Tài liệu Physics exercises_solution: Chapter 43 pdf

So if the spin magnetic moment of the proton is parallel to the magnetic field, U 0,and if they are antiparallel, U 0.. So the parallel case has lower energy.. b For electrons, the neg

81 has 81 protons and 124 neutrons.

43.2: a) Using R(1.2fm)A 3, the radii are roughly 3.6 fm, 5.3 fm, and 7.1 fm b) Using 4 R 2for each of the radii in part (a), the areas are 163 fm2,353fm2 and

7928.2(2

Hz)1027.2(s)J1063.6(2so,But

27

7 34

TeV1015245.3)(

9130.1(

b)   66.9MHz, λ 4.48m

f

c h

E

f

43.5: a) U  μBz B.N andS point in the same direction for a proton So if the spin magnetic moment of the proton is parallel to the magnetic field, U 0,and if they are antiparallel, U 0 So the parallel case has lower energy.

The frequency of an emitted photon has a transition of the protons between the two states given by:

Hz

1002.7)

sJ1063.6(

)T65.1)(

TJ10051.5)(

7928.2(2

2

7 34

E E h

E

Hz1002.7

sm1000.3

This is a radio wave

b) For electrons, the negative charge means that the argument from part (a) leads to the

m state (antiparallel) having the lowest energy, since N and point in  S

opposite directions So an emitted photon in a transition from one state to the other has a frequency

h

B h

E E h

sm1000.3λ

so

Hz1062.4

kg)1011.9(4

T)C)(1.6510

60.1)(

00232.2(4

)00232.2(

4

)00232.2(2

)00232.2(

3 10

8 10

31 19

π πm

eB f

m

e S

m

e μ

e

e

z e z

eV10224.2(

s)m10998.2)(

sJ10626.6(

19 6

8 34

43.8: a) 7(mn mH)mN 0.112u, which is 105 MeV, or 7.48 MeV per nucleon b) Similarly, 2(mH mn)mHe 0.03038u 28.3MeV,or7.07MeVper nucleon, slightly lower (compare to Fig (43.2)).

43.9: a) For 11B

5 the mass defect is:

5 e

MeVu)(931.5081815

.0(energy binding

The

u0.081815

u009305

11)u000549

0(5)u008665

1(6)u007277

Z Z C A C A C E

2 4

3 2

1 B

)2()1(

11

)4(5)MeV(0.7100MeV)(11)

80.17(11)MeV75.15(

 E

)11(

)1011()MeV69.23(

2



  EB 76.68MeV

So the percentage difference is 100 0.62%

MeV76.21

MeV76.21MeV

43.10: a) 34mn 29mH mCu 34(1.008665)u29(1.007825)u62.929601u 

andu

MeV931.5(usingnucleon per

MeV8.75orMeV,551iswhich

)63(

)28)(

29()MeV(0.7100MeV)(63)

80.17()63)(

MeV75.15(





(The fifth term is zero since the number of neutrons is even while the number of protons

is odd, making the pairing term zero.)

This result differs from the binding energy found from the mass deficit by 0.86%,

a very good agreement comparable to that found in Example 43.4

43.11: Z is a magic number of the elements helium (Z = 2), oxygen (Z = 8), calcium (Z =

20), nickel (Z =28), tin (Z = 50) and lead (Z = 82) The elements are especially stable,

with large energy jumps to the next allowed energy level The binding energy for these elements is also large The protons’ net magnetic moments are zero

43.12: a) 146mn 92mH mU 1.93u,which is

b) 1.80103 MeV,orc)7.56MeVper nucleon (using 931.5MeVu and 238 nucleons)

43.13: a) αdecay:Z decreases by 2,Adecreasesby 4

b) βdecay:Z decreases by 1,Aremains thesame:

c)  decay:Z decreases by 1,Aremains thesame:

43.14: a)The energy released is the energy equivalent of 8.40 10 4 u,

e p

m

,b)

uMeVu)(931.510

8.1(So:u101.68

0.0005491u))

7(0.000549u

(14.003074u))

6(0.000549u

C

(

4 4

e

14 7

M

m

43.18: a) As in the example, (0.000898 u)(931.5 MeV u)0.836MeV

b) 0.836MeV0.122MeV0.014MeV0.700MeV

43.19: a) If tritium is to be unstable with respect to  decay, then the mass of the 

products of the decay must be less than the parent nucleus

u3.014932u)

