Chapter28 Student

8/10/2019 Chapter28 Student

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PHYSICS CHAPTER 28

1

is defined as the spontaneousdisintegration of certain atomicnuclei accompanied by theemission of alpha particles,beta particles or gammaradiation.

CHAPTER 28: Radioactivity(2 Hours)


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PHYSICS CHAPTER 28Learning Outcome:

At the end of this chapter, students should be able to:

Explain , + , and decays.State decay law and use

Define and determine activity, A and decay constant, .Derive and use

Define and use half-life

2

28.1 Radioactive decay

OR


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PHYSICS CHAPTER 28

28.1 Radioactive decayis defined as the phenomenon in which anunstable nucleus disintegrates to acquire a morestable nucleus without absorb an external energy .The radioactive decay is a spontaneous reactionthat is unplanned, cannot be predicted andindependent of physical conditions (such aspressure, temperature) and chemical changes.This reaction is random reaction because the

probability of a nucleus decaying at a given instant isthe same for all the nuclei in the sample.Radioactive radiations are emitted when an unstablenucleus decays. The radiations are alpha particles,

beta particles and gamma-rays . 3


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PHYSICS CHAPTER 28

28.1.1 Alpha particle ( ) An alpha particle consists of two protons and two neutrons .It is identical to a helium nucleus and its symbol is

It is positively charged particle and its value is +2e with massof 4.002603 u.When a nucleus undergoes alpha decay it loses four nucleons,two of which are protons, thus the reaction can be representedby general equation below:

Examples of decay :

4

He42 42OR

Q HePbPo 42214

82218

84

(Parent) ( particle)

(Daughter) QHe4

2

Q HeRaTh 422268823090

Q HeRnRa 42222

86226

88

Q HeThU 422349023892


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PHYSICS CHAPTER 28

28.1.2 Beta particle ( )Beta particles are electrons or positrons (sometimes is calledbeta-minus and beta-plus particles).The symbols represent the beta-minus and beta-plus (positron)are shown below:

Beta-minus particle is negatively charged of 1e and its massequals to the mass of an electron .Beta-plus (positron) is positively charged of +1e (antiparticleof electron) and it has the same mass as the electron .In beta-minus decay, an electron is emitted, thus the massnumber does not changed but the charge of the parentnucleus increases by one as shown below:

5

e0

1 OR

e0

1 ORBeta-minus

(electron) :

Beta-plus

(positron) :

(Parent) ( particle)

(Daughter)

Qe01


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PHYSICS CHAPTER 28Examples of minus decay:

In beta-plus decay, a positron is emitted, this time the charge of

the parent nucleus decreases by one as shown below:

For example of plus decay is

6

Q ePaTh 0

1

234

91

234

90Q eUPa

01

23492

23491

Q ePoBi 0

1214

84214

83

(Parent) (Positron)

(Daughter)

X A Z Y1 A

Z Qe01

Qv en p 0110

11

Neutrino is unchargedparticle with negligiblemass .


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PHYSICS CHAPTER 28

7

Neutrino (and antineutrino) was introduce to make sure betadecay follow the conservation of relativistic energy (kinetic and

rest energies)

and conservation of linear momentum:

More information about neutrino and antineutrino :http://hyperphysics.phy-astr.gsu.edu/hbase/particles/neutrino.html#c1

momentumlinear reactionafter
http://hyperphysics.phy-astr.gsu.edu/hbase/particles/neutrino.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/particles/neutrino.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/particles/neutrino.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/particles/neutrino.html


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PHYSICS CHAPTER 2828.1.3 Gamma ray ( )

Gamma rays are high energy photons (electromagnetic

radiation ).Emission of gamma ray does not change the parent nucleusinto a different nuclide , since neither the charge nor thenucleon number is changed.

