bai-tap-hat-nhan-nguyen-tu-giai-chi-tiet.pdf

Luyn thi i hc mn Vt L: Bi tp phn ng ht nhn c p n BI TP V HT NHN NGUYN T P - 1

Cu 1. Ngi ta dng ht proton bn vo ht nhn Li73 ng yn, gy ra phn ng

H11 + Li7

3 2 . Bit phn ng ta nng lng v hai ht c cng ng nng. Ly

khi lng cc ht theo n v u gn bng s khi ca chng. Gc to bi hng ca

cc ht c th l: A. C gi tr bt k. B. 600 C. 1600 D. 1200 Gii: Theo L bo ton ng lng

PP = P1 + P2 P2 = 2mK K l ng nng

cos2

=

P

PP

2=

2

1

Km

Km PP

2

2 =

Km

Km PP

2

1=

Km

Km PP

2

1=

K

K P

.4

.1

2

1

cos2

=

K

K P

4

1

KP = 2K + E -----> KP - E = 2K ------> KP > 2K

cos2

=

K

K P

4

1 >

4

22

4

1

K

K ------>

2

> 69,3

0 hay > 138,60

Do ta chn p n C: gc c th 1600

Cu 2. ng v Si3114 phng x . Mt mu phng x Si3114 ban u trong thi gian 5 pht

c 190 nguyn t b phn r nhng sau 3 gi trong thi gian 1 pht c 17 nguyn t b phn r. Xc nh chu k bn r ca cht . A. 2,5 h. B. 2,6 h. C. 2,7 h. D. 2,8 h.

Gii: 11 0 0 1(1 )

tN N e N t (t1

Luyn thi i hc mn Vt L: Bi tp phn ng ht nhn c p n

Ta c ln 2

22 2ln 2 1

4

TT Te e e

(3). Thay (1), (3) vo (2) ta c t l cn tm:

2

11 4 3

1 1

1 4

k k

k

. Chn p n C

Cu 4: C hai mu cht phng x A v B thuc cng mt cht c chu k bn r T = 138,2 ngy v c khi 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 A nhiu hn mu B l

A. 199,8 ngy B. 199,5 ngy C. 190,4 ngy D. 189,8 ngy

Gii Ta c NA = N0 1te ; NB = N0 2

te 2 1( ) 1 2ln 2

2,72 ( ) ln 2,72t tB

A

Ne t t

N T

----- t1 t2 = ln 2,72

199,506 199,5ln 2

T ngy Chn p n B : 199,5 ngy

Cu 5: Mt bnh nhn iu tr bng ng v phng x, dng tia dit t bo bnh. Thi

gian chiu x ln u l 20t pht, c sau 1 thng th bnh nhn phi ti bnh vin khm bnh v tip tc chiu x. Bit ng v phng x c chu k bn r T = 4 thng (coi t T ) v vn dng ngun phng x trong ln u. Hi ln chiu x th 3 phi tin hnh trong bao lu bnh nhn c chiu x vi cng mt lng tia nh ln u?

A. 28,2 pht. B. 24,2 pht. C. 40 pht. D. 20 pht.

Gii: Lng tia phng x ln u: 1 0 0(1 )

tN N e N t

( p dng cng thc gn ng: Khi x


ln 2 ln 2

2 20 0 0

T

t TN N e N e N e

. Thi gian chiu x ln ny t

ln2 ln2

'2 20 0' (1 ) '

tN N e e N e t N

Do t= 22ln

e t = 14,1 pht Chn p n

C

Cu 7: ta dng prtn c 2,0MeV vo Nhn 7Li ng yn th thu hai nhn X c cng

ng nng. Nng lng lin kt ca ht nhn X l 28,3MeV v ht khi ca ht 7Li l 0,0421u. Cho 1u = 931,5MeV/c

2; khi lng ht nhn tnh theo u xp x bng s khi. Tc ca ht nhn X bng: A. 1,96m/s. B. 2,20m/s. C. 2,16.10

7m/s. D. 1,93.10

7m/s.

Gii: Ta c phng trnh phn ng: H11 + Li7

3 2 X4

2

mX = 2mP + 2mn mX -----> mX = 2mP + 2mn - mX vi mX =5,931

3,28 = 0,0304u

mLi = 3mP + 4mn mLi ------>mLi = 3mP + 4mn - mLi

M = 2mX (mLi + mp) = mLi - 2mX = - 0,0187u < 0; phn ng ta nng lng E

E = 0,0187. 931,5 MeV = 17,42MeV 2WX = E + Kp = 19,42MeV -----> WX = 2

2mv

= 9,71 MeV

v = m

WX2 = u

WX

4

2 =

25,931.4

71,9.2

c

MeV

MeV = c

5,931.4

71,9.2 = 3.10

8.0,072 = 2,16.10

7 m/s

Chn p n C.

Cu 8: Cho chm ntron bn ph ng v bn 55

25Mn ta thu c ng v phng x 56

25Mn . ng v phng x 56Mn c chu tr bn r T = 2,5h v pht x ra tia -. Sau qu

trnh bn ph 55Mn bng ntron kt thc ngi ta thy trong mu trn t s gia s

nguyn t 56Mn v s lng nguyn t 55Mn = 10-10. Sau 10 gi tip th t s gia

nguyn t ca hai loi ht trn l: A. 1,25.10

-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 5625Mn gim,

c s nguyn t 55

25Mn khng i, Sau 10 gi = 4 chu k s nguyn t ca 56

25Mn gim 24

= 16 ln. Do th t s gia nguyn t ca hai loi ht trn l:

55

56

Mn

Mn

N

N=

16

10 10= 6,25.10

-12 Chn p n C


Cu 9 . Dng ht Prtn c ng nng pK = 5,45 MeV bn vo ht nhn Beri ng yn

to nn phn ng: H11 + Be9

4 eH4

2 + Li6

3 . H li sinh ra bay theo phng vung gc

vi phng chuyn ng ca Prtn. Bit ng nng ca Hli l K = 4MeV v khi

lng cc ht tnh theo n v u bng s khi ca chng. ng nng ht nhn Liti c gi tr:

A. 46,565 MeV ; B. 3,575 MeV C. 46,565 eV ;

D. 3,575 eV.

Gii:Phng trnh phn ng: LiHeBep6

3

4

2

9

4

1

1

Theo L bo ton ng lng

Pp = P + PLi

Do ht hli bay ra theo phng vung gc vi ht Proton

PLi2 = P

2 + Pp

2 (1)

ng lng ca mt vt: p = mv ng nng ca vt K = mv2/2 = P2/2m----> P2 = 2mK T (1)

2mLiKLi = 2mK + 2mpKp

----> 6 Kli = 4K + Kp

------> KLi = (4K + Kp )/6 = 21,45/6 = 3,575(MeV)

Chn p n B

Cu 10. Cho prtn c ng nng KP = 2,25MeV bn ph ht nhn Liti 7

3 Li ng yn.

Sau phn ng xut hin hai ht X ging nhau, c cng ng nng v c phng chuyn ng hp vi phng chuyn ng ca prtn gc nh nhau. Cho bit mp = 1,0073u; mLi = 7,0142u; mX = 4,0015u; 1u = 931,5 MeV/c

2.Coi phn ng khng km theo phng x gamma gi tr ca gc l A. 39,45

0 B. 41,35

0 C. 78,9

0. D. 82,7

0.

Gii: Cng thc lin h gia ng lng v ng nng ca vt

K =

22 2

2

PP mK

m Phng trnh phn ng:

1 7 4 4

1 3 2 2H Li X X mP + mLi = 8,0215u ; 2mX = 8,0030u.Nng lng phn ng to ra :

E = (8,0215-8,0030)uc2 = 0,0185uc2= 17,23MeV

2KX = KP + E = 19,48 MeV--- KX =9,74 MeV. Tam gic OMN:

2 2 2 2 osX X P X PP P P P P c

N

M

O

PX

PX

PH

P

PLi

Pp


Cos = 21 1 2.1,0073.2,25

0,12062 2 2 2 2.4,0015.9,74

P P P

X X X

P m K

P m K Suy ra = 83,070

Cu 11 : Ht c ng nng K = 3,1MeV p vo ht nhn nhm ng yn gy ra

phn ng nPAl 30152713 , khi l-ng ca cc ht nhn l m = 4,0015u, mAl = 26,97435u,

mP = 29,97005u, mn = 1,008670u, 1u = 931,5Mev/c2. Gi s hai ht sinh ra c cng vn

tc. ng nng ca ht n l

A. Kn = 0,8716MeV. B. Kn = 0,9367MeV.

C. Kn = 0,2367MeV. D. Kn = 0,0138MeV.

Gii Nng lng phn ng thu : E = (m + mAl - mP - mn ) uc2 = - 0,00287uc

2 = -

2,672 MeV KP + Kn = K + E = 0,428 MeV KP = 2

2

P Pm v ;

Kn = 2

2

n nm v m vP = vn

1 1

30 30 1

n n n

P P P n

K m K

K m K K

0,4280,0138

31 31

P nn

K KK MeV

. p n D

Cu 12 : . Tim vo mu bnh nhn 10cm3 dung dch cha Na2411 c chu k bn r T = 15h

vi nng 10-3mol/lt. Sau 6h ly 10cm3 mu tm thy 1,5.10-8 mol Na24. Coi Na24 phn b u. Th tch mu ca ngi c tim khong: A. 5 lt. B. 6 lt. C. 4 lt. D. 8 lt.

Gii: S mol Na24 tim vo mu: n0 = 10-3

.10-2

=10-5

mol.

