Nuclear Energy and Elementary Particles

596

Chapter 30

Nuclear Energy and Elementary Particles

Problem Solutions

30.1 The number of fissions required is

10

20

13

3.30 10 J 1 MeV9.92 10

208 MeV 1.60 10 J

EN

Q

The mass of a single 235U atom is 27 25

atom (235 u)(1.66 10 kg u) 3.90 10 kgm , so the

total mass of 235U required is

20 25 4

total atom 9.92 10 3.90 10 kg 3.87 10 kg 0.387 gm Nm

30.2 The energy released in the reaction 1 235 98 135 1

0 92 40 52 0n U Zr + Te + 3 n is

235 98 13592 40 52

2 2

nU Zr Te2

235.043 923 u 2 1.008 665 u 97.912 0 u 134.908 7 u 931.5 MeV u

192 MeV

Q m c m m m m c

30.3 The energy released in the reaction 1 235 88 136 1

0 92 38 54 0n U Sr + Xe + 12 n is

235 88 136 92 38 54

2 2

nU Sr Xe11

235.043 923 u 11 1.008 665 u 87.905 614 u 135.907 220 u 931.5 MeV u

126 MeV

Q m c m m m m c

Nuclear Energy and Elementary Particles 619

30.4 Three d ifferent fission reactions are possible:

1 235 90 144 1

0 92 38 54 0n U Sr Xe 2 n 144

54 Xe

1 235 90 143 1

0 92 38 54 0n U Sr Xe 3 n 143

54 Xe

1 235 90 142 1

0 92 38 54 0n U Sr Xe 4 n 142

54 Xe

30.5 (a) With a specific gravity of 4.00, the density of soil is 34 000 kg m . Thus, the mass

of the top 1.00 m of soil is

2

2 7

3

kg 1 m4 000 1.00 m 43560 ft 1.62 10 kg

3.281 ftmm V

At a rate of 1 part per million, the mass of uranium in this soil is then

7

U 6 6

1.62 10 kg16.2 kg

10 10

mm

(b) Since 0.720% of naturally occurring uranium is 235

92 U , the mass of 235

92 U in the soil of

part (a) is

23592

3 3

UU7.20 10 7.20 10 16.2 kg 0.117 kg 117 gm m

30.6 At 40.0% efficiency, the useful energy obtained per fission event is

13 11

event 0.400 200 MeV event 1.60 10 J MeV 1.28 10 J eventE

The number of fission events required each day is then

9 4

24

11

event

1.00 10 J s 8.64 10 s d6.75 10 events d

1.28 10 J event

tN

E

P

Each fission event consumes one 235

92 U atom. The mass of this number of 235

92 U atoms is

24 27

atom 6.75 10 events d 235 u 1.66 10 kg u 2.63 kg dm Nm

In contrast, a coal-burning steam plant producing the same electrical power uses more

than 66 10 kg d of coal.

618 CHAPTER 30

30.7 The mass of 235 U in 1.0-kg of fuel is 0.017 kg, and the num ber of 235 U nuclei is

22

27atom

0.017 kg4.36 10

235 u 1.66 10 kg u

mN

m

At 208 MeV per fission event and 20% efficiency, the useful energy available from this

number of fission events is

22 13 114.36 10 events 208 MeV event 1.60 10 J Mev 0.20 2.9 10 JE

From dragWork F s E , the d istance the ship can travel on this 1.0-kg of fuel is

11

6 3

5

drag

2.9 10 J2.9 10 m 2.9 10 km

1.0 10 N

Es

F (or about 1800 miles)

30.8 (a) The mass of 235U in the reserve is

235

6 3 3 10

U

0.70 kg g4.4 10 metric ton 10 10 3.1 10 g

100 ton kgm

(b) The number of moles in the quantity of 235U found above is

10

83.1 10 g1.3 10 mol

235 g mol

mn

M

The number of 235U atoms in this reserve is

8 23 311.3 10 mol 6.02 10 atoms mol 7.8 10 atomsAN nN

(c) Assuming all atoms undergo fission and all released energy captured , the total

energy available is

13

31 21MeV MeV 1.6 10 J208 7.8 10 events 208 2.6 10 J

event event 1 MeVE N

(d) At a consumption rate of 131.5 10 J s , the maximum time this energy supply could

last is

21

13 7

2.6 10 J 1 yr5.5 yr

1.5 10 J s 3.156 10 s

EtP


(e) Fission is not sufficient to supply the world with energy at a price of $130 or less per

kilogram of uranium.

30.9 The total energy required for one year is

6 102 000 kWh month 3.60 10 J kWh 12.0 months 8.64 10 JE

The number of fission events needed will be

10

21

13event

8.64 10 J2.60 10

208 MeV 1.60 10 J MeV

EN

E

and the mass of this number of 235 U atoms is

21 27

atom 2.60 10 235 u 1.66 10 kg um Nm

31.01 10 kg 1.01 g

30.10 (a) At a concentration of 3 3 33 mg m 3 10 g mc , the mass of uranium dissolved in

the oceans covering two-thirds of Earth’ s surface to an average depth of av 4 kmh

is 22 2U av av3 3

( ) [ (4 )]Em cV c A h c R h , or

2

3 6 3 15

U 3

g 23 10 4 6.38 10 m 4 10 m 4 10 g

3mm

(b) Fissionable 235U makes up 0.7% of the mass of uranium computed above. If we

assume all of the 235U is collected and caused to undergo fission, with the release of

about 200 MeV per event, the potential energy supply is

235

atom

235 U0.7number of U atoms 200 MeV 200 MeV

100U

mE

m

and at a consumption rate of 131.5 10 J sP , the time this could supply the

world’ s energy needs is t E P , or

618 CHAPTER 30

235atom

U

15 13

3 13 727

3

200 MeV0.7

100

0.7 4 10 g 1 kg 200 MeV 1.6 10 J 1 yr

100 1 MeV10 g 1.5 10 J s 3.156 10 s235 u 1.66 10 kg u

5 10 yr

U

mt

m P

(c) The uranium comes fro d issolving rock and minerals. Rivers carry such solutes into

the oceans, so the ocean’ s supply of uranium is steadily replenished. Further, if

breeder reactors are used , the current ocean supply can last half a million years!

30.11 (a) 4 4 8

2 2 4He He Be

(b) 8 4 12

4 2 6Be He C

(c) The total energy released in this pair of fusion reactions is

4 12

2 2

He C3Q m c m m c

3 4.002 602 u 12.000 000 u 931.5 MeV u 7.27 MeV

30.12 The energy released in the reaction 1 2 3

1 1 2H H He is

1 2 31 1 2

2 2

H H He

1.007 825 u 2.014 102 u 3.016 029 u 931.5 MeV u 5.49 MeV

Q m c m m m c

30.13 The energy released in the reaction 2 3 4 1

1 1 2 0H H He n is

2 3 41 1 2

2 2

nH H HeQ m c m m m m c

2.014102 u 3.016 049 u 4.002 603u 1.008 665 u 931.5 MeV u

13 1217.6 MeV 1.60 10 J MeV 2.82 10 J

The total energy required for the year is

6 102 000 kWh month 12.0 months yr 3.60 10 J kWh 8.64 10 J yrE


so the number of fusion events needed for the year is

10

22

12

8.64 10 J yr3.06 10 events yr

2.82 10 J event

EN

Q

30.14 (a) 1 12 13

1 6 7H C N (b) 13 13 0

7 6 1N C e

(c) 13 1 14

6 1 7C H N (d) 14 1 15

7 1 8N H O

(e) 15 15 0

8 7 1O N e (f) 15 1 12 4

7 1 6 2N H C He

30.15 With the deuteron and triton at rest, the total momentum before reaction is zero. To

conserve momentum, the neutron and the alpha particle must move in opposite

d irections with equal magnitude momenta after reaction, or np p . Neglecting

relativistic effects, we use the classical relationship between momentum and kinetic

energy , 2 2KE p m , and write n n2 2m KE m KE , or

n n( )KE m m KE .