A gamma ray photon is emitted when a nucleus in an excited

state makes a transition to a ground state .Examples of decay are :

It is uncharged (neutral) ray and zero mass .The differ between gamma-rays and x-rays of the samewavelength only in the manner in which they are produced ;gamma-rays are a result of nuclear processes , whereas x-rays originate outside the nucleus . 8

HePbPo 42214

82218

84

eUPa 0

1234

92234

91

TiTi 20881208

81

Gamma ray


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PHYSICS CHAPTER 2828.1.4 Comparison of the properties between alpha

particle, beta particle and gamma ray.Table 15.1 shows the comparison between the radioactiveradiations.

9

Alpha Beta Gamma

Charge

Deflection byelectric andmagnetic fields

Ionization power

Penetration power

Ability to affect aphotographic plate

Ability to producefluorescence

+2e 1e OR +1e 0 (uncharged)

Yes Yes No

Strong Moderate Weak

Weak Moderate Strong

Yes Yes Yes

Yes Yes YesTable 28.1


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PHYSICS CHAPTER 28Figures 28.1 and 28.2 show a deflection of , and in electricand magnetic fields.

10

Figure 28.1

B

E

Figure 28.2

Radioactivesource


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PHYSICS CHAPTER 2828.1.5 Decay constant ( )

Law of radioactive decay states:For a radioactive source, the decay rate is directly

proportional to the number of radioactive nuclei N remaining in the source .i.e.

Rearranging the eq. (15.1):

Hence the decay constant is defined as the probability that a

radioactive nucleus will decay in one second . Its unit is s 1.11

dt dN

N dt

dN

Negative sign means the number ofremaining nuclei decreases with time

Decay constant

(28.1)

N dt dN

nucleieradioactivremainingof number

ratedecay


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PHYSICS CHAPTER 28The decay constant is a characteristic of the radioactive nuclei.Rearrange the eq. (15.1), we get

At time t= 0 , N=N 0 (initial number of radioactive nuclei in thesample) and after a time t , the number of remaining nuclei is

N . Integration of the eq. (28.2) from t= 0 to time t :

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dt N dN (28.2)

t N

N dt

N dN

00

t N N t N 00ln

t N N

0

ln

Exponential law ofradioactive decay

(28.3)


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PHYSICS CHAPTER 28From the eq. (28.3), thus the graph of N , the number ofremaining radioactive nuclei in a sample, against the time t isshown in Figure 28.3.

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t e N N 0

20

N

0 N

40 N

160

N 80

N

2/1T 2/12T 2/13T 2/14T 2/15T 0 t ,time

N

lifehalf :2/1 T

Figure 28.3

Simulation 28.1

Note:

From the graph (decay curve),the life of any radioactivenuclide is infinity , therefore totalk about the life of radioactivenuclide, we refer to its half-life .
http://localhost/var/www/apps/conversion/tmp/scratch_7/AF_4409.swfhttp://localhost/var/www/apps/conversion/tmp/scratch_7/AF_4409.swfhttp://localhost/var/www/apps/conversion/tmp/scratch_7/AF_4409.swf


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PHYSICS CHAPTER 2831.1.6 Half-life ( T 1/2 )

is defined as the time taken for a sample of radioactive

nuclides disintegrate to half of the initial number of nuclei .From the eq. (28.3), and the definition of half-life,

when , thus

The half-life of any given radioactive nuclide is constant , itdoes not depend on the number of remaining nuclei .

14

t e N N 0

2/1T t 2

; 0 N

N

2/10

0

2T e N N

2/12 T e 2/1

21 T e

2/1ln2ln T e

Half-life (28.4)


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PHYSICS CHAPTER 28The units of the half-life are second (s), minute (min), hour (hr), day (d) and year (y). Its unit depend on the unit of decay

constant .Table 28.2 shows the value of half-life for several isotopes.

