C
c.TRC NGHIM:Cu 1: Urani ( 23892U) c chu k bn r l 4,5.10
9nm. Khi phng x, urani bin thnh thri ( 23490Th ). Khi lng th
to thnh trong 23,8 g urani sau 9.109 nm l bao nhiu?A. 17,55g B. 18,66g C. 19,77g D. Phng n khc
Cu 2: Chu k bn r 21184 Po l 138 ngy. Ban u c 1mmg21184 Po . Sau 276 ngy, khi lng
21184 Po b phn r l:
A. 0,25mmg B. 0,50mmg C. 0,75mmg D. p n khc
* Cht phng x210
84 Poc chu k bn r 140 ngy, bin thnh ht nhn ch(Pb). Ban u c 42mg.Tr li cc cu 3,4,5Cu 3 : S prtn v ntron ca Pb nhn gi tr no sau y.A. 80notron v 130 proton B. 84 notron v 126 protonC. 84notron v 124 proton D. 82 notron v 124 proton
Cu 4 : phng x ban u ca 21084 Po nhn gi tr no ?
A. 6,9.1016 Bq B. 6,9.1012 Bq C. 9,6.1012 Bq D. 9,6.1016 BqCu 5 : Sau 280 ngy m phng x, khi lng ch trong mu l ?A. 10,5mg B. 21mg C. 30,9mg D. 28mg
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Dng 4: Xc nh chu k bn r T.a.Phng php:1) Tm chu k bn r khi bit khi cho bit m & m0 ( hoc N & N0 ; H&H0 ):- Bit sau thi gian t th mu vt c tl m/m0 ( hay N/N0 ) . Tm chu k bn r T ca mu vt ?a) T s s nguyn t ban u v s nguyn t cn li sau thi gian phng x t
N= N0t
e. => T=
N
N
t
0ln
2ln.Hoc m=m0
te . => T=t ln
mln
m
0
2
Nu xN
N=0 2 => x =
t
THoc: x
m
m=0 2 => x =
t
T
Nu0m
m=
0N
N=
n2
1(vi n N * )
n
tTn
T
t== .
Nu t s :0m
m=
0N
Nkhng p th: m T
t
m
= 2.0
==
02
0
log2m
m
T
t
m
mT
t
T=.
Tng t cho s nguyn t v phng x:
N Tt
N
= 2.0
==
02
0
log2N
N
T
t
N
NT
t
T =.
H Tt
H
= 2.0
==
02
0
log2H
H
T
t
H
HT
t
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-t
e . ) =>
0N
N=1- te . =>T= -
)1ln(
2ln.
0N
N
t
2)Tm chu k bn r khi bit sht nhn(hay khi lng) cc thi im t1 v t2
-Theo s ht nhn: N1= N01.te
; N2=N0
2.te
;2
1
N
N=
).( 12 tte
=2 1ln2.( )t t
Te
=>T =
2
1
12
ln
2ln)(
N
N
tt
-Theo s khi lng: m1= m01.te
; m2= m0
2.te
=>1
2
m
m=
).( 12 tte
=2 1
ln2.( )t t
Te
=>T = 2 11
2
( ) ln 2
ln
t t
m
m
3)Tm chu k bn khi bit sht nhn bphn r trong hai thi gian khc nhau1N 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=1
1
t
N; -Sau t(s) H=
2
2
t
Nm H=H0
te . => T=
2
1ln
2ln.
N
N
t
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b. Sdng lnh SOLVE trong my tnh Fx-570ES tm nhanh mt i lng cha bit :
-My Fx570ES Ch dng trong COMP: MODE 1 ) SHIFTMODE 1 Mn hnh: Math
Cc bc Chn ch Nt lnh ngha- Kt quDng COMP Bm: MODE 1 COMP l tnh ton chungChnhdng nhp / xut ton Math Bm: SHIFTMODE 1 Mn hnh xut hin MathNhp bin 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: SHIFT CALC = hin th kt qu X= .....
V d: Mt mu Na2411 ti t=0 c khi lng 48g. Sau thi gian t=30 gi, mu Na2411 cn li 12g. Bit Na
2411 l cht
phng x - to thnh ht nhn con l Mg2412 .Chu k bn r ca Na2411 l
A: 15h B: 15ngy C: 15pht D: 15giy
Ta dng biu thc0
0 .2 :
2
t
Tt
T
mm m Hay m
= = Vi i lng cha bit l: T ( T l bin X)
Nhp my :
30
12 48.2X
= Bm: SHIFT CALC = (chkhong thi gian 6s)Hin th:X= 15 .Chn ATv 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 nm B. 4,5 nm C. 9 nm D. 48 nm
HD Gii : Ta c0m
m=
n2
1=
42
1
16
1=
n
tTn
T
t== . =
4
12= 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 l
A. 128t. B. 128
t. C. 7
t. D. 128 t.
