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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