15

Table 28.2

Isotope Half-life

4.5 10 9 years

1.6 10 3 years

138 days

24 days

3.8 days

20 minutes

U23892

Po210884

Ra22688

Bi21483

Rn22286

Th23490


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PHYSICS CHAPTER 2831.1.7 Activity of radioactive sample ( A )

is defined as the decay rate of a radioactive sample.

Its unit is number of decays per second .Other units for activity are curie (Ci ) and becquerel (Bq ) S.I.unit.

Unit conversion:

Relation between activity ( A ) of radioactive sample and time t :

From the law of radioactive decay :

and definition of activity :

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second perdecays1073Ci1 10 .

dt dN

second perdecay1Bq1

N dt dN

dt dN

A


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PHYSICS CHAPTER 28Thus

17

00 N A

N A and t e N N 0 t e N A 0

Activity at time t Activity at time, t =0

and t e N 0(28.5)


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PHYSICS CHAPTER 28

18

A radioactive nuclide A disintegrates into a stable nuclide B. The

half-life of A is 5.0 days. If the initial number of nuclide A is 1.0 10 20 ,calculate the number of nuclide B after 20 days.Solution :

The decay constant is given by

The number of remaining nuclide A is

The number of nuclide A that have decayed is

Therefore the number of nuclide B formed is

Example 28.1 :

Q B A

0.52ln

days20;101.0days;0.5 2002/1 t N T

2/1

2lnT

1days139.0

t

e N N

0 20139.020

100.1

e N nuclei102.6 18

1820 102.6100.1 nuclei1038.9 19


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PHYSICS CHAPTER 28

19

a. Radioactive decay is a random and spontaneous nuclearreaction. Explain the terms random and spontaneous.

b. 80% of a radioactive substance decays in 4.0 days. Determinei. the decay constant,ii. the half-life of the substance.

Solution :a. Random means that the time of decay for each nucleus

cannot be predicted . The probability of decay for eachnucleus is the same .

Spontaneous means it happen by itself without externalstimuli . The decay is not affected by the physical conditionsand chemical changes .

Example 28.2 :


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PHYSICS CHAPTER 28

20

Solution :b. At time

The number of remaining nuclei is

i. By applying the exponential law of radioactive decay, thus thedecay constant is

ii. The half-life of the substance is

days,0.4t

00 10080 N N N

nuclei2.0 0 N

t e N N 0 0.4002.0 e N N 0.42.0 e

0.4ln2.0ln e

eln0.42.0ln

2ln

2/1 T 402.0

2ln2/1 T


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PHYSICS CHAPTER 28

21

Phosphorus-32 is a beta emitter with a decay constant of 5.6 107

s 1. For a particular application, the phosphorus-32 emits 4.0 10 7beta particles every second. Determinea. the half-life of the phosphorus-32,b. the mass of pure phosphorus-32 will give this decay rate.

(Given the Avogadro constant, N A =6.02 10 23 mol 1)

Solution :

a. The half-life of the phosphorus-32 is given by

Example 28.3 :

2ln

2/1 T

1717 s104.0;s106.5 dt

dN

7106.5

2ln


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PHYSICS CHAPTER 28

22

Solution :

b. By using the radioactive decay law, thus

6.02 10 23 nuclei of P-32 has a mass of 32 g

7.14 10 13 nuclei of P-32 has a mass of

1717 s104.0;s106.5 dt

dN

0 N dt

dN

077 106.5100.4 N nuclei1014.7 130 N

32

1002.6

1014.723

13


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PHYSICS CHAPTER 28

23

A thorium-228 isotope which has a half-life of 1.913 years decaysby emitting alpha particle into radium-224 nucleus. Calculatea. the decay constant.b. the mass of thorium-228 required to decay with activity of

12.0 Ci.c. the number of alpha particles per second for the decay of 15.0 g

thorium-228.(Given the Avogadro constant, N A =6.02 10 23 mol 1)Solution :

a. The decay constant is given by

Example 28.4 :

2ln

2/1 T

6060243651.913y913.12/1 T

2ln

1003.6 7

s1003.6 7


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PHYSICS CHAPTER 28

24

Solution :

b. By using the unit conversion ( Ci decay/second ),

the activity is

Since then

If 6.02 10 23 nuclei of Th-228 has a mass of 228 g thus

3.86 10 19 nuclei of Th-228 has a mass of

10107.30.12Ci0.12 Adecays/s1044.4 11

second perdecays1073Ci1 10 .