HD Gii:Ta c =0H
Hn2
1=
72
1
128
1=
77
tT
T
t== 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 khChu k bn r ca cht phng x l
A. 12 gi. B. 8 gi. C. 6 gi. D. 4 gi.Tm tt Gii :
?
24
%5,870
=
=
=
T
ht
m
m
Ta c : 300
0 2
1
88
7
8
7
100
5,87=====
mm
mm
m
m
Hay . ht
TT
t8
3
24
33 ==== . Chn B
Bi 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 kbn r ca ng v l:
A. 1h B. 3h C. 4h D. 2h
HD: ht
TT
tk
N
Nkk
22
24
1
2
175.0
2
11
0
=======
Bi 5. Phng trnh phng x ca Plni c dng: 21084 PoA
ZPb + .Cho chu k bn r ca Plni T=138 ngy.Kh
lng ban u m0=1g. Hi sau bao lu khi lng Plni ch cn 0,707g?
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A: 69 ngy B: 138 ngy C: 97,57 ngy D: 195,19 ngy
Hd gii: Tnh t:0m
m=
te . => t=2ln
ln. 0m
mT
=2ln
707,0
1ln.138
= 69 ngy (Chn A)
Bi 6. Vo u nm 1985 phng th nghim nhn mu qung cha cht phng x Cs17355 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 .
HD Gii : a/ Ta bit H0 = N0 , vi N0 =A
mNA => m =A
0
A
0
N.693,0
ATH
N.
AH=
Thay s m = 5,6.108g
b/ Sau 10 nm : H = H0 te ; t = 231,030
10.693,0= => H = 1,4.105 Bq .
c/ H = 3,6.104Bq =>H
H0 = 5 => t = ln5 =T
t.693,0=> t =
693,0
5lnT= 69 nm .
Bi 7.ng v Cacbon 146C phng x v bin thnh nito (N). Vit phng trnh ca s phng x. Nu cu to c
ht nhn nito. Mu cht ban u c 2x10-3 g Cacban 146C . Sau khong thi gian 11200 nm. Khi lng ca Cacb
146C trong mu cn li 0.5 x 10
-3 g . Tnh chu k bn r ca cacbon 146C .
HD Gii: Phng trnh ca s phng x : 14 o 146 1 7C e N +
-Ht nhn nit147 N gm Z = 7 prtn V N = A Z = 14 7 = 7 ntrn
- Ta c:
t toT T
om
m m 2 2m
= = (1)
Theo bi:3
2o3
m 2 104 2
m 0.5 10
= = =
(2) T (1) v (2)
t t 112002 T 5600
T 2 2 = = = = nm
Bi 8. Ht nhn C146 l cht phng x
- c chu k bn r l 5730 nm. Sau bao lu lng cht phng x ca mt mu c
cn bng8
1lng cht phng x ban u ca mu .
HD gii .Ta c: N = N0 Tt
2 0N
N= T
t
2 ln0N
N= -
T
tln2 t =
2ln
ln.0
N
NT
= 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 =
H
H0 => t = ln(H
H0 )
= )ln(1 0
H
H
tm = =
T
2ln)ln(
1 0H
H
t
T =07,1ln
.2ln t=
067658,0
693,0.100ngy 1023 ngy.
Bi 10. Bit ng v phng x 146 C c chu k bn r 5730 nm. Gi s mt mu g c c phng x 200 phn r/pht
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. Ttui ca mu g c.
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HD gii . Ta c: H = H0. Tt
2 =T
t
H
2
0 T
t
2 =H
H0 = 8 = 23T
t= 3 t = 3T = 17190 (nm).
Bi 11. Silic 3114 Si l cht phng x, pht ra ht v bin thnh ht nhn X. Mt mu phng x 3114 Si ban u trong th
gian 5 pht c 190 nguyn t b phn r, nhng sau 3 gicng trong thi gian 5 pht ch c 85 nguyn t b phn r. Hxc 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=
H
Ht
0ln
2ln. =
85
190ln
2ln.3 = 2,585 gi
Bi 12. Mt mu phng x Si3114 ban u trong 5 pht c 196 nguyn t b phn r, nhng sau 5,2 gi(k t lc t =
cng trong 5 pht ch c 49 nguyn t b phn r. Tnh chu k bn r ca Si3114 .
HD gii . Ta c: H = H0 Tt
2 =T
t
H
2
0 Tt
2 =H
H0 = 4 = 22T
t= 2 T =
2
t= 2,6 gi.
Bi 13. Ht nhn Plni l cht phng x ,sau khi phng x n trthnh 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:Po
Pb
mm =
mm' =
t
A
t
emNAeN
.0
..0 ')1(
=AA' (1- te . )
T=-
)'.
.1ln(
2ln.
Am
Am
t
Po
Pb
=
)206
210.1595,01ln(
2ln.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% hnhn cha b phn r. n thi im t2 = t1 + 100 (s) s ht nhn X cha b phn r ch cn 5% so vi s ht nhn bu. Tnh chu k bn r ca cht phng x.
