Chapter 43

\(\textbf{We Are Stardust.}\) In 1952 spectral lines of the element technetium-99 (\(^9$$^9\)Tc) were discovered in a red giant star. Red giants are very old stars, often around 10 billion years old, and near the end of their lives. Technetium has no stable isotopes, and the half-life of \(^9$$^9\)Tc is 200,000 years. (a) For how many halflives has the \(^9$$^9\)Tc been in the red giant star if its age is 10 billion years? (b) What fraction of the original \(^9$$^9\)Tc would be left at the end of that time? This discovery was extremely important because it provided convincing evidence for the theory (now essentially known to be true) that most of the atoms heavier than hydrogen and helium were made inside of stars by thermonuclear fusion and other nuclear processes. If the \(^9$$^9\)Tc had been part of the star since it was born, the amount remaining after 10 billion years would have been so minute that it would not have been detectable. This knowledge is what led the late astronomer Carl Sagan to proclaim that "we are stardust".

Measurements indicate that 27.83% of all rubidium atoms currently on the earth are the radioactive \(^8$$^7\)Rb isotope. The rest are the stable \(^8$$^5\)Rb isotope. The half-life of \(^8$$^7\)Rb is 4.75 \(\times\) 10\(^1$$^0\) y. Assuming that no rubidium atoms have been formed since, what percentage of rubidium atoms were \(^8$$^7\)Rb when our solar system was formed 4.6 \(\times\) 10\(^9\) y ago?

The nucleus \(^{15}_{8}O\) has a half-life of 122.2 s; \(^{19}_{8}O\) has a half- life of 26.9 s. If at some time a sample contains equal amounts of \(^{15}_{8}O\) and \(^{19}_{8}O\), what is the ratio of \(^{15}_{8}O\) to \(^{19}_{8}O\) (a) after 3.0 min and (b) after 12.0 min?

A \(^6$$^0\)Co source with activity 2.6 \(\times\) 10\(^-$$^4\) Ci is embedded in a tumor that has mass 0.200 kg. The source emits \(\gamma\) photons with average energy 1.25 MeV. Half the photons are absorbed in the tumor, and half escape. (a) What energy is delivered to the tumor per second? (b) What absorbed dose (in rad) is delivered per second? (c) What equivalent dose (in rem) is delivered per second if the RBE for these g rays is 0.70? (d) What exposure time is required for an equivalent dose of 200 rem?

\(\textbf{An Oceanographic Tracer.}\) Nuclear weapons tests in the 1950s and 1960s released significant amounts of radioactive tritium (\(^{3}_{1}H\), half- life 12.3 years) into the atmosphere. The tritium atoms were quickly bound into water molecules and rained out of the air, most of them ending up in the ocean. For any of this tritium-tagged water that sinks below the surface, the amount of time during which it has been isolated from the surface can be calculated by measuring the ratio of the decay product, \(^{3}_{2}He\), to the remaining tritium in the water. For example, if the ratio of \(^{3}_{2}He\) to \(^{3}_{1}H\) in a sample of water is 1:1, the water has been below the surface for one half-life, or approximately 12 years. This method has provided oceanographers with a convenient way to trace the movements of subsurface currents in parts of the ocean. Suppose that in a particular sample of water, the ratio of \(^{3}_{2}He\) to \(^{3}_{1}H\) is 4.3 to 1.0. How many years ago did this water sink below the surface?

In the 1986 disaster at the Chernobyl reactor in eastern Europe, about \\(\frac{1}{8}\\) of the \(^{137}Cs\) present in the reactor was released. The isotope \(^{137}Cs\) has a half-life of 30.07 y for \(\beta\) decay, with the emission of a total of 1.17 MeV of energy per decay. Of this, 0.51 MeV goes to the emitted electron; the remaining 0.66 MeV goes to a \(\gamma\) ray. The radioactive \(^{137}Cs\) is absorbed by plants, which are eaten by livestock and humans. How many \(^{137}Cs\) atoms would need to be present in each kilogram of body tissue if an equivalent dose for one week is 3.5 Sv? Assume that all of the energy from the decay is deposited in 1.0 kg of tissue and that the RBE of the electrons is 1.5.

Consider the fusion reaction \(^{2}_{1}H\) + \(^{2}_{1}H \rightarrow ^{3}_{2}He + ^{1}_{0}n\). (a) Estimate the barrier energy by calculating the repulsive electrostatic potential energy of the two \(^{2}_{1}H\) nuclei when they touch. (b) Compute the energy liberated in this reaction in MeV and in joules. (c) Compute the energy liberated \(per\) \(mole\) of deuterium, remembering that the gas is diatomic, and compare with the heat of combustion of hydrogen, about \(2.9 \times 10^{5} J/mol\).

Radioisotopes are used in a variety of manufacturing and testing techniques. Wear measurements can be made using the following method. An automobile engