NUCLEAR CHEMISTRY

Discovery of Radiation Roentgen (1895) Discovered a mysterious form of radiation was given off even without electron beam. This radiation could pass through paper and other objects but not dense materials (lead, bone). Called them X-rays Studied fluorescent materials that glowed when hit with a beam of electrons.

Becquerel (1896) Studied fluorescent minerals containing uranium. Discovered radioactivity by accident on a cloudy day: Thought that an external source was needed to produce the mysterious radiation. Found that uranium emits radiation without external source.

Thought radioactivity was a property of heavy elements. Pierre and Marie Curie During study, discovered new radioactive elements: Polonium and Radium. Wondered how small mass can give off large amount of energy: Explained by Einstein with E=mc 2.

Studied radioactivity and named types of nuclear radiation. Rutherford Discovered that elements decay into other elements after emitting nuclear radiation. Called it Nuclear Decay. Gold foil experiment revealed that the mass of an atom is concentrated in the nucleus (atom is mostly space)

Atomic Structure Recall: Atoms – consist of a positively charged nucleus, which has protons and neutrons. IsotopeSymbol# protons# neutronsAtomic Mass Mo-101 H-2 C-14 U-238 Bi-210 He Isotope – atoms of the same chemical element that have a different number of neutrons. Each isotope of a given element is designated by the total number of its protons plus its neutrons.

Difference between the mass of an atom and the mass of its individual particles amu amu Mass Defect Nuclear Forces

Nuclear Binding Energy Energy released when a nucleus is formed from nucleons. High binding energy = stable nucleus. E = mc 2 E:energy (J) m:mass defect (kg) c:speed of light (3.00×10 8 m/s)

Nuclear Binding Energy Unstable nuclides are radioactive and undergo radioactive decay.

Alpha Emission parent nuclide daughter nuclide alpha particle Numbers must balance!! occurs when the nucleus has too many protons which cause excessive repulsion.

Alpha Emission Ex. Plutonium-239 undergoes alpha decay Atomic Mass: = Atomic #: = Masses must be equal = Conservation of mass

Ex. Polonium-210 undergoes alpha decay to produce this daughter nuclide + Alpha Emission Atomic Mass: 210 = A4 + Atomic #: 84 = Z = A4 - = = Z2 - = 82 =

Beta Emission electron occurs when the neutron to proton ratio is too great.

Ex. Polonium-210 undergoes beta decay to produce this daughter nuclide + Beta Emission Atomic Mass: 210 = A0 + Atomic #: 84 = Z = A0 - = 84 = Z1 + = 85 =

positron Positron Emission Occurs when the neutron to proton ratio is too small.

Ex. Polonium-210 undergoes positron emission to produce this daughter nuclide + Positron Emission Atomic Mass: 210 = A0 + Atomic #: 84 = Z = A0 - = 84 = Z1 - = 83 =

Electron Capture electron occurs when the neutron to proton ratio in the nucleus is too small.

Ex. Polonium-210 captures an electron to produce this daughter nuclide + Electron Capture Atomic Mass: 210 = A0 + Atomic #: 84 = Z + A = 210 Z = 83 =

Gamma Emission –Emission of high energy electromagnetic wave. occurs when the nucleus is at too high an energy.

Ex. Polonium-210 undergoes gamma decay to produce this daughter nuclide + Gamma Emission Atomic Mass: 210 = A0 + Atomic #: 84 = Z 0 + A = 210 Z = 84 =

Nuclear Decay Why nuclides decay –need stable ratio of neutrons to protons Transmutation- One element becomes another. More than 83 protons means that the nuclei is unstable (radioactive)

Types of Radiation Alpha particle (  ) –helium nucleus paper 2+ Beta particle (  -) –electron 1- lead Positron (  +) –positron 1+ Gamma (  ) –high-energy photon 0 concrete ChargeShielding

Penetrating Ability of Radiation

Half-life Half-life (t ½ ) –Time required for half the atoms of a radioactive nuclide to decay. –Shorter half-life = less stable.

Half-life m f :final mass m i :initial mass n:# of half-lives

Half-life  Fluorine-21 has a half-life of 5.0 seconds. If you start with 25 g of fluorine-21, how many grams would remain after 60.0 s? GIVEN: t ½ = 5.0 s m i = 25 g m f = ? total time = 60.0 s n = 60.0s ÷ 5.0s =12 WORK : m f = m i (½) n m f = (25 g)(0.5) 12 m f = g