HD gii . . Ta c: N = N0 Tt
2 Tt
2 =0N
N. Theo bi ra: T
t1
2
=0
1
N
N= 20% = 0,2 (1); T
t2
2
=0
2
N
N= 5% = 0,05 (2).
T (1) v (2) suy ra:T
t
T
t
2
1
2
2
= Ttt 12
2
=05,0
2,0= 4 = 22
T
tt 12 = 2 T =2
100
21112 tttt +=
= 50 s.
Bi 15.o chu k ca mt cht phng x ngi ta cho my m xung btu m tthi im t0=0.n thi it1=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 ccht phng x ny.
HD Gii:-S xung m c chnh l s ht nhn b phn r: N=N0(1-t
e.
)
-Ti thi im t1: N1= N0(1- 1.t
e
)=n1
-Ti thi im t2 : N2= N0(1-2.te
)=n2=2,3n1
1- 2.t
e
=2,3(1- 1.t
e
) 1- 1.3 t
e
=2,3(1- 1.t
e
) 1 + 1.t
e
+ 1.2 t
e
=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 0.2g 6027Co 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 s NA = 6.032 x 10
23mol-1.
HD gii .a) Phng trnh ca s phng x: +60 - 6027 28Co N i
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Thnh phn ca ht nhn 6028Ni l: 28 proton v 60 28 = 32 neutron.
b) phng x: = t t0 0H = H .e H .e ; vi = = =23
210 A0m N 0.2x6.022x10
N 2.0073x10M 60
ht.
T (1) suy ra: = =0 0 0ln 2
H l.N N .T
=> = = = =21
8010
0
N ln2 2.0073x10 ln2T 1.67x10 s 5.3H 225x3.7x10
nm.
Theo nh lut phng x: = =lt 00m
m m .e
t2 T
suy ra = = = =
t20 0T
0
m m2 4 2
m (1 0.75)m
t = 2T = 2x 5.3=10.6 nm.
Bi 17. Cban( )6027Co phng 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 ( )6027Co phn r ht?
HD Cch 1: a.Phng trnh phn r: 60270 60
Co 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% =1/4 Hay 0
0
mm 14
m 4 m
= =
nh lut phng x:
ln 2 t.tt T T
0 0 0m m .e m e m 2
= = = Hay
t0T m2 4 t 2T 10,54
m
= = = = nm
HD Cch 2 . Ta c: m = m0 - m = m0 Tt
2 t =2ln
'ln.
0
0
m
mmT
= 10,54 nm.
Bi 18 : C 0,2(mg) Radi Ra22688 phng ra 4,35.108 ht trong 1 pht. Tm chu k bn r ca Ra ( cho T >> t).
Cho x
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S ht nhn it b phn r l: 23 21Am 1,31
N .N x6,022x10 6,022x10N 131
= = = = ht
Mt ht nhn phn r, phng x 1 ht - nn s ht -c phng x cng l N = 6,022 x 1021 ht.
Bi 20. Mt bnh nhn iu tr bng ng v phng x, dng tia dit t bo bnh. Thi gian chiu x ln u 20t = pht, c sau 1 thng th bnh nhn phi ti bnh vin khm bnh v tip tc chiu x. Bit ng v phng
c chu k bn r T = 4 thng (coi t T T = t2ln
2ln= 2t = 2.2 = 4 gi.
Bi 22. xc nh chu k bn r T ca mt ng v phng x, ngi ta thng o khi lng ng v phngx trong mu cht khc nhau 8 ngy c cc thng so l 8g v 2g.Tm chu k bn r T ca ng v?
A. 4 ngy. B. 2 ngy. C. 1 ngy. D. 8 ngy.HDGii: Tm chu k bn r khi bit s ht nhn( hay khi lng) cc thi im t1 v t2
m1= m01.te
; m2=m0
2.te
=>1
2
m
m
=).( 12 tte
=>T = 2 1
1
2
( ) ln 2
ln
t t
mm
Th s :T = 2 11
2
( ) ln 2
ln
t t
m
m
=
(8 0) ln 28
ln2
=
8ln24 y
ln 4ng=
Bi 23:(H-2011) : Cht phng x poolooni Po21084 pht ra tia v bin i thnh ch Pb206
82 . Cho chu k c
Po21084 l 138 ngy. Ban u (t = 0) c mt mu plni chuyn cht. Ti thi im t1, t s gia s ht nh
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plni v s ht nhn ch trong mu l31
. Ti thi im t2 = t1 + 276 ngy, t s gia s ht nhn plni v
ht nhn ch trong mu l
A.91
. B.161
. C.151
. D.251
.
