Matriculation Physics ( Radioactivity )
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Transcript of Matriculation Physics ( Radioactivity )
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PHYSICS CHAPTER 15
is defined as the spontaneous is defined as the spontaneous disintegration of certain disintegration of certain atomic nuclei accompanied by atomic nuclei accompanied by the emission of alpha the emission of alpha particles, beta particles or particles, beta particles or gamma radiation.gamma radiation.
CHAPTER 15: RadioactivityCHAPTER 15: Radioactivity(3 Hours)(3 Hours)
PHYSICS CHAPTER 15
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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: ExplainExplain αα, , ββ++, , βˉβˉ and and γγ decays. decays. StateState decay law and use decay law and use
DefineDefine activity, activity, AA and decay constant, and decay constant, . . Derive and use Derive and use
DefineDefine half-life and half-life and useuse
Learning Outcome:
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15.1 Radioactive decay (2 hours)
Ndt
dN
teNN 0teAA 0OROR
2ln
2/1 T
PHYSICS CHAPTER 15
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Definition (refer to section 14.1.3). The radioactive decay is a spontaneousspontaneous reaction
that is unplannedunplanned, cannot be predictedcannot be predicted and independentindependent of physical conditionsphysical conditions (such as pressure, temperature) and chemical changeschemical changes.
This reaction is randomrandom reaction because the probabilityprobability of a nucleus decayingnucleus decaying at a given instant is the samesame for all the nuclei in the sample.
Radioactive radiations are emitted when an unstable nucleus decays. The radiations are alpha particles, alpha particles, beta particles and gamma-raysbeta particles and gamma-rays.
15.1 Radioactive decay
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An alpha particle consists of two protons and two neutronstwo protons and two neutrons. It is identical to a helium nucleusa helium nucleus and its symbol is
It is positively chargedpositively 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 :
15.1.1 Alpha particle ()
He42 α4
2OR
Q HePbPo 42
21482
21884
(Parent)(Parent) (( particle) particle)(Daughter)(Daughter)
XAZ Y4
2
AZ QHe4
2
Q HeRaTh 42
22688
23090
Q HeRnRa 42
22286
22688
Q HeThU 42
23490
23892
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Beta particles are electrons or positronselectrons 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-minusBeta-minus particle is negatively chargednegatively charged of 1e and its mass equals to the mass of an electronmass of an electron.
Beta-plus (positron)Beta-plus (positron) is positively chargedpositively charged of +1e (antiparticle of electron) and it has the same mass as the electronmass as the electron.
In beta-minus decay, an electron is emitted, thus the mass mass number does not charge but the charge of the parent number does not charge but the charge of the parent nucleus increasesnucleus increases by oneby one as shown below:
15.1.2 Beta particle ()
e01
βOR e01
βORBeta-minus Beta-minus (electron) :(electron) :
Beta-plus Beta-plus (positron) :(positron) :
(Parent)(Parent) (( particle) particle)(Daughter)(Daughter)
XAZ Y1
AZ Qe0
1
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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)(Parent) (Positron)(Positron)(Daughter)(Daughter)
XAZ Y1
AZ Qe0
1
Qv enp 01
10
11
NeutrinoNeutrino is uncharged uncharged particle with negligible particle with negligible massmass.
PHYSICS CHAPTER 15
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Gamma rays are high energy photonshigh energy photons (electromagnetic electromagnetic radiationradiation).
Emission of gamma ray does not change the parent nucleus does not change the parent nucleus into a different nuclideinto 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 nucleus in an excited state makes a transition to a ground statestate makes a transition to a ground state.
Examples of decay are :
It is uncharged (neutral) ray and zero massuncharged (neutral) ray and zero mass. The differ between gamma-rays and x-rays of the same differ between gamma-rays and x-rays of the same
wavelengthwavelength only in the manner in which they are producedthey are produced; gamma-raysgamma-rays are a result of nuclear processesresult of nuclear processes, whereas x-x-rays originate outside the nucleusrays originate outside the nucleus.
15.1.3 Gamma ray ()
γ HePbPo 42
21482
21884
γ eUPa 0
123492
23491
γ TiTi 20881
20881
Gamma rayGamma ray
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Table 15.1 shows the comparison between the radioactive radiations.
15.1.4 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+2e 1e OR +1e1e OR +1e 0 (uncharged)0 (uncharged)
YesYes YesYes NoNo
StrongStrong ModerateModerate WeakWeak
WeakWeak ModerateModerate StrongStrong
YesYes YesYes YesYes
YesYes YesYes YesYesTable 15.1Table 15.1
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Figures 15.1 and 15.2 show a deflection of , and in electric and magnetic fields.