To conserve energy, it is necessary that the kinetic energies of the reaction products

satisfy the relation n 17.6 MeVKE KE Q . Then, using the result from above, we

have n n n( ) 17.6 MeVKE m m KE , or the kinetic energy of the emerging neutron must

be

n

17.6 MeV14.1 MeV

1.008 665 u1

4.002 603 u

KE

30.16 (a) The energy released in the reaction 1 11 4

1 5 4H B 3( He) is

1 11 41 5 2

2 2

H B He3

1.007 825 u 11.009 306 u 3 4.002 603 u 931.5 MeV u

8.68 MeV

Q m c m m m c

(b) The proton and the boron nucleus both have positive charges. Thus, they must have

enough kinetic energy to overcome the repulsive Coulomb force one exerts on the other

and approach each other very closely.

618 CHAPTER 30

30.17 Note that pair production cannot occur in a vacuum. It must occur in the presence of a

massive particle which can absorb at least some of the momentum of the photon and

allow total linear momentum to be conserved.

When a particle-antiparticle pair is produced by a photon having the minimum possible

frequency, and hence minimum possible energy, the nearby massive particle absorbs all

the momentum of the photon, allowing both components of the particle-antiparticle pair

to be left at rest. In such an event, the total kinetic energy afterwards is essentially zero,

and the photon need only supply the total rest energy of the pair produced.

The minimum photon energy required to produce a proton -antiproton pair is

13 10

photon proton2 2 938.3 MeV 1.60 10 J MeV 3.00 10 JRE E

Thus, 10

photon 23

34

3.00 10 J4.52 10 Hz

6.63 10 J s

Ef

h

and 8

16

23

3.00 10 m s6.64 10 m 0.664 fm

4.52 10 Hz

c

f

30.18 The total kinetic after the pair production is

3

total photon proton2 2.09 10 MeV 2 938.3 MeV 213 MeVRKE E E

The kinetic energy of the antiproton is then

total 213 MeV 95.0 MeV 118 MeVp pKE KE KE

30.19 The total rest energy of the 0 is converted into kinetic energy of the photons. Since the

total momentum was zero before the decay, the two photons must go in opposite

d irections with equal magnitude momenta (and hence equal energies). Thus, the rest

energy of the 0 is split equally between the two photons, giving for each photon

0,

photon

135 MeV67.5 Mev

2 2

RE

E

photon

photon 67.5 MeVE

p cc

and

13

photon 22

34

67.5 MeV 1.60 10 J1.63 10 Hz

1 MeV6.63 10 J s

Ef

h


30.20 Observe that the given reactions involve only mesons and baryons. With no leptons

before or after the reactions, we do not have to consider the conservation law s

concerning the various lepton numbers. All interactions always conserve both charge

and baryon numbers. The strong interaction also conserves strangeness. Conservation

of strangeness may be violated by the weak interaction but never by more than one un it.

With these facts in mind consider the given interactions:

0K

0K total before total after

Charge 0 0 +1 1 0

Baryon number 0 0 0 0 0

Strangeness +1 +1 0 0 0

This reaction conserves both charge and baryon number, but does violate strangeness by

one unit. Thus, it can occur via the weak interaction but not other interactions.

0

0 total before total after

Charge 0 0 +1 1 0

Baryon number +1 +1 0 0 0

Strangeness 1 1 0 0 0

This reaction fails to conserve baryon number and cannot occur via any interaction.

