Phy 310 chapter 8

is defined as the spontaneous

disintegration of certain atomic

nuclei accompanied by the

emission of alpha particles,

beta particles or gamma

radiation.

CHAPTER 8: Radioactivity

(3 Hours)

Dr Ahmad Taufek Abdul RahmanSchool of Physics & Material Studies, Faculty of Applied Sciences, Universiti Teknologi MARA Malaysia, Campus of Negeri Sembilan

1

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

Explain α, β+, βˉ and γ decays.

State decay law and use

Define activity, A and decay constant, .

Derive and use

Define half-life and use

Learning Outcome:

8.1 Radioactive decay (2 hours)

Ndt

dN

teNN 0teAA 0

OR

2ln2/1 T

2

DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY

The radioactive decay is a spontaneous reaction

that is unplanned, cannot be predicted and

independent of physical conditions (such as

pressure, temperature) and chemical changes.

This reaction is random reaction because the

probability of a nucleus decaying at a given instant

is the same for all the nuclei in the sample.

Radioactive radiations are emitted when an unstable

nucleus decays. The radiations are alpha particles,

beta particles and gamma-rays.

8.1 Radioactive decay

3


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 mass

of 4.002603 u.

When a nucleus undergoes alpha decay it loses four nucleons,

two of which are protons, thus the reaction can be represented

by general equation below:

Examples of decay :

8.1.1 Alpha particle ()

He42 α4

2OR

Q HePbPo 42

21482

21884

(Parent) ( particle)(Daughter)

XAZ Y4

2

AZ QHe4

2

Q HeRaTh 42

22688

23090

Q HeRnRa 42

22286

22688

Q HeThU 42

23490

23892 4


Beta particles are electrons or positrons (sometimes is called

beta-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 mass equals to the mass of an electron.

Beta-plus (positron) is positively charged of +1e (antiparticle of electron) and it has the same mass as the electron.

In beta-minus decay, an electron is emitted, thus the mass number does not charge but the charge of the parent nucleus increases by one as shown below:

8.1.2 Beta particle ()

e01

βOR e01

βORBeta-minus

(electron) :

Beta-plus

(positron) :

(Parent) ( particle)(Daughter)

XAZ Y1

AZ Qe0

15


Examples 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

Q ePaTh 01

23491

23490

Q eUPa 01

23492

23491

Q ePoBi 01

21484

21483

(Parent) (Positron)(Daughter)

XAZ Y1

AZ Qe0

1

Qv enp 01

10

11

Neutrino is uncharged

particle with negligible

mass.

6


Gamma rays are high energy photons (electromagnetic radiation).

Emission of gamma ray does not change the parent nucleus into a different nuclide, since neither the charge nor the nucleon 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 same wavelength only in the manner in which they are produced; gamma-rays are a result of nuclear processes, whereas x-rays originate outside the nucleus.

8.1.3 Gamma ray ()

γ HePbPo 42

21482

21884

γ eUPa 0

123492

23491

γ TiTi 20881

20881

Gamma ray

7


Table 8.1 shows the comparison between the radioactive

radiations.

8.1.4 Comparison of the properties between alpha

particle, beta particle and gamma ray.

Alpha Beta Gamma

Charge

Deflection by

electric and

magnetic fields

Ionization power

Penetration power

Ability to affect a

photographic plate

Ability to produce

fluorescence

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

Yes Yes No

Strong Moderate Weak

Weak Moderate Strong

Yes Yes Yes

Yes Yes YesTable 8.1 8


Figures 8.1 and 8.2 show a deflection of , and in electric

and magnetic fields.

Figure 8.1

B

E

αγ β γ

β

α

Figure 8.2

Radioactive

source

9


Law of radioactive decay states:

For a radioactive source, the decay rate is directly

proportional to the number of radioactive nuclei Nremaining in the source.

i.e.

Rearranging the eq. (8.1):

Hence the decay constant is defined as the probability that a

radioactive nucleus will decay in one second. Its unit is s1.

8.1.5 Decay constant ()

dt

dN

Ndt

dN

Ndt

dN

Negative sign means the number of

remaining nuclei decreases with time

Decay constant

(8.1)

N

dt

dN

nuclei eradioactiv remaining ofnumber

ratedecay

10


The decay constant is a characteristic of the radioactive nuclei.

Rearrange the eq. (8.1), we get

At time t=0, N=N0 (initial number of radioactive nuclei in the

sample) and after a time t, the number of remaining nuclei is

N. Integration of the eq. (8.2) from t=0 to time t :

dtN

dN (8.2)

tN

Ndt

N

dN

00

tN

N tN 00ln

λtN

N

0

ln

λteNN 0

Exponential law of

radioactive decay(8.3)

11


From the eq. (8.3), thus the graph of N, the number of remaining

radioactive nuclei in a sample, against the time t is shown in

Figure 8.3.

teNN 0

2

0N

0N

4

0N

16

0N8

0N

2/1T 2/12T 2/13T2/14T

2/15T0t,time

N

lifehalf:2/1 T

Figure 8.3

Stimulation 8.1

Note:

From the graph (decay curve),

the life of any radioactive

nuclide is infinity, therefore to

talk about the life of radioactive

nuclide, we refer to its half-life.