582(0.000548u

3.016029)

He(

u3.015500u

0.00054858u

016049

3)H(

2 3 2

3 1

e

3 2

so the decay is possible

b) The energy of the products is just

keV

19MeV0.019u)

MeV5.931)(

u100.2

E

43.20: Note that Eq 43.17 can be written as follows

2 / 2

0

T t

2(ln

A A T

X has 39 protons and 90 protons plus neutrons, so it must be Y.90

(b) Use base 2 because we know the half life

yr1902

log

01.0log)yr28(

2log

01.0log

201

.0

2

2

0 0

0

2 2

A A

A A

T t

T t

43.23: a) H e 3He

2

0 1

3

0e ,N 0.100N andλ (ln2) T N

)100.0

43.24: a) 500Ci(50010 6)(3.701010s 1)

dt dN

sdec1085



dt dN

s106.69d)

s(86,40012

2ln2

lnλλ

2

2 2

T

nuclei10

77.2s

1069.6

s/dec1085.1λ

1 7

1066.1131(nuclei10

77

400,86

ds1029.9

s1069.6

)5001(lnλ

)5001(ln

λ)500/1(ln

)Ci500(Ci1

6

1 7 λ

e μ

43.25: ( ln 2 / 1 / 2

0

λ 0

T t

e A

(ln 2) ln( 0)

2

A A

T t 



days80.2)83183091(ln

)days00.4)(

2ln()(ln

)2ln(

0

A A t T

s1036.1λ

yr1

s1015.3yr1620

2ln2

lnλ

665.2g

226

atoms10

022.6g

s

dec1062.3)s1036.1)(

10665.2(λ

10

10 1

11 25

Ci1Bq

1062

10807.1λ2)

ln

(

)s(10836.3)(

λ

;λ

sdecays3.00

mindecays180

atoms14-carbon10

82.7)10016.6()103.1(soatoms,10

016

6

molkg10(12.011mol)

atoms10

022.6()kg100.12(

;

11 2

/

1

1 12

11 23

12 23

tot

3 23

3 A

yeardays365()y27.5(

2ln2

g1060.3u

17 24

)60()yrs27.5(Ci0408.0λ

Ra 1/2Ra

Co 1/2Co Co

A T N N

43.29: a) (0) 7.561011 Bq7.561011decays s

dt dN

and

min)s60(min)8.30(

693.0693

.0

103.75

sdecays10

7.56)

0(λ

1)

0

1 4

b) The number of nuclei left after one half-life is 1.01 1015

2

)0(





N

nuclei, and the

activity is half: 3.781011decays s

dt dN

c) After three half lives (92.4 minutes) there is an eighth of the original amount

14

1053

45

43.30: The activity of the sample is 102Bq kg,

kg)(0.500min)

sec(60

mindecays

activity of atmospheric carbon is 255 Bq kg (see Example 43.9) The age of the sample

is then

.y7573y

1021.1

)255102(lnλ

)255102(ln

693.0693

.0

7 9

kg1063

Bq421

decays10

43.33: a) 4.91 10 s

)ys10156.3(y1047.4(

693.0693

.0

7 9

decays10

70.3)Ci1020.1(Ci1020

sdecays10

44.41

N dt

dN





N m

N 9.041022 nuclei (238u)



kg

0357.0)10kg)(9.0410

66.1)(

1 18

23 27

107.46nuclei)10

52.1(s1091.4(λ

nuclei10

52.1)kg1066.1(238

kg0600.0

dN

N

alpha particles emitted each second

43.34: (a) rem = rad  RBE

43.35: 1 rad = 102 Gy, so 1 Gy = 100 rad and the dose was 500 rad

rem=(rad) (RBE) = (500 rad) (4.0) = 2000 rem

kgJ5.0so,kgJ1Gy

43.36: a) 5.4Sv(100rem Sv)540rem b) The RBE of 1 gives an absorbed dose of

540 rad c) The absorbed dose is 5.4 Gy, so the total energy absorbed is (5.4Gy)(65kg)J

351 The energy required to raise the temperature of 65 kg0.010o C is (65kg)(4190

kJ

3C)(0.01

K)

kg

43.37: a) We need to know how many decays per second occur.

.s1079.1)ys10156.3()y3.12(

693.0693

0

7 2

)CiBq10(3.70Ci)(0.35λ

1)0

18

102540

s 3600 ( 24 ( 7 ( s 10 79 1 ( 18

)0()

43.38: a) From Table (43.3), the absorbed dose is 0.0900 rad b) The energy absorbed

is (9.0010 4J kg)(0.150kg)1.3510 4J; each proton has energy 1.28210 13J, so the number absorbed is 1.05109c) The RBE for alpha particles is twice that for

protons, so only half as many, 5.27108, would be absorbed

43.39: a)

m102.00

)sm10(3.00s)J10(6.63)10(6.50

11

8 34

rad100)kgJ10(1.08kg

0.600

J106.46

The rem dose for x rays (RBE = 1) is just 0.108 rem

43.40: (0.7210 6Ci)(3.71010Bq Ci)(3.156107s)8.411011 α particles The absorbed dose is

rad

108Gy1.08kg)

(0.50

)eVJ10(1.602eV)10(4.0)10

43.41: a) H Be X4 He So29 4 7and14 2

2

9 4

9 4

u002603

4u016003

7u012182

9u014102

m)102.1(

m101.5(2)

m)102.1(

15 3

15 9

15 3

15 2

C)1060.1(44

15 0

2 19 Be

H 0

q q πε U

MeV

1.4eV101.4 6 



 U

H He H

2

M

u,010882

0,

u012937

10)B(u,

M

reaction  Q(0.010882u)(931.5MeV u)10.14MeV

m M

M K

u01.2u0.14

43.44:(200106 eV)(1.60210 19J eV)(6.0231023molecules mol)1.93

,

mol

J

1013 which is far higher than typical heats of combustion

43.45: The mass defect is ( U) (236U*)

92 n

0

u045562

236u008665

1u043923

)uMeV(931.5u)

(0.007025)

43 46: a) Z 3205and A47110 b) The nuclide is a boron nucleus,

B n Li

43.47: The energy liberated will be

MeV

586.1

)vMeV(931.5

v)7.016929v

4.002603v

(3.016029Be)

()He(He)

4

4 2

Z z A Z

Add the mass of Z electrons to each side and we find:     

 Y)(X)

2

A Z

A

M m

43.50: Denote the reaction as

.Y

The mass defect is related to the change in the neutral atomic masses by

),(

])1([][mX Zme  mY  Z me me  mX mY

where m and X m are the masses as tabulated in, for instance, Table (43.2).Y

YX

e 1

e 1

e

e 1

m M

M

m M

m M

m M

M

m

A Z

A

Z

A Z

A

Z

A Z

])1([]

[mXZme me mY  Z  me  mX mY

where mX and mY are the masses as tabulated in, for instance, Table (43.2)

43.53: a) Only the heavier one (25 Al)

13 can decay into the lighter one (25 Mg)

)u00054858

0(2u985837

24u990429

24

)12)Mg(()13)Al((

3

e e

25 12 e

25 13

u985837

24u990429

24Mg)()Al

(

3

25 12

10592.4()

82 210

alpha particle is (206 210)times this, or 5.30MeV(see Example 43.5)

H Bi

83 210

n Po

84 210

d) At 210Po,

84 210

m  so the decay is not possible (see Problem (43.50))

e) Bi 2 e 210Po,

84 210

ZM M

A Z

%

022.010023.990963

23.985823.990963

error

%

u9858.23uMeV931.5

MeV)(198.31u)

.00866513(1

u)511(1.00782Na)

(

MeV

31.198(24)

MeV)39(24

))11(224()

.13

24

11

3 2

3

3 B

23

990963

231987

43.56: The -particle will have

43.57: Au Hg    ( Au) (198Hg)197.968225u197.966752u

80

198 79

198 80

198

MeV

372.1uMeV(931.5

u)

10

473.1()

( wasavailableenergy

totaltheAndu

10

1.473

3

2 3

5 11 6

u)582(0.00054813.003355

u005739

13

2C)()N(

3

e

13 6

13 7

Therefore the decay is possible because the initial mass is greater than the final mass

43.60: a) A least-squares fit to log of the activity vs time gives a slope of

,hr

Bq)1000.2

N dt

t dN dt

t dN t A

N dt

t dN e

N t

0

λ 0

)0(λ

)()

.)

λ

)(

t t

e N t N

2

1)

(So

2 0

) 2

1 ( ln 0 2)

(ln 0

2 2

T

t n N

t N e

N e

N

n T

t

T t

43.63: 19 1

7 10

2

s1062.4)ys10156.3(y1075.4(

693.0693

0 y) s 10 156 3 ( y 10 6 4 ( s 10 62 4 ( 0 0

87 0

0 87

0694.1

7 9

1 19 87

87

87 87

N N

e N N

e N N e

)2783.01(

2783.02783

N N N

.2920.0))3856.0(0694.11(

)3856.0(0694.1So

.3856

0

87 85

87 85

0 0

So the original percentage of 87 0 85

85

(29%

2

T A

m N

d)

mCi

Bq1017.1

1025

about an hour and a half Note that this time is

so small in comparison with the half-life that the decrease in activity of the source may be neglected

43.65: a) 2.610 4 Ci(3.701010 decays sCi)9.6106 decays s

dt

dN

so in one second there is an energy delivered of

s

J1062

9

)eVJ10(1.60eV)10(1.25s)00.1(s106.9(2

12

1

7

19 6

1 6

b) Absorbed dose

kg500.0

sJ106

9   7





m E

1.9 10 rad

skgJ

rad100skgJ109

srem10

2

2 ( 240 ) 261.9 1221.2

9 26

240

2 122

13000y

693.0

5730)21.0ln(

λ

)21.0

8

(

Ci)Bq10(3.70Ci)102

C)1060.1(4

3 15

0

2 19 2

u103.51

u1.008665u

3.016029u)

014102

2(2

3 2

5.23110 13 J c) A mole of deuterium has 6.0221023 molecules, so the energy per mole is

J

103.150J)

10231.5(10

7 16

8 16

kg1052

054.210

28.1

)10793.3(693.0(

dN

So in 50 years the energy absorbed is: E (0.50MeV decay)(50y)(2.054109 decay y)

J

108.22MeV

43.72: In terms of the number Nof cesium atoms that decay in one week and the mass

Sv)(3.5kg)0.1(

soJ),10283.2(

MeV))(0.51

(1.5)MeV)

66.0(1((

)E(RBE)E

RBE)((

13

e e

N

γ γ

The number N0 of atoms present is related to by λ,so

N

.10536.1)

10

535.1(sec

1030.7)yrsec10(3.156yr)

13 days)

sec 10 (8.64 days) 7 ( sec 10

4 1

m v

M M

m

m v v

2 2

2 2

2

2 2

2

)(

21

)(

2

1)

(2

12

12

1

v M m

m M m

mM M

m M

v M m

Mm v

M m

mM v

M v

M K

mv M

M

m M K

K m M

M K

140 54

u1.008665u

93.915360u

139.921636u

230439.235

43.76: a) A least-squares fit of the log of the activity vs time for the times later than 4.0 hr

gives a fit with correlation 12106 and decay constant of 0.361hr1, corresponding

to a half-life of 1.92 hr Extrapolating this back to time 0 gives a contribution to the rate of

about 2500/s for this longer-lived species A least-squares fit of the log of the activity vs

time for times earlier than 2.0 hr gives a fit with correlation = 0.994, indicating the

presence of only two species

b) By trial and error, the data is fit by a decay rate modeled by

5000Bqe t 1 733 hr  2500Bqe t 0 361 hr 

This would correspond to half-lives of 0.400 hr and 1.92 hr

c) In this model, there are 1.04107 of the shorter-lived species and 2.49107 of the longer-lived species

d) After 5.0 hr, there would be 1.80103 of the shorter-lived species and 4.10106

of the longer-lived species

43.77: (a) There are two processes occurring: the creation of 128I by the neutron

irradiation, and they decay of the newly produced I128 So K N K

dt

dN

whereλ

K

N d

λlnλ

ln

λλ

t K N

K

t N

at various times is:

Bq;

105.1min)180(Bq;

103.1min)75(

Bq;

101.1min)50(Bq;

105.7min)25(

Bq;

106.3min)10(Bq;

101.4min)1(

6 6

6 5

5 4

9 6

d) The maximum activity is at saturation, when the rate being produced equals that decaying and so it equals 1.5106 decays s

43.78: The activity of the original iron, after 1000 hours of operation, would be

Bq108306.12

)CiBq103.7(Ci)

The activity of the oil is 84 Bq, or 4.5886104 of the total iron activity, and this must

be the fraction of the mass worn, or mass of 4.5910 2 g The rate at which the piston rings lost their mass is then 4.5910 5 g hr