30.21

Reaction Conservation Law Violated

(a) p p e : 0 0 0 1 ; and : 0 0 1 0eL L

(b) p p Charge, : 1 1 1 1Q

(c) p p p Baryon Number, : 1 1 1 0B

(d) p p p p n Baryon Number, : 1 1 1 1 1B

(e) 0p n Charge, : 0 1 0 0Q

30.22 The relevant conservation laws are 0eL , 0L , and 0L .

(a) 0 e ? 0 0 0 1e eL L , so 1eL e

(b) ? p p 0 0 1 0 0L L , so 1L

618 CHAPTER 30

(c) 0 p ? 0 0 0 1L L , so 1L

(d) ? ? 0 0 1L L , so 1L

0 1 0L L , so 1L

One particle must be with 1L , while the other is with 1L

30.23 ? p n

Conservation of charge 0Q e e or 0Q

Conservation of Baryon number 1 1 0B or 0B

Conservation of electron-lepton number 0 0 0eL or 0eL

Conservation of muon-lepton number 0 0 1L or 1L

Conservation of tau -lepton number 0 0 0L or 0L

The particle having these properties is an antimuon -neutrino. It is the .

30.24 (a) p K

Baryon number, B : 0 1 0 1 0B

Charge, Q : 1 1 1 1 0Q

p

Baryon number, B : 0 1 0 1 0B

Charge, Q : 1 1 1 1 0Q


(b) Strangeness is conserved in the first reaction:

Strangeness, S : 0 0 1 1 0S

The second reaction does not conserve strangeness:

Strangeness, S : 0 0 0 1 1S

The second reaction cannot occur via the strong or electromagnetic interactions .

(c) If one of the neutral Kaons were also produced in the second reaction, giving 0p K , then strangeness would no longer be violated:

Strangeness, S : 0 0 0 1+1 0S

Because the total mass of the product particles in this reaction would be greater

than that in the first reaction [see part (a)], the total incident energy of the reacting

particles would have to be greater for this reaction that for the first reaction .

30.25 (a) 0

Charge: 1 0 1 0 0Q

Baryon number: 1 1 0 0 0B

Lepton numbers, :eL 0 0 0 0 0eL

:L 0 0 1 1 0L

:L 0 0 0 0 0L

Strangeness: 2 1 0 0 0S

618 CHAPTER 30

(b) 0 0K 2

Charge: 0 0 0 0Q

Baryon number: 0 0 0 0B

Lepton numbers, :eL 0 0 0 0eL

:L 0 0 0 0L

:L 0 0 0 0L

Strangeness: 1 0 0 0S

(c) 0K p n

Charge: 1 1 0 0 0Q



:L 0 0 0 0 0L

:L 0 0 0 0 0L



(d) 0 0

Charge: 0 0 0 0Q

Baryon number: 1 1 0 0B

Lepton numbers, :eL 0 0 0 0eL

:L 0 0 0 0L

:L 0 0 0 0L

Strangeness: 1 1 0 0S

(e) e e

Charge: 1 1 1 1 0Q



:L 0 0 1 1 0L

:L 0 0 0 0 0L


618 CHAPTER 30

(f) 0p n

Charge: 1 0 0 1 0Q



:L 0 0 0 0 0L

:L 0 0 0 0 0L


30.26

proton u u d total neutron u d d total

strangeness 0 0 0 0 0 strangeness 0 0 0 0 0

baryon

number 1

1

3

1

3

1

3 1

baryon

number 1

1

3

1

3

1

3 1

charge e 2 3e 2 3e 3e e charge 0 2 3e 3e 3e 0


30.27 The number of water molecules in one liter ( mass 1 000 g ) of water is

23 251000 g

6.02 10 molecules mol 3.34 10 molecules18.0 g mol

N

Each molecule contains 10 protons, 10 electrons, and 8 neutrons. Thus, there are

2610 3.34 10 electronseN N , 2610 3.34 10 protonspN N ,

and 268 2.68 10 neutronsnN N

Each proton contains 2 up quarks and 1 down quark, while each neutron has 1 up quark

and 2 down quarks. Therefore, there are

262 9.36 10 up quarksu p nN N N , and

262 8.70 10 down quarksd p nN N N

30.28

0K Particle 0 Particle

d s total u d s total

strangeness 1 0 1 1 strangeness 1 0 0 1 1

baryon number 0 1 3 1 3 0 baryon number 1 1 3 1 3 1 3 1

charge 0 3e 3e 0 charge 0 2 3e 3e 3e 0

30.29 Compare the given quark states to the entries in Table 30.4:

(a) suu (b) ud

(c) 0sd K (d) ssd

30.30 (a) uud : 2 2 1

charge3 3 3

e e e e . This is the antiproton .

