1 nuclear reactions and radioactivity part i. 2 antoine-henri becquerel (1896) while experimenting...
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Nuclear Reactionsand
Radioactivity
Part I
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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
Discovery of Radioactivity
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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
Discovery of Radioactivity
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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.
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Radioactive Emissions (Radiation)
Penetrating Power
SymbolEquivalent
Description
Type
He Stopped by thick paper ()
Helium nucleus
Dense (+) charged particle
42
-1
-1
Stopped by 6mm of Al
High speed electron
(-) charged particle
e0
0
Alpha
Beta
Gamma
Stopped by several cm of Pb
High energy photons
Type of energy 0
0
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Element SymbolK
39
19
Mass number
Atomic number
Number of protons (p+) =
19Number of electrons (e-)
=19Number of neutrons (n0) =39 – 19 = 10
From this notation we can determine:
Nuclear Terminology
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Nuclear Terminology (cont.)
Nuclide - a nuclear species with specified numbers of protons and neutrons
Reactant Nuclide - Parent NuclideProduct 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:
92
238U
Parent nuclide
2
4He
Radiation
90
234Th
Daughter Nuclide
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Balancing Nuclear Equations
Total mass (A) andTotal charge (Z) are conservedX
AZ
Mass: 234 = 234 + 0
Example:
90234Th 91
234Pa 10e
Is this Nuclear Equation balanced?
Charge: 90 = 91 + (-1)
Are mass and charge conserved?
yes
yes
The nuclear equation is balanced.
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Radioactive Emissions (Radiation)
Penetrating Power
SymbolEquivalent
Description
Type
He Stopped by thick paper ()
Helium nucleus
Dense (+) charged particle
42
-1
-1
Stopped by 6mm of Al
High speed electron
(-) charged particle
e0
0
Alpha
Beta
Gamma
Stopped by several cm of Pb
High energy photons
Type of energy 0
0
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Penetrating Power of Radioactive Emissions
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Types of Radioactive DecayAlpha 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.
88226
86222
24Ra Rn 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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Alpha Decay ?
95
241
2
4Am He
Example: Balancing Nuclear Equations
Mass No. (A): 241 = A + 4 241 = 237 + 4
Atomic No. (Z): 95 = Z + 295 = 93 + 2
What element corresponds to an atomic number of 93?
XA
Z
From the periodic table, Np corresponds to Z = 93
Np23793
Final Answer:
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Types of Radioactive DecayBeta Decay (): 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.
2863
2963
10Ni Cu e
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.
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Beta Decay ?
90
234
-1
0Th e
Example: Balancing Nuclear Equations
Mass No. (A): 234 = A + 0 234 = 234 + 0
Atomic No. (Z): 90 = Z + (-1)90 = 91 + (-1)
What element corresponds to an atomic number of 91?
XA
Z
From the periodic table: Pa corresponds to Z = 91
Pa23491
Final Answer:
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Gamma ray emission (): 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.
92238
24
90234
002U He Th
Types of Radioactive Decay
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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Gamma Decay ?
82
209
0
0Pb*
Example: Balancing Nuclear Equations
Mass No. (A): 209 = A + 0 209 = 209 + 0
Atomic No. (Z): 82 = Z + 082 = 82 + 0
What element corresponds to an atomic number of 82?
XA
Z
From the periodic table, Pb corresponds to Z = 82 Pb209
82
Final Answer:
(The * in the equation indicates the nucleus is in an excited state)
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Positron Decay:
1122
10
1022Na e Ne
80201
10
79201
00Hg e Au
Other Types of Radioactive Decay
Electron Capture: (inner-orbital electron is captured by the nucleus)
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Nuclear Stability
Determined by:Mass Number (A): number of protons + neutronsAtomic Number (Z): number of protonsNumber of neutrons (N): where N=A-ZRatio of neutrons to proton: N/Z
XAZ
Stable:If 0 < Z < 20 & N/Z Ratio = 1.0or 20 < Z < 83 & Z is even
Unstable:If Z > 83
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Zone of Stability
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Sample Problems: Predicting Stability
N/Z = 0.8 UNSTABLE
N/Z = 1.0 & Z<20 STABLE
Z>83 UNSTABLE
N/Z= 1.20 & Z is even STABLE
Ne1810
(a)
S3216
(b)
Th23690
(c)
Ba12356
(d)
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Decay Series
A radioactive nucleus reaches a stable state by a series of steps.
90232
82208Th Pb
series of decays
Example 1: Thorium (Th) decay into Lead (Pb).
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.
Units: SI unit of activity: becquerel (Bq) Bq = 1 disintegration/second (d/s) 1 curie (Ci) = 3.7 x 1010 d/s
Activity = number of decays = λ Ntime
Where: λ = Decay constant N = Number of nuclei
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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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Example:Decay of a 10.0g sample of C-14
Half-life (t1/2)
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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.
• By incorporating 14C and 32P into foods, metabolic pathways can be studied.
• 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.
• The thyroid gland can be monitored by a scanner after patients drink a solution containing Na131I.
Examples:
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Examples of Radioactive Tracers
Nuclide Half-life Area of body studied
131 I 8.05 days Thyroid 59Fe 45.1 days Red Blood
Cells 87Sr 2.8 hours Bones
133Xe 5.3 days Lungs
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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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Known(s): λ = 1.16 x 10-1 d/hr
Unknown(s): t1/2
Half-life(t1/2) of technetium-99 = 5.97 hr
Equation(s): t1/2 =
λln 2
1.16 x 10-1 d/hr0.693 d=
Calculating Half-life (Example Problem)
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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 hrInitial Activity = 2.5 x 109 d/sTime elapsed = 4.0 days
Unknown(s): Activity after 4.0 days Decay constant (λ)
Equation(s): N = Ni e-λt
N =
Activityλ
λ = ln 2t1/2
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Calculating Activity (Problem Solution)
Solve:
0.693λ = ln 2t1/2
= 15 hr
= 0.046 hrs-1
Ni =Activityi
λ =2.5 x 109 d/s
λ
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Calculating Activity (Problem Solution)
Solve: N = Ni e-λt
Activityλ
= 2.5 x 109 d/sλ
e-(0.046 hrs-1)(4 days)x
Activity = 2.5 x 109 d/s e-(0.046 hrs-1)(96 hrs)x
Activity = 2.5 x 109 d/s 0.012x
Activity of Na-24 after 4 days = 3.0x107 d/s