CHAPTER 31 RADIOACTIVITY AND NUCLEAR PHYSICS

Suppose a particle of ionizing radiation deposits 0.500 MeV of energy in this Geiger tube.. a Confirm that the annihilation equation e+ +e− →γ +γ conserves charge, electron family number

CHAPTER 31: RADIOACTIVITY AND

NUCLEAR PHYSICS

31.2 RADIATION DETECTION AND DETECTORS

1 The energy of 30.0 eV is required to ionize a molecule of the gas inside a Geiger tube,

thereby producing an ion pair Suppose a particle of ionizing radiation deposits 0.500 MeV of energy in this Geiger tube What maximum number of ion pairs can it create?

eV/pair30.0

eV/MeV10

1.00MeV0.500pairs

2 A particle of ionizing radiation creates 4000 ion pairs in the gas inside a Geiger tube

as it passes through What minimum energy was deposited, if 30.0 eV is required to create each ion pair?

Solution (4000pairs)(30.0eV/pair) =120keV

3 (a) Repeat Exercise 31.2, and convert the energy to joules or calories (b) If all of this

energy is converted to thermal energy in the gas, what is its temperature increase, assuming 50.0cm3 of ideal gas at 0.250-atm pressure? (The small answer is

consistent with the fact that the energy is large on a quantum mechanical scale but small on a macroscopic scale.)

Solution

(a)

J1092.1eV

1

J1060.1)eV1020.1

T nR

nRT PV

nR

E T

T nR E

nRT E

3

21

Also 3

2so

,2

32

)cm101m1)(

cm50)(

atmPa1001.1)(

atm250.0(

)J1092.1)(

K293)(

32(

13

3 6 3

3 5

4 Suppose a particle of ionizing radiation deposits 1.0 MeV in the gas of a Geiger tube,

all of which goes to creating ion pairs Each ion pair requires 30.0 eV of energy (a) The applied voltage sweeps the ions out of the gas in 1.00µs What is the current?

(b) This current is smaller than the actual current since the applied voltage in the Geiger tube accelerates the separated ions, which then create other ion pairs in subsequent collisions What is the current if this last effect multiplies the number of ion pairs by 900?

Solution

(a)

A10071s10001

C10601eV030

eV100012

6

19 6

t

Ne t

Q I

(b) I =(900)(1.067×10−8)=9.601.00×10−6 A=9.60 μA

31.3 SUBSTRUCTURE OF THE NUCLEUS

5 Verify that a 2.3×1017kg mass of water at normal density would make a cube 60 km

on a side, as claimed in Example 31.1 (This mass at nuclear density would make a cube 1.0 m on a side.)

Solution

km61m101.6kg/m

1000

103

3

17 3

6 Find the length of a side of a cube having a mass of 1.0 kg and the density of nuclear

matter, taking this to be

103.2

kg0

3 17

7 What is the radius of an α particle?

Solution 3 (1.2 10 15m) ( )4 3 1.9 10 15m 1.9fm

r

8 Find the radius of a 238Pu

nucleus Pu238 is a manufactured nuclide that is used as a power source on some space probes.

, one of the most tightly bound stable nuclei (b) What

is the ratio of the radius of N i5 8 to that of 258Ha, one of the largest nuclei ever made? Note that the radius of the largest nucleus is still much smaller than the size of

an atom.

Solution

Ni 0

m10645.4

thatso ,fm7.6m106.7258m102.1

15 15

Ha Ni

15 3

15 Ha

10 The unified atomic mass unit is defined to be 1u=1.6605×10− 27kg Verify that this

amount of mass converted to energy yields 931.5 MeV Note that you must use digit or better values for c and q e

four-Solution E =mc2 =(1.6605×10− 27 kg)(2.998×108 m/s)2 =1.49246×10− 10J

J106022.1

eV1J

1049246

11 What is the ratio of the velocity of a β particle to that of an α particle, if they have

the same nonrelativistic kinetic energy?

27

31 2

2 2

2 2

1017.1kg

106605.1002.4

kg1011.92

12

12

1KE

α

α

β β

α α

α β

β

m

m v

v

m

m v

v v

m v

m mv

1 to4.8510

17.1

12 If a 1.50-cm-thick piece of lead can absorb 90.0% of the γ rays from a radioactive

source, how many centimeters of lead are needed to absorb all but 0.100% of the γ

rays?