(b) udd : 2 1 1

charge 03 3 3

e e e . This is the antineutron .

618 CHAPTER 30

30.31 The reaction is 0 p X , or on the quark level, uds uud uus 0 ?

The left side has a net 3u, 2d, and 1s . The right side has 2u, 0d, and 1s , leaving

1 and 2u d missing.

The unknown particle is a neutron, udd . Baryon and strangeness numbers are conserved.

30.32 (a) 0p is forbidden by conservation of charge

(b) e is forbidden by conservation of electron-lepton number

conservation of energy , and conservation of muon-lepton number

(c) p is forbidden by conservation of baryon number

30.33 To the reaction for nuclei, 1 3 4 0

1 2 2 1H He He e e , we add three electrons to both

sides to obtain 1 3 4 0 0

1 atom 2 atom 2 atom 1 1H He He e e e . Then we use the masses of the

neutral atoms to compute

1 3 41 2 2

2 2

eH He He2Q m c m m m m c

1.007 825 u 3.016 029 u 4.002 603 u 2 0.000 549 u 931.5 MeV u

18.8 MeV

30.34 For the particle reaction, 2e , the lepton numbers before the event are 1L

and 1eL . These values must be conserved by the reaction so one of the emerging

neutrinos must have 1L while the other has 1eL . The emerging particle are

and e

30.35 (a) e Violates conservation of muon-lepton number

and also conservation of electron-lepton number

(b) en p e Violates conservation of electron-lepton number


(c) 0 0p Violates conservation of charge

(d) 0p e Violates conservation of electron-lepton number

and also conservation of baryon number

(e) 0 0n Violates conservation of strangeness number by 2 units

Even weak interactions only violate this rule by 1 unit.

30.36 Assuming a head -on collision, the total momentum is zero both before and after the

reaction p p p X . Therefore, since the proton and the pion are at rest after

reaction, particle X must also be left at rest.

Particle X must be a neutral baryon in order to conserve charge and baryon number in

the reaction. The rest energy this particle is

0 0 0 00 02 70.4 MeV 140.8 MeVX p p pE E E E E E

or 0 938.3 MeV 139.6 MeV 140.8 MeV 939.5 MeVXE

Particle X is a neutron .

30.37 If a neutron starts with kinetic energy 2.0 MeViKE and loses one half of its kinetic

energy in each collision with a moderator atom, its kinetic energy after n collisions will

be 2n

f iKE KE .

The kinetic energy associated with particle in a gas at temperature 20.0 C 293 KT

(see Chapter 10 of the textbook) is

23

19

3 3 1 eV1.38 10 J K 293 K 0.0379 eV

2 2 1.60 10 Jf BKE k T

Thus, the number of collisions the neutron must make before it reaches the energy

associated with a room temperature gas is ln2 ln( )i fn KE KE , or

61 2.0 10 eV

ln 26ln 2 0.0379 eV

n

618 CHAPTER 30

30.38 (a) The number of deuterons in one kilogram of deuterium is

26

27atom

1 kg3.00 10

2.01 u 1.66 10 kg u

mN

m

Each occurrence of the reaction 2 2 3 1

1 1 2 0D D He n consumes two deuterons and

releases 3.27 MeVQ of energy. The total energy available from the one kilogram

of deuterium is then

26 13

133.00 10 1.60 10 J3.27 Mev 7.85 10 J

2 2 1 MeV

NE Q

(b) At a rate of eight cents per kilowatt-hour, the value of this energy is

13 6

6

$0.08 1 kWh7.85 10 J $1.74 10 $1740 000

kWh 3.60 10 Jvalue E rate

(c) Deuterium makes up four-twentieths or one-fifth of the mass of a heavy water

molecule. Thus, five kilograms of heavy water is necessary to obtain one kilogram

of deuterium. The cost for this water (5 kg)($300 kg) $1500 .

