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Example 19.1 Writing Nuclear Equations for Alpha Decay

Example 19.1 Writing Nuclear Equations for Alpha Decay. Write the nuclear equation for the alpha decay of Ra-224. Solution. Begin with the symbol for Ra-224 on the left side of the equation and the symbol for an alpha particle on the right side .

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Example 19.1 Writing Nuclear Equations for Alpha Decay

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  1. Example 19.1Writing Nuclear Equations for Alpha Decay Write the nuclear equation for the alpha decay of Ra-224. Solution Begin with the symbol for Ra-224 on the left side of the equation and the symbol for an alpha particle on the right side. Equalize the sum of the mass numbers and the sum of the atomic numbers on both sides of the equation by writing the appropriate mass number and atomic number for the unknown daughter nuclide. Refer to the periodic table to deduce the identity of the unknown daughter nuclide from its atomic number and write its symbol. Since the atomic number is 86, the daughter nuclide is radon (Rn). For Practice 19.1 Write the nuclear equation for the alpha decay of Po-216.

  2. Example 19.2Writing Nuclear Equations for Beta Decay, Positron Emission, and Electron Capture Write the nuclear equation for each type of decay. a. beta decay in Bk-249 b. positron emission in O-15 c. electron capture in I-111 Solution a.In beta decay, the atomic number increases by 1 and the mass number remains unchanged. The daughter nuclide is element number 98, californium. b. In positron emission, the atomic number decreases by 1 and the mass number remains unchanged. The daughter nuclide is element number 7, nitrogen. c. In electron capture, the atomic number also decreases by 1 and the mass number remains unchanged. The daughter nuclide is element number 52, tellurium.

  3. Example 19.2Writing Nuclear Equations for Beta Decay, Positron Emission, and Electron Capture Continued For Practice 19.2 a. Write three nuclear equations to represent the nuclear decay sequence that begins with the alpha decay of U-235 followed by a beta decay of the daughter nuclide and then another alpha decay. b. Write the nuclear equation for the positron emission of Na-22. c. rite the nuclear equation for electron capture in Kr-76. For More Practice 19.2 Potassium-40 decays to produce Ar-40. What is the method of decay? Write the nuclear equation for this decay.

  4. Example 19.3Predicting the Type of Radioactive Decay Predict whether each nuclide is more likely to decay via beta decay or positron emission. a. Mg-28 b. Mg-22 c. Mo-102 Solution a. Magnesium-28 has 16 neutrons and 12 protons, so N/Z = 1.33. However, for Z = 12, you can see from Figure 19.5 that stable nuclei should have an N/Z of about 1. Alternatively, you can see from the periodic table that the atomic mass of magnesium is 24.31. Therefore, a nuclide with a mass number of 28 is too heavy to be stable because the N/Z ratio is too high and Mg-28 undergoes beta decay, resulting in the conversion of a neutron to a proton. b. Magnesium-22 has 10 neutrons and 12 protons, so N/Z = 0.83 (too low). Alternatively you can see from the periodic table that the atomic mass of magnesium is 24.31. A nuclide with a mass number of 22 is too light; the N/Z ratio is too low. Therefore, Mg-22 undergoes positron emission, resulting in the conversion of a proton to a neutron. (Electron capture would accomplish the same thing as positron emission, but in Mg-22, positron emission is the only decay mode observed.) Figure 19.5 Stable and Unstable Nuclei A plot of N (the number of neutrons) versus Z (the number of protons) for all known stable nuclei—represented by green dots on this graph—shows that these nuclei cluster together in a region known as the valley (or island) of stability. Nuclei with an N/Z ratio that is too high tend to undergo beta decay. Nuclei with an N/Z ratio that is too low tend to undergo positron emission or electron capture.

  5. Example 19.3Predicting the Type of Radioactive Decay Continued c. Molybdenum-102 has 60 neutrons and 42 protons, so N/Z = 1.43. However, for Z = 42, you can see from Figure 19.5 that stable nuclei should have an N/Z ratio of about 1.3. Alternatively you can see from the periodic table that the atomic mass of molybdenum is 95.94. A nuclide with a mass number of 102 is too heavy to be stable; the N/Z ratio is too high. Therefore, Mo-102 undergoes beta decay, resulting in the conversion of a neutron to a proton. Figure 19.5 Stable and Unstable Nuclei A plot of N (the number of neutrons) versus Z (the number of protons) for all known stable nuclei—represented by green dots on this graph—shows that these nuclei cluster together in a region known as the valley (or island) of stability. Nuclei with an N/Z ratio that is too high tend to undergo beta decay. Nuclei with an N/Z ratio that is too low tend to undergo positron emission or electron capture. For Practice 19.3 Predict whether each nuclide is more likely to decay via beta decay or positron emission. a.Pb-192 b.Pb-212 c.Xe-114

  6. Example 19.4Radioactive Decay Kinetics Plutonium-236 is an alpha emitter with a half-life of 2.86 years. If a sample initially contains 1.35 mg of Pu-236, what mass of Pu-236 is present after 5.00 years? Sort You are given the initial mass of Pu-236 in a sample and asked to find the mass after 5.00 years. Given: mPu-236(initial) = 1.35 mg; t = 5.00 yr; t1/2 = 2.86 yr Find: mPu-236 (final) Strategize Use the integrated rate law (Equation 19.3) to solve this problem. You must determine the value of the rate constant (k) from the half-life expression (Equation 19.1). Use the value of the rate constant, the initial mass of Pu-236, and the time along with integrated rate law to find the final mass of Pu-236. Since the mass of the Pu-236 (mPu-236) is directly proportional to the number of atoms (N), and since the integrated rate law contains the ratio (Nt/N0), the initial and final masses can be substituted for the initial and final number of atoms.