Solution If there is a 90.0% reduction for each 1.50 cm, then (0.100)x

will remain, where x =

number of distances of 1.50 cm

(0.100) 3.00 (1.50cm) 4.50cmln

0.00100ln

100.0ln0.00100

ln100

.00.00100

x

x

13 The detail observable using a probe is limited by its wavelength Calculate the energy

of a γ-ray photon that has a wavelength of 1×10− 16 m, small enough to detect

details about one-tenth the size of a nucleon Note that a photon having this energy is difficult to produce and interacts poorly with the nucleus, limiting the practicability of this probe.

J1060.1

eV1m/s

10989.1

J10989.1m

1000.1

m/s1000.3sJ1063.6

7 16

9

9 16

8 34

14 (a) Show that if you assume the average nucleus is spherical with a radius 1 / 3

0A r

0

3 0 3

3 0

4

u33

4

u3

4

,

r A r

A r

M V

M A

r r

ππ

3

1 3

m1000.1

kg106605.110381.1fm

u10381.1fm2.14

u3

3

17kg/m10

3

2 ×

=

15 What is the ratio of the velocity of a 5.00-MeV β ray to that of an α particle with the

same kinetic energy? This should confirm that β s travel much faster than α s even when relativity is taken into consideration (See also Exercise 31.11.)

γ

−

=c v

For the β particle: KE =5.00MeV=(γ −1)(0.511MeV), so that (γ −1)=9.785 or

785.10

=

sm10985.2)785.10(

11

)sm10998.2(

uMeV5.931)u0026.4)(

1(MeV00.5

=

11

)sm10998.2

1

to3.19sm1055.1

sm10985.2

16 (a) What is the kinetic energy in MeV of a β ray that is traveling at 0.998c ? This

gives some idea of how energetic a β ray must be to travel at nearly the same speed as a γ ray (b) What is the velocity of the γ ray relative to the β ray?

Solution

MeV57.7)MeV511.0)(

819.14()

1(KE

,819.15)

998.0(11

2

2 2 2

2 2

γγ

(b)

002.1998

=

c

c v

c

31.4 NUCLEAR DECAY AND CONSERVATION LAWS

3

1H He

18 β− decay of 40K

, a naturally occurring rare isotope of potassium responsible for some

of our exposure to background radiation.

40 20 21

106

23 α decay of 210Po

, the isotope of polonium in the decay series of 238U that was discovered by the Curies A favorite isotope in physics labs, since it has a short half-life and decays to a stable nuclide.

2 124

206 82 126

222 86 138

137 56 82

is also radioactive.)

90 39 52

2 140

228 88 142

Solution 4 2

2 126

208 82 128

212

29 When an electron and positron annihilate, both their masses are destroyed, creating

two equal-energy photons to preserve momentum (a) Confirm that the annihilation equation e+ +e− →γ +γ conserves charge, electron family number, and total number

of nucleons To do this, identify the values of each before and after the annihilation (b) Find the energy of each γ ray, assuming the electron and positron are initially nearly

at rest (c) Explain why the two γ rays travel in exactly opposite directions if the center

of mass of the electron-positron system is initially at rest.

Solution (a) charge :( ) ( )+1 + −1 =0 ;electron family number:( ) ( )+1 + −1 =0 ;A:0+0=0

2and

,0since

P

2 2 2

2 1

2 2

1 2

1 tot

2 1

m E E

c m E E E

E

e

e

P P

(c) The two γ rays must travel in exactly opposite directions to conserve momentum, since initially there is zero momentum if the center of mass is initially at rest

30 Confirm that charge, electron family number, and the total number of nucleons are all

conserved by the rule for α decay given in the equation 2

4 2 2

A

this, identify the values of each before and after the decay.

Solution Z =(Z−2)+2before/after;efn :0+0=0;A=(A−4)+4;

31 Confirm that charge, electron family number, and the total number of nucleons are all

conserved by the rule for β− decay given in the equation N e

A Z N

A

Z → +β +− ν

− + 1Y 1

do this, identify the values of each before and after the decay.

Solution Z =(Z +1)−1before/after;efn :0=( ) ( )+1 + −1;A= A

32 Confirm that charge, electron family number, and the total number of nucleons are all

conserved by the rule for β− decay given in the equation N e

A Z N

do this, identify the values of each before and after the decay.

Solution Z =(Z −1)+1before/after;efn :0=( ) ( )1 + −1;A= A

33 Confirm that charge, electron family number, and the total number of nucleons are all

conserved by the rule for electron capture given in the equation

e N

A Z N

Solution Z −1=(Z −1 )before/after;efn :( ) ( )+1 = +1;A= A

34 A rare decay mode has been observed in which 222Ra

emits a 14C nucleus (a) The decay equation is 222 Ra →A X + 14 C Identify the nuclide AX

(b) Find the energy emitted in the decay The mass of 222Ra is 222.015353 u.