S mol Na24 cn li sau 6h: n = n0 e- t

= 10-5

. Tt

e

.2ln

= 10-5 15

6.2ln

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

Cu 13.Cho phn ng ht nhn 21 D +

2

1 D He3

2 + n1

0 . Bit ht khi ca D2

1 l ( mD

= 0,0024u, mHe = 0,0505u v 1u = 931,5Mev/c2, NA = 6,022.10

23 mol

-1. Nc trong t

nhin c cha 0,015% D2O, nu ton b D2

1 c tch ra t 1kg nc lm nhin liu

dng cho phn ng trn th to ra nng lng l A. 3,46.10

8KJ B.1,73.10

10KJ C.3,46.10

10KJ D. 30,762.10

6 kJ

Gii: ht khi: m = Zmp + (A-Z)mn m -----> m = Zmp + (A-Z)mn m Nng lng mt phn ng to ra

E = (2mD mHe mn ) c2 = [2(mP + mn - mD) (2mp + mn - mHe ) - mn]c

2 = (mHe -

2mD)c2

= 0,0457uc2 = 42,57MeV = 68,11.10

-13J

Khi lng D2O c trong 1000g H2O = 0,015x 1000/100 = 0,15 g.


S phn t D2 cha trong 0,15 g D2O : N=20

AN 0,15 = 20

15,0.10.022,6 23 = 4,5165.10

21

Nng lng c th thu c t 1 kg nc thng nu ton b tri thu c u dng lm nhin liu cho phn ng nhit hch l

E = N.E = 4,5165.1021. 68,11.10-13 = 307,62.108 J = 30,762.106 kJ . p n D

Cu 14: Mt hn hp 2 cht phng x c chu k bn r ln lt l T1= 1 gi v T2 =2 gi. Vy chu k bn r ca hn hp trn l bao nhiu? A. 0,67 gi. B. 0,75 gi. C. 0,5 gi. D. p n khc. Gii: Sau t = T1 = 1h s ht nhn ca cht phng x th nht gim i mt na, cn s ht

nhn ca cht phng x th hai cn 2

1

02

2

N=

2

02N > 2

02N . Nh vy chu k bn r cu hn hp

T > 1h. Chn p n D

Cu 15 :Bn mt hat anpha vo ht nhn nito N147 ang ng yn to ra phn ng

NHe 1474

2 H1

1 + O17

8 . Nng lng ca phn ng l E =1,21MeV.Gi s hai ht sinh ra c

cng vecto vn tc. ng nng ca ht anpha:(xem khi lng ht nhn tnh theo n v u gn bng s khi ca n) A1,36MeV B:1,65MeV C:1.63MeV D:1.56MeV

Gii:

Phng trnh phn ng NHe 1474

2 H1

1 + O17

8 . Phn ng thu nng lng E = 1,21

MeV

Theo L bo ton ng lng ta c;

mv = (mH + mO )v (vi v l vn tc ca hai ht sau phn ng) ----> v = OH mm

vm

=

9

2v

K = 2

2

vm = 2v2

KH + KO = 2

)( 2vmm OH = 2

)( OH mm (9

2)

2 v

2 =

9

4v

2 =

9

2K

K = KH + KO + E --------> K - 9

2K =

9

7K = E

------> K =7

9E = 1,5557 MeV = 1,56 MeV. Chn p n D


Cu 16: Mc nng lng ca ng t hidro c biu thc En= -13.6/n2 eV. Khi kch thch ng t hidro t qu o dng m ln qu o n bng nng lng 2.55eV, thy bn knh qu o tng 4 ln .bc sng nh nht m ng t hidro c th pht ra l: A:1,46.10

-6 m B:9,74.10

-8 m C:4,87.10

-7 m D:1,22.10

-7 m

Gii: rm = m2r0; rn = n

2r0 ( vi r0 bn knh Bo)

m

n

r

r=

2

2

m

n = 4----> n = 2m----> En Em = - 13,6 ( 2

1

n-

2

1

m) eV = 2,55 eV

-----> - 13,6 (24

1

m-

2

1

m) eV = 2,55 eV------>

24

3

m13,6. = 2,55------> m = 2; n = 4

bc sng nh nht m ng t hidro c th pht ra l:

hc = E4 E1 = -13,6.( 2

1

n - 1) eV = 13,6

16

15,1,6.10

-19 = 20,4. 10

-19 (J)

-----> = 14 EE

hc

=

19

834

10.4,20

10.310.625,6

= 0,974.10-7

m = 9,74.10-8m . Chn p n B

Cu 17 : Bn ht nhn c ng nng 18 MeV vo ht nhn 147N ng yn ta c phn

ng 14 177 8N O p . Bit cc ht nhn sinh ra cng vc t vn tc. Cho m = 4,0015u;

mp= 1,0072u; m N = 13,9992u; m O =16,9947u; cho u = 931 MeV/c

2. ng nng ca ht

prtn sinh ra c gi tr l bao nhiu?

