Transcript Nuclear Reactions and Radioactivity

Nuclear Reactions
and
Radioactivity
Part I
1
Discovery of Radioactivity
Antoine-Henri Becquerel (1896)
While experimenting with uranium compounds, he
discovered that:
• The compounds emit penetrating radiation that
produces images on photographic film
• This phenomenon occurs even when wrapped in
paper and stored in the dark
• Radiation creates an electric discharge in air,
providing a way to measure its intensity
2
Discovery of Radioactivity
Marie & Pierre Curie (Early 1900s)
• Found that thorium minerals also emit radiation
• Showed that the intensity of radiation is directly
proportional to the concentration of the element in
the mineral, not the nature of the compound in
which element occurs
• Named this behavior radioactivity
• Discovered the elements polonium and radium
3
Discovery of Radioactivity
Rutherford & Colleagues (1902)
• Discovered that elements other than radium formed
when radium emitted radioactive emissions
• Proposed that radioactive emissions cause one
element to change into another
• This proposal was met with skepticism (sounded
similar to alchemy)
• Led to an understanding of the three types of
radioactive emissions:
alpha, beta, and gamma
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Radioactivity
The spontaneous breakdown of the nuclei
of atoms accompanied by a release of
some type of radiation. (The atom’s nuclei
are trying to become more stable.
5
Radioactive Emissions (Radiation)
Type
Alpha
Beta
Description
Equivalent
Dense (+)
charged
particle
(-) charged
particle
Helium
nucleus
Gamma Type of
energy
High
speed
electron
High
energy
photons
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Symbol
Penetrating
Power
He
Stopped by
thick paper
4
2
()
0
-1
e
0
b
-1
0
0
g
Stopped by
6mm of Al
Stopped by
several cm
of Pb
Nuclear Terminology
Mass number 
Atomic number 
39
K
19
 Element Symbol
From this notation we can determine:
Number of protons (p+) = 19
Number of electrons (e-) = 19
Number of neutrons (n0) = 39 – 19 = 10
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Nuclear Terminology (cont.)
Nuclide - a nuclear species with specified
numbers of protons and neutrons
Reactant Nuclide - Parent Nuclide
Product Nuclide - Daughter Nuclide
When a reactant nuclide decays, a lower
energy product nuclide is formed and the
excess energy is emitted as radiation.
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Nuclear Terminology (cont.)
The reactant nuclide decay can be summarized by
writing a NUCLEAR EQUATION:
238
92
U
Parent nuclide


4
He
+
234
Radiation
+
Daughter Nuclide
2
9
Th
90
Balancing Nuclear Equations
A
Z
X
Total mass (A) and
Total charge (Z) are conserved
Example:
Is this Nuclear Equation balanced?
234
 234Pa + -0e
Th
90
91
1
Are mass and charge conserved?
yes
Mass: 234 = 234 + 0
Charge: 90 = 91 + (-1)
yes
The nuclear equation is balanced.
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Radioactive Emissions (Radiation)
Type
Alpha
Beta
Description
Equivalent
Dense (+)
charged
particle
(-) charged
particle
Helium
nucleus
Gamma Type of
energy
High
speed
electron
High
energy
photons
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Symbol
Penetrating
Power
He
Stopped by
thick paper
4
2
()
0
-1
e
0
b
-1
0
0
g
Stopped by
6mm of Al
Stopped by
several cm
of Pb
Penetrating Power of Radioactive Emissions
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Types of Radioactive Decay
Alpha Decay (): emits an alpha particle.
An alpha particle is composed of 2 protons and 2
neutrons bound together, which is the same as a
helium nucleus.
226
222
4
+

Ra
Rn
88
86
2
He
Application:
Home smoke alarms use Americium-241 which emits
alpha particles. Particulates in the air (smoke) prevent
the particles from reaching a detector, which sets off
the alarm.
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Example: Balancing Nuclear Equations
Alpha Decay
241
95
A
?
Z

Am
X
+
4
2
Mass No. (A):
241
241
= A + 4
= 237 + 4
Atomic No. (Z):
95
95
= Z + 2
= 93 + 2
He
What element corresponds to an atomic number of 93?
From the periodic table, Np corresponds
to Z = 93
Final Answer:
237
93
Np
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Types of Radioactive Decay
Beta Decay (b): emits a beta particle (an electron). In
beta decay a neutron in the nucleus changes into a
proton, an electron and a neutrino and ejects the high
speed electron (beta particle) from the nucleus.
63
28
Ni 
63
29 Cu
+
0
-1e
Application:
Carbon 14 Dating - By examining the change in carbon
due to the loss of beta particles we can determine the
age of a biological substance.
15
Example: Balancing Nuclear Equations
Beta Decay
234
Th
90

A
?
Z
X
+
0
e
-1
Mass No. (A):
234
234
= A + 0
= 234 + 0
Atomic No. (Z):
90
90
= Z + (-1)
= 91 + (-1)
What element corresponds to an atomic number of 91?
From the periodic table:
Pa corresponds to Z = 91
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Final Answer:
234
91
Pa
Types of Radioactive Decay
Gamma ray emission (g): occurs when an excited
nucleus releases a high energy photon. It can result
from the spontaneous fission (splitting) of an atom. In
this process the excess energy is emitted as a gamma
ray.
238
92 U
 42 He +
234
90Th
+ 2 00 g
Application:
Food Preservation – Due to the high penetration of
gamma rays, they can be directed into a food product
to kill bacteria without inducing measurable radiation
in the food or affecting its nutritional value.
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Example: Balancing Nuclear Equations
Gamma Decay 209
Pb*
82

