CHAPTER 28 Nuclear Chemistry

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Presentation transcript:

CHAPTER 28 Nuclear Chemistry Radioactive Decay

Nuclide = atom of an isotope A. Nuclear Stability Nuclide = atom of an isotope

A. Nuclear Stability Nuclear stability – stable nuclei always have at least as many neutrons as protons.

A. Nuclear Stabiity For an odd/even or even/odd nucleus, if the mass number is different by more than 1 amu from the rounded atomic mass, the nuclide is unstable. Ex:

A. Nuclear Stability For an even/even nucleus, if the mass number is different by more than 3 amu from the rounded atomic mass, the nuclide is unstable. Ex:

A. Nuclear Stability For odd/odd nuclei, only four stable isotopes are found in nature:

B. Nuclear Decay 2+ 1- 1+ Alpha particle () helium nucleus paper 2+ Beta particle (-) electron 1- lead Positron (+) positron 1+ concrete Gamma () high-energy photon

B. Nuclear Decay Top and bottom numbers must balance!! Alpha Emission parent nuclide daughter nuclide alpha particle Top and bottom numbers must balance!!

B. Nuclear Decay Beta Emission electron Positron Emission positron

B. Nuclear Decay Electron Capture electron Gamma Emission Usually follows other types of decay. Transmutation Atom of one element changes into an atom of another element.

B. Nuclear Decay Why nuclides decay… need stable ratio of neutrons to protons DECAY SERIES TRANSPARENCY

C. Half-life Half-life (t½) Time required for half the atoms of a radioactive nuclide to decay. Shorter half-life = less stable.

C. Half-life mf: final mass mi: initial mass n: # of half-lives

C. Half-life t½ = 5.0 s mf = mi (1/2)n mi = 25 g mf = (25 g)(0.5)12 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: t½ = 5.0 s mi = 25 g mf = ? total time = 60.0 s n = 60.0s ÷ 5.0s =12 WORK: mf = mi (1/2)n mf = (25 g)(0.5)12 mf = 0.0061 g

C. Half-life Nt: final mass N0: initial mass t: elapsed time

C. Half-life k: rate constant t1/2: half-life

C. Half-life t½ = 11.4 days k = 0.693 / t1/2 k = 0.693 / 11.4 days A sample of radium-223 has a half-life of 11.4 days. What is the rate constant for this isotope? GIVEN: t½ = 11.4 days WORK: k = 0.693 / t1/2 k = 0.693 / 11.4 days k = 0.0607 days -1

C. Half-life k = 0.035 yr -1 k = 0.693 / t 1/2 t ½ = 0.693 / k The rate constant for gold-200 is 0.035 /year. What is the half-life of gold-200? GIVEN: k = 0.035 yr -1 WORK: k = 0.693 / t 1/2 t ½ = 0.693 / k t ½ = 0.693 / 0.035 yr -1 t ½ = 19.8 yr ≈ 20. years

C. Half-life t ½ = 1.7 x 107 yr ln (N0/Nt) = kt N0 = 3.75 g The half-life of iodine 129 is 1.7 x 107 years. If a nuclear bomb explosion resulted in 3.75 g of iodine-129, how much time would have to elapse for the amount of iodine-129 to be 0.75 g? GIVEN: t ½ = 1.7 x 107 yr N0 = 3.75 g Nt = 0.75 g t = ? WORK: ln (N0/Nt) = kt ln (3.75/0.75) = 0.693/1.7 x 107 t t = 3.9 x 107 years (39, 000, 000 years!)

D. Radiocarbon Dating Carbon-14 is in all living things through the carbon cycle. Amount of carbon-14 stays constant until organism dies, then it begins to decay.

D. Radiocarbon Dating Amount of carbon-14 can be expressed as either a percentage or as a decimal number. Example: amount of carbon-14 in a dead tree could be expressed as 38% or 0.38 of the original amount.

Half-life of carbon-14 : 5730 years D. Radiocarbon Dating Half-life of carbon-14 : 5730 years

D. Radiocarbon Dating The remnants of an ancient canoe are found in a cave in northern Australia. The amount of carbon-14 is 6.28 counts per minute, and the amount of carbon-14 in a tree today is 13.6 counts per minute. What is the approximate age of the canoe?

D. Radiocarbon Dating t ½ = 5730 yr k = 0.693 / 5730 yr GIVEN: t ½ = 5730 yr N0 = 13.6 cts/min Nt = 6.28 cts/min t = ? WORK: k = 0.693 / 5730 yr k = 1.20942 x 10-4 yr -1 ln (N0/Nt) = kt ln (13.6/6.28) = 1.20942 x 10-4 t t = 6390 years 2011 – 6390 = 4379 BC

E. Fission Occurs when isotopes are bombarded with neutrons and split the nucleus into smaller fragments, accompanied by the release of neutrons and a large amount of energy. (Each atom can capture 1 neutron.)

E. Fission Chain reaction – occurs when atomic nuclei that have split release energetic neutrons that split more nuclei.

E. Fission Two steps in controlling fission: Neutron moderation – water or carbon slows down the neutrons Neutron absorption – decreases the number of slow neutrons through the use of control rods made of neutron-absorbing materials (usually cadmium)

F. Fusion Occurs when two light nuclei combine to produce a nucleus of heavier mass, accompanied by the release of a large amount of energy.

F. Fusion Occurs in all stars High temperatures are necessary to initiate fusion (no cold fusion yet) Possible future energy source Hydrogen bomb is a fusion reaction (fusion of two deuterium nuclei).

G. Methods of Detection Geiger Counters (primarily beta) Scintillation counter – coated screen detects radiation particles. Film badge – several layers of photographic film encased in a holder. Detects beta and gamma.

H. Radioisotopes in Medicine X-rays: Useful in imaging soft-tissue organs. Tracers: Iodine-131 is used to check for thyroid problems Radiation treatment: Some cobalt isotopes are used as radiation sources to treat cancer.