A. 0,111 MeV B. 0,555MeV C. 0,333 MeV D. p s khc

Gii: Nng lng phn ng thu : E = (m + mN - mO mp ) uc2 = - 0,0012uc

2 = -

1,1172 MeV

KO + Kp = K + E = 16,8828 MeV

KO = 2

2

O Om v ; Kp = 2

2

p pm v m vO = vp --

1 1

17 17 1

p p p

O O O p

K m K

K m K K

16,8828

0,937918 18

O p

p

K KK MeV

Chn p n D

Cu 18 ng v phng x Na24 pht ra phng x - vi chu k bn r T v ht nhn con l Mg24. Ti thi im ban u t s khi lng Mg24 v Na24 l . Sau thi gian 2T th t s l: A. 1. B. 2. C. 3. D. 4.

Gii: Phng trnh phng x: eMgNa0

1

24

12

24

11 Sau mi phn ng khi lng Mg24 c to thnh ng bng khi lng Na24 b phn r.

Gi m0 l khi lng ban u ca Na24. Khi lng Mg24 lc u: m1 = m0/4 Sau t = 2T: Khi lng Na24 cn m = m0/2

2 = m0/4


Khi lng Mg24 c to thnh: m2 = m = m0 m = 3m0/4 Lc khi lng Mg24 m = m1 + m2 = m0 Do t s m/m = 4. Chon p n D,

Cu 19. Phn tch mt mu g c v mt khc g va mi cht c ng v phng x C14 vi chu k bn r 5600 nm. o phng x ca hai khc g th thy phng x ca khc g va mi cht gp 1,2 ln ca khc g c vi khi lng ca mu g c gp i khi lng khc g mi cht. Tui ca mu g c l: A. 4903 nm. B. 1473 nm. C. 7073 nm. D. 4127 nm Gii: Gi H l phng x ca mt na khi lng (m/2) ca khc g c, H0 l phng x ca khc g mi. Theo bi ra m = 2m0 -----> 2H = 1,2H0 ---> H = 0,6H0 (*)

Theo L phng x ta c: H = H0 te (**) T (*) v (**) suy ra: te = 0,6 ------> -

T

2ln t = ln0,6 ------>

t = -T2ln

6,0ln = 4127 nm. Chn p n D

Cu 20 . Mt khi cht phng x hn hp gm hai ng v vi s lng ht nhn ban u nh nhau .ng v th nht c chu k T1 = 2,4 ngy ngy ng v th hai c T2 = 40 ngy ngy.Sau thi gian t1 th c 87,5% s ht nhn ca hn hp b phn r,sau thi gian

t2 c 75% s ht nhn ca hn hp b phn r.T s 2

1

t

t l. A. t1 = 1,5 t2. B. t2 = 1,5 t1

C. t1 = 2,5 t2 D. t2 = 2,5 t1

Gii: Gi T l khong thi gian m mt na s ht nhn ca hn hp hai ng v b phn r ( chu k bn r ca hn hp, ta c th tnh c T = 5,277 ngy).

Sau thi gian t1 s ht nhn ca hn hp cn li N1 = N0 1te = 8

0N = .3

0

2

N----> t1 = 3T

(*) Sau thi gian t2 s ht nhn ca hn hp cn li N2 =

N0 2te = 4

0N .=2

0

2

N----> t2 = 2T. (**). T (*) v (**) suy ra

2

1

t

t=

2

3

hay t1 = 1,5t2 Chn p n A

Cu 21 : Bit U235 c th b phn hch theo phn ng sau : nYIUn 1094

39

139

53

235

92

1

0 3

Khi lng ca cc ht tham gia phn ng: mU = 234,99332u; mn = 1,0087u; mI = 138,8970u; mY = 93,89014u; 1uc

2 = 931,5MeV. Nu c mt lng ht nhn U235

nhiu, gi s ban u ta kch thch cho 1010 ht U235 phn hch theo phng trnh trn v sau phn ng dy chuyn xy ra trong khi ht nhn vi h s nhn ntrn l k = 2. Coi phn ng khng phng x gamma. Nng lng to ra sau 5 phn hch dy chuyn u tin (k c phn hch kch thch ban u): A. 175,85MeV B. 11,08.10

12MeV C. 5,45.10

13MeV D. 8,79.10

12MeV

Gii: Nang lng ta ra sau mi phn hch:


E = ( mU + mn - mI - mY - 3mn )c2

= 0,18878 uc2 = 175,84857 MeV

= 175,85 MeV

Khi 1 phn hch kch thch ban u sau 5 phn hach dy chuyn s phn hch xy ra l 1 + 2 + 4 + 8 + 16 = 31

Do s phn hch sau 5 phn hch dy chuyn t 1010 phn hch ban u N = 31.1010

Nng lng ta ra E = N E = 31.1010 x175,85 = 5,45.1013 MeV Chn p n C Cu 22: Ngy nay t l ca U235 l 0,72% urani t nhin, cn li l U238. Cho bit chu k bn r ca chng l 7,04.108 nm v 4,46.109 nm. T l ca U235 trong urani t nhin vo thi k tri t c to thnh cch y 4,5 t nm l: A.32%. B.46%. C.23%. D.16%.