HDGii cch 1:Ti thi im t1, t s gia s ht nhn plni v s ht nhn ch trong mu l31
.Suy ra 3 ph
b phn r ,( cn li 1 phn trong 4 phn) -> cn 21 1 14 22
t
T
= = Hay 2tT
=
=> t1 = 2T=2.138=276 ngy . Suy ra t2 = t1 + 276 = 4T
Ta c :4 4
2 02 24 4
2 2 0 2 0
.2 2 1
(1 2 ) 1 2 15Po
Pb
N NN N
N N N N N
= = = = =
HDGiicch 2:Phng trnh phng x ht nhn: Po21084 + Pb20682
S ht nhn ch sinh ra bng s ht Poloni b phn r: Popb NN =
thi im t1: 276223
1
)21(
2.111
0
10
10
1
1
1
1
1 ====
=
=
=
TtkN
N
NN
N
N
N
N
Nk
k
Pb
Po ngy
thi im t2 = t1 + 276 = 552 ngy k2 = 4 151
212
)21(2.
4
4
20
20
20
2
2
2
2
2 =
=
=
=
=
k
k
Pb
Po
N
N
NN
N
N
N
N
N
Bi 24: Gi s ban u c mt mu phng x X nguyn cht, c chu k bn r T v bin thnh ht nhn bn Y. Tthi im 1t t l gia ht nhn Y v ht nhn X l k. Ti thi im 2 1 2t t T= + th t l l
A. k + 4. B. 4k/3. C. 4k+3. D. 4k.
HDGii: p dng cng thc L phng x ta c:1
1 1
1
1
01
1 1 0
(1 ) 1
1
tY t
t
X
N N eNk e
N N N e k
= = = =
+(1)
2 12
2 1 1
2
( 2 )02
2 ( 2 ) 21 2 0
(1 ) (1 ) 11
t t TY
t t T t T
X
N N eN ek
N N N e e e e
+
+
= = = = = (2)
Ta c:
ln 222 2 ln 2 1
4
TT Te e e
= = = (3).
Thay (1), (3) vo (2) ta c t l cn tm: 21
1 4 31 1
1 4
k k
k
= = +
+
. Chn p n C
*Dng my o xung phng x pht ra:a.Phng php: Mt mu vt cht cha phng x. ti thi im t1 my o c H1 xung phng x v sau 1khong t ti t2 o c H2 xung phng x. Tm chu k bn r ca ng v phng x l ?
Chn thi im ban u ti t1. Khi : t0 t1 c H0 H1 v t t2 c H H2 .Suy ra c :
teHH
.0 .
= 0
.
H
He t =
=
0
ln
2ln.
H
H
tT
Hoc Tt
HH
= 2.0 0
2H
HT
t
=
=
02log
H
H
T
t
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b. Bi tp:Bi 25: Magi Mg2712 phng x vi chu k bn r l T, lc t1 phng x ca mt mu magie l 2,4.10
6Bq. Vo lc t2phng x ca mu magi l 8.105Bq. S ht nhn b phn r t thi im t1n thi im t2 l 13,85.10
8 ht nhn. Tichu k bn r T
A. T = 12 pht B. T = 15 pht C. T = 10 pht D.T = 16 phtGii
H0 = H1 = N0H2 = H = N H1 H2 = H0 H = (N0 N)
HHNT
= 0.2ln sNHH
T 600.2ln0
=
= = 10 pht
Bi 26: Mt lng cht phng x Radon( Rn222 ) c khi lng ban u l m0 = 1mg. Sau 15,2 ngy th phng x cn gim 93,75%. Tnh chu k bn r v phng x ca lng cht phng x cn li.
HDGii: + T0 0
0 0
11 93,75%
164 3,8
42 2
t t
T T
H H
H H t tT ngay
TH H
H H
= =
= = =
= =
=> BqAT
NmH
k
A 110 10.578,3.
2..693,0==
c.TRC NGHIM:Cu1: Mt lng cht phng x Rn22286 ban u c khi lng 1mg. Sau 15,2 ngy phng x gim 93,75%. Chu bn r ca Rn l
A. 4,0 ngy B. 3,8 ngy C. 3,5 ngy D. 2,7 ngy
Cu2: Cht phng x Po21084 pht ra tia v bin i thnh Pb20682 . Chu k bn r ca Po l 138 ngy. Ban u c 10
Po th sau bao lu lng Po ch cn 1g?A. 916,85 ngy B. 834,45 ngy C. 653,28 ngy D. 548,69 ngy
Cu3: Sau thi gian t, phng x ca mt cht phng x- gim 128 ln. Chu k bn r ca cht phng x l
A. 128t. B.128
t. C.
7
t. D. 128 t.