(d) Whether it would be cost-effective depends on how much it cost to fuse the

deuterium and how much net energy was produced. If the cost is nine-tenths of the

value of the energy produced, each kilogram of deuterium would still yield a profit

of $174 000 .


30.39 (a) The number of moles in 1.0 gal of water is

3 3

3

2

g 3.786 L 10 cm1.0 1.0 gal

1 gal 1 Lcm2.1 10 mol

18 g mol

m Vn

M M

so the number of hydrogen atoms (2 per water molecule) will be

2 23 1 26

H 2 2 2.1 10 mol 6.02 10 mol 2.5 10AN nN

and one of every 6 500 of these is a deuteron. Thus, the number of number of

deuterons contained in 1.0 gal of water is

26 22

D H 6 500 2.5 10 6 500 3.9 10 deuteronsN N

and the available energy is

22 13 103.9 10 deuterons 1.64 MeV deuteron 1.6 10 J MeV 1.0 10 JE

(b) At a consumption rate of 41.0 10 J sP , the time that this could supply a

person’ s energy needs is

10

4

1.0 10 J 1 d12 d

86 400 s1.0 10 J s

EtP

30.40 The total energy input required is required

input

E tE

efficiency efficiency

P

or

3

15

input

100 000 10 J s 100 d 86 400 s d2.9 10 J

0.30E

At 208 MeV per fission event, the number of fissions of 235 U nuclei needed is

15

input 25

-13

2.9 10 J 1 MeV8.7 10

208 Mev 208 Mev 1.60 10 J

EN

The mass of this number of 235 U atoms will be

25

23 3

8.7 10 atoms g 1 kg235 34 kg

mol6.02 10 atoms mol 10 gA

Nm M

N

618 CHAPTER 30

30.41 Conserving energy in the decay , with the initially at rest gives

,RE E E , or

139.6 MeVE E [1]

Since the total momentum was zero before the decay, conservation of momentum

requires the muon and antineutrino go off in opposite d irections with equal magnitude

momenta, or p p . The relativistic relation between total energy and momentum of a

particle (Equation 26.10 in the textbook) then gives for the antineutrino: E p c , or

p E c . Applying the same equation to the muon, we obtain

2 22 2 2 2 2

, , ,R R RE p c E p c E E E

or

22 2 2

, 105.7 MeVRE E E E E E E

and

2105.7 MeV

E EE E

[2]

Substituting Equation [1] into [2] gives 2(105.7 MeV) 139.6 MeVE E , or

80.0 MeVE E [3]

Subtracting Equation [3] from Equation [1] yields 2 59.6 MeVE , and

29.8 MeVE

30.42 The reaction 0 0p K is ud uud ds uds at the quark level. There is

a net 1u and 2d quarks both before and after the reaction . This reaction conserves the net

number of each type quark.

For the reaction 0p K n , or ud uud ds udd , there is a net

1u and 2d before the reaction and 1u, 3d, and 1 antistrange quark afterwards . This reaction

does not conserve the net number of each type quark.

30.43 To compute the energy released in each occurrence of the reaction

4p 2


we add four electrons to each side to produce neutral atoms and obtain 1 4

1 atom 2 atom4( H ) He 2e 2 . Then, recalling that the neutrino and the photon both

have zero rest mass, and using the atomic masses from Appendix B in the t extbook gives

1 41 atom 2 atom

2 2

eH He

1312

4 2

4 1.007 825 u 4.002 603 u 2 0.000 549 u 931.5 MeV u

1.60 10 J25.7 Mev 4.11 10 J

1 MeV

Q m c m m m c

With each occurrence of this reaction consuming four protons, the rate of proton

consumption required to produce the power output of the Sun is

26 26

38

12

3.85 10 J s 3.85 10 J s3.74 10 proton s

Q 4 1.03 10 J protonrate

energy per proton

P

Nuclear Energy and Elementary Particles

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