  7. Example 19.4Radioactive Decay Kinetics Continued Conceptual Plan Solve Follow your plan. Begin by determining the rate constant from the half-life. Solve the integrated rate law for Nt and substitute the values of the rate constant, the initial mass of Pu-236, and the time into the solved equation. Calculate the final mass of Pu-236.

  8. Example 19.4Radioactive Decay Kinetics Continued Solution Check The units of the answer (mg) are correct. The magnitude of the answer (0.402 mg) is about one-third of the original mass (1.35 mg), which seems reasonable given that the amount of time is between one and two half-lives. (One half-life would result in one-half of the original mass and two half-lives would result in one-fourth of the original mass.)

  9. Example 19.4Radioactive Decay Kinetics Continued For Practice 19.4 How long will it take for the 1.35 mg sample of Pu-236 in Example 19.4 to decay to 0.100 mg?

  10. Example 19.5Radiocarbon Dating A skull believed to belong to an ancient human being has a carbon-14 decay rate of 4.50 disintegrations per minute per gram of carbon (4.50 dis/min · g C). If living organisms have a decay rate of 15.3 dis/min · g C, how old is the skull? (The decay rate is directly proportional to the amount of carbon-14 present.) Sort You are given the current rate of decay for the skull and the assumed initial rate. You are asked to find the age of the skull, which is the time that passed in order for the rate to have reached its current value. Given: ratet = 4.50 dis/min · g C; rate0= 15.3 dis/min · g C; Find: t Strategize Use the expression for half-life (Equation 19.1) to find the rate constant (k) from the half-life for C-14, which is 5730 yr (Table 19.3). Use the value of the rate constant and the initial and current rates to find t from the integrated rate law (Equation 19.4).

  11. Example 19.5Radiocarbon Dating Continued Conceptual Plan Solve Follow your plan. Begin by finding the rate constant from the half-life. Substitute the rate constant and the initial and current rates into the integrated rate law and solve for t.

  12. Example 19.5Radiocarbon Dating Continued Solution Check The units of the answer (yr) are correct. The magnitude of the answer is about 10,000 years, which is a little less than two half-lives. This value is reasonable given that two half-lives would result in a decay rate of about 3.8 dis/min  g C.

  13. Example 19.5Radiocarbon Dating Continued For Practice 19.5 A researcher claims that an ancient scroll originated from Greek scholars in about 500 b.c. A measure of its carbon-14 decay rate gives a value that is 89% of that found in living organisms. How old is the scroll and could it be authentic?

  14. Example 19.6Using Uranium/Lead Dating to Estimate the Age of a Rock A meteor contains 0.556 g of Pb-206 to every 1.00 g of U-238. Assuming that the meteor did not contain any Pb-206 at the time of its formation, determine the age of the meteor. Uranium-238 decays to lead-206 with a half-life of 4.5 billion years. Sort You are given the current masses of Pb-206 and U-238 in a rock and asked to find its age. You are also given the half-life of U-238. Given: mU-238 = 1.00 g; mPb-206 = 0.556 g; t1/2 = 4.5 × 109yr Find: t Strategize Use the integrated rate law (Equation 19.3) to solve this problem. To do so, you must first determine the value of the rate constant (k) from the half-life expression (Equation 19.1). Before substituting into the integrated rate law, you also need the ratio of the current amount of U-238 to the original amount (Nt/N0). The current mass of uranium is simply 1.00 g. The initial mass includes the current mass (1.00 g) plus the mass that has decayed into lead-206, which can be determined from the current mass of Pb-206. Use the value of the rate constant and the initial and current amounts of U-238 along with the integrated rate law to find t.

  15. Example 19.6Using Mass Percent Composition as a Conversion Factor Continued Conceptual Plan

  16. Example 19.6Using Mass Percent Composition as a Conversion Factor Continued Solve Follow your plan. Begin by finding the rate constant from the half-life. Solution Determine the mass in grams of U-238 that is required to form the given mass of Pb-206. Substitute the rate constant and the initial and current masses of U-238 into the integrated rate law and solve for t. (The initial mass of U-238 is the sum of the current mass and the mass that is required to form the given mass of Pb-206.)

  17. Example 19.6Using Mass Percent Composition as a Conversion Factor Continued Solution Check The units of the answer (yr) are correct. The magnitude of the answer is about 3.2 billion years, which is less than one half-life. This value is reasonable given that less than half of the uranium in the meteor has decayed into lead. For Practice 19.6 A rock contains a Pb-206 to U-238 mass ratio of 0.145 : 1.00. Assuming that the rock did not contain any Pb-206 at the time of its formation, determine its age.

  18. Example 19.7Mass Defect and Nuclear Binding Energy Calculate the mass defect and nuclear binding energy per nucleon (in MeV) for C-16, a radioactive isotope of carbon with a mass of 16.014701 amu. Solution Calculate the mass defect as the difference between the mass of one C-16 atom and the sum of the masses of 6 hydrogen atoms and 10 neutrons. Calculate the nuclear binding energy by converting the mass defect (in amu) into MeV. (Use 1 amu = 931.5 MeV.) Determine the nuclear binding energy per nucleon by dividing by the number of nucleons in the nucleus. For Practice 19.7 Calculate the mass defect and nuclear binding energy per nucleon (in MeV) for U-238, which has a mass of 238.050784 amu.

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