Hoa phong xa

7/30/2019 Hoa phong xa

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BI 1. HO PHNG X

1. NH LUT CHUYN DCH PHNG X:(1) Khi phn r s khigim 4 cn s th t gim 2 n v (A'=A- 4; Z'=Z-2),

(2) Khi phn r - s khi khng thay i, s th t tng 1 n v.Cc ng v thuc cng h phng x c s khi khc nhau 4n (u).

Bng 2.1(L5.1): H Thori (A=4n)

Nng l- ng bc x cc i

(MeV)232Th 1,41.1010 n m 4,01228Ra(MsTh1) 5,57 n m

0,014228Ac(MsTh2) 6,13 h

2,11228Th(RdTh) 1,91 n m 5,42224Ra(ThX) 3,66 ngy 5,69220Rn(Tn) 55,6 s 6,29216Po(ThA) 0,15 s 6,78212Pb(ThB) 10,64h 0,57212Bi(ThC) 60,6 min , : 6,09; : 2,25212Po(ThC') 3,05.107s 8,79208Tl(ThC") 3,07 min 1,80208Pb(ThD) Bn

Ht nhn

Thi gian b n

hu Dng phn r


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Bng 2.2 (L5.3): H urani-radi (A=4n+2)

Nng l- ng bc x cc i(MeV)

238U(UI) 4,47.109 n m 4,20234Th(UX1) 24,1 ngy

0,199234mPa(UX2) 1,17 min

2,30234Pa(UZ) 6,7 h 1,2234U(UII) 2,44.105n m 4,78230Th(Io) 7.7.104 n m 4,69226Ra 1600 n m 4,78

222Rn 3,82 ngy 5,49218Po(RaA) 3,05 min , 1) : 6,00214Pb(RaB) 2,68 min 1,02218At 2s , 1) : 6,76218Rn 0,035s 7,13214Bi(RaC) 19,8 min 1), : 5,51;: 3,27214Po(RaC') 1,64.104s 7,69210Tl(RaC") 1,3 min 2,34210Pb(RaD) 22,3 n m 1), : 3,72;: 0,061206Hg 8,15 min

1,31210Bi(RaE) 5,01 ngy 1), : 4,69;: 1,16206Tl(RaE") 4,2 min 1,53210Po(RaF) 138,4 ngy 5,31206Pb(RaG) Bn

Ht nhn Thi gian b nhu

Dng phn r

1) < 0,1%


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Bng 2.3.(L5.4.): H actini (A=4n+3)

1) < 5%


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Bng 2.4. (L5.2.): H neptuni (A=4n+1)

Nng l- ng bc x cc i(MeV)

237Np 2,14.106n m 4,87233Pa 27,0 ngy 0,25233U 1,59.105n m 4,82229Th 7,34.103n m 4,89225Ra 14,8 ngy 0,32225

Ac 10,0 ngy 5,83221Fr 4,8 min 6,34217At 0,032 s 7,07213Bi 45,65 min 1), : 5,87;: 1,42213Po 4,2.106s 8,38209Tl 2,2 min 1,83209Pb 3,3 h 0,64209Bi Bn

Ht nhnThi gian b n

hu Dng phn r

1) < 2,2%

2. NNG LNG HC CA PHN R PHNG X V PHN NG

HT NHN

Trn c s nguyn l 2 ca nhit ng lc hc, ta bit rng mt qu

trnh ho hc ch c th t din ra khi n lm cho h chuyn sang trng thibn vng hn v mt nng lng, ngha l trong chuyn ho y, h gii

phng mt nng lng dng cho mi trng. Quy lut y cng p dng cho

s phn r phng x.


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S phn r phng x c th biu din bi phng trnh phn ng tng

qut:

AB + x + E . (2.17)

Phng trnh ny cho bit rng mt nguyn t A chuyn ho thnhnguyn t B pht ra mt ht x v gii phng nng lng E. S tnh E cho

bit kh nng t din ra phn ng (2.17). E>0 ngha l s phn r l c kh

nng t xy ra. Cn E


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(1) Nng lng E ca phn r tnh theo ht khi da vo

phng trnh Einstein:

E = (m1 - m2 - m)c2 (2.88)

trong m1, m2, m ln lt l khi lng ca ht nhn m, con, ht

. tnh E ngi ta cng thng s dng nguyn t khi

(M = m + Zme) ca cc nuclit m, con v hli:

E = (M1 - M2 - MHe)c2 (2.89)

(2) Nng lng ca s phn r - cng c tnh da vo phng

trnh Einstein:

E = (m1 - m2 - me)c2 (2.97)

Trong m1, m2, me ln lt l khi lng ca ht nhn m, con v

electron. Khi lng ca phn ntrino c th b qua (< 2.10-7.v.C.). Khi

thay khi lng ht nhn bng nguyn t khi, (2.97) tr thnh:

E = [ M1 - Z1 me - M2 + (Z1 + 1) me - me] = (M1 - M2) c2 . (2.98)

(3) Phng x +. Khi y, mt proton trong ht nhn bin i thnh

mt ntron, mt pozitron v mt ntrino, s th t gim mt n v cn skhi khng thay i. Nng lng phn r c tnh tng t nh trng hp

phn r -, nhng v

Z2 = Z1 - 1

nn ta c :

E = [ M1 - Z1 me - M2 + (Z1 - 1) me - me] = (M1 - M2 - 2me) c2 . (2.99)

Nh vy, nu chnh lch nguyn t khi ca m v con khng lnhn 2 ln khi lng electron (tnh theo u) th phng x + khng t din

bin c

Nhng ngay c khi E > 0, s phn r c din ra hay khng li cn lvn khc. Nng lng hc ca phn ng (2.17) c m t bi s


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hnh 2.1, s chnh lch v nng lng ca ht nhn m (A) v sn

phm phn r (B+x) l E. Cng ging nh trong phn ng ho hc, cc htnhn khng bn (A) phi vt qua mt hng ro th c chiu cao ES

chuyn ho thnh sn phm phn r (B+x). Ch nhng ht nhn m no c

nng lng cao hn mt lng ES so vi nng lng trung bnh thng k EA

ca tp hp cc ht nhn A mi vt qua c hng ro th v phn r

c. Chiu cao ca hng ro th cng thp, xc sut phn r cng cao, tc

l tc ca s phn r phng x cng ln.

Tuy nhin, s phn r phng x khng ging hon ton vi phn ng

ho hc. Trong phn r , ht nhn c th khng cn phi vt qua nh

hng ro th m xuyn qua hng ro nh hiu ng ng hm. Xc sut cavic xuyn qua hng ro th nh vy s cng cao khi E cng ln.

