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Transcript of RADIOACTIVITY and RADIOACTIVE DECAY REYNALDO S. JIMENEZ Nuclear Training Center Philippine Nuclear...
![Page 1: RADIOACTIVITY and RADIOACTIVE DECAY REYNALDO S. JIMENEZ Nuclear Training Center Philippine Nuclear Research Institute.](https://reader036.fdocuments.in/reader036/viewer/2022062421/56649e615503460f94b5cd76/html5/thumbnails/1.jpg)
RADIOACTIVITY and
RADIOACTIVE DECAY
REYNALDO S. JIMENEZNuclear Training Center
Philippine Nuclear Research Institute
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What is RADIATION ?
Has been around since the earth was formed 4500 million years ago.
Can be detected, measured and controlled.
87% of radiation dose comes from natural sources, e.g... cosmic, food we eat, our homes
13% result of man’s activities..
- Medical Applications (diagnosis and treatment of disease)- Industrial application (inspection of welds, detection of cracks in or cast metal)- Research application (dating of antiquities, food preservation)
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Radiation – any form energy that
travels and dissipates.
Ionizing Radiation – Its interaction with matter may result to creation of charged particles.
Non-Ionizing Radiation – its interaction with matter will not result to creation of charged particles.
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Everything in nature, every creature
and every material contains, and
always has contained, radioactive
substances.
You are radioactive yourself,
and so is your dog, your coffee,
your seatmate,and your mother-in-
law.
Radiation is all around us.
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Sunshine is one of the most familiar forms of radiation.
IONIZING RADIATION
Potentially harmful or beneficial to Humans…depending on how it is used.
Short wavelength = high energy
Long wavelength =
low energy
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What is IONIZING RADIATION?
- the kind of radiation which is a result of the radioactive process
- that which changes the physical state of atoms which it strikes causing them to be electrically charged or “ionized”
e-
Neutral atom
e-
e-
Ionized atom
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SOURCES OF RADIOACTIVITY
Natural Sources Artificial (Man-made)
accounts for 15% of the total
radiation burden 97% of man-made
radiation is due to diagnostic medical exposures
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Natural Sources• Terrestrial• Floors and
walls of our homes, schools or offices
• Food, water and air
• Muscles, bones and tissues
• Cosmic radiation or rays
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Background Radiation• Value of background radiation is not stable, may vary widely from place to place and from time to time, depending, for instance, on the structure and wetness of the soil, the seasons, changes in weather, wind direction and the level above sea• Radiation emitted from natural radioactive substances in our environment and from the cosmos
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Artificial Sources– Dental and other
medical x-rays– Radiation used to
diagnose diseases and for cancer therapy
– Industrial uses of nuclear techniques
– Consumer products such as luminous wrist watches, ionization smoke detectors
– Fallout from nuclear weapons testing
– Small quantities of radioactive materials released to the environment from coal and nuclear power plants
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What is RADIOACTIVITY ?
RADIOACTIVTY
is the process by
which certain
atoms
spontaneously
emit high energy
particles or ra
ys
from th
eir
nucleus.
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Where does it come from? How does it happen??
Radiation comes from the nucleus of the atom.
. . . RADIATION . . . radiation . . .
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•Everything in the world is composed of different types of matter (chemical elements).
•Each element consists of very small parts called the "Atoms".
ATOM
electron
NUCLEUS
ATOM
Typical diameter: ~ 10-10 m
Typical diameter: ~ 10-15 m
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NUCLEUS
neutron
proton
particle mass charge location
proton 1 amu + in nucleus
neutron 1 amu no charge
in nucleus
electron 1/1850 amu
- around nucleus in various energy
levels
Subatomic Particles
1 amu = 1.675 x 10-27 kg
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Radioactivity depends on the structure of the nucleus.
A nucleus should contain “appropriate” # of neutrons to become stable non-radioactive
C-12 C-13C-10 C-11 C-14 C-15
Stable configuration (For Z ≤ 20) : # of neutrons = or a bit higher than # of protons
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An unstable nucleus has too much energy in it. An atom cannot hold this energy forever.
Unstable nuclei make substances radioactive.
Sooner or later, the atom must get rid of the excess energy
. . . and return to its normal (stable) state.
radiation
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WHY, again, do certain ATOMS DECAY?
Atoms with too much energy in their nuclei are called "radioactive".
Some nuclear arrangements are less stable than others.
A radioactive isotope decays to form a more stable
nucleus.
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… get rid of their excess energy
(DECAY) by emitting radiation.