A
?
Z
X
+
0
0
g
(The * in the equation indicates the nucleus is in an excited state)
Mass No. (A):
209
209
= A + 0
= 209 + 0
Atomic No. (Z):
82
82
= Z + 0
= 82 + 0
What element corresponds to an atomic number of 82?
From the periodic table,
Pb corresponds to Z = 82
Final Answer:
209
82
Pb
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Other Types of Radioactive Decay
Positron Decay:
22
Na
11
 0e +
1
22
Ne
10
Electron Capture:
(inner-orbital electron is captured by the nucleus)
201
Hg
80
+
0
- 1e

19
201
Au
79
+ 0g
0
Nuclear Stability
A
Z
X
Determined by:
Mass Number (A): number of protons + neutrons
Atomic Number (Z): number of protons
Number of neutrons (N): where N=A-Z
Ratio of neutrons to proton: N/Z
Unstable:
If
Z > 83
Stable:
If
or
0 < Z < 20 & N/Z Ratio = 1.0
20 < Z < 83 & Z is even
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Zone of Stability
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Sample Problems: Predicting Stability
(a)
18
10
Ne
N/Z = 0.8 UNSTABLE
(b)
32
16
S
N/Z = 1.0 & Z<20 STABLE
(c)
236
90
(d)
123
56
Th
Ba
Z>83 UNSTABLE
N/Z= 1.20 & Z is even
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STABLE
Decay Series
A radioactive nucleus reaches a stable state by a
series of steps.
Example 1: Thorium (Th) decay into Lead (Pb).
series of decays 208
232


Th
Pb
90
82
This decay series consists of 10 decays
(6 alpha decays and 4 beta decays)
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Decay Series
Example 2: Uranium to Lead
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Rate of Nuclear Decay
Radioactive nuclei decay at a characteristic rate,
regardless of the chemical substance in which they
occur. A measure of this decay is activity.
Activity = number of decays = λ N
time
Where:
λ = Decay constant
N = Number of nuclei
Units:
SI unit of activity: becquerel (Bq)
Bq = 1 disintegration/second (d/s)
1 curie (Ci) = 3.7 x 1010 d/s
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Half-life (t1/2)
The time it takes for half the nuclei present to decay.
Half the number of nuclei remain after each half-life.
Half-life for a nuclear change and a chemical change
are the same.
Half-life is related to the activity constant:
t 1/2 = ln 2 = 0.693
λ
λ
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Half-life (t1/2)
Example:
Decay of
a 10.0g
sample
of C-14
27
Half-life (t1/2)
Example:
Decay of
a 10.0g
sample
of Co-60
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Medical Applications of Radioactive
Nuclides as Radioactive Tracers
Radiotracers: radioactive nuclides that are introduced
into organisms via food or drugs; the pathway of the
radiotracer can be “traced” by monitoring their
radioactivity.
Examples:
• By incorporating 14C and 32P into foods, metabolic
pathways can be studied.
• The thyroid gland can be monitored by a scanner
after patients drink a solution containing Na131I.
• 201Th can be used to assess damage to heart caused
by a heart attack by determining the amount of Th
present in heart muscle tissue because Th is
concentrated in healthy muscle tissue.
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Examples of Radioactive Tracers
Nuclide
Half-life
Area of body
studied
I
8.05 days
Thyroid
59Fe
45.1 days
Red Blood Cells
87Sr
2.8 hours
Bones
133Xe
5.3 days
Lungs
131
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Calculating Half-life (Example Problem)
Technetium-99 is used to form images of internal organs
in the body and is often used to determine heart damage.
This nuclide, 99Tc decays to ground state by gamma
emission. The rate constant for decay is 1.16 x 10-1 d/hr.
What is the half-life of this nuclide?
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Calculating Half-life (Example Problem)
Known(s): λ = 1.16 x 10-1 d/hr
Unknown(s): t1/2
Equation(s): t1/2 =
ln 2
λ
0.693 d
=
1.16 x 10-1 d/hr
Half-life(t1/2) of technetium-99 = 5.97 hr
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Calculating Activity (Example Problem)
Sodium-24 has a half-life of 15 hours and is used to
study blood circulation. If a patient is injected with a
24NaCl solution whose activity is 2.5 x 109 d/s, how
much of the activity is present in the patient’s body
and excreted fluids after 4.0 days?
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Calculating Activity (Problem Solution)
Known(s): t1/2 = 15 hr
Initial Activity = 2.5 x 109 d/s
Time elapsed = 4.0 days
Unknown(s): Activity after 4.0 days
Decay constant (λ)
Equation(s): N = Ni e-λt
Activity
N=
λ
ln 2
λ= t
1/2
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Calculating Activity (Problem Solution)
Solve:
ln 2
0.693 =
λ=
0.046 hrs-1
=
t1/2
15 hr
Activityi
Ni =
λ
=
35
2.5 x 109 d/s
λ
Calculating Activity (Problem Solution)
Solve:
N = Ni e-λt
Activity
2.5 x 109 d/s x e-(0.046 hrs-1)(4 days)
=
λ
λ
Activity = 2.5 x
109
d/s x e
-(0.046 hrs-1)(96 hrs)
Activity = 2.5 x 109 d/s x 0.012
Activity of Na-24 after 4 days = 3.0x107 d/s
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