Gii: N1 = N01 te 1 ; N2 = N01 te 2 ------> 2

1

N

N=

02

01

N

Nt

e)( 12

--------> 02

01

N

N=

2

1

N

Nt

e)( 21 =

28,99

72,0 2ln)11

(21 TT

t

e

=28,99

72,0 2ln)46,4

1

704,0

1(5,4

e = 0,303

02

01

N

N= 0,3 ------>

0201

01

NN

N

=

3,1

3,0= 0,23 = 23%. Chn p n C

Cu 23: cho chu k bn r T ca mt cht phng x, ngi ta dng my m xung. Trong t1 gi u tin my m c n1 xung; trong t2 = 2t1 gi tip theo my m c n2

= 64

9n1 xung. Chu k bn r T c gi tr l bao nhiu?

A. T = t1/2 B. T = t1/3 C. T = t1/4 D. T = t1/6

Gii Ta c n1 = N1 = N0(1- 1t

e ) n2 = N2 = N1(1- 2

te

) = N0 1t

e (1- 12 te )

2

1

n

n=

)1(

111

1

2 tt

t

ee

e

=

)1(

12XX

X

(Vi X = 1te do ta c phng trnh: X2 + X

=2

1

n

n=

64

9 hay X

2 + X

64

9= 0. Phng btrnh c cc nghim X1 = 0,125 v X2 = - 1,125


mCe = 58mp + 82mn - mCe mNb = 41mp + 52mn - mNb

M = mCe + mNb - mU + 7mn 7mp 7me mCe + mNb - mU

WLKR = A

WLK -----> Wlk = WLKR.A = mc2

-----> m = 2c

AWLKR

mU = 235 . 7,7 2c

MeV = 1809,5

2c

MeV

mCe = 140 . 8,43 2c

MeV = 1180,2

2c

MeV

mNb = 93 . 8,7 2c

MeV = 809,1

2c

MeV

Do E = Mc2

= 1180,2 + 809,1 1809,5 = 179,8 MeV. Chn p n A

Cu 25 .Trong phn ng dy chuyn ca ht nhn U235 , phn ng th nht c 100 ht

nhn U235 b phn r v h s nhn notron l 1,6. Tnh tng s ht nhn b phn r n phn ng th 101. A. 5,45.10

23 B.3,24.10

22 C. 6,88.10

22 D. 6,22.10

23

Gii: Phn ng th nht c 100 ht nhn U235 b phn r, phn ng th hai c 100x1,6

=160 ht nhn U235 ; phn ng th ba c 100 x (1,6)2 ht nhn U235 ;..... phn ng th 100 c 100x (1,6)

99

Tng s ht nhn b phn r n phn ng th 101

N = 100( 1,60 + 1,6

1 + 1,6

2 +.... +1,6

100) =

16,1

)1.6,1(100 101

= 6,88.10

22 ht . p n C


gian chiu x ln u l t = 30 pht, c sau 1 thng th bnh nhn phi ti bnh vin khm bnh v tip tc chiu x. Bit ng v phng x c chu k bn r T = 4 thng (coi t T ) v vn dng ngun phng x trong ln u. Hi ln chiu x th 3 phi tin hnh trong bao lu bnh nhn c chiu x vi cng mt lng tia nh ln u?

A. 40 pht. B. 20 pht C. 28,2pht. D. 42,42 pht

Gii:

Lng tia phng x ln u: 1 0 0(1 )

tN N e N t



ln2 ln2

'2 20 0' (1 ) '

tN N e e N e t N

Do t = 22ln

e t = 2 .30 = 42,42 pht. Chn p n D Cu 27: Mt hn hp gm hai cht phng x X v Y ban u s ht phng x ca hai cht l nh nhau. Bit chu k phng x ca hai cht ln lt l T1 v T2 vi T2 = 2T1. Sau thi gian bao lu th hn hp trn cn li mt phn hai s ht ban u? A. 1,5T2 B. 2T2 C. 3T2 D. 0,69T2

Gii: T2 = 2T1 ------> 1 = 22 Sau thi gian t s ht nhn ca X v Y cn li:

N1 = N01t

e 1

; N2 = N02t

e 2

vi N01 = N02 =2

0N ; N0 l s ht nhn ban u ca hn hp

S ht nhn cn li ca hn hp: N = N1 + N2 =N01(t

e 1 + te 2 ) =

2

0N ( te 22 + te 2 )

Gi T l khong thi s ht nhn ca hn hp gim i mt na: N = 2

0N

khi t = T th Te 22 + Te 2 =1. t Te 2 =X >0 ta c : X2 + X 1 = 0 (*)

Phng trnh (*) c nghim X = 2

51; loi nghim m X =

2

15 = 0,62

---> T

e 2 = 0,62-----> -

2T

Tln2 = ln0,62 ------> T = 0,69T2 p n D

Cu 28. xc nh lng mu trong bnh nhn ngi ta tim vo mu mt ngi mt lng nh dung dch cha ng v phng x Na24( chu k bn r 15 gi) c phng x

2Ci. Sau 7,5 gi ngi ta ly ra 1cm3 mu ngi th thy n c phng x 502 phn r/pht. Th tch mu ca ngi bng bao nhiu? A. 6,25 lt B. 6,54 lt C. 5,52 lt D. 6,00 lt

Gii: H0 = 2,10

-6.3,7.10

10 = 7,4.10

4Bq; H = 502V phn r/pht = 8,37V Bq (V th tch ca mu tnh theo cm

3 )

H = H0 2-t/T

= H0 2-0,5

-------> 2-0,5

= 0H

H =

410.4,7

37,8 V ------> 8,37 V = 7,4.10

4.2

-0,5

V = 37,8

210.4,7 5,04 = 6251,6 cm

3 = 6,25 dm

3 = 6,25 lit. Chn p n A

Cu 29: Ngi ta trn 2 ngun phng x vi nhau. Ngun phng x c hng s phng x l 1 , ngun phng x th 2 c hng s phng x l 2 . Bit 12 2 . S ht nhn ban


u ca ngun th nht gp 3 ln s ht nhn ban u ca ngun th 2. Hng s phng x ca ngun hn hp l A.

12,1 B. 15,1 C. 15,2 D. 13

GII. Gi N01 l s ht nhn ban u ca ngun phng x 1 Gi N02 l s ht nhn ban u ca ngun phng x 2. Th N02 = N01/2. Sau thi gian t s ht nhn cn li ca mi ngun l:

11 01.

tN N e v 2 .012 12 02. .3

t tNN N e e .

Tng s ht nhn cn li ca 2 ngun:

. 2011 2 1 11 2 01

1( . ) (3. )

3 3

t t t tNN N N N e e e e (1)

Khi t = T(T l chu k bn r ca hn hp) th N = (N01 +N02)=2/3 N01. (2)

T (1) v (2) ta c : . 21 13. 2t te e

t .1 te = X ta c : 2 3 2 0X X (*) Phng trnh (*) c nghim X = 0,5615528.

Do : .1 te = 0,5615528. T

1 1

1

1 1 ln 2 ln 2.ln . 1,20.

10,5615528 ln0,5615528

t TT

.

P N A

Cu 30. Ht nhn Na24 phng x vi CKBR 15 g, to thnh ht nhn X. Sau thi

gian bao lu th mt mu cht px Na24 nguyn cht lc u s c t s s nguyn t ca X v ca Na c trong mu bng 0,75 Bi gii: Theo L phng x ta c:

N = N0e-t. S nguyn t ca X c to thnh bng s nguyn t Na24 phn r

NX = N = N0 N = N0(1- e-t

)

NX/N = (1- e-t

)/ e-t

= 0,75. Suy ra et

=1,75 - t = (ln1,75/ln2) T = 0,8074T =12,1 h p s t = 12,1h

Cu 31. Mt khi cht phng x. Trong t1 gi u tin phng ra n1 tia phng x trong t2

= 2t1 gi tip theo phng ra n2 tia phng x. Bit 2 19

64n n . Chu k bn r l:

A. 16

tT B. 1

2

tT C. 1

4

tT D. 1

3

tT

Gii: Ta c n1 = N1 = N0(1- e-t

1 )

n2 = N2 = N1(1- e-t

2 ) = N0e

-t1

(1- e

-2t1

)


n1 /n2 =(1- e-t

1 )/e

-t1

(1- e

-2t1

) =(1-X)/X(1-X

2) = 1/X(1+X) Vi X = e-t1

do ta c phng trnh: X2 + X = n2/n1 =9/64 hay X2 + X 9/64 = 0. Phng btrnh c

cc nghim X1 = 0,125 v X2 = - 1,125 t = ln2,697 = 0,99214 ------> = 0,19843

= T

2ln -----> T =

2ln = 3,493 pht = 3,5 pht. p n A


gian chiu x ln u l t = 30 pht, c sau 1 thng th bnh nhn phi ti bnh vin khm bnh v tip tc chiu x. Bit ng v phng x c chu k bn r T = 4 thng (coi t T ) v vn dng ngun phng x trong ln u. Hi ln chiu x th 3 phi tin hnh trong bao


lu bnh nhn c chiu x vi cng mt lng tia nh ln u?

A. 40 pht. B. 20 pht C. 28,2pht. D. 42,42 pht

Gii:


tN N e N t



-------> 2

1

E

E =

210.235

10

6,17.2

A

A

N

N= 3,939 4 -------> E1 = 4E2 , Chn p n C

Cu 37. Mt ngi bnh phi chy thn bng phng php phng x. Ngun phng x uc s dng c chu k bn r 40T ngy. Trong ln khm u tin ngi bnh c chp trong khong thi gian 12pht. Do bnh giai on u nn trong 1 thng ngi ny 2 ln phi ti bnh vin chp c th lch hn vi bc s nh sau:

Thi gian: 08h Ngy 05/11/2012 PP iu tr: Chp phng x (BS. V Ngc Minh)

Thi gian: 08h Ngy 20/11/2012 PP iu tr: Chp phng x (BS. V Ngc Minh)

Hi ln chp th 3 ngi ny cn chp trong khong thi gian bng bao nhiu nhn c liu lng phng x nh cc ln trc: Coi rng khong thi gian chp rt nh so vi thi gian iu tr mi ln. A. 15,24pht B. 18,18pht C. 20,18pht D. 21,36pht.