Radioactive isotopes…
They decay by spitting out:
- mass (alpha particles)- charge (beta particles)- energy (gamma rays)
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PARENT and DAUGHTER
PARENT NUCLIDE – the original nuclide which undergoes radioactive decay
DAUGHTER NUCLIDE (or progeny) - the more stable nuclide which
results from radioactive decay
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NUCLIDE - any atomic species characterized by the number of protons and number of neutrons
notations: AzXN X-A AX
where X: symbol of element Z: atomic number = no. of protons N: number of neutrons
A: atomic mass = Z + N Examples:
6027Co33 Co-60 60Co
32
15P17 P-32 32P
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ALPHA, •A helium nucleus, 4
2He- Consists of 2 protons and 2 neutrons
•heavy ( Mass : 4 units or 7340 times beta particle )•Charge : +2•High energy ( Energy range : 4 to 8 MeV )•Slow moving ( Speed : 2 X 107 m/s)•Emitted when the nucleus is too big•Least penetrating but much damage where it penetrates
- Limited range : < 10 cm in air ; 60 microns in tissue•Easily Shielded (e.g., paper, skin)
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(1) alpha decay - emission of an alpha particle (a He nucleus), resulting in a decrease in both mass and atomic number.
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Example: Alpha Decay of Americium-241 to Neptunium-237
αNp Am 4
2
237
93
241
95
- decay•Heavy nuclei more massive than Pb decay by this method•Atomic Number, Z, decreases by 2•Atomic Mass Number, A, decreases by 4•Products are a new element and an particle
αYX 42
4A2Z
AZ
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Beta,
• A fast moving electron originating from the nucleus• Emitted when the nucleus has too many neutrons• Comes from a neutron which has changed into a p+
and an e-• Very light ( Mass : 0.00055 amu )• Charge : -1• Energy dependent on radionuclide - ( Energy range : several KeV to 5 MeV ) Range : ~ 12' / MeV in air ; few mm in tissue• Shielding (aluminum and other light (Z<14) materials,
plastics)
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(2) Beta decay - emission of a beta particle (an electron from the nucleus), resulting in an increase in atomic number.
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- DECAY•A radioactive nucleus that undergoes - decay has a neutron in its nucleus convert into a p+ and an e-
•Atomic Number, Z, increases by 1
•Atomic Mass Number, A, remains the same
Example: Beta Decay of Hydrogen-3 to Helium-3. He H 3
2
3
1
YX A
1ZAZ
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Gamma, •Not a particle but a burst of very high energy as electromagnetic radiations of very high frequency•Results from the transition of nuclei from excited state to their ground state•No mass, 0 charge•Have highest energy of all EM radiations•Very dangerous (can do a lot of biological damage)•Energies well defined and characteristics of the emitting radionuclide (up to several MeV)•Speed : speed of light• Long range : km in air ; m in body•Shielding : large amounts of lead or concrete
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(3) Gamma decay - This is the photon that carries the energy that is emitted. The wavelength is in the order of 10-11 to 10-14 m (higher energy than x-rays).
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DECAY•Only energy is released•Parent and daughter atoms are the same chem'l element•Atomic Number, Z, remains the same•Atomic Mass Number, A, remains the same
XX A
Z
A
Z
Example: Gamma Decay of Helium-3.
HeHe 3
2
3
2
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POSITRON, +
• Similar to e- but opposite charge
• Comes from a proton which has changed into a neutron and a positron
p+ n + +
• The neutron stays in the nucleus and the positron ejected at high speed
• Charge : +1
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POSITRON DECAY•Occurs in nuclei which have an excess of protons•Atomic Number, Z, reduces by 1•Atomic Mass Number, A, remains the same
1A
1-ZAZ YX
•Example: Positron Decay of Carbon-11 to Boron-11. B C 11
5
11
6
Comes from a proton which has changed into a neutron and a
positron
p+ n + +
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ELECTRON CAPTURE•Occurs in atoms of excess protons•e- in the innermost shell (K-shell) is captured by a p+ in the nucleus to form a neutron• always accompanied by emission of x-rays• x-rays emitted are characteristics of the progeny nuclide•Atomic Number, Z, reduces by 1•Atomic Mass Number, A, remains the same
ray- x YX A1-Z
AZ
Example: Electron Capture of Beryllium-7. It decays to Lithium-7.
LiBe 7
3
7
4
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(4) positron emission - emission of a positively charged electron (positron) from the nucleus, resulting in a decrease in the atomic number. A positron has the same mass as an electron, but opposite in charge. In other words, inside the nucleus, a proton is being converted into a neutron.
(5) electron capture - This happens in heavy atoms in which an inner shell (1s) electron is captured by the nucleus, resulting in a decrease in atomic number. This process has the same effect as positron emission.
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Summary of Radioactive Decay Modes
Decay Mode Symbol Common Source
Change in Z
Change in N
Change in A
Alpha Heavy Nuclei - 2 - 2 - 4
Beta - Excess Neutrons
+ 1 - 1 0
Gamma Excited Nuclei
0 0 0
Positron + Excess Protons
- 1 + 1 0
Electron Capture
Excess Protons
- 1 + 1 0
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Decay Chains / Decay Series
A radionuclide may decay to a nuclide which is also radioactive.
