Download Ionizing Radiation

Study Guide 7: Ionizing Radiation
Text: Chapter 6, sections 1-11 (more than
described in Study Guide), plus text 2.5 and lab
manual section 7A-1 (on inverse-square law).
Upcoming quizzes:
Quiz 4 (final day, Friday, March 14, 2008) you
should know:
Material from SG 5 and SG 6.
Quiz 5 (final day, Friday, March 28, 2008) you
should know:
Material from Computer lab exp. 7 (Radiation) and
exp. 8 (Electricity): in particular
From semilog and log-log graphs of provided data
(exp. 7);
1. How to calculate the linear attenuation
coefficient.
2. Determine half-time T1/2.
3. Study the exponent in power-law decay from a
point radioactive source.
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4. Apparatus used in exp. 8 (bottom of page 8-1 of
lab manual) and how to arrange this to study
parallel and series circuits.
5. Using 4 to calculate current and resistances.
NUCLEAR EQUATIONS
Nuclear equations are derived from conservation
laws:
• mass/energy
• charge
• angular momentum (spin)
Radioactive decay is how UNSTABLE NUCLEI can
increase their binding energy (i.e., how strongly
nuclear particles – protons and neutrons – stick to
each other).
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e.g. The α-decay of Radium into Radon.
226
222
4
Ra →
Rn + He + γ
88
86
2
Radium
(Parent)
Radon
(Daughter)
A gamma ray,
sometimes required
for energy
conservation
He-4 nucleus (also called
an α particle)
The numbers in the above equation have important
meanings:
The mass number A = (# of protons + # of neutrons).
Defines the ISOTOPE.
226
Ra →
88
222
4
Rn + α + γ
86
2
The CHEMICAL ELEMENT is defined by the atomic
number Z = (# of protons)
In general …
AX =
Z
# of protons + # of neutrons
X
# of protrons
3
Little things you should know…
1. THE MASS NUMBER ≈ MASS OF 1.0 MOLE ( NA =
Avogadro’s number) OF THE ISOTOPE IN GRAMS.
2. Symbol X = Fe, H, Au, etc., is redundant (already
determined by the atomic number Z).
3. Alternative symbols are occasionally used, e.g., α
(instead of He) for alpha particles.
4. A nucleus with extra energy (excited) is designated as
follows…
222Rn*
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This extra energy can be transferred to emitted
radiation as kinetic energy in an α or β particle or as a γ
photon.
5. Energies can be given in electron volts, 1 eV = 1.60 ×
10-19 Joules.
6. Masses can be given as REST ENERGIES using E=mc2
law. E.g., Eproton=(1.60×10-27)×(3.0×108)2 ≅ 939MeV.
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TYPES OF RADIATION
• different types create different decays.
• we focus on 3 types
1) Alpha (α) Radiation: emission of an α particle
(Helium-4 nucleus) from a larger nucleus.
e.g.
226
222
4
Ra →
Rn + α + γ
88
86
2
222
218
4
Rn →
Po + α + γ
86
84
2
CASCADE,
SERIES
OR
CHAIN
e.g.
(
251
247
4
Fm →
Cf + α (K = 6.833 MeV ) + γ 55 keV
100
98
2
α
This denotes the kinetic energy of the α particle
The α and γ rays emitted by an isotope have a specific
energy spectrum that can be used to identify it. E.g., in
the decay above, 87% of the α particles produced have
6.833 MeV of kinetic energy and 1.5% have 7.305 MeV.
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)
GENERAL FORMULA FOR α DECAY
AX
Z
→
A - 4Y + 4α + (γ )
Z-2
2
WHY ALPHA PARTICLES?
Other nuclei can be emitted (besides 4He) but is rare.
For example,
223Ra
88
→
209Pb + 14C
82
6
4
He emission gets rid of more energy per nucleon.
All these emissions of smaller nuclei from the parent
are really examples of fission because the nucleus
splits into two more stable parts.
6
In the table, the numbers in the upper left corners
are the atomic numbers (not mass numbers). You
don’t have to memorize this table!
