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Nuclear Chemistry

Nuclear Chemistry. Chapter 21 Nuclear Chemistry. Some isotopes of elements are unstable and spontaneously emit particles and high energy electromagnetic radiation. This spontaneous emission of particles and energy is known as radioactive decay. An example of radioactive decay is:.

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Nuclear Chemistry

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  1. Nuclear Chemistry

  2. Chapter 21 Nuclear Chemistry Some isotopes of elements are unstable and spontaneously emit particles and high energy electromagnetic radiation. This spontaneous emission of particles and energy is known as radioactive decay. An example of radioactive decay is: The particle being emitted here is a helium nucleus or alpha particle. So this is referred to as alpha decay. Notice the sum of the mass numbers is the same on both sides. This is also true for the sum of the atomic numbers. Type of Radiation alpha a +2 helium nucleus 1 beta b -1 an electron -1e 100 gamma g 0 high energy photons 0g 1000

  3. Electric or Magnetic Field + Radioactive material Beta particle = e-1 Gamma radiation Alpha particle He nucleus Lead shielded container

  4. 1. alpha decay 2. beta decay (equivalent to converting a neutron to a proton) positron (positive electron) 4. positron emission (antimatter) (equivalent to converting a proton to a neutron) 5. electron capture (also equivalent to converting a proton to a neutron) 3. gamma radiation usually accompanies all radioactive decay. Gamma photons are usually not written

  5. Examples Write nuclear equations for the following processes: 1. radium-226 undergoes alpha decay 2. thorium-231 undergoes beta emission (A neutron changes to a proton and electron) 3. mercury-201 undergoes electron capture (equivalent to converting a proton to a neutron) 4. oxygen-15 undergoes positron emission (converting a proton to a neutron)

  6. Rates of Radioactive Decay All radioactive decay is 1st order ln[A]t = -kt + ln[A]o One way of indicating rate of radioactive decay is half-life Example:The half-life of cobalt-60 is 5.3 yr. How much of a 1.000 mg sample of cobalt-60 is left after a 15.9 yr period? Two ways of solving this 15.9 yrs is 3 half-lifes 1.000mg a 0.500mg a 0.250mg a 0.125mg ln[Co-60]t = -kt + ln[1.000]o ln[Co-60]t = -(0.13/yr)(15.9yr)t + 0 ln[Co-60]t = -2.07 + 0 [Co-60]t = e-2.07 = .126

  7. Energy Changes in Nuclear Reactions Matter is “destroyed” in nuclear reactions. Actually, the “destroyed” matter is converted to energy. The amount of energy produced is calculated by using Einstein’s famous E = mc2. The missing matter is referred to as the mass defect.

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