Es

A

B + x

E

Trng th i

Nnglng

Hnh 2.1. (L5.2) Hng ro th trong phn r phng x

Phn r phng x l mt trng hp ring ca phn ng ht nhn:

A + x B + y + E

E = (mA + mx mB my)c2 (m l kh lng ht nhn)

Thay m = M Zme ta c:

E = (MA + Mx MB My)c2

Khi kh lng nguyn t c biu din qua u (.v.C) th:


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E = (MA + Mx MB My).931,5 MeV = (MA + Mx MB My).1,602.10-13. 931,5 J

3. NG HC PHNG X

Phn r phng x tun theo quy lut ng hc bc nht

N=Noe- t ; (2.2)

N l s nguyn t ca nuclit phng x ang kho st, l hng s tc phn r, No l s nguyn t ca nuclit phng x thi im t=0.

Thi im mt na s nguyn t ban u b phn r (N=No/2),

gi l thi gian bn hu t1/2, c th tnh c bng cch ly lgarit 2 v ca

biu thc:N/No=1/2= e-

t1/2 (2.3)

v thu c:

t1/2=ln2/ =0.69315/ (2.4)

hoc:

=ln2/ t1/2 . (2.5)

a (2.5) vo (2.2) ta c:

N=No(1/2)t/ t1/2 . (2.6)

T phng trnh (2.6) d thy rng s nguyn t phng x sau 1ln

thi gian bn hu cn li 1/2, sau 2 ln t1/2 cn 1/4, sau 7 ln t1/2 cn 1/128

(tc l t hn 1%), sau 10 t1/2 cn 1/1024 (t hn 1 phn nghn) so vi lng

ban u.

Mt i lng cng thng c s dng l i sng trung bnh ca htnhn phng x , c nh ngha theo cch thng thng ca cc gi trtrung bnh:

(2.8NdtN1

00

=


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a (2.2) vo (2.8) ta c:

(2.91

dte0

t

==

So snh cc biu thc (2.9) v (2.4) d thy rng bng 1,443 ln thi gian

bn hu.

t gi tr t==1/ vo (2.2) ta thu c N = N0/e v a ra nhn xt

sau y: thi gian sng trung bnh l khong thi gian cn thit s

nguyn t phng x gim i e ln.

S khc bit quan trng gia ng hc ca qu trnh phn r phng x

vi cc qu trnh ho hc l ch hng s tc phn r, thi gian bnhu hoc thi gian sng trung bnh ca cc ng v phng x ni chung

khng ph thuc vo cc iu kin bn ngoi nh nhit , p sut, trng

thi vt l hoc lin kt ho hc.

4.Hot v khi lng

Tc phn r tnh bng s phn r, tc l s bin i ht nhn, trong

1 giy cng c gi l hot phng xA:

A=-dN/dt= N. (2.10)

V th, quy lut thay i hot phng x theo thi gian cng chnh

l quy lut ng hc kho st mc 3.

A=A0.e- t=A0(1/2)t/t1/2, (2.11)

Trong A0 l hot phng x ban u.

Trong h SI n v hot phng x l Becquerel, vit tt l Bq, c

nh ngha l 1phn r trong 1giy, ngha l:

1Bq=1s-1 .

Trong thc t, o hot phng x ngi ta thng s dng n v

curi, cc c s v c cc bi s ca n.


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1 Ci = 3,7.1010 Bq

Phng trnh (2.10) cng cho bit quan h gia hot v khi lng

cht phng x, n cho php xc nh c khi lng cht phng x khi o

hot phng x ca n, hoc lng cht phng x cn dng t cmt hot phng x cho trc. T cc biu thc (2.5) v (2.10) rt ra:

(2.12t.2ln

N 2/1AA

=

=

hay:

(2.13t.2ln.NN

N.Mm 2/1

AvAv

A.M==

vi M l nguyn t gam, NAv l s Avogadro.

L v d minh ho ta th tnh khi lng 32P cn thit c hot

phng x 1Ci, cho t1/2 ca ng v ny bng 14,3 ngy.

Gii: S nguyn t 32P cn thit c hot phng x 1Ci l:

10163,7.10

.14,3.24.3600 6,6.10ln 2

N = =

Suy ra khi lng 32P cn c l:

g5,3g10.5,310.02,6

10.6,6.32m 6

23

16===

Mt i lng quan trng khc l hot ringAs ca mt nguyn

t phng x, c nh ngha l hot phng x ca 1 n v khi lng,

thng l 1g, nguyn t ( bao gm c khi lng cc ng v phng x v

khng phng x:

(2.14gCi

hocg

Bqs

=m

AA


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i khi hot phng x ring c quy v mt mol hp cht ho

hc cha nguyn t phng x:

Cihoc (2.15)

mols

A BqA

n mol

=

Chng hn hot phng x ring ca benzen c nh du bi 14C

thng c cho theo n v mCi/mmol=Ci/mol.

S thay i hot phng x ring theo thi gian cng tun theo

phng trnh (2.11):

(2.16

2

1.e

1/2

00

t/t

st-

ss

== AAA

Trong As0 l hot phng x ring ti thi im t=0 (hot

phng x ring ban u).

Trong ho hc thng thng ngi ta ch quan tm n khi lng

cc cht c mt trong h, nhng trong ho phng x, cng nh trong cc ng

dng cht phng x, bn cnh khi lng, hot phng x ring l thng

tin rt quan trng. Ngoi ra, bng cch ng thi xc nh khi lng v

hot phng x ngi ta c th nhn c nhng thng tin quan trng vcc qu trnh bin i vt cht trong h kho st.

4. CN BNG PHNG X

4.1. Khi nim v cn bng phng x

Khi nim cn bng phng x v thc cht khng ng nht vi khi

nim cn bng ho hc. hiu r khi nim ny chng ta kho st trng

hp quan trng v thng gp trong ho phng x, mt ng v m

phn r thnh ng v con, ri ng v con ny li phn r tip tc. Nhng

bin i nh vy c biu din bng s :

Nuclit 1Nuclit 2Nuclit 3 (2.21)


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Tc tch lu nuclit con (2) l hiu gia tc hnh thnh ng v

ny do s phn r ca nuclit m (1) v tc phn r ca con:

dN2/dt = -dN1/dt - 2N2 = 1N1- 2N2 (2.22)

Thay vo (2.22) biu thc ca N1 rt ra t (2.2) ta c:

dN2/dt + 2N2 - 1N10e-1t = 0 (2.23)

Gii phng trnh vi phn tuyn tnh (2.23) (xem ph lc 1) ngi ta

thu c:

(2.24eNeeNN t02tt0

112

12

221 +

=

Gi nh rng thi im t=0 nuclit con c tch hon ton khinuclit m, tc l N20=0 th (2.24) tr thnh:

( ) (2.25eeNN tt0112

12

21

=

Rt ra:

[ ] (2.26e1eNN t)(t0112

12

121

=

hay:

[ ] (2.27e1NN t)(112

12

12

=

T (2.27) dng nhn thy rng trong trng hp 2>1 sau mt

thi gian t ln c th chp nhn :0)( 12 te (2.28)

v (2.27) tr thnh:


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(2.29NN 112

12

=

Ngha l:

(2.30constNN

12

1

1

2 =

=

Trng thi t s nng nuclit m v nuclit con trung gian

khng thay i theo thi gian gi trng thi cn bng phng x. S khc

nhau cn bn gia cn bng phng x vi cn bng ho hc nm ch cn

bng phng x khng phi l trng thi ca mt qu trnh thun nghch.

T iu kin c cc biu thc (2.29) v (2.30) c th a ra 4

trng hp sau y:

(1) 2>>1 cng c ngha l thi gian bn hu ca nuclit m t1/2(1) rtln so vi thi gian bn hu ca nuclit con t1/2(2), h s nhanh chng t

c cn bng phng x. y l trng hp cn bng th k.

(2) 2>1 ngha l thi gian bn hu ca nuclit m t1/2(1) tuy ln so vi

thi gian bn hu ca nuclit con t1/2(2) nhng tc phn r ca m cngkhng th b qua. l trng hp cn bng tm thi.

(3) 2


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4.2. Cn bng phng x th k

Khi t1/2(2) 1 c th chp nhn 2-1 2 v

phng trnh (2.27) tr thnh:

N2/ N1 = 1/ 2 = t1/2(2)/ t1/2(1) . (2.32)

T (2.32) rt ra:

2 N2 = 1 N1 (2.33)

hay:

A2 =A1 (2.34)

yA2 = 2 N2; A1 = 1 N1 l hot phng x .

Nh th khi t n cn bng phng x, t s gia s nguyn t ca

nuclit con v m lun lun l hng s v hot phng x ca m v con

lun lun bng nhau. Cn bng phng x nh vy c gi l cn bng

th k.

V 1


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con chu bt k ca mt dy phng x, nu cc iu kin c cn bng

phng x c tho mn.

(1). Tnh thi gian bn hu ca cc nuclit c thi gian bn hu qu

di, khi m vic xc nh thi gian bn hu gp kh khn do s thay i hot phng x khng th o c bng thc nghim.

1

1/2 1/ 2

2

(1) (2) (2.37)N

t tN

=

(2). Tnh hm lng ca cc nuclit nm trong cn bng phng x ca

mt dy.

(2.38)1(t

)2(t.

MM

NN

.MM

mm

2/1

2/1

1

2

1

2

1

2

1

2 ==

trong M1, M2 l nguyn t lng.

(3).ng dng trong phn tch, chng hn xc nh hm lng ng v

m trong khong vt thng qua o hot phng x ca nuclit con. xc

nh hm lng urani trong qung ngi ta c th tin hnh o hot ca

Th-234 hoc Pa-234m (Pa l k hiu ca nguyn t protactini).Hm lng rai trong mu c th c xc nh vi nhy rt cao

nh o raon nm cn bng phng x vi rai.

Cng thc tnh khi lng ca nuclit m t hot phng x ca

nuclit con c th rt ra trc tip t cc phng trnh (2.10) v (2.34):

(2.39)1(t.2ln

.NM

m 2/12

Av

11

A=

4.3. Cn bng phng x tm thi

Cn bng phng x tm thi xy ra khi 2>1 ngha l thi gian bnhu ca nuclit m t1/2(1) tuy ln so vi thi gian bn hu ca nuclit con

t1/2(2) nhng tc phn r ca m cng khng th b qua.


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tin lp lun chng ta nhc li vi gi thit ti t=0 nuclit con c

tch hon ton khi nuclit m, tc l N20 = 0 v tr li vi phng trnh

(2.27)

[ ] (2.27e1NN t)(11212 12 =

Khi t l ln, trong thc t thng ly)2(t)1(t

)2(t).1(t10t

2/12/1

2/12/1

> ,

e-(2 -1)t tr thnh nh so vi 1, ta c:

(2.40NN 112

12

=

v rt ra:

(2.41(2)t-(1)t

(2)t

NN

1/21/2

1/2

12

1

1

2 =

=

Nh vy t s gia s nguyn t (cng l t s khi lng) ca hai

nuclit m v con tr thnh hng s, khng thay i theo thi gian, h t

c cn bng phng x.

Da vo nh ngha hot phng x cho bi phng trnh (2.10) v

phng trnh (2.41) d dng tm thy:

(2.42)1(t

)2(t11

NN

2/1

2/1

2

1

22

11

2

1 =

=

=A

A

C th thy rng khc nhau c bn ca cn bng tm thi vi cn

bng th k l ch khi t n cn bng tm thi hot ca nuclit m

lun nh hn hot phng x ca nuclit con, trong khi cn bng th

k hai hot phng x ny lun lun bng nhau.

Cc biu thc rt ra c t vic nghin cu trng thi cn bng

phng x tm thi cng c cc ng dng tng t nh trng hp cn bng


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th k, s khc nhau ch dng c th ca cc phng trnh tnh ton m

thi. Thay cho cc phng trnh (2.37), (2.38), (2.39), y ta c:

(2.431

N

N)2(t)1(t

2

12/12/1

+=

(2.44)2(t)1(t

)2(t.

MM

NN

.MM

mm

2/12/1

2/1

1

2

1

2

1

2

1

2

==

[ ] (2.45)2(t)1(t.2ln

.NM

m 2/12/12

Av

11 =

A

0 1 2 3 4 5 6 7 8 9 10

A=A + A1 2A1

2A

A2

Thi gian t/t1/2

HotA

1

10

102

Hnh 2.2.( L5.9) S ph thuc thi gian ca hot phng x tng cng v

hot phng x ring r ca cc nuclit trong cn bng th k.


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HotA

1/2Thi gian t/t

2A

21A=A + A

109876543210

A2max

maxA

1A

210

10

1

Hnh 2.3.(L5.10) S ph thuc thi gian ca hot phng x tng

cng v hot phng x ring r ca cc nuclit trong cn bng tm thi.

Hnh 2.2 v 2.3 cho thy r s khc nhau cn bn ca cn bng th k

v cn bng tm thi. Khi t n cn bng th k hot phng x ca cc

ng v m v con lun lun bng nhau v khng thay i. Trong trnghp ca cn bng tm thi, ng bin thin hot A1 ch ct A2 ti 1 im

A2max , cn khi t ti cn bng, cc hot ny khng bng nhau v lun

lun gim. (Ch : Trc tung ca cc th c chia theo thang logarit)

4.4. Phn r ni tip trong trng hp tng qut


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i vi trng hp mt dy phng x c n nuclit, phn r theo s

tng qut sau:

Nuclit 1Nuclit 2Nuclit 3Nuclit 4...Nuclit n (2.55).

Nu thi gian bn hu ca nuclit m l rt ln hn so vi cc nuclitcon chu, tc l:

1


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S tch phn phng trnh vi phn (2.70) cho ta:

NA = NA0e-(ab + ac)t (2.71)

A c th phn r theo nhiu nhnh khc nhau vi cc tc ring r

khc nhau, nhng A ch c mt thi gian bn hu t1/2(A):

(2.722ln2ln

)A(tacabA

2/1 +=

=

Tc tch lu ca nuclit B v C bng hiu s gia tc hnh thnh

(do s phn r ca A) vi tc phn r ca chng:

(2.73NN

dt

dNBBAab

B =

Vi nuclit C ta cng c phng trnh tng t:

(2.74NNdt

dNCCAac

C =

Thay (2,71) vo (2.74) ta c phng trnh :

(2.75NeN

dt

dNBB

t)(0Aab

B acab = +

S tch phn phng trnh vi phn (2.75) vi cc iu kin u NB = 0

khi t=0 cho ta:

[ ] (2.76eeN)(

N tt)(0AacabB

abB

Bacab + +

=

Phng trnh (2.76) c dng hon ton tng t vi phng trnh

(2.25) ca trng hp phn r khng r nhnh kho st mc 2.3.4.

Vi nuclit C ta cng c phng trnh tng t.

Khi nuclit m c i sng di hn nhiu so vi nuclit con, tc l khi

ab + ac = A


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[ ] (2.77e1NN tAB

abB

B

=

Sau mt thigian t ln, e-Bt > B v ab + ac = A >> C , phngtrnh (2.76) c th rt gn thnh:

[ ] (2.831eNN t)(Aacab

abB

acab +

= +

hoc tng t, i vi nuclit C:

[ ] (2.841eNNt)(

Aacab

acC

acab

+

=+

Chia 2 v ca (2.83) cho (2.84) ta c:

NB/NC = ab/ ac (2.85)

t


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e(ab + ac)t = 1+(ab + ac)t (2.86)

v t (2.83) v (2.84) ngi ta thu c:

NB/NA=abt v NC/NA=act (2.87)

NHNG CNG THC CN GHI NH

1. Quan h gia hng s tc phn r v thi gian bn hu

t1/2=ln2/ =0.69315/ (2.4)

hoc:

=ln2/ t1/2 . (2.5)2, S ht nhn cn li sau thi gian t:

N=Noe- t

N=No(1/2)t/ t1/2 . (2.6)

3. nh ngha hot phng x

A=-dN/dt=N. (2.10)

4. S thay i hot phng x theo thi gian:

A=A0.e-t=A0(1/2)t/t1/2, (2.11)

trong A0 l hot phng x ban u.

5. Cn bng phng x (1


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yA2 = 2 N2; A1 = 1 N1 l hot phng x .

Nh th khi t n cn bng phng x, t s gia s nguyn t ca

nuclit con v m lun lun l hng s v hot phng x ca m v con

lun lun bng nhau. Cn bng phng x nh vy c gi l cn bngth k.

V 1


24/45

- Phn r - v EC (electron capture):

E = (M1 - M2) c2 . (2.98)

- Phn r +

E = (M1 - M2 - 2me) c2 . (2.99)

- Phn r :

E = E

- T phn hch:

E = [MA - (MB + Mx)]c2 . (2.19)

SE = [MA - (MB + Mx)]c2 . (2.19)

Ch rng 1u(.v.C) = 1,660566.10-24g; c = 2,997925.108ms-1, nn

theo (2.19), s ht khi 1u pht sinh mt nng lng E = 1,49244.10-10J.

Trong khoa hc ht nhn ngi ta thng s dng n v nng lng

eV,

1eV = 1,60219.10-19J,

rt ra : Ht khi 1u sinh ra 931,5 MeV. (2.20)


25/45

BI 2. BI TP HO PHNG X

I. MT S BI TP N GINBI TP 1.Chui phn r ca U-238 kt thc Pb-206. Trong chui ny phi c bao

nhiu phn r v bao nhiu phn r -?

Gii BT18 phn r v 6 phn r -

Bi tp 2.Triti (3H) phn r - vi thi gian bn hu ca t1/2(3H) = 12,33 nm). Mtmu triti c hot phng x 1 MBq.- Vit phng trnh biu din s phn r phng x ca triti- i hot phng x ni trn ra Ci,- Tnh s nguyn t v khi lng triti ca mu,- Tnh hot phng x ring ca triti (ch cha triti)

Gii BT2- Phng trnh biu din s phn r phng x ca triti:

31H 32He + -

- Hot phng x tnh ra Ci,106/3,7x1010 27Ci

- S nguyn t triti trong muN =A/=A/ (0,693/t1/2) = 106/s /(0,693/ 12, 33 x 24x3600 x 365 s)

= 5,59 x 1014 nguyn t.

- Khi lng triti ca mum = 3.N/6,02 x 1023 = 2,78 x 10 -9 g

- Hot phng x ring ca triti (ch cha liti) As = (106/s)/(2,78 x 10 -9 g)


26/45

BT 3. Triti phn r theo quy lut bc nht vi chu k bn r l 12, 5 nm.Mt bao nhiu nm hot ca mu triti gim i cn li 15% so vi banu?

Gii

T phng trnh ng hc ca s phn r phng x: A = A0. et

rt ra t =1

ln 0

A

A= 1/2

t

ln 2.ln 0

A

A=

12,5

ln 2.ln

100

15= 34, 2 nm

BT 4. ng v phng x 13N c chu k bn r l 10 pht, thng c dng chp cc b phn trong c th. Nu tim mt mu 13N c hot phngx l 40 Ci vo c th, hot phng x ca n trong c th sau 25 phts cn li bao nhiu?

Gii

Hot phng x l s phn r phng x trong mt n v thi gian. nv o hot thng l Becquerel (Bq) v Curie (Ci).1 Bq = 1 phn r/giy = 1s-11Ci = 3,7. 1010 Bq.

A =dN

dt= . N0. e

t = . N

A0 = . N0A = A0. et = A0.

12

ln 2.t

t

e

= 40. e2,5.ln2 = 7,01 Ci.

BT 5. Gadolini-153 l nguyn t c dng xc nhbnh long xng, c chu k bn r l 242 ngy. Tnh phntrm Gd-133 cn li trong c th bnh nhn sau 2 nm (730ngy) k t khi cho vo c th?

Gii

Qu trnh phng x tun theo nh lut: N = N0.et

12ln 2

.