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2) Beta (β) Radiation – Actually 3 different processes;
electron emission, positron emission and electron
capture. We only consider electron (β) emission from
the nucleus. Can also occur with γ radiation (like α
radiation).
FOR EXAMPLE
90Sr
38
→
90 Y + 0β + ν
39
-1
(No γ)
Anti-neutrino
17 N
7
17
0
O + β +ν + γ
→
8
-1
3H →
1
3He + 0β + ν
2
-1
(No γ)
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Little things you should know…
You ONLY have to remember that a γ MAY be produced
by β and α decays. When a decay will produce a γ is
beyond the scope of this course.
Remember that ALL electron emissions (the only β decays
we study) produce an anti-neutrino.
The anti-neutrino is required to conserve momentum.
Unlike α’s, β’s are not emitted only at specific energies
but over a range from 0 to a maximum value because the
energy is shared with the anti-neutrino.
GENERAL FORMULA FOR β DECAY
A
A
0
X→
Y + β +ν
Z
Z +1
-1
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3) Gamma (γ) Radiation: Composed of high energy
photons (light) which have no rest mass, therefore
parent nucleus = daughter. Often occurs in
conjunction with α and β radiation (which do cause the
daughter to change from the parent). For example,
Excited state
60 Co →
27
STEP 1
60Ni*
28
60 Ni* + 0β + ν
28
-1
→60
Ni
+
2
γ
28
STEP 2
It is also possible for a nucleus to simply emit a gamma
photon. In such cases the general form is,
A *
X
Z
→
A
X+γ
Z
10
Some examples:
242Pu
94
→
65
Ni
28
→
24 Na
11
→
238U + ?
92
?
0
+ β +ν
-1
24Mg
12
+ ?
RADIOACTIVITY AND HALF LIVES
Over time, a parent sample will gradually turn into a
daughter. In this section we will learn how to determine
the rate at which a given sample emits radiation and the
rate at which radioactive substances are removed from
physical and biological systems.
Radioactive decay is a random process. Each parent atom
has the same probability, λp, ("p" for physical) of
decaying per unit time, Δt. If Δt is short compared to
the half life (see below for def’n of half life), the number
of events that will occur (= the change in the number of
parent nuclei, ΔN) is,
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ΔN = − NλpΔt
(1)
the "-" indicates that N decreases
The parent number change, ΔN, per unit time Δt is then
(2)
ΔN/Δt = −Nλp
For tiny changes, Eq. (1) can be expressed as,
dN = - Nλp dt .
Therefore with a little integral calculus,
⎛ N ⎞
ln ⎜
⎟=
⎜N0 ⎟
⎝ ⎠
−λ t
p ,
(3)
where N0 = initial # of parent nuclei and N = final # of
parent nuclei. Rearranging Equation (3) a bit we get the:
Physical Decay Law
N = N 0 e
− λp t
This equation only applies in the absence of
biological decay.
(4)
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1. Number of nuclei decreases exponentially over time.
2. Probability/unit time, λp, (dimension: time-1) = the
DECAY CONSTANT.
3. The decay constant is related to the half-life, T1/2.
The half life is the time it takes half of the sample to
undergo a particular decay. When t = pT1/2 then N =
N0/2. Putting this in Eq. (4) gives
N =
N0
2
= N0 e
(
− λp pT
1/2
⎛1⎞
∴ − ln ⎜ ⎟ = 0.693 = λp
⎝2⎠
)
'
( )
T
p
1/2
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∴ Physical Half Life
p T1/2
⎛ 0.693 ⎞
=⎜
⎟
⎜ λp ⎟
⎝
⎠
(5)
Note from the equation N = N0e−λpt, by differentiating
with respect to time t, we get
-λpt
dN
= -λpN e
= -λpN
0
dt
The quantity |dN/dt| is the count-rate, the number of
disintegrations per unit time, which is sometimes denoted
A (for “activity”). We see that A also decreases
exponentially with time, i.e.
−λpt
A= A e
0
(6)
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Sample problem 1
232
U α decays with a half life of 72 years into
Polonium 228. If a sample of 232U has an initial
count-rate of 2.0 × 104 counts/s (or C/s), what is its
count-rate after 50 years?