Cu4: Sau khong thi gian 1 ngy m 87,5% khi lng ban u ca mt cht phng x b phn r thnh cht khChu k bn r ca cht phng x l
A. 12 gi. B. 8 gi. C. 6 gi. D. 4 gi.Cu5: Mt gam cht phng x trong 1s pht ra 4,2.1013 ht. Khi lng nguyn t ca cht phng x ny 58,933 u; lu1,66.10-27 kg. Chu k bn r ca cht phng x ny l:
A. 1,78.108s. B.1,68.108s. C.1,86.108s. D.1,87.108 s.
Cu6: Mt mu phng x Si3114 ban u trong 5 pht c 196 nguyn t b phn r, nhng sau 5,2 gi(K t t = 0)
cng trong 5 pht ch c 49 nguyn t b phn r. Chu k bn r ca Si3114 lA. 2,6 gi B. 3,3 gi C. 4,8 gi D. 5,2 gi
Cu 7:o chu k ca cht phng x, ngi ta dng mt my m xung. trong t1 giu tin my m c
xung; trong t2 = 2t1 gitip theo my m c 2 19
64n n= xung. Chu k bn r T c g tr l :
A. 13tT= B. 1
2tT= C. 1
4tT = D. 1
6tT =
Cu8: ng v Na 24 phng x vi chu k T = 15 gi, to thnh ht nhn con l Mg. Khi nghin cu mt mu chngi ta thy thi im bt u kho st th t s khi lng Mg24 v Na 24 l 0.25, sau mt thi gian t th t sbng 9. Tm t ?
A. t =4,83 gi B. t =49,83 gi C. t =54,66 gi D. t = 45,00 giCu9: Mt cht phng x pht ra tia , c mt ht nhn b phn r cho mt ht . Trong thi gian 1 pht u chtphng x pht ra 360 ht , nhng 6 gisau, k t lc bt u o ln th nht, trong 1 pht cht phng x ch pht ra 4ht . Chu k bn r ca cht phng x ny l:
A. 1 gi B. 2 gi C. 3 gi D. 4 gi
Tm ttt1 : H1 = 2,4.10
6Bq
t2 : H2 = 8.10
5
BqN= 13,85.108 T = ?
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Dng 5: Xc nh thi gian phng x t, tui th vt cht.a.Phng php: Tng tnhdng 4 :Lu : cc i lng m & m0, N & N0, H &H0 phi cng n v..
Tui ca vt c: 0 0ln lnln 2 ln 2
N mT Tt
N m= = hay 0 0
1 1ln ln
N mt
N 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 trong khong thi gian bng ba ln s ht nhn cn li ca ng vy?
A. 2T. B. 3T. C. 0,5T. D. T.
Gii : m=3m. Theo , ta c : 3
2.
)21(
0
0 =
=
T
t
T
t
m
m
m
m 42312 == T
t
T
t
t = 2T. Chn p n : A
Bi 2: 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 :
A. 75 ngy B. 11,25 gi C. 11,25 ngy D. 480 ngy
Gii: T = 360h ;32
1
0=
m
m. t? Ta c
32
1
0=
m
m52
1= 5=
T
t t = 5T t = 1800 gi= 75 ngy. Chn A.
Bi 3: Lc u mt mu Plni21084 Po 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 10
23 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 = 206Ap dng nh lut bo ton in tch : 84 = 2 + Z Z = 82
Vy 210 1 20684 2 82Po He Pb + . Ht nhn21084 Po c cu to t 82 prtn v 124 ntrn
b) Ta c : - S ht Plni ban u : o Aom N
N
A
= ; - S Plni cn li : toN N .e=
-S ht Plni b phn r : oN N N = ;t
oN N (1 e ) = ;- S ht ch sinh ra : tPb oN N N (1 e )
= =
- Khi lng ch to thnh : Pb PbPbA
N .Am
N= (1); - Khi Plni cn li : ( )tom m e 2
=
( )
( )
( ) ( )
( )
t tPbPb Pb Pb
t t tA o
t
A 1 e 1 e1 m N .A 2060,6
2 m N .m e A e 210 e
e 0,62 t 95,19
= = =
= ngay
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. Bichu k ca 14C l 5600 nm. Tui ca tng g l :
A. 1900 nm B. 2016 nm C. 1802 nm D. 1890 nm
Gii : cho:H= 0,8H0 v m nh nhau. Theo ta c : 32,08,0log8,02 20
====
T
t
H
HT
t
.
t = 0,32T = 0,32.5600 = 1802 nm Chn p n C
Bi 5: Plni 21084 Po l cht phng x to thnh ht nhnAZX 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 .
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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 206
84 2 Z Z 82
= + =
= + = A 206Z 82X Pb = .
Vy X l Pb. 20682 Pb c 82 ht prtn v 206 82 = 124 ht ntrn
Theo nh lut phng x ta c:ot m
o Tmt
T
m 1m 2 8
0,1252
= = = hay
t3T2 2= t = 3T = 3 x 138 = 414 ngy 2
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 sht ch to ra
Theo bi:
o
Pb oA
Po
A
N N.206
m N NN 206. 0, 406
Nm N 210.210N
= = =
o oN N N 85,561N N 206
= =
oN 1 0,4138 1,4138 2N
= + = = Vy
1 1oT 2N T 1382 2 t 69
N 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)
m37
mth 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) oA
Am N (1 e )
N; Khi lng U 238:
= =t
(U) o(U) (U)
A A
A N em .N A .