Gii: Liu lng phng x mi ln chiu: tNeNNt

00 )1( Vi t = 12 pht



N = 238

mN A ; NPb = 206

PbAmN ;

Theo L phng x: N = N0 e-t

-----------------------> 238

mN A = (238

mN A +206

PbAmN )e-t

-------> et

= 206

2381

238

206238

m

m

mN

mNmN

Pb

A

PbAA

= 1,0525

--------> 0525,1ln2ln

tT

------> t = 3,3 .108 nm. Chn p n C

Cu 39 . Tnh cng cn thit tng tc mt electron t trng thi ngh n vn tc 0,50c. A.0,144m0c

2. B.0,225m0c

2. C.0,25m0c

2. D.0,5m0c

2.

Gii:

E0 = m0c2; E = E0 + W = m0c

2 +

2

2mv= m0c

2 +

2

2

0

1c

v

m

2

)5,0( 22 c

E = m0c2 +

4

3

0m

2

)5,0( 22 c = m0c

2 + 0,144m0c

2

Do A = E E0 = 0,144m0c2 Chn p n A

Cu 40 Mt ng h chuyn ng vi tc v = 0,8c. Sau 1h tnh theo ng h chuyn ng th ng h ny chy chm so vi ng h gn vi quan st vin ng yn mt lng l bao nhiu? A. 20 pht B. 30 pht C. 40 pht D. 50 pht

ng h gn vi quan st vin chuyn ng chy chm hn ng h gn vi quan st

vin ng yn t =

2

2

0

1c

v

t

t0 l khong thi gian gn vi quan st vin ng yn

Thi gian ng h chuyn ng chm hn ng h gn vi quan st vin ng yn

t = t - t0 = t0 (

2

2

1

1

c

v

- 1) = 60( 6,0

1- 1) = 60.

6,0

4,0 = 40 pht. Chn p n C

Cu 41: o chu k bn r ca 1 cht phng x - ngi ta dng my m electron. K t thi im t=0 n t1= 2 gi my m ghi dc N1 phn r/giy. n thi im t2 = 6 gi


my m dc N2 phn r/giy. Vi N2 = 2,3N1. tm chu k bn r. A 3,31 gi. B 4,71 gi C 14,92 gi D 3,95 gi

Gii: H1 = H0 (1- 1t

e ) -----> N1 = H0 (1- 1

te

) H2 = H0 (1- 2t

e ) -----> N2 = H0 (1-

2te ) -----> (1- 2te ) = 2,3(1- 1te ) ----> (1- 6e ) = 2,3 ( 1 - 2e ) t X = 2e ta c: (1

X3) = 2,3(1-X) ------> (1-X)( X

2 + X 1,3) = 0.

Do X 1 0 -----> X2 + X 1,3 = 0 -----. X = 0,745

2e = 0,745 ------> - T

2ln2 = ln0,745 ------> T = 4,709 = 4,71 h Chn p n B

Cu 42: Cho mt hat nhn khi lng A ang ng yn thi phn ra thanh 2 ht nhn c khi lng B va D ( vi B < D ). Cho tc anh sang trong chn khng la c. ng nng ca B ln hn ng nng ht D l:

A.(D B)(A B D)c^2 / (B + D) A. DB

cDBABD

2))((

B.(B + B A)(A + B D)*c^2 / (B + D) B. DB

cDBAABB

2))((

C. B(A B D)*c^2 / D C. D

cDBAB 2)(

D. D(A B D)*c^2 / B D. B

cADBD 2)(

Gii: Gi ng nng ca B v D l KB v KD

KB = 2

2

BBv ; KD = 2

2

DDv . theo L bo ton ng lng ta c BvB = DvD---> D

B

v

v=

B

D

Nng lng phn ng ta ra E = (A - B - D)c2 = KB + KD (*)

D

B

K

K =

2

2

D

B

Dv

Bv =

B

D ----->

D

DB

K

KK =

B

BD (**)

v D

DB

K

KK =

B

BD (***)

T (**) v (***) DB

DB

KK

KK

=

BD

BD

----->

KB KD = BD

BD

(KB + KD) =

BD

cDBABD

2))((. p n A


Cu 43. .mt gia nh s dng ht 1000kwh in trong mt thng. Cho tc nh sng l 3.10

8 m/s. nu c cch chuyn mt chic mng tay nng 0,1g thnh in nng th s

cho gia nh s dng trong bao lu A. 625 nm B.208 nm 4 thng C. 150 nm 2 thng D. 300 nm trn Gii: in nng gia inh s dng trong 1 thng W = 1000kWh = 3,6.109J Nng lng ngh ca 0,1g mng tay: E = mc2 = 9.1012J

Thi gian gia nh s dng t = W

mc 2 =

9

164

10/6,3

10.9.10 = 2500 thng = 208 nm 4 thng.

p n B

Cu 44:Dng p c ng nng 1K bn vo ht nhn 9

4 Be ng yn gy ra phn ng: 9 6

4 3p Be Li . Phn ng ny ta ra nng lng bng W=2,1MeV . Ht nhn 6

3 Li v ht

bay ra vi cc ng nng ln lt bng 2 3,58K MeV v 3 4K MeV . Tnh gc gia

cc hng chuyn ng ca ht v ht p (ly gn ng khi lng cc ht nhn, tnh theo n v u, bng s khi).