0

tt

tN e eN

= = =ln 2

.730242e

= 12,25%.


27/45

BT 7.

1. Di tc ng ca ntron nng lng cao trong tia v tr, ht nhn

Nit-14 bin i thnh ht nhn C-12 cng vi s to thnh ht nhn triti.Hy vit phng trnh ca phn ng ht nhn ni trn.

2. Di tc ng ca ntron nhit trong tia v tr, ht nhn Nit-14

bin i thnh ht nhn C-14 cng vi s to thnh ht nhn 1H. Hy vit

phng trnh ca phn ng ht nhn ni trn. 14N(n,p)14C

Gii

14

7N +1

0n 12

6C +3

1H. Phn ng c th vit tm tt:14

N(n,t)12

C

147N + 10n 146C + 11p. Phn ng c th vit tm tt: 14N(n,p)14C

BT8

2 g 2964Cu c chu k bn hu 12,7 h c lu gi trong mt bung ch, cho

n khi thu c 0,39 g 2864Ni v 0,61 g 3064Zn, c hai u l cc ng v bn.

Vit phng trnh biu din s phn r ca 2964Cu.

Mu 2964Cu c lu gi bao lu? (Gi nh rng cc php cn PTN ny

khng nhy pht hin c s ht khi trong qu trnh phn r phng

x).

Tnh hng s tc ca cc qu trnh phn r ca 2964Cu to thnh 2864Ni v

3064Zn.

Gii

2964Cu 3064Zn.+ _

2964Cu 2864Ni + +


28/45

Cc phn r khng thay i khi lng ca h (khi khng k n s htkhi). Khi lng ca Ni v Zn c to thnh bng gim khi lng ca

ng: mZn + mNi = 1 g

Khi lng ca 2964

Cu gim i mt na. Thi gian lu gi mu ng bngchu k bn hu: 12,7h.

(64Cu) = ln2/12,7 h = 5,46.10-2.h-1

(64Cu) = + + _ = + + (39/61).+

+ = 3,33.10-2.h-1; _= 2,13.10-2.h-1

BT 9.

1. Vit phng trnh biu din s phn r - ca ht nhn triti.

2. Vit phng trnh ca cc qu trnh phn r phng x:222Rn 3,82d

218Po 3,1min 214Pb 26,8min

214Bi 19,9min 214Po 164 s

3. Vit phng trnh ca cc qu trnh phn r phng x sau:

Phn r - ca Sr-90 Phn r ca Th-232

Phn r +

ca Cu-62 Phn r -

ca C-144. Chui phn r ca U-238 kt thc Pb-206. Trong chui ny phi c

bao nhiu phn r v bao nhiu phn r -?

Gii

1. 31H 32He + -

2.

22286Rn 21884Po + 42He

21884Po 21482Pb + 42He214

82Pb 21483Bi + -

21483Bi 21484Po + -

21484Po 21082Pb +


29/45

3.

9038Sr 9039Y + -232

90Th 22888Ra + 42He

6229Cu 6228Ni + +

146C 147N + -

4. 8 phn r v 6 phn r -

BT 10. Thi gian bn hu ca triti 3H t1/2(3H) = 12,33 nm). Mt mu triti chot phng x 1 MBq.- i hot phng x ni trn ra Ci,

- Tnh s nguyn t v khi lng triti ca mu,- Tnh hot phng x ring ca triti

Gii106/3,7x1010 27Ci

N =A/=A/ (0,693/t1/2) = 106/s /(0,693/ 12, 33 x 24x3600 x 365 s) = 5,59 x1014 nguyn t.M = N/6,02 x 1023 = 2,78 x 10 -9 gAs = (106/s)/(2,78 x 10 -9 g)

BT 11

Thi gian bn hu ca 14C l t1/2(14C) = 5730 nm. 2 gam mt mu cha14C c hot phng x 3,7 Bq.- i hot phng x ni trn ra Ci,- Tnh s nguyn t 14C c trong mu,- Tnh hot phng x ring ca mu .

Gii

3,7 Bq = 3,7 /3,7 x 1010 Ci = 10-10 Ci.N =A x t1/2/0,693 = 3,7 x 5730 x 365 x 24 x 3600/0,6935 = 9,64 x 1011 htnhn.

As = 3,7 Bq /2g = 1,85 Bq/g


30/45

BT 11.

Cho dy phng x sau:

222Rn 3,82d 218Po 3,1min

214Pb 26,8min 214Bi 19,9min

214Po 164 s

Gi thit rng ban u ch c mt mnh radon trong mu nghin cu vi

hot phng x 3,7.104 Bq,

a) Vit cc phng trnh biu din cc phn r phng x trong dy trn.

b) Ti t = 240 min (pht) hot phng x ca 222Rn bng bao nhiu?

c) Cng ti t = 240 min hot phng x ca 218Po bng bao nhiu?

d) Ti t = 240 min hot phng x chung ln hn, nh hn hay bng

hot phng x ban u ca 222Rn.

Giia)

22286Rn 21884Po + 42He

21884Po 21482Pb + 42He214

82Pb 21483Bi + -

21483Bi 21484Po + -

21484Po 21082Pb + 3,7.104 Bq = 1Ci , 240 min = 4 h

b) A1=A01e-t = 1Ci.e-ln2.4/24.3,82 = 0,97 Ci

c) t = 240 min > 10 t1/2(Po), h t c cn bng phng x tm

thi, nn


31/45

A1/A2 = 1 t1/2(2)/t1/2(1) A2 =A1/[1 3,1/(3,82.24.60)] = 0,9705 Ci

Nu quan nim gn ng rng c cn bng th k (1


32/45

II. NG DNG PHNG PHP HO PHNG X TRONG

PHN TCH

Bi tp 1. Tnh thi gian bn hu ca cc nuclit c thi gian bn hu

qu di, khi m vic xc nh thi gian bn hu gp kh khn do sthay i hot phng x khng th o c bng thc nghim.

V d: Trong 1kg urani cn bng phng x c cha 0,34mg 226Ra c

t1/2 = 1600 nm. C th tnh c thi gian bn hu ca 238U:

(2.37nm10.5,41600.238226

.34,0

10)2(t

NN

)1(t 96

2/12

12/1 ===

Bi tp 2. Tnh hm lng ca cc nuclit nm trong cn bng phngx ca mt dy.

(2.38)1(t

)2(t.

MM

NN

.MM

mm

2/1

2/1

1

2

1

2

1

2

1

2 ==

trong M1, M2 l nguyn t lng.

V d: Tnh lng 228Ra c t1/2(2) l 5,75 nm c trong 1g 232Th c

t1/2(1) l 1,41.1010 nm:

-102 1/ 22 1 10

1 1/ 2

(2) 228 5,75. . 4,01.10 g

(1) 232 1,42.10

M tm m

M t= = =

Nhng tnh ton nh vy c tm quan trng ln trong cng ngh x l

qung urani v thori, n cung cp thng tin v lng b thi phng x cn

c x l v qun l.