N N
Gi thit =(U)
pb
m37
m
= =
t
t
e 37 20632,025
2381 e =t t(1 e )32,025.e 1 = =t
33,025e 1,03
32,025
= t ln1,031 0.03 = 9 80.03t 4,6 10 2 10 nam0.693
Bi 7: Tnh tui ca mt ci tng c bng g, bit rng phng x ca C14 trong tng g bng 0.707 ln phnx 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 hi
nay l Ho.Do ta c= = = = =
t(t ) t T
o
H 1 Te 2 0,707 t 2800
H 22nm.
Bi 8: C hai mu cht phng x A v B thuc cng mt cht c chu k bn r T = 138,2 ngy v c kh
lng ban u nh nhau . Ti thi im quan st , t s s ht nhn hai mu cht 2,72B
A
N
N
= .Tui ca mu
nhiu hn mu B lA. 199,8 ngy B. 199,5 ngy C. 190,4 ngy D. 189,8 ngy
Gii : NA = N01te ; NB = N0 2
te . 2 1( ) 1 2
ln 22,72 ( ) ln 2,72t tB
A
Ne t t
N T
= = =
=> t1 t2 =ln2,72
199,506 199,5ln 2
T= = ngy. p n B : 199,5 ngy
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Bi 9: Mt pho tng c bng g bit rng phng x ca n bng 0,42 ln phng x ca mt mu gti cng loi va mi cht c khi lng bng 2 ln khi lng ca pho tng c ny. Bit chu k bn r cng v phng x C146 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 bi ta 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
Bi 10: Trong qung urani t nhin hin nay gm hai ng v U238 v U235. U235 chim t l 7,143 000
. Gi s lcu trI t mi hnh thnh 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 nm B: 6,04 triu nm C: 604 t nm D: 60,4 t nm
Gii S ht U235 v U238 khi tri t mi hnh thnh l N0 . S ht U238 by gi 1Tt
2.01
=NN
S ht U235 by gi 2Tt
2.02
=NN => 9
2
1 10.04,61000
143,7== t
N
N(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 = 1,67.10
11Bq.
2. Sau thi gian bao lu phng x ca n gim i 16 ln.3. Tm nng lng ta ra khi cht phng x trn phn r ht.
HD: 1. XHePo AZ+42
21084
=
=
82
206
Z
A
mggN
TAHm
AT
Nm
A
NmH
A
AA 110.1.693,0.
..693,0 300
000 =====
2. T ngayTt
H
H
H
H
T
t
T
t552422
2
2164
0
40
===
=
==
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
Trong m0 = 1mg c N0 =18
323
10.867,2210
10.10.022,6=
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 thnh206Pb84, chu k bn r l 138 ngy. Sa
bao lu th t s s ht gia Pb v Po l 3 ?A. 276 ngy B. 138 ngy C. 179 ngy D. 384 ngy
Gii cch 1: Ti thi im t, t s gia s ht nhn ch v s ht nhn plni trong mu l 3.Suy ra 3 phn 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 = ? th 3 3Pb Pb PoPo
NN N
N= = (1)
S ht nhn ch sinh ra l NPb: .206
PbPb A
mN N= S ht nhn Po cn li NPo:
0 o o. .
2 .2 210.2
oP oPt A At t t
T T Tme
N m mN N N
A
= = =
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Thay vo ( 1) ta c:o o
o. 3. . 210. 3 .206 210. .2 3.206 (2)206210.2 2
t
Pb oP oP TA A Pb Pb oPt t
T T
m m mN N m m m= = =
vi. .206
210Po Pb Po
Pb
Po
m A mm
A
= = M:
oPo
oPo
1.(1 ).206
1 2.(1 )210
2
t
T
Po Pbt
T
m
m m m
= =
Thay vo (2) ta c:
oPo
oPo oPo oPo
2
1.(1 ).206
12210. .2 3.206. .(1 ).2 3.210
2
1(1 ).2 3 2 1 3 2 4 2 2 2 2.138 276 y
2
tt t
TT T
t
T
t t t
T T T
t
T
m
m m m
tt T ng
T
= =
= = = = = = = =
Bi 13: Plni Po21084 l cht phng x v bin thnh ch Pb20682 .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 =Tt
N/
0
2=
30
2
Nvi N0 =
A
Nm t0 = 2323
10.004816,0210
10.02,6.168,0= ngt.
N= 193
23
10.02,62
10.004816,0=
S nguyn t b phn r: N = N0 N = 48,16.1019 6,02.1019 = 42,14.1019 ngt
b, S nguyn t Pb c to thnh bng s nguyn t Po b phn r bng N.
Khi lng Ch c to thnh: mPb =
N
.
N
AN= g1442,0
10.02,6
206.10.14,4223
19
=
Bi 14: xc nh hng s phng x ca Co55 . Bit rng s nguyn t ca ng vy c mi gigim i 3,8%.Gii: p dng nh lut phng x: N = N0 e
- tSau t = 1h s nguyn t b mt i: N = N0 N = N0( 1 - e
- t ) (1)
theo :0N
N= 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 ca Co55 l: = 0,04 (h-1).