This radionuclide may in turn give rise to another radionuclide…
… and this will be repeated until the atom finally reaches a stable form.
This path to stability is called a DECAY CHAIN or a DECAY SERIES.
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CHART OF NUCLIDES• a plot of Z vs. N of all known nuclides
• provides several important data concerning the isotopes
• may be used to determine how a particular nuclide will decay• may be used to determine the progeny of a parent nuclide• decay modes are given in order of abundance
• From the chart it is possible to find nuclides which are stable.
• Stable nuclides form a rough band running diagonally up and to the right on the chart.
• General Rule : The closer a nuclide is to the line of stability, the more stable it is.
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238 234 230 226 222 218 214 210 206
U-92Pa-91Th-90Ac-89Ra-88Fr-87Rn-86At-85Po-84Bi-83Pb-82Tl-81
U
Th Th
Ra
Rn
Po
Pb
Bi
Po
Tl
Pb
Bi
Po
Pb
U
Pa
Atomic Mass NumberA
tom
ic e
lem
en
t an
d n
um
ber
Uranium-238 Series
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Alpha in
Beta-
out
n
out
Original nucleus
n
in
Beta+
out
Alpha
out
Ato
mic
Mass N
um
ber
Atomic element and number
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NUCLEAR EQUATIONS
•Nuclear equations show how atoms decay.
•Similar to chemical equations
- must still balance mass and charge
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NUCLEAR EQUATIONS
Example: A patient is injected with radioactive phosphorus. What happens to the phosphorus?
Is this equation balanced?
You must see if the mass and charge are the same on both sides.
Charge +15 (protons) + 16 (protons) - 1 ()-------------------------------------- +15 total charge +15 total charge
Yes, it is balanced.
Mass 15 protons 16 protons 17 neutrons 16 neutrons-------------------------------------32 total mass 32 total mass
3215P17 32
16S16 +
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DECAY PARAMETERS: characteristics of specific radionuclide
•ACTIVITY, A - number of disintegrations of a nucleus occurring per second
•DECAY CONSTANT, - the fractions of atoms which undergo decay per unit time
•HALF-LIFE, T1/2 - time taken for half the atoms of a radionuclide to undergo radioactive decay
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RADIOACTIVE DECAY LAW
N = No e - T
where :
No : original number of nuclei present
T : time which passed
: radioactive decay constant
N : remaining nuclei after time T
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UNITSACTIVITY
- described by the number of nuclear disintegrations per unit time.
1 Becquerel (Bq) = 1 disintegration per second - named after Henri Becquerel, discoverer of radioactivity
1 Curie (Ci) = 37 billion disintegrations per second
- named after Marie Curie who discovered and named radium and polonium
1 Ci = 3.7 x 1010 Bq
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Activity: how much is present?Activity – tells how many unstable nuclei decay in a second and emit radiation
Activity of an object need not be in any proportion to its size.
High activity Low activity
Radiography isotope
Low level waste
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ACTIVITY, AProportional to the number of unstable nuclei
A = N
Can be written in the form:
A = Ao e - T
where: Ao: original activity of radionuclide
T : time which passed : radioactive decay constant A : remaining activity after decay time
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Half-Life
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Each radioisotope has its own half-life.
Half-life values can range from milliseconds to billions of years.
Half-life examples:
•Molybdenum-99 67 hours• Iodine-131 8 days• Phosphorus-32 14.3 days• Iron-59 45 days• Cobalt-60 5.3 years• Carbon-14 5760 years• Uranium-235 710 million years
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HALF-LIFE, T1/2 - time taken for half the atoms of a
radionuclide to undergo radioactive decay
To determine relationship between and T1/2 :
at T=T1/2 , N = No / 2
substituting to N = No e -T
taking ln of both sides:
21T
e21
21T
693.0
21T
oo eN
2N
ln (1/2) = ln e(-T1/2)
- 0.693 = - T1/2
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CALCULATING ACTIVITY
Activity is also defined as:
Where A : activity at time T
Ao : initial activity
n : number of half-lives which
have elapsed, i.e.
no
2A
A
21T
Tn
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EXAMPLE OF ACTIVITY CALCULATION
PROBLEM:
P-32 has a half-life of 14.3 days. On Jan. 10, 2006, the activity of the P-32 sample was 10 mCi. What will the activity be on February 6, 2006?
no
2A
A METHOD 1: USE Eqn. 1
SOLUTIONS:
Given: T = 27 days : Time interval bet Jan. 10, Feb. 6, 2006
T1/2 = 14.3 days : Half-life of P-32
Ao = 10 Ci : Activity of source on Jan. 10, 2006
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Activity of P-32 on Feb. 6, 2006 is 2.698 mCi.
substituting to Eqn. 1 :noAA
2
91.117.30
5.57
21
y
y
T
Tn
27 days
14.3 days1.89
mCimCimCi
A 698.2706.3
10
2
1089.1
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METHOD 2 : USET
oeAA Eqn. 2
Substituting to Eqn. 2 :
A = 10 e – (0.048 / day) x (27 days)
= 10 e - 1.308
= 10 x 0.27 = 2.7023 Ci
Activity of P-32 on Feb. 6, 2006 is 2.7023 Ci.Note: Difference is due to rounding adjustments.
