1 Nuclear Reactions and Radioactivity Part I

2 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

3 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

4 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

5 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.

6 Radioactive Emissions (Radiation) Penetrating Power SymbolEquivalentDescriptionType He Stopped by thick paper (  ) Helium nucleus Dense (+) charged particle 4 2 Stopped by 6mm of Al High speed electron (-) charged particle e 0  0 Alpha Beta Gamma Stopped by several cm of Pb High energy photons Type of energy  0 0

7  Element Symbol K Mass number  Atomic number  Number of protons (p + ) = 19 Number of electrons (e - ) =19 Number of neutrons (n 0 ) =39 – 19 = 10 From this notation we can determine: Nuclear Terminology

8 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.

9 Nuclear Terminology (cont.) The reactant nuclide decay can be summarized by writing a NUCLEAR EQUATION: U Parent nuclide 2 4 He Radiation   Th Daughter Nuclide  

10 Balancing Nuclear Equations Total mass (A) and Total charge (Z) are conserved X A Z Mass: 234 = Example: Th Pa  1 0 e  Is this Nuclear Equation balanced? Charge: 90 = 91 + (-1) Are mass and charge conserved? yes The nuclear equation is balanced.

11 Radioactive Emissions (Radiation) Penetrating Power SymbolEquivalentDescriptionType He Stopped by thick paper (  ) Helium nucleus Dense (+) charged particle 4 2 Stopped by 6mm of Al High speed electron (-) charged particle e 0  0 Alpha Beta Gamma Stopped by several cm of Pb High energy photons Type of energy  0 0

12 Penetrating Power of Radioactive Emissions

13 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 RaRnHe   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.

14 Alpha Decay ? Am He  Example: Balancing Nuclear Equations Mass No. (A):241= A = Atomic No. (Z):95= Z = What element corresponds to an atomic number of 93? X A Z From the periodic table, Np corresponds to Z = 93 Np Final Answer:

15 Types of Radioactive Decay Beta 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 NiCue   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.

16 Beta Decay ? Th e  Example: Balancing Nuclear Equations Mass No. (A):234= A = Atomic No. (Z):90= Z + (-1) 90 = 91 + (-1) What element corresponds to an atomic number of 91? X A Z From the periodic table: Pa corresponds to Z = 91 Pa Final Answer:

17 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 UHeTh  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.

18 Gamma Decay ? Pb *   Example: Balancing Nuclear Equations Mass No. (A):209= A = Atomic No. (Z):82= Z = What element corresponds to an atomic number of 82? X A Z From the periodic table, Pb corresponds to Z = 82 Pb Final Answer: (The * in the equation indicates the nucleus is in an excited state)

19 Positron Decay: NaeNe  HgeAu    Other Types of Radioactive Decay Electron Capture: (inner-orbital electron is captured by the nucleus)

20 Nuclear Stability 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 X A Z Stable: If 0 < Z < 20 & N/Z Ratio = 1.0 or 20 < Z < 83 & Z is even Unstable: If Z > 83

21 Zone of Stability

22 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 (a) (b)(c)(d)

23 Decay Series A radioactive nucleus reaches a stable state by a series of steps ThPb seriesofdecays  Example 1: Thorium (Th) decay into Lead (Pb). This decay series consists of 10 decays (6 alpha decays and 4 beta decays)

24 Decay Series Example 2: Uranium to Lead

25 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 d/s Activity = number of decays = λ N time Where: λ = Decay constant N = Number of nuclei

26 Half-life (t 1/2 ) 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 = λ λ

27 Example: Decay of a 10.0g sample of C-14 Half-life (t 1/2 )

28 Half-life (t 1/2 ) Example: Decay of a 10.0g sample of Co-60

29 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 14 C and 32 P into foods, metabolic pathways can be studied. 201 Th 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 Na 131 I. Examples:

30 Examples of Radioactive Tracers NuclideHalf-lifeArea of body studied 131 I8.05 daysThyroid 59 Fe45.1 daysRed Blood Cells 87 Sr2.8 hoursBones 133 Xe5.3 daysLungs

31 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, 99 Tc decays to ground state by gamma emission. The rate constant for decay is 1.16 x d/hr. What is the half-life of this nuclide?

32 Known(s): λ = 1.16 x d/hr Unknown(s): t 1/2 Half-life(t 1/2 ) of technetium-99 = 5.97 hr Equation(s): t 1/2 = λ ln x d/hr d = Calculating Half-life (Example Problem)

33 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 24 NaCl solution whose activity is 2.5 x 10 9 d/s, how much of the activity is present in the patient’s body and excreted fluids after 4.0 days?

34 Calculating Activity (Problem Solution) Known(s): t 1/2 = 15 hr Initial Activity = 2.5 x 10 9 d/s Time elapsed = 4.0 days Unknown(s): Activity after 4.0 days Decay constant ( λ) Equation(s): N = N i e - λt N = Activity λ λ = ln 2 t 1/2

35 Calculating Activity (Problem Solution) Solve: λ = ln 2 t 1/2 = 15 hr = hrs -1 N i = Activity i λ = 2.5 x 10 9 d/s λ

36 Calculating Activity (Problem Solution) Solve:N = N i e - λt Activity λ = 2.5 x 10 9 d/s λ e -( hrs -1 )(4 days) x Activity = 2.5 x 10 9 d/s e -( hrs -1 )(96 hrs) x Activity = 2.5 x 10 9 d/s x Activity of Na-24 after 4 days = 3.0x10 7 d/s