From equation (5),
λp = 1/(pT1/2) = 0.693/72 = 0.00963 year-1
Note, we can keep this unit for λp if time is also
expressed in units of years. From eq. (6),
A = A0 e−λpt
∴A = (2.0 × 104 C/s) exp(−0.00963(50))
= (2.0 × 104 C/s) exp(−0.48) = (2× 104 C/s) (0.62)
= 1.24 × 104 C/s
BIOLOGICAL ACTIVITY
Foreign materials in living organisms are expelled due to
the biological activity (e.g. urination, breathing, sweating)
of the organism.
Biological activity = # of parent atoms ejected
by biological processes/unit time.
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Things to remember…
1. Random. Don’t know which atoms are excreted.
2. Biological decay depends on BOTH the isotope AND
the ORGANISM.
3. Biological decay is governed by exponential laws. If
the physical decay = 0, the # of foreign atoms N in a
biological system at time t is,
Biological Decay Law…
N = N e
0
− λb t
This equation only applies in the absence of
radioactive decay.
The symbols have the same meaning as for the physical
decay equation. The biological decay constant, λb,
replaces the λp.
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The biological half life bT1/2 is
T
b 1/2
⎛
⎞
0.693 ⎟
⎜
= ⎜
λb ⎟
⎝
⎠
Combining Physical and Biological Activities
Radioactive isotopes in living organisms are affected by
both physical and biological processes. The effective
decay constant, λe, is given by,
λe = λp + λb
In terms of half-lives,
1
T
e 1/2
=
1
T
p 1/2
+
1
T
b 1/2
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The number of atoms N present at time t is
N = N e
0
− λe t
= N e
0
(
)
− λp + λb t
This equation ALWAYS applies
This equation can always be used because if there is no
biological decay, then λb = 0, while if there is no physical
decay then λp = 0.
Sample Problem 2: Celtic Remains?
THE FACTS:
1) The natural occurrence of carbon-14 on earth (created
by cosmic neutron bombardment from nitrogen in the
air) is responsible for 15.3 disintegrations per minute
per gram of total carbon.
2) Carbon-14 β-decays with a half-life of 5370 years.
3) Charcoal samples from ancient (sacrificial?) fire pits
at Stonehenge are found to have an activity of 9.65
disintegrations per minute per gram of carbon.
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THE QUESTIONS:
A) What are the end products of this decay?.
B) Based on these findings, how old is Stonehenge?
(B) Ans. The trees died 3571 years ago.
SAMPLE PROBLEM 3 (Similar to “killer question” 6-8
in text)
A radioactive isotope has a physical half-life of 65 min.
An initial measurement of it shows a count-rate of 15,500
Cs−1. Thirty-nine minutes later the sample is administered
to a mouse. One hour after this, the (unfortunate) mouse
has a total body count of 1025 Cs−1. What is the biological
half-life of the isotope in the mouse?
Ans. bT1/2 = 25.1 min
ABSORPTION OF RADIATION
Absorption in living tissue (Bad!):
• Ionizing radiation damages molecules: can result in cell
death.
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• Damaged DNA causes genetic mutations (AND
superheros!).
REDUCTION OF RADIATION EXPOSURE:
1)
DISTANCE AND TIME
In the absence of an absorptive barrier, activity A (C in
lab manual) obeys an inverse square law:
⎛K⎞
A∝⎜
2⎟
⎝r ⎠
r = distance from source
A person three times further away from the source,
receives 1/9 as much exposure. This does not include
absorption by air (or other material) that intervenes.
2)
ABSORPTION BY MATTER
Alpha, beta and gamma radiation each behave differently
in the presence of barriers. Let's assume the incident
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radiation has intensity I0 or activity A0 (text and lab
manual use N).
GAMMA RAY (OR X-RAY)
absorption is similar to
radioactive decay or absorption of light. Each γphoton can be absorbed at any point along the path
through the barrier. The probability of
absorption/unit length/photon is μ (linear attenuation
coefficient).