N A

A

N 8

11

1015.11044.4

N

nuclei1086.3 19

2281002.61086.3

23

19


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PHYSICS CHAPTER 28

25

Solution :c. If 228 g of Th-228 contains of 6.02 10 23 nuclei thus

15.0 g of Th-228 contains of

Therefore the number of emitted alpha particles per second is

given by

231002.62280.15

nuclei1096.3 22 N

228 1096.31015.1 N

dt dN

A Ignored it.


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PHYSICS CHAPTER 28Learning Outcome:

At the end of this chapter, students should be able to:

Explain the application of radioisotopes as tracers.

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28.2 Radioisotope as tracers


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PHYSICS CHAPTER 2828.2 Radioisotope as tracers28.2.1 Radioisotope

is defined as an isotope of an element that is radioactive .It is produced in a nuclear reactor, where stable nuclei arebombarded by high speed neutrons until they becomeradioactive nuclei .Examples of radioisotopes:

a.

b.

c.

27

Q PnP 3215103115

Q eSP 0

13216

3215

Q Nan Na 241110

2311

Q eMg Na 0

12412

2411

Q AlnlA 281310

2713

Q eSiAl 0128142813

(Radio phosphorus)

(Radio sodium)

(Radio aluminum)


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PHYSICS CHAPTER 2828.2.2 Radioisotope as tracers

Since radioisotope has the same chemical properties as the

stable isotopes then they can be used to trace the path madeby the stable isotopes .Its method :

A small amount of diluted radioisotope solution is prepared .The solution is either swallowed by the patient or injectedinto the body of the patient .

After a while certain part of the body will have absorbedeither a normal amount, or an amount which is larger thannormal or less than normal of the radioisotope. A detector(such as Geiger counter ,gamma camera , etc..) thenmeasures the count rate at the part of the bodyconcerned .

It is used to investigate organs in human body such as kidney,thyroid gland, heart, brain, and etc..It also used to monitor the blood flow and measure the bloodvolume . 28


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PHYSICS CHAPTER 28

29

A small volume of a solution which contained a radioactive isotope

of sodium had an activity of 12000 disintegrations per minute whenit was injected into the bloodstream of a patient. After 30 hours theactivity of 1.0 cm 3 of the blood was found to be 0.50 disintegrationsper minute. If the half-life of the sodium isotope is taken as 15hours, estimate the volume of blood in the patient.

Solution :The decay constant of the sodium isotope is

The activity of sodium after 30 h is given by

Example 28.5 :

h30;min12000h;15 1

02/1

t AT

2ln

2/1 T 2ln

1512 h1062.4

t e A A 0 301062.4 212000 e

1

min3000

A


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PHYSICS CHAPTER 28

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Solution :

In the dilution tracing method, the activity of the sample, A isproportional to the volume of the sample present, V .

thus the ratio of activities is given by

Therefore the volume of the blood is

h30;min12000h;15 102/1 t AT

V A11 kV A 22 kV A then and

initial final

(28.6)


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PHYSICS CHAPTER 28

28.2.3 Other uses of radioisotope

In medicineTo destroy cancer cells by gamma-ray from a high-activitysource of Co-60.To treat deep-lying tumors by planting radium-226 or caesium-137 inside the body close to the tumor.

In agricultureTo enable scientists to formulate fertilizers that will increase theproduction of food.To develop new strains of food crops that are resistant todiseases, give high yield and are of high quality.To increase the time for food preservation.