Solution

14 6 134

222

88Ra →AXN+ C

82688208;

14

222− = = − =

u105485.3u003241

14u976627

207u015353

222

CPb

Ra

2

2 2

2

2 8

14 6 126

208 82 134

222 88

E

m m

m m

35 (a) Write the complete α decay equation for 226Ra

(b) Find the energy released in the decay.

4 2 136

222 86 138

u10229.5u002603

4017570

222025402

226

RnRa

2

2 3

2

3

222 226

E

m m

m

36 (a) Write the complete α decay equation for 249Cf

(b) Find the energy released in the decay.

Solution (a) 2

4 2 149

245 96 151

u10758.6u002603

4065483

245074844

249

2

2 3

E m

37 (a) Write the complete β− decay equation for the neutron (b) Find the energy

released in the decay.

u104042.8u00054858

0007276

1008665

1

2

2 4

E

m m m

38 (a) Write the complete β− decay equation for 90Sr

, a major waste product of nuclear reactors (b) Find the energy released in the decay.

Solution

51

90 39 52

u1086.5u907152

89907738

89YSr

2

2 4

2

4 90

E

m m

m

39 Calculate the energy released in the β+ decay of 22Na

, the equation for which is given

in the text The masses of 22Na and 22Ne are 21.994434 and 21.991383 u,

u10954.1u00054858

0(2991383

21994434

21

2NeNa

2

2 3

2

3

22 22

E

m m

m

40 (a) Write the complete β+ decay equation for 11C

(b) Calculate the energy released in the decay The masses of 11C and 11B

are 11.011433 and 11.009305 u, respectively.

11 5 5

u10031.1u00054858

02u009305

11u011433

11

2BC

2

2 3

2

3

11 11

E

m m

m

41

(a) Calculate the energy released in the α decay of 238U

(b) What fraction of the mass of a single 238U is destroyed in the decay? The mass of 234Th

is 234.043593 u (c) Although the fractional mass loss is large for a single nucleus, it is difficult to

observe for an entire macroscopic sample of uranium Why is this?

u10588.4u002603

4043593

234050784

238

ThU

HeTh

U

2

2 3

2

3

234 238

2

4 2 144

234 90 146

238 92

E

m m

238

u10588

×

=

×

(c) Since 238U is a slowly decaying substance, only a very small number of nuclei decay

on human timescales; therefore, although those nuclei that decay lose a noticeablefraction of their mass, the change in the total mass of the sample is not detectable for a macroscopic sample

42 (a) Write the complete reaction equation for electron capture by 7Be

(b) Calculate the energy released.

7

(b) The energy released in electron capture is given by ( ) 2

1 c m m

7.0160034u

8

15 7 7

15.000108u

2 1

c m m

E Z,A Z ,A

31.5 HALF-LIFE AND ACTIVITY

44 An old campfire is uncovered during an archaeological dig Its charcoal is found to

contain less than 1/1000 the normal amount of 14C Estimate the minimum age of the charcoal, noting that 210 =1024.

Solution

1024

1 to

ln2

source is labeled 4.00 mCi, but its present activity is found to be 1.85×107 Bq.

(a) What is the present activity in mCi? (b) How long ago did it actually have a mCi activity?

4.00-Solution

Bq103.7

1CiBq

46 (a) Calculate the activity R in curies of 1.00 g of 226Ra

(b) Discuss why your answer is not exactly 1.00 Ci, given that the curie was originally supposed to be exactly the activity of a gram of radium.

Solution (a) First we must determine the number of atoms for radium We use the molar mass

mol

atoms10

022.6g226

molg

t

N

R=

, where we know the half life of 226Ra is

y106

1 × 3 ,

s10156.3

y1y

106.1

106646.2

7 3

47 Show that the activity of the 14C

in 1.00 g of 12C found in living tissue is 0.250 Bq.

Bq250.0)ys10156.3)(

y5730(

)105217.6)(

693.0(693

.0

;105217.6g00.1g0.12

1002.61030.1

7 10

2

10

23 12

N

48 Mantles for gas lanterns contain thorium, because it forms an oxide that can survive

being heated to incandescence for long periods of time Natural thorium is almost 100% 232Th

, with a half-life of 1.405×1010 y If an average lantern mantle contains

300 mg of thorium, what is its activity?