For a beam of intensity, I, traversing a distance dx along
its path, the change in intensity will be (compare to the
expression for decay) dI = - μIdx
With a little calculus, we see…
⎛ I⎞
ln⎜⎜ ⎟⎟ = − μx
⎝ I0 ⎠
or…
I = I0e
− μx
μ depends on:
1)
Energy of the photon - The more energetic the
photon is, the more deeply it penetrates.
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2)
Nature of the absorbing material.
However, if we define the mass attenuation coefficient,
μ m,
μ = μmρ
where ρ is the density (mass per volume), we find that μm
is largely independent of the material. It still depends on
the gamma ray (or x-ray) energy.
For a 100 keV x-ray, 3.0 cm of iron cuts the intensity by
a factor of two. Given that human tissue is about 1/10 as
dense as iron, these x-rays would penetrate about 30 cm
of human tissue. As one might expect, this is about the
energy of a medical x-ray.
ABSORPTION OF CHARGED PARTICLES (α AND β) occurs
everywhere along the path (rather than at just one point
as for γ-rays). The mean distance traveled by the
particles is represented by the “penetration depth”,
defined below. The activity drops sharply beyond this
distance for particle beams of a single energy. Alpha and
beta rays are much less penetrative than γ-rays.
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ABSORPTION OF α PARTICLES
1)
High mass means very little deflection with each
collision.
2)
Short intense tracks in target material.
3)
Intense ionization (damage) along path.
4)
A 5.3 MeV alpha has a penetration depth of 30 mm
in air and about 30 micrometers in water or human
tissue.
5)
A major hazard if adjacent to the skin (α emitter
was swallowed), but can be easily shielded.
ABSORPTION OF β PARTICLES
1)
Low mass means beta particles are strongly
deflected with each collision.
2)
Leave long and tortuous (twisted) tracks in target
material.
3)
Mean “penetration depth” of β particles of energy,
E > 0.6 MeV is given by:
Penetration depth = [5.42 E(MeV) – 1.33]/ρ
where ρ = density (kg/m3)
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4)
Note that that higher energy beams penetrate more
deeply.
5)
Can penetrate several meters of air (see text
example 6-2) and are hazardous at a distance.
Sample Problem 4: Lead Shielding:
A researcher working at some distance from a γemitting radioisotope would be exposed to 100 times
the allowable yearly limit of radiation over one hour
(lethal!). What thickness of lead shielding is required
to reduce the exposure to one tenth the yearly limit
for an hour-long exposure at the same distance? The
mass attenuation coefficient of lead is 0.0121 m2/kg
for the γ-photons involved and the density of lead is
1.14 × 104 kg/m3.
Ans. 5.0 cm
Sample Problem 5: Beta radiation in air and with
shielding:
What is the penetration depth in air of β radiation
from Bismuth-210? The E for 210Bi β’s is 1.17 MeV
and the density of air is 1.3 kg/m3. Would 1.0 mm
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of Pb (density 1.14 × 104 kg/m3) stop this radiation?
How about 1.0 mm of Al (density 2.7 × 103 kg/m3)?
Ans. D = 3.88 m
Yes
No
Some final things
Radiation damage at the molecular level: most radiation
causes ionization of molecules, disrupting chemical bonds
(such as in DNA and proteins), producing “free radicals”
and high-speed electrons which can cause other
ionizations.
Qualitative only: (text Sections 6.10 and 6.11 are not very
clear):
Effects of radiation on an organism can be
described by the following concepts: Exposure,
Absorbed Dose and Equivalent Dose.
(1) Exposure = amount of radiation energy which is
incident per unit mass of “victim”.
(DOES NOT DEPEND ON ABSORBER)
(Ignore “mysterious” question 1, especially 1(c), on SelfTest IV. This mixes up exposure, absorbed dose and
equivalent dose. Also skip text exercise 6-4)
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(2) Absorbed Dose = amount of radiation energy which is
absorbed per unit mass of “victim”.
(DOES DEPEND ON THE PROPERTIES OF THE
ABSORBER −E.G., IONIZATION ENERGIES)
(3) Equivalent Dose: measures damage produced by
absorbed dose. Depends on the interaction of the
radiation type (e.g., α) with the “victim” material.
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