Bi tp 3.Xc nh hm lng ng v m trong khong vt thngqua o hot phng x ca nuclit con.

Cng thc tnh khi lng ca nuclit m t hot phng x ca

nuclit con c th rt ra trc tip t cc phng trnh (2.10) v (2.34):


33/45

(2.39)1(t.2ln

.NM

m 2/12

Av

11

A=

xc nh hm lng urani trong qung ngi ta c th tin hnh

o hot ca Th-234 hoc Pa-234m.Hm lng rai trong mu c th c xc nh vi nhy rt cao

nh o raon nm cn bng phng x vi rai.

Bi tp 4. Phng php nh du bng ng v phng x trong phntch xc nh hm lng axit aspatic trong sn phm thu phn mt protein,ngi ta thm vo dung dich thu phn 5,0 mg axit aspatic nh du c hot phng x ring 0,46 Ci/mg. Sau , ngi ta tch ra 0,21 mg axitaspatic nguyn cht c hot phng x ring 0,01 Ci/mg. Tnh lng axitaspatic c trong mu dung dch thu phn ban u.Ch thch: Axit aspatic l mt amino axit c trong c th ng thc vt, cnhiu trong mt ma, c ci ng, cng thc phn t C4H7NO4.

Gii :Gi x l khi lng axit aspatic (mg) c trong dung dch thu phn,

y l lng axit (nh du) a thm vo,D l hot phng x,As1 l hat phng x ring ca cht nh du ban u,

As2 l hot dung dch sau khi nh du, ta c:As1 = D/y (1)

As2 = D/(x+y) (2).Chia (1) cho (2) v bin i mt cch n gin:

x = y(As1/ As2 - 1). (3)Thay s vo (3), thu c: x = 225 mg


34/45

III. NH TUI BNG PHNG PHP PHNG X1. TNH t KHI C N0/N

N = N0e

-t

0lnN

Nt =

Bi tp 1. .Khi nghin cu mt mu c vt ngun gc hu c cha 1 mg C, ngi tathy rng t l ng v 14C/12C ca mu l 1,2 x 10-14.

a. C bao nhiu nguyn t 14C c trong mu?b. Tc phn r ca 14C trong mu bng bao nhiu?c. Tui ca mu nghin cu bng bao nhiu?Cho t1/2(14C) = 5730 nm, hot phng x ring ca cacbon thi chac cc hot ng ht nhn ca con ngi l 227 Bq/kgC.

Gii14

7N + 10n 126C + 31H. Phn ng c th vit tm tt: 14N(n,t)12C(ntron nhanh)


(ntron nhit)a. Tng s nguyn t C trong mu c vt = (10-3g/12g/ngtg) x 6,02 x

10

23

ngt/ngtg = 5,02 x 10

19

ngtS nguyn t 14C l N (1,2 x 10-14)(5,02 x 1019) = 6,02 x 105 ngt.b. A = (ln2/5730 x 365 x 24 x 3600 s) x 6,02 x 105 = 2,3 x 10-6 Bqc. tui t = [ln(227 x 10-6/2,3 x 10-6)]/(ln2)/5730 nm = 38 000 nm

2. TNH t KHI C Dt/PtKhi khng c thng tin v N0 vic nh tui s tnh theo t s Dt/PtTrong Dtl s ht nhn thi im t ca mt ng v con chu bn, Pt l s ht nhn ca m thi im t.Con khng c mt khi t = 0 v khng mt i (do khuch tn, bay hi...)

Dt + Pt = P0 (1)

Pt = P0 e-t (2)Chia 2 v cho Pt ;

Dt/ Pt + 1 = et (3)

1

ln 1 t

t

Dt

P

= +

(4)


35/45

Bi tpHy tnh tui ca loi c t s nguyn t 206Pb so vi 238U bng 0,60. Chot1/2 ca 238U l 4,5.109 nm.

1 ln 1 tt

DtP

= +

= [1/(ln2/4,5.109 nm)].ln(1 + 0,6) = 3,1.109 nm

2.2. Trng hp ng v con c mt ti t = o

Dt + Pt = P0 + D0 (5) nh c tui trong trng hp ny cn c thng tin v mt ng v bnkhc ca con m ng v ny khng c to ra do phn r ca m.

Dst = Dso = Ds (6)Chia c 2 v ca (5) cho Ds :

Dt/ Ds + Pt/ Ds = D0/ Ds + P0/ Ds (7)Hay:Dt/ Ds = D0/ Ds + P0/ Ds - Pt/ Ds (8)

Thay P0 = Pt et (9)

Ta c:Dt/ Ds = D0/ Ds + ( e

t - 1) ( Pt/ Ds) (10)y = b + ax (11)C th v ng thng y = b + ax v thu c h s gc l ( et - 1). Cngc th tnh a khi c 2 cp gi tr ca y v x.Bi tpTui ca mt trng, do tu Apollo 16 thu lm c, c xc nh davo t s nguyn t ca cc ng v 87Rb/87Sr v 87Sr/86Sr trong mt skhong vt c trong mu:Khong vt 87Rb/86Sr 87Sr/86SrA 0,004 0,699B 0.180 0,709

a) 87Rb phng x - . Hy vit phng trnh biu din qu trnh phn rht nhn ny. t1/2(87Rb) = 4,8.1010 nm.

b) Tnh tui ca mu . Bit rng 87Sr v 86Sr l cc ng v bn v ban

u (t = 0) t s 87Sr/86Sr trong cc khong A v B l nh nhau.Gii:37

87Rb 3887Sr + -

Phng trnh (10) c th vit nh sau:87Srnow/86Sr = 87Sr0/86Sr + (e

t - 1) 87Rbnow/86Sr (12)Trong mu A:


36/45

0,699 = 87Sr0/86Sr + (et -1)0,004 (a)

Trong B:0,709 = 87Sr0/86Sr + (e

t - 1)0,180 (b)(b) - (a) v bin i ta c:

et

= (0,709 0,699)/(0,180 0,004) +1 = 1,0568t = (ln2)t/t1/2 = ln1,0568t = (4,8.1010.ln1,0568)/ln2 = 3,8.109 nm

Trong bi tp trn ngi ta c th i tnh thm 87Sr0/86Sr t = 0.

Ngi ta c th cho cc gi tr khc nhau ca 87Rb/86Sr v 87Sr/86Sr trong

nhiu khong vt khc nhau. Nu a ln th m thu c mt ng

thng th l bng chng cho thy t = 0, t s 87Sr0/86Sr trong cc khong

ny nh nhau.