Solution

(300 10 g) 7.784 10 ;g

32.2

1002

y10405.1(

)10784.7)(

693.0(693

7 10

49 Cow’s milk produced near nuclear reactors can be tested for as little as 1.00 pCi of 131I

per liter, to check for possible reactor leakage What mass of 131I

has this activity?

Solution

)1002(0.693)(6

1002.6693.0

693.0

23 2

23 2

m M

Rt N t

N R

g1007.8)

1002.6)(

693.0(

)ds1064.8)(

d04.8)(

Bq/Ci10

70.3)(

Ci1000.1)(

131

23

4 10

50 (a) Natural potassium contains 40K

, which has a half-life of 1.277×109 y What mass

of 40K

in a person would have a decay rate of 4140 Bq? (b) What is the fraction of 40K

in natural potassium, given that the person has 140 g in his body? (These numbers are typical for a 70-kg adult.)

Solution

(a)

mg0.16g1060.1)

1002.6)(

693.0(

)ys10156.3)(

y10277.1)(

Bq/Ci4140)(

g0.40(

)1002.6)(

693.0(

1002.6693.0

693.0

2 23

7 9

23 2

23 2

m M

Rt N t

N R

(b)

4 2

2

1014.1g1060.1g140

g1060

×

51 There is more than one isotope of natural uranium If a researcher isolates 1.00 mg of

the relatively scarce 235U and finds this mass to have an activity of 80.0 Bq, what is its half-life in years?

) m

)(

( t

N

y1003.7s10156.3

y1s

10219.2

)Bq0.80)(

g235(

g)10)(1.0010

02(0.693)(6

)1020.693)(6.0(

8 7

16

3 23

23 2

has one of the longest known radioactive half-lives In a difficult experiment, a researcher found that the activity of 1.00 kg of 50V is 1.75 Bq What is the half-life in years?

02.6()693.0(6930

t

m M t

N

From the periodic table, M =50.94g/mol, so

y1048.1s10156.3

y1s

10681.4

)Bq75.1)(

molg94.50(

)g1000)(

molatoms10

02.6)(

693.0(

17 7

24

23 2

53 You can sometimes find deep red crystal vases in antique stores, called uranium glass

because their color was produced by doping the glass with uranium Look up the natural isotopes of uranium and their half-lives, and calculate the activity of such a vase assuming it has 2.00 g of uranium in it Neglect the activity of any daughter nuclides.

Bq10468.2)ys10156.3)(

y10468.4(

)Bq10689.3)(

693.0(

Bq10151.1)ys10156.3)(

y1004.7(

)Bq10689.3)(

693.0(

;693.0

10022.5)99274.0)(

g00.2(g238

1002.6

;10689.3)00720.0)(

g00.2(g235

1002.6

4 7

9

3 238

3 7

8

3 235

2

21 23

238

19 23

N R

N

N

Bq1058

1002.6)103.1(10

02.6)103.1

[ ] 5.6 10 y

693.0

)y5730()s3600d1()sd10499.2(ln

693.0

lnln

4 1

2 0 0

λ

λ

55 What fraction of the 40K

that was on Earth when it formed 4.5×109 years ago is left

today?

Solution 0 693 [( 0 693 )( 4 5 10 y )]( 1 28 10 y ) 0.087

0

9 9

=

e e

e N

56 A 5000-Ci 60Co

source used for cancer therapy is considered too weak to be useful when its activity falls to 3500 Ci How long after its manufacture does this happen?

Solution ( )

693.0

)y2714.5()Ci3500Ci5000ln(

693.0

ln693

.0

2

0 0

t t

R

R e

and 238U in natural uranium when Earth formed 4.5×109 years ago?

Solution Assume we have a 10,000-particle sample and take it backwards in time

y105.4693.072

8 9

2 1/

t -0.693/

235 0

235 2

1 693 0 235 235

e N

%77

%73.76

%100)()(

)(

%

;950,19)y10468.4()y105.4(693.09927)

(

;

%23

%27.23

%100)()(

)(

%

0 238 0

235

0 238 238

8 9

0 238

0 238 0

235

0 235 235

=

=

×+

N N

e N

N N

N N

58 The β− particles emitted in the decay of 3H

(tritium) interact with matter to create light in a glow-in-the-dark exit sign At the time of manufacture, such a sign contains 15.0 Ci of 3H

(a) What is the mass of the tritium? (b) What is its activity 5.00 y after manufacture?

Solution

(a)

2 2

693.0693

.0

t

M m t

1

kg106605.1u016050

2M Rt

or