Figure 15.1Figure 15.1
B
E
α
γ β γβ
α
Figure 15.2Figure 15.2
Radioactive source
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Law of radioactive decay states:
For a radioactive source, the decay rate is directly For a radioactive source, the decay rate is directly
proportional to the number of radioactive nuclei proportional to the number of radioactive nuclei NN remaining in the sourceremaining in the source.
i.e.
Rearranging the eq. (15.1):
Hence the decay constantdecay constant is defined as the probability that a the probability that a radioactive nucleus will decay in one secondradioactive nucleus will decay in one second. Its unit is ss11.
15.1.5 Decay constant ()
dt
dN
Ndt
dN
Ndt
dN
Negative signNegative sign means the number of number of remaining nuclei decreases with timeremaining nuclei decreases with time
Decay constantDecay constant
(15.1)(15.1)
NdtdN
nuclei eradioactiv remaining ofnumber
ratedecay
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The decay constant is a characteristic of the radioactive nuclei. Rearrange the eq. (15.1), we get
At time t=t=00, , N=NN=N00 (initial number of radioactive nucleiinitial number of radioactive nuclei in the
sample) and after a time time tt, the number of remaining nucleinumber of remaining nuclei is
NN. Integration of the eq. (15.2) from t=0 to time t :
dtN
dN (15.2)(15.2)
tN
Ndt
N
dN00
tNN tN 00
ln
λtN
N
0
ln
λteNN 0Exponential law of Exponential law of radioactive decayradioactive decay
(15.3)(15.3)
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From the eq. (15.3), thus the graph of N, the number of
remaining radioactive nuclei in a sample, against the time t is shown in Figure 15.3.
teNN 0
20N
0N
40N
160N8
0N
2/1T 2/12T 2/13T 2/14T 2/15T0 t,time
N
lifehalf:2/1 T
Figure 15.3Figure 15.3
Stimulation 15.1
Note:Note:
From the graph (decay curve), the life of any radioactive life of any radioactive nuclide is infinitynuclide is infinity, therefore to talk about the life of radioactive nuclide, we refer to its half-lifehalf-life.
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is defined as the time taken for a sample of radioactive the time taken for a sample of radioactive nuclides disintegrate to half of the nuclides disintegrate to half of the initial number of initial number of nucleinuclei.
From the eq. (15.3), and the definition of half-life,
when , thus
The half-life of any given radioactive nuclidehalf-life of any given radioactive nuclide is constantconstant, it does not depend on the number of remaining nucleidoes not depend on the number of remaining nuclei.
15.1.6 Half-life (T1/2)
teNN 0
2/1Tt 2
; 0NN
2/10
0
2TeN
N 2/12 Te
2/1
2
1 Te
2/1ln2ln Te
λλT
693.02ln2/1 Half-lifeHalf-life (15.4)(15.4)
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The units of the half-life are secondsecond (s), minuteminute (min), hourhour (hr), day day (d) and yearyear (y). Its unit depend on the unit of decay unit depend on the unit of decay constantconstant.
Table 15.2 shows the value of half-life for several isotopes.
Table 15.2Table 15.2
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
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secondper decays1073Ci1 10 .
is defined as the decay rate of a radioactive sample.the decay rate of a radioactive sample.
Its unit is number of decays per secondnumber of decays per second. Other units for activity are curie curie (CiCi) and becquerelbecquerel (BqBq) – 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 :
15.1.7 Activity of radioactive sample (A)
dt
dN
secondper decay 1Bq 1
Ndt
dN
dt
dNA
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Thus
00 NA
NA andteNN 0
teNA 0
λteAA 0
Activity at time Activity at time tt Activity at time, Activity at time, tt =0=0
and teN 0
(15.5)(15.5)
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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 :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 eNnuclei 102.6 18
1820 102.6100.1 nuclei 1038.9 19
nuclei 1038.9 19
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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 :Solution :
a. Random means that the time of decay for each nucleus time of decay for each nucleus
cannot be predictedcannot be predicted. The probability of decay for each probability of decay for each
nucleus is the samenucleus is the same.
Spontaneous means it happen by itself without external happen by itself without external
stimulistimuli. The decay is not affected by the physical conditions decay is not affected by the physical conditions
and chemical changesand chemical changes.
Example 2 :
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Solution :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
2ln
2/1 T402.0
2ln2/1 T
days 72.12/1 T
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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 :Solution :
a. The half-life of the phosphorus-32 is given by
Example 3 :
2ln
2/1 T
1717 s 104.0 ;s 106.5 dt
dN
7106.5
2ln
s 1024.1 62/1 T
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Solution :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
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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 15.0 g
thorium-228.