C khi ngi ta cho bit tui khong vt (t), tnh hoc t1/2.

Ngi ta thng nh tui da vo phn r 40K thnh 40Ar (php nh tui

K/Ar) hoc 235U v 207Pb; 238U v 206Pb.

3. MT S DNG BI TP KHCBi tp 1.

Cho dy phng x sau:

222Rn 3,82d 218Po 3,1min

214Pb 26,8min 214Bi 19,9min

214Po 164 s

Gi thit rng ban u ch c mt mnh radon trong mu nghin cu vi

hot phng x 3,7.104 Bq,

e) Vit cc phng trnh biu din cc phn r phng x trong dy trn.f) Ti t = 240 min (pht) hot phng x ca 222Rn bng bao nhiu?

g) Cng ti t = 240 min hot phng x ca 218Po bng bao nhiu?

h) Ti t = 240 min hot phng x chung ln hn, nh hn hay bng

hot phng x ban u ca 222Rn.


37/45

Li gii bi tp 6

a)

22286Rn 21884Po + 42He21884Po 21482Pb + 42He214

82Pb 21483Bi + -

21483Bi 21484Po + -

21484Po 21082Pb +

3,7.104 Bq = 1Ci , 240 min = 4 h

b) A1=A01e-t = 1Ci.e-ln2.4/24.3,82 = 0,97 Ci

c) t = 240 min > 10 t1/2(Po), h t c cn bng phng x v

+ Quan nim gn ng rng c cn bng th k (1


38/45


Bi tp 2.

4. Vit phng trnh biu din s phn r - ca ht nhn triti.

5. Vit phng trnh ca cc qu trnh phn r phng x:222Rn 3,82d

218Po 3,1min 214Pb 26,8min

214Bi 19,9min 214Po 164 s

6. Vit phng trnh ca cc qu trnh phn r phng x sau:

Phn r - ca Sr-90 Phn r ca Th-232

Phn r +

ca Cu-62 Phn r -

ca C-144. Chui phn r ca U-238 kt thc Pb-206. Trong chui ny phi c

bao nhiu phn r v bao nhiu phn r -?

Gii

1. 31H 32He + -

2.

22286Rn 21884Po + 42He

21884Po 21482Pb + 42He214

82Pb 21483Bi + -

21483Bi 21484Po + -

21484Po 21082Pb +

3.

9038

Sr 90

39

Y +

-

232

90Th 22888Ra + 42He

6229Cu 6228Ni + +

146C 147N + -

4. 8 phn r v 6 phn r -


39/45

Bi tp 3. Thi gian bn hu ca triti 3H t1/2(3H) = 12,33 nm). Mt mu tritic hot phng x 1 MBq.- i hot phng x ni trn ra Ci,- Tnh s nguyn t v khi lng triti ca mu,- Tnh hot phng x ring ca triti

Gii106/3,7x1010 27Ci

N =A/=A/ (0,693/t1/2) = 106/s /(0,693/ 12, 33 x 24x3600 x 365 s) = 5,59 x1014 nguyn t.M = N/6,02 x 1023 = 2,78 x 10 -9 gAs = (106/s)/(2,78 x 10 -9 g)

Bi tp 4

Thi gian bn hu ca 14C l t1/2(14C) = 5730 nm. 2 gam mt mu cha14C c hot phng x 3,7 Bq.- i hot phng x ni trn ra Ci,- Tnh s nguyn t 14C c trong mu,- Tnh hot phng x ring ca mu .

Gii

3,7 Bq = 3,7 /3,7 x 1010 Ci = 10-10 Ci.N =A x t1/2/0,693 = 3,7 x 5730 x 365 x 24 x 3600/0,6935 = 9,64 x 1011 htnhn.

As = 3,7 Bq /2g = 1,85 Bq/g


40/45

Bi tp 9

Cacbon 14 c to thnh t nit do tc dng ca cc ntron

(chm) trong cc tia v tr, ri i vo c th sinh vt qua quang

hp v lu chuyn thc phm ca ng thc vt. 14C phn r -

vi thi gian bn hu t1/2 = 5730 nm. S phn tch cacbon

phng x trong cc c th sng cho gi tr hot phng x

ring ca cacbon l

230 Bq/kg cacbon.

a). Vit cc phng trnh phn ng ht nhn biu din qu trnhhnh thnh v phn r ca 14C trong t nhin.

b) T l ng v 14C/12C trong c th sng bng bao nhiu?

c) Mt nh kho c ly c mt mu, c cho l ca mt ho

thch hu c, ti mt kim t thp Ai-cp v thy rng t l

ng v ca cacbon trong mu ny, xc nh bng phng php

khi ph, l 14C/12C = 6. 10-13 . ng s cho rng tui ca mu

ni trn l bao nhiu?

Gii

1.

14 1 14 1

7 0 6 1

14 14

6 7

N C H

C N

n

+ +


41/45

3. Hot 230 Bq/kg tng ng vi t s ng v 14C/12C sau y:

14

12 12

1/ 2

1/ 2

ln 2. ln2

= . . = .C

A A

C C

NA m tAs N w N w

m m M m M t

= =

(khi b qua hm lng ca 13C). Trong , w l t s ng v 14C/12C

121/ 2 12

23

A

. 230 5730 365 24 3600 12= = = 1.20 10

.ln 2 6.02 10 1000 ln 2

CAs t M

wN

Ch : Khi thay s cn i 230 Bq/kg ra 230/1000 (Bq/g), v mol nguyn t

tnh ra gam.

V 6.0 1013 / 1.20 1012 = 1/2, mt khong thi gian bng thi gian bn hu

tri qua (chng ta s dng gi tr thi gian bn hu 5730 nm xc nh

tui). Nh kho c hc cho rng cht bt ny c lm ra vo nm 3560

trc CN.

4. Thc ra, nhm phenoxyacetyl c hnh thnh t axit phenoxyacetic

c tng hp trong cng nghip t cc sn phm ch bin than v du m.

N khng cha cacbon phng x. Ch c 8 trong s 16 nguyn t cacbon l

c ngun gc t nhin (to thnh t c th sng). Nh th, trong phn cngun gc t nhin, hm lng 14C phi gp i w = 1.2 1012, ngha l cht

bt ny l sn phm ca thi nay.

Bi tp 11.1. Khi phn tch qung urani (uranium) ngi ta tm thy 3 ng v ca

urani l 238U, 235U v 234U, u c tnh phng x. Hai ng v 235U v 234Uc phi l ng v con chu ca 238U khng? Ti sao? (Ngi ta quan stc cc nguyn t phng x t nhin tnh phng x v tnh phng

x ).2. Khi thu luyn mt mu qung urani ly t m Nng Sn (Qung Nam),

ngi ta thu c dung dch c nng UO2SO4 (uranyl sunfat) l 0,01Mcn nng Fe2(SO4)3 ln ti 0,05M. S tch urani khi st v cc tpcht khc c th thc hin bng phng php chit hoc trao i ion,nhng cng c th bng kt ta phn on.