Bi 15: U238 phn r thnh Pb206 vi chu k bn r 4,47.109 nam .Mt khi cha 93,94.10-5 Kg v 4,27.10-5 Kg Pb.Gi s khi lc u hon ton nguyn cht ch c U238.Tui ca khi l:
A.5,28.106(nm) B.3,64.108(nm)C.3,32.108(nam) B.6,04.109(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 =238
mNA ; NPb =206
PbAmN ;
Theo L phng x: N = N0 e-t
------>238
mNA = (238
mNA +206
PbAmN )e-t
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=> et =206
2381
238
206238m
m
mN
mNmN
Pb
A
PbAA
+=+
= 1,0525 => 0525,1ln2ln
=tT
=> t = 3,3 .108 nm. Chn p n C
Bi 16: Tim vo mu bnh nhn 10cm3 dung dch cha Na2411 c chu k bn r T = 15h vi nng 10-3mol/l
Sau 6h ly 10cm3 mu tm thy 1,5.10-8 mol Na24. Coi Na24 phn bu. Th tch mu ca ngi c timkhong:
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. T
t
e
.2ln
= 10-5 156.2ln
e = 0,7579.10-5 mol.
Th tch mu ca bnh nhn V = litl 505,55,1
578,7
10.5,1
10.10.7579,08
25
==
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 lng1,5M mi cht l 15 Bq. Xc nh tui ca bc tng c. Bit chu k bn r ca C14 l T= 5600 nmA 1800 nm B 2600 nm C 5400 nm D 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 trN= 109 ht/s (ht nhn con khng phng x)A: 9,5 ngy B: 5,9 ngy C: 3,9 ngy D: Mt gi tr khc
Gii Cu 2: Tc phn r trong thi gian t l: Tt
2.01
=NN ; Tc to thnh trong thi gian t l N0= q.t
Tc to thnh ht nhn trong thi gian t l )21( Tt
0
=NN =109 .Thu c t 0,667.T= 9,5 ngy
* Poloni 21084 Pophng xbin thnh ht nhn Pb vi chu k bn r 138 ngy. Lc u c 1g Po cho NA= 6,02.10
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 ngy B. 110 ngy C. 85 ngy D. 105 ngyCu 4 Sau 2 nm th tch kh He c gii phng KTC .A. 95cm3 B. 103,94 cm3 C. 115 cm3 D.112,6 cm3
Cu 5(H- C-2010)Bit ng v phng x 146 C c chu k bn r 5730 nm. Gi s mt mu g c c phng x 2
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 phr/pht. Tui ca mu g c cho l
A. 1910 nm. B. 2865 nm. C. 11460 nm. D. 17190 nm.
Dng 6: XC NH PHNG X Ha.Phng php:
p dng cng thc:H = H0t
e. 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 bnr bng 8 ngy m) theo n v Bq v Ci.
Gii:S ht nhn ngt sau t = 1,5T: N = N0t
e .
=22220
5,10
/0 NNNTt
== => N = 2020
10.5,222
10.07.7= ngt.
phng x ti thi im t.: H = . N = CiBqNT
310
1420 10.77,6
10.7,3
10.056,2506,210.2.
3600.24.8
693,0.
2ln===
Bi 2: Cht Plni Po210 c chu k bn r T = 138 ngy m.
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a, Tm phng x ca 4g Plni.b, Hi sau bao lu phng x ca n gim i 100 ln.
Gii: a, phng x ban u ca 4g Po. H0 = .N0 (1)
vi3600.24.138
693,02ln==
T (j-1) v
210
10.02,6.4 2300 ==
A
NmN A thay s vo (1) =>: H = 6,67.1014 Bq.
b, Tm thi gian: H = H0t
e .
H
He t 0. = t = 916100ln.
693,0ln
1 0 ==
T
H
H
ngy.
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 phn trm so vi phng x ca lncht phng x ban u?
A. 25%. B. 75%. C. 12,5%. D. 87,5%.Cu 2 : Mt lng cht phng x Rn22286 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. 3,58.1011Bq D. 5,03.1011Bq
III. PHNNG HT NHN:
1. Phng trnh phn ng:31 2 4
1 2 3 4
AA A A
Z Z Z ZA B C D+ + Trng hp phng x: 31 4
1 3 4
AA A
Z 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 4m m m mp p p p hay v v v v+ = + + = +
- Bo ton nng lng:1 2 3 4X X X X
K K E K K + + = + ;
Trong : E l nng lng phn ng ht nhn; 21
2X 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:2 2X 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 =
=> 2 2 21 2 1 22p p p p p cos= + +
hay 2 2 21 1 2 2 1 2 1 2( ) ( ) ( ) 2mv m v m v m m v v cos= + +
hay 1 1 2 2 1 2 1 22mK m K m K m m 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 2
2 2 1 1
K v m A
K 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
p
1p
2p
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Lu : - Nu M0 > M th phn ng to nng lng |E| = |E0-E| di dng ng nng ca cc ht C, D hophtn . Cc ht sinh ra c ht khi ln hn nn bn vng hn.
