Chuyen de VATLY Hat NhanDH

7/31/2019 Chuyen de VATLY Hat NhanDH

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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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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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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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* 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.


9/58



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.


10/58



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)


11/58



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.


12/58



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


13/58



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.


14/58



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


15/58



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 :


16/58



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%.


17/58



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.


18/58



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


19/58



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


20/58



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?


21/58



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.


22/58



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


23/58



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


24/58



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


25/58



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


26/58



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 = ?


27/58



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 .


28/58



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


29/58



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

= = =


30/58



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


31/58



=> 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.


32/58



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


33/58



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= = =


34/58



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

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