12
21
103.217.30
693.0693.0 yxyT
Where
0.048 / day 14.3 days
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EXAMPLE OF ACTIVITY CALCULATION
PROBLEM:
A Cs-137 source had an activity of 800 MBq on Jan. 1, 1973. What will its activity be on July 1, 2030?
no
2A
A METHOD 1: USE Eqn. 1
SOLUTIONS:
Given: T = 57.5 y : Time interval bet Jan. 1, 1973 and July 1, 2030
T1/2 = 30.17 y : Half-life of Cs-137 (from Table of Nuclide)
Ao = 800 MBq : Activity of source on Jan. 1, 1973
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Computation for T
year , mo. , day
Tf = July 1, 2030 2030 , 7 , 1
Ti = Jan 1, 1973 1973 , 1 , 1---------------------------------- 57 yrs , 6 mo. , 0 days
57 yrs + 6 mo / 12 mo/yr + 0 days
57 yrs + 0.5 yrs -------------------------------------
T = Tf - Ti = 57.5 yrs
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91.117.30
5.57
21
y
y
T
Tn
MBqMBqMBq
A 21376.3
800
2
80091.1
Activity of Cs-137 on July 1, 2030 is 213 MBq.
subsituting to Eqn. 1 :noAA
2
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METHOD 2 : USET
oeAA Eqn. 2
Substituting to Eqn. 2 :
A = 800 e – (0.023 / y) x (57.5 y)
= 800 e - 1.32
= 800 x 0.267 = 214 MBq
Activity of Cs-137 on July 1, 2030 is 214 MBq.Note: Difference of 1 MBq is due to rounding adjustments.
12
21
103.217.30
693.0693.0 yxyT
Where
0.023 / y
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Sample computation for T
Given: Tf = 3:20 pm
Ti = 10:00 am
---------------------------------
5 hrs + 20 mins
Convert hrs to mins:
(5 hrs) X (60 mins/hr) = 300 mins
Add the 20 mins:
300 mins + 20 mins = 320 mins = T
3 + 12 = 15 hrs , 20 mins = 10 hrs, 0 mins
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Seatwork: Sample computation for TGiven: Tf = Sept. 12, 1997 Ti = Jan. 1, 1966
--------------------------------- T = Tf – Ti 31 yrs + 8mos. + 11 days
Convert 8 mos. to yrs: (8 mos.) X (1 yr/12 mos.) = 0.6666 yrConvert 11 days to yrs: (11 days) X (1 yr / 365 days) = 0.0301 yrAdd all the numbers converted to yrs: (31 + 0.6666 + 0.0301) yrs = 31.6967 yrs or 31.70 yrs = T
1997 , 9 , 12 1966 , 1 , 1
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Seatwork: 32P has a half life of 14.3 days. At 0 day it
has an activity of 1mCi. Compute for the activity of 32P after the:
1) 1st half life : 2) 2nd half life: 3) 3rd half life: 4) 4th half life:
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EXAMPLE OF HALF-LIFE CALCULATION
PROBLEM : A sample is counted and found to have 952 counts per minute. Seven minutes later it is measured again and has a count of 148 counts per minute. A background measurement gave 6 counts per minute. What is the half-life of the sample?
SOLUTIONS :
GIVEN:
Initial Activity : Ao = 952 - 6 = 946 cpm
Activity 7 minutes after : A = 148 - 6 = 142 cpm
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This means that the number of half-lives (n) in 7 minutes is 2.74.
METHOD 1 : UsenoAA
2
66.6142
9462
A
Aon
n log 2 = log 6.66
74.23010.0
8235.0
2log
66.6logn
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74.2
72
1 T
Using2
1T
Tn
n
TT
21
Therefore, the half-life of the sample is 2.55 mins.
T1/2 = 2.55 minutes
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UseT
o
eA
A
TA
Aln
o
TAA
ln o
TA
Aln o
METHOD 2 :
A = Ao e -T
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7142946
ln
7
66.6ln
7
89.1
271.0
Then use:
21
693.0
T
693.0
21 T
271.0
693.02
1 T
T1/2 = 2.55 minutes
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THANK YOU for your attention.