(Given the Avogadro constant, NA =6.02 1023 mol1)
Solution :Solution :
a. The decay constant is given by
Example 4 :
2ln
2/1 T
6060243651.913y 913.12/1 T
2ln
1003.6 7 18 s 1015.1
s 1003.6 7
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Solution :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 Adecays/s 1044.4 11
secondper decays1073Ci1 10 .
NA
A
N 8
11
1015.1
1044.4
N
nuclei 1086.3 19
2281002.6
1086.323
19
g 1046.1 2
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Solution :Solution :
c. If 228 g of Th-228 contains of 6.02 1023 nuclei thus
15.0 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.
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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to:
ExplainExplain the application of radioisotopes as tracers. the application of radioisotopes as tracers.
Learning Outcome:
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15.2 Radioisotope as tracers (1 hour)
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15.2.1 Radioisotope is defined as an isotope of an element that is radioactivean isotope of an element that is radioactive. It is produced in a nuclear reactor, where stable nuclei are stable nuclei are
bombarded by high speed neutrons until they become bombarded by high speed neutrons until they become radioactive nucleiradioactive nuclei.
Examples of radioisotopes:
a.
b.
c.
15.2 Radioisotope as tracers
Q PnP 3215
10
3115
Q eSP 01
3216
3215
Q NanNa 2411
10
2311
Q eMgNa 01
2412
2411
Q AlnlA 2813
10
2713
Q eSiAl 01
2814
2813
(Radio phosphorus)(Radio phosphorus)
(Radio sodium)(Radio sodium)
(Radio aluminum)(Radio aluminum)
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Since radioisotoperadioisotope has the same chemical properties as the same chemical properties as the stable isotopesstable isotopes then they can be used to trace the path made trace the path made by the stable isotopesby the stable isotopes.
Its method : A small amount of suitable radioisotopesmall amount of suitable radioisotope is either
swallowedswallowed by the patient or injected into the body of the injected into the body of the patientpatient.
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 detector (such as Geiger counter ,gamma cameraGeiger counter ,gamma camera, etc..) then measures the count rate at the part of the body measures the count rate at the part of the body concernedconcerned.
It is used to investigate organs investigate organs in human body such as kidney, thyroid gland, heart, brain, and etc..
It also used to monitor the blood flowmonitor the blood flow and measure the blood measure the blood volumevolume.
15.2.2 Radioisotope as tracers
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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 15 hours, estimate the volume of blood in the patient.
Solution :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
2ln
2/1 T
2ln15
12 h 1062.4 teAA 0
301062.4 2
12000 e
1min 3000 A
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Solution :Solution :
In the dilutiondilution tracing method, the activityactivity of the sample, AA is
proportionalproportional to the volumevolume of the sample present, VV.
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
initialinitial finalfinal
(15.6)(15.6)
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In medicineIn 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 agricultureIn 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.
15.2.3 Other uses of radioisotope
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In industryIn 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 geologyIn 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.
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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?
(Given the Avogadro constant, NA =6.02 1023 mol1)
Example 6 :
I13153
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Solution :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 52/1 T
Bq 100.8 50 A
2ln
2/1 T
2ln1091.6 5
16 s 1000.1
00
dt
dNA
065 1000.1100.8 N
00 NA
nuclei 100.8 110 N
1311002.6
100.823
11
g 1074.1 10
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Solution :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 52/1 T
Bq 100.8 50 A
s 108.64360024hr 24 4tg 1074.1 10
teNN 0 46 1064.81000.10
101074.1
eN 46 1064.81000.110
0 1074.1
eNg 1090.1 10
0N
s 108.64360024hr 24 4t
teAA 0 46 1064.81000.15100.8
eABq 1034.7 5A
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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 :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
2ln
2/1 T
2ln5730
14 y 1021.1
teNN 0 teNN
41021.100
21025.9
et ln1021.11025.9ln 42 years 19674t
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Exercise 15.1 :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. :ANS. : 8754 years8754 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 drug’s activity.(Physics, 3(Physics, 3rdrd edition, James S. Walker, Q27&28, p.1107) edition, James S. Walker, Q27&28, p.1107)
ANS. :ANS. : 0.115 h0.115 h11; 1.7; 1.7101099 nuclei; 1.19 nuclei; 1.19 CiCi
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PHYSICS CHAPTER 15
Good luckFor
2nd semester examination