31


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PHYSICS CHAPTER 28In industry

To measure the wear and tear of machine part and the

effectiveness of lubricants.To detect flaws in underground pipes e.g. pipes use to carrynatural petroleum gas.To monitor the thickness of metal sheet during manufacture bypassing it between gamma-ray and a suitable detector.

In archaeology and geologyTo estimate the age of an archaeological object found byreferring to carbon-14 dating.To estimate the geological age of a rock by referring to

potassium-40 dating.

32


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PHYSICS CHAPTER 28

33

Radioactive iodine isotope of half-life 8.0 days is used forthe treatment of thyroid gland cancer. A certain sample is requiredto have an activity of 8.0 10 5 Bq at the time it is injected into thepatient.a. Calculate the mass of the iodine-131 present in the sample to

produce the required activity.b. If it takes 24 hours to deliver the sample to the hospital, what

should be the initial mass of the sample?c. What is the activity of the sample after 24 hours in the body of the

patient?(Given the Avogadro constant, N A =6.02 10 23 mol 1)

Example 28.6 :

I13153


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PHYSICS CHAPTER 28

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Solution :

The decay constant of the iodine isotope is

a. From the relation between the decay rate and activity,

If 6.02 10 23 nuclei of I-131 has a mass of 131 g thus

8.0 10 11 nuclei of I-131 has a mass of

s;1091.66060240.8 52/1 T Bq100.8 50 A

2ln

2/1 T

2ln

1091.6 5

16 s1000.1

00

dt dN A

0

65 1000.1100.8 N 00 N A

nuclei100.8 110 N

1311002.6100.8

23

11


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PHYSICS CHAPTER 28

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Solution :

b. GivenLet N : mass of I-131 after 24 hours

N 0 : initial mass of I-131By applying the exponential law of radioactive decay, thus

c. GivenThe activity of the sample is

s;1091.66060240.8 52/1 T Bq100.8 50 A

s108.64360024hr 24 4t g1074.1 10

t

e N N

0 46 1064.81000.10

101074.1

e N 46 1064.81000.1100 1074.1

e N

s108.64360024hr 24 4t

t e A A 0 46 1064.81000.15100.8 e A


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PHYSICS CHAPTER 28

36

An archeologist on a dig finds a fragment of an ancient basket

woven from grass. Later, it is determined that the carbon-14 contentof the grass in the basket is 9.25% that of an equal carbon samplefrom the present day grass. If the half-life of the carbon-14 is 5730years, determine the age of the basket.

Solution :The decay constant of carbon-14 is

The age of the basket is given by

Example 28.7 :

years5730;1025.9100

25.91/20

20

T N N N

2ln

2/1 T

2ln

573014

y1021.1

t e N N 0

t e N N 41021.1

0021025.9

et ln1021.11025.9ln 42


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PHYSICS CHAPTER 28

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Exercise 28.1 :Given N A =6.02 10

23 mol 1

1. Living wood takes in radioactive carbon-14 from theatmosphere during the process of photosynthesis, the ratio ofcarbon-14 to carbon-12 atoms being 1.25 to 10 12 . When thewood dies the carbon-14 decays, its half-life being 5600years. 4 g of carbon-14 from a piece of dead wood gave atotal count rate of 20.0 disintegrations per minute. Determinethe age of the piece of wood.

ANS. : 8754 years2. A drug prepared for a patient is tagged with Tc-99 which has a

half-life of 6.05 h.a. What is the decay constant of this isotope?b. How many Tc-99 nuclei are required to give an activity of

1.50 Ci?c. If the drug of activity in (b) is injected into the patient 2.05 h

after it is prepared, determine the drugs activity.(Physics, 3 rd edition, James S. Walker, Q27&28, p.1107)

ANS. : 0.115 h 1; 1.7 10 9 nuclei; 1.19 Ci


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PHYSICS CHAPTER 28

Good luckFor

2nd semester examination

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