A. 045 . B. 090 . C. 075 . D. 0120 .

Gii; ng nng ca proton: K1 = K2 + K3 - E = 5,48 MeV

Gi P l ng lng ca mt vt; P = mv; K = 2

2mv =

m

P

2

2

P12 = 2m1K1 = 2uK1; P2

2 = 2m2K2 = 12uK2 ; P3

2 = 2m3K3 = 8uK3

P1 = P2 + P3

P22 = P1

2 + P3

2 2P1P3cos

cos = 31

2

2

2

3

2

1

2 PP

PPP =

31

231

162

1282

KK

KKK = 0

---> = 900 Chn p n B

Cu 45: Thnh phn ng v phng x C14 c trong kh quyn c chu k bn r l 5568 nm. Mi thc vt sng trn Tri t hp th cacbon di dng CO2 u cha mt lng cn bng C14. Trong mt ngi m c, ngi ta tm thy mt mnh xng nng 18g vi phng x 112 phn r/pht. Hi vt hu c ny cht cch y bao nhiu lu, bit phng x t C14 thc vt sng l 12 phn r/g.pht. A. 5734,35 nm B. 7689,87nm C. 3246,43 nm D. 5275,86 nm. Gii: phng x ca 18g thc vt sng H0 = 18.12 phn r/pht = 216 phn r/pht

Ta c H = H0 te ------> te = 0H

H = 216

112 = 27

14 -----> - t = ln

27

14

P2

P3

P1


t = - 2ln

T ln27

14 = 5275,86 nm

Cu 1: C hai mu cht phng x A v B thuc cng mt cht c chu k bn r T = 138,2 ngy v c khi 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 A nhiu hn mu B l

A. 199,8 ngy B. 199,5 ngy C. 190,4 ngy D. 189,8 ngy

Gii Ta c NA = N0 1te ; NB = N0 2

te

2 1( ) 1 2ln 2

2,72 ( ) ln 2,72t tB

A

Ne t t

N T

----- t1 t2 = ln 2,72

199,506 199,5ln 2

T

ngy

Chn p n B : 199,5 ngy


gian chiu x ln u l 20t pht, c sau 1 thng th bnh nhn phi ti bnh vin khm bnh v tip tc chiu x. Bit ng v phng x c chu k bn r T = 4 thng (coi t T ) v vn dng ngun phng x trong ln u. Hi ln chiu x th 3 phi tin hnh trong bao lu bnh nhn c chiu x vi cng mt lng tia nh ln u?

A. 28,2 pht. B. 24,2 pht. C. 40 pht. D. 20 pht.

Gii:


tN N e N t



Gii: Theo bi ra 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

Cu 4. Tim vo mu bnh nhn 10cm3 dung dch cha Na2411 c chu k bn r T = 15h

vi nng 10-3mol/lt. Sau 6h ly 10cm3 mu tm thy 1,5.10-8 mol Na24. Coi Na24 phn b u. Th tch mu ca ngi c tim khong: A. 5 lt. B. 6 lt. C. 4 lt. D. 8 lt.

Gii: S mol Na24 tim vo mu: n0 = 10-3

.10-2

=10-5

mol.

S mol Na24 cn li sau 6h: n = n0 e- t

= 10-5

. Tt

e

.2ln

= 10-5 15

6.2ln

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

Cu 5. xc nh lng mu trong bnh nhn ngi ta tim vo mu mt ngi mt lng nh dung dch cha ng v phng x Na24( chu k bn r 15 gi) c phng x

2Ci. Sau 7,5 gi ngi ta ly ra 1cm3 mu ngi th thy n c phng x 502 phn r/pht. Th tch mu ca ngi bng bao nhiu? A. 6,25 lt B. 6,54 lt C. 5,52 lt D. 6,00 lt

Gii: H0 = 2,10

-6.3,7.10

10 = 7,4.10

4Bq; H = 502V phn r/pht = 8,37V Bq (V th tch ca mu tnh theo cm

3 )

H = H0 2-t/T

= H0 2-0,5

-------> 2-0,5

= 0H

H =

410.4,7

37,8 V ------> 8,37 V = 7,4.10

4.2

-0,5

V = 37,8

210.4,7 5,04 = 6251,6 cm

3 = 6,25 dm

3 = 6,25 lit. Chn p n A

Documents

bai-tap-hat-nhan-nguyen-tu-giai-chi-tiet.pdf