- Nu M0 < M th phn ng thu nng lng |E| =|E0-E| di dng ng nng ca cc ht A, B hophtn . Cc ht sinh ra c ht khi nh hn nn km bn vng.
+ Trong phn ng ht nhn 31 2 41 2 3 4
AA A A
Z Z Z ZA B C D+ + Cc ht nhn A, B, C, D c:
-Nng lng lin kt ring tng ng l 1, 2, 3, 4.-Nng lng lin kt tng ng l E1, E2, E3, E4
- ht khi tng ng l m1, m2, m3, m4-Nng lng ca phn ng ht nhn
E = A33 +A44 - A11 - A22E = E3 + E4 E1 E2E = (m3 + m4 - m1 - m2)c
2
3. Quy tc dch chuyn ca sphng x+Phng x ( 42He ):
4 42 2
A A
Z ZX He Y
+ : So vi AZX, ht nhn con
42
A
Z Y
li 2 (BngTH) v s khi gim 4
+Phng x- ( 10 e ): 01 1
A A
Z ZX e Y
+ + : So vi AZX, ht nhn con 1
A
Z Y+ tin 1 (BngTH) v c cng s kh
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 + +
Lu : - Bn cht (thc cht) ca tia phng x- l ht electrn (e-)- Ht ntrin (v) khng mang in, khng khi lng (hoc rt nh) chuyn ng vi vn tc c
nh sng v hu nh khng tng tc vi vt cht.+Phng x+ ( 10 e
+ ): 01 1A A
Z ZX e Y+ + :So viA
ZX, ht nhn con 1A
Z Y+ li 1 (BngTH) 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 ntrinp n e v+ + +
Lu : Bn cht (thc cht) ca tia phng x+ l ht pzitrn (e+)+ Phng x (ht phtn): Ht nhn con sinh ra trng thi kch thch c mc nng lng E1 chuyn xun
mc nng lng E2ng thi phng ra mt phtn c nng lng:
1 2
hchf E E
= = =
Lu : Trong phng xkhng c s bin i ht nhn phng xthng i km theo phng x v .
4. ng dng cc nh lut bo ton gii mt bi ton vt l ht nhn.
Xt phn ng: 31 2 41 2 3 41 2 3 4
AA A A
Z Z Z ZX X X X E+ +
Gi: *1 2 3 4; ; ;X X X XK K K K : L ng nng ca cc ht nhn X1 ; X2 ; X3 ;X4
Vi 21 ; :2X x x
K m v dv J = Nu ht nhn ng yn th K = 0
Trong : m: l khi lng tng ht nhn. v: kg , uv: l vn tc tng ht nhn. v: m/s
* 1 2 3 4; ; ;p p p p
: L ng lng ca cc ht nhn X1 ; X2 ; X3 ; X4
Vi pX= mX.vX v: kg.m/s- Mi quan h gia ng lngpXv ng nng KXca htXl:
2 22 ( . ) 2 . 2X X X X X X X X X X Xp m K m v m K m v m K= = =
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a.Cc nh lut bo ton:
+ Bo ton ng lng: 1 2 3 4 1 1 2 2 4 3 4 4m m m mp p p p hay v v v v+ = + + = +
+ Bo ton nng lng:1 2 3 4X X X X
K K E K K + = + (1)
Trong : E l nng lng phn ng ht nhn .- Nu phn ng ta nng lng th phng trnh (1) ly +E- Nu phn ng thu nng lng th phng trnh (1)ly E
Lu : - Khng c nh lut bo ton khi lng.
b. Dng bi tp tnh gc gia cc ht to thnh.Cho htX1 bn ph htX2(ng ynp2 = 0) sinh ra htX3 vX4 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 vvectng 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 ht X3 v 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 ht X1 v 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 sdng+ S Avgar: NA = 6,022.10
23 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:i) Xc nh tn ht nhn cha bit ( AZX cn thiu) :
- p dng nh lut bo ton skhi v in tch .Ch : nn hc thuc mt vi cht c sin tch Z thng gp trong phn ng ht nhn (khng cn quan tm n skhi 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 vin tch, skhi ca chng :
Ht 42 He , ht ntron
10 n , ht proton
11p , tia
01 e , tia
+
01.+ e , tia c bn cht l sng in t.
ii) Xc nh scc ht ( tia ) phng x pht ra ca mt phn ng :- Thng thng th loi bi tp ny thuc phn ng phn r ht nhn . Khi ht nhn m sau nhiu ln phng x to raht v y ht ( ch l cc phn ng ch yu to loi v ngun phng x+ l rt him ) . Do khi gii bi tp lony c cho l , nu gii h h