42/45

C th kt ta 99% lng st (Fe3+) c trong dung dch ni trn pH bngbao nhiu m khng lm mt lng urani c mt trong dung dch? Gi nhrng s hp ph urani trn b mt kt ta Fe(OH)3 l khng ng k. Bitrng trong iu kin nhit tin hnh th nghim, tch s tan ca UO2(OH)2l 10-22 ca Fe(OH)3 l 3,8.10-38.3. Nc thi ca dung dch thu luyn qung urani c cha ng v phngx 226Ra c thi gian bn hu 1600 nm. bo v mi trng, ngi ta cth ng kt ta rai vi BaSO4 v lu gi khi cht thi ny trong kho thiht nhn. Cn lu gi chng bao lu hot phng x ca khi cht thiny ch cn li


43/45

A B CD

d. Hy ngh mt phng n chit c > 95% lng urani c trong100ml dung dch nc vo 100 ml pha hu c. Nng uranyl nitrat trongnc v thnh phn h dung mi chit cho php chp nhn h s phn b Dkhng i v bng 6.Gii1. Khi xy ra phn r , nguyn t khi khng thay i. Khi xy ra 1 phn

r , nguyn t khi thay i 4 .v. khi lng nguyn t (u). Nh th, skhi ca cc ng v con chu phi khc s khi ca ng v m 4n(u),vi n l s nguyn. Ch 234U tho mn iu kin ny vi n = 1. Trong 2ng v 234U, 235U, ch 234U l ng v con chu ca 238U.

2. Nng Fe3+ cn li trong dung dch sau khi 99% st trong dung dchFe2(SO4)3 0,05 M b kt ta l:

[Fe3+] = 2.0,05.10-2 M = 10-3 M.[OH-] cn c trong dung dch [Fe3+] ch cn li trong dung dch vi nng10-3 M l:

( )31 1

38 13 3Fe(OH) 1233 3

T 3,8x10OH 38 x10Fe 10

+

= = =

[OH-] = (38)1/3.10-12 ng vi gi tr pH l:pH = -log{10-14/(38)1/3.10-12} = 2 + (1/3)log38 = 2,53. pH , tch s ion ca UO2(OH)2 trong dung dch 0,01M l:[UO22+][OH-]2 = 0,01.[(38)1/3.10-12]2 = 1,13.10-25< 10-22

V tch s ion ni trn rt nh so vi tch s tan ca UO2(OH)2 (TUO2(OH)2) nnurani khng kt ta trong iu kin trn.3.

a. Sau n chu k bn hu ca rai, hot phng x ca thng cht thi chcn li 1/2n. Hot phng x ch cn 103 hay:nlog2 > 3. Rt ra: n > 3/0,301 10. Thi gian cn lu gi hot phngx ca khi cht thi rai cn


44/45

TN1 TN2 TN3 TNnX x1 x2 x3 ... xnY y1 = Co-x1 y2=2(Co-x2) y3=3(Co-x3) yn=n(Co-xn)b.

0 1 0 2 0 n

1 2 n

C x 2(C x ) n(C x )... D const

x x x

= = = = = (1)

T biu thc trn rt ra:0

n

nCx

D n=

+(2)

Ly o hm xn theo n, ta c:(xn)' = CoD/(D + n)2 > 0 (3)

Nh th , xn tng khi n tng: x1 < x2 < x3. (4)y1 = x1.D ; y2 = x2.D ; y3 = x3.DT cc bt ng thc (4) rt ra: y1 < y2 < y3 . (5)

A. x1y3 saiC. x1y3 saiD. x1y3 sai

c. Khi n tng, nng uranyl nitrat trong pha nc v pha hu c nm cnbng, xn v yn, u tng dn ln. Nhng yn khng th vt qua nng bo ho ca uranyl nitrat trong pha hu c. V th ng ng nhit chits tim cn vi ng thng nm ngang

y = ybh (ybh l nng bo ho uranyl nitrat trong pha hu c). Ch c th A c dng iu nh vy.d. Nu chit 1 ln th nng uranyl nitrat trong pha nc sau khi chit

c tnh nh sauD = y/x = (C0 - x)/x (6)

Rt ra: x/ C0 = 1/ (D+1) = 1 / 7 = 0,143 > 0,1Nh th lng uranyl nitrat cn li trong dung dch nc s ln hn 10%nng ban u. chit c > 95% urani vo pha hu c ta c th chia 100 ml dung michit thnh n phn bng nhau, ri chit thnh n bc. Dung dch nc sau khichit vi phn dung mi th nht ( bc chit 1), tip tc a vo chit vi

phn dung mi th 2 (bc chit 2)...c th cho n bc n.Khi 100 ml dung mi hu c c chia thnh n phn bng nhau, t l thtch pha nc (Vaq) vi th tch pha hu c (Vo) trong mi bc chit s l:

Vaq/Vo = 100 ml/ (100 ml/n) = nBiu thc ca h s phn b cho bc chit 1 s l:


45/45

D = y1/x1 = n(C0 - x1)/x1 (7)Rt ra: x1 = n C0 / (D+n) = C0/[(D/n) +1] (8)Vo bc chit th 2, nng ban u ca pha nc l x1. Tng t nh biuthc (7) ta c:

x2 = x1 / [(D/n) +1] (9).Thay x1 bng biu thc (8) ta thu c:

x2 = C0 / [(D/n) +1] 2 (10)Tng t nh vy, vi bc chit th 3,..., th n ta c:

x3 = C0 / [(D/n) +1] 3 (11)xn = C0 / [(D/n) +1] n (12)

> 95% urani c chit vo pha hu c th:xn/C0 = 1 / [(D/n) +1] n < 5x10-2 (13)

Lp bng bin thin 1 / [(D/n) +1] n theo n:n = 1 2 3

1/[(D/n) +1] n = 1/7 > 5.10-2 1/16 > 5.10-2 1/27 < 5.10-2Kt lun: C th chia 100 ml dung mi hu c thnh 3 phn bng nhau vchit 3 bc a c > 95% urani vo pha hu c.Ch : Phn d cn c th gii theo nhiu cch khc nhau. Cc cch gii ngkhc u c cho im. Cc hc sinh c bit gii c th a ra phngn chit lin tc ngc dng v tnh s bc l thuyt. C hc sinh lm theo

phng n loi tr dn (th phng n chit 2 bc, ri sang phng n chit3 bc...). tnh n t phng trnh (13), hc sinh c th logarit ho ...

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