12


..\SSP1223\AF_4409.html

is defined as the time taken for a sample of radioactive

nuclides disintegrate to half of the initial number of nuclei.

From the eq. (8.3), and the definition of half-life,

when , thus

The half-life of any given radioactive nuclide is constant, it

does not depend on the number of remaining nuclei.

8.1.6 Half-life (T1/2)

teNN 0

2/1Tt 2

; 0NN

2/1

00

2

TeN

N

2/12T

e

2/1

2

1 Te

2/1ln2lnT

e

λλT

693.02ln2/1 Half-life (8.4)

13


The 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 8.2 shows the value of half-life for several isotopes.

Table 8.2

Isotope Half-life

4.5 109 years

1.6 103 years

138 days

24 days

3.8 days

20 minutes

U23892

Po210884

Ra22688

Bi21483

Rn22286

Th23490

14


secondper decays1073Ci1 10 .

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 :

8.1.7 Activity of radioactive sample (A)

dt

dN

secondper decay 1Bq 1

Ndt

dN

dt

dNA

15


Thus

00 NA

NA andteNN 0

teNA 0

λteAA 0

Activity at time t Activity at time, t =0

and teN 0

(8.5)

16


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 1.01020,

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 1 :

QBA

0.5

2ln

days 20;101.0 days; 0.5 2002/1 tNT

2/1

2ln

T

1days 139.0

teNN 0 20139.020100.1 eN

nuclei 102.6 18

1820 102.6100.1 nuclei 1038.9 19

nuclei 1038.9 1917


a. Radioactive decay is a random and spontaneous nuclear

reaction. Explain the terms random and spontaneous.

b. 80% of a radioactive substance decays in 4.0 days. Determine

i. 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 each

nucleus is the same.

Spontaneous means it happen by itself without external

stimuli. The decay is not affected by the physical conditions

and chemical changes.

Example 2 :

18


Solution :

b. At time

The number of remaining nuclei is

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

decay constant is

ii. The half-life of the substance is

days, 0.4t

00

100

80NNN

nuclei 2.0 0N

teNN 0 0.4

002.0 eNN 0.42.0 e

0.4ln2.0ln e

1day 402.0

eln0.42.0ln

2ln2/1 T

402.0

2ln2/1 T

days 72.12/1 T 19


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

s1. For a particular application, the phosphorus-32 emits 4.0 107

beta particles every second. Determine

a. the half-life of the phosphorus-32,

b. the mass of pure phosphorus-32 will give this decay rate.

(Given the Avogadro constant, NA =6.02 1023 mol1)

Solution :

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

Example 3 :

2ln2/1 T

1717 s 104.0 ;s 106.5 dt

dN

7106.5

2ln

s 1024.1 62/1 T

20


Solution :

b. By using the radioactive decay law, thus

6.02 1023 nuclei of P-32 has a mass of 32 g

7.14 1013 nuclei of P-32 has a mass of

1717 s 104.0 ;s 106.5 dt

dN

0Ndt

dN

077 106.5100.4 N

nuclei 1014.7 130 N

321002.6

1014.723

13

g 1080.3 9

21


A thorium-228 isotope which has a half-life of 1.913 years decays

by emitting alpha particle into radium-224 nucleus. Calculate

a. 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 8.0 g

thorium-228.


Solution :

a. The decay constant is given by

Example 4 :

2ln2/1 T

6060243651.913y 913.12/1 T

2ln1003.6 7

18 s 1015.1

s 1003.6 7

22


Solution :

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

the activity is

Since then

If 6.02 1023 nuclei of Th-228 has a mass of 228 g thus

3.86 1019 nuclei of Th-228 has a mass of

10107.30.12Ci 0.12 A

decays/s 1044.4 11

secondper decays1073Ci1 10 .

NA

AN

8

11

1015.1

1044.4

N

nuclei 1086.3 19

2281002.6

1086.323

19

g 1046.1 223


Solution :

c. If 228 g of Th-228 contains of 6.02 1023 nuclei thus

8.0 g of Th-228 contains of

Therefore the number of emitted alpha particles per second is

given by

231002.6228

0.15

nuclei 1096.3 22N

228 1096.3 1015.1

Ndt

dNA

secondparticles/ 1055.4 14 αA

Ignored it.

24


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

Explain the application of radioisotopes as tracers.

Learning Outcome:

8.2 Radioisotope as tracers (1 hour)

25


8.2.1 Radioisotope

is defined as an isotope of an element that is radioactive.

It is produced in a nuclear reactor, where stable nuclei are bombarded by high speed neutrons until they become radioactive nuclei.

Examples of radioisotopes:

a.

b.

c.

8.2 Radioisotope as tracers

Q PnP 32

15

1

0

31

15

Q eSP 0

1

32

16

32

15

Q NanNa 24

11

1

0

23

11

Q eMgNa 0

1

24

12

24

11

Q AlnlA 28

13

1

0

27

13

Q eSiAl 0

1

28

14

28

13

(Radio phosphorus)

(Radio sodium)

(Radio aluminum)

26


Since radioisotope has the same chemical properties as the

stable isotopes then they can be used to trace the path made

by the stable isotopes.

Its method :

A small amount of suitable radioisotope is either

swallowed by the patient or injected into the body of the

patient.

After a while certain part of the body will have absorbed

either a normal amount, or an amount which is larger than

normal or less than normal of the radioisotope. A detector

(such as Geiger counter ,gamma camera, etc..) then

measures the count rate at the part of the body

concerned.

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 blood

volume.

8.2.2 Radioisotope as tracers

27


A small volume of a solution which contained a radioactive isotope

of sodium had an activity of 12000 disintegrations per minute when

it was injected into the bloodstream of a patient. After 30 hours the

activity of 1.0 cm3 of the blood was found to be 0.50 disintegrations

per minute. If the half-life of the sodium isotope is taken as 8 hours,

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 5 :

h 30;min 12000 h; 15 102/1 tAT

2ln2/1 T

2ln15

12 h 1062.4

teAA 0

301062.4 2

12000 e

1min 3000 A28


Solution :

In the dilution tracing method, the activity of the sample, A is

proportional to the volume of the sample present, V.

thus the ratio of activities is given by

Therefore the volume of the blood is

h 30;min 12000 h; 15 102/1 tAT

VA

11 kVA 22 kVA then and

2

1

2

1

V

V

A

A

2

1

3000

5.0

V

32 cm 6000V

initial final

(8.6)

29


In medicine

To destroy cancer cells by gamma-ray from a high-activity

source of Co-60.

To treat deep-lying tumors by planting radium-226 or caesium-

137 inside the body close to the tumor.

In agriculture

To enable scientists to formulate fertilizers that will increase the

production of food.

To develop new strains of food crops that are resistant to

diseases, give high yield and are of high quality.

To increase the time for food preservation.

31.2.3 Other uses of radioisotope

30


In 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 carry

natural petroleum gas.

To monitor the thickness of metal sheet during manufacture by

passing it between gamma-ray and a suitable detector.

In archaeology and geology

To estimate the age of an archaeological object found by

referring to carbon-14 dating.

To estimate the geological age of a rock by referring to

potassium-40 dating.

31


Radioactive iodine isotope of half-life 8.0 days is used for

the treatment of thyroid gland cancer. A certain sample is required

to have an activity of 8.0 105 Bq at the time it is injected into the

patient.

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?


Example 6 :

I131

53

32


Solution :

The decay constant of the iodine isotope is

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

If 6.02 1023 nuclei of I-131 has a mass of 131 g thus

8.0 1011 nuclei of I-131 has a mass of

s; 1091.66060240.8 5

2/1 TBq 100.8 5

0 A

2ln2/1 T

2ln1091.6 5

16 s 1000.1

0

0

dt

dNA

0

65 1000.1100.8 N00 NA

nuclei 100.8 11

0 N

1311002.6

100.823

11

g 1074.1 10 33


Solution :

b. Given

Let N : mass of I-131 after 24 hours

N0 : initial mass of I-131

By applying the exponential law of radioactive decay, thus

c. Given

The activity of the sample is

s; 1091.66060240.8 5

2/1 TBq 100.8 5

0 As 108.64360024hr 24 4t

g 1074.1 10

teNN 0 46 1064.81000.1

0

101074.1

eN

46 1064.81000.110

0 1074.1

eN

g 1090.1 10

0

N

s 108.64360024hr 24 4t

teAA 0 46 1064.81000.15100.8

eA

Bq 1034.7 5A34


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

woven from grass. Later, it is determined that the carbon-14 content

of the grass in the basket is 9.25% that of an equal carbon sample

from the present day grass. If the half-life of the carbon-14 is 5730

years, determine the age of the basket.

Solution :

The decay constant of carbon-14 is

The age of the basket is given by

Example 7 :

years 5730;1025.9100

25.91/20

20

TNNN

2ln2/1 T

2ln5730

14 y 1021.1

teNN 0

teNN41021.1

0021025.9

et ln1021.11025.9ln 42

years 19674t 35


Exercise 8.1 :Given NA =6.021023 mol1

1. Living wood takes in radioactive carbon-14 from the atmosphere during the process of photosynthesis, the ratio of carbon-14 to carbon-12 atoms being 1.25 to 1012. When the wood dies the carbon-14 decays, its half-life being 5600 years. 4 g of carbon from a piece of dead wood gave a total count rate of 20.0 disintegrations per minute. Determine the age of the piece of wood.

ANS. : 8754 years

2. 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 drug’s activity.

(Physics, 3rd edition, James S. Walker, Q27&28, p.1107)

ANS. : 0.115 h1; 1.7109 nuclei; 1.19 Ci 36


Next Chapter…

CHAPTER 9 : Nuclear Reactor

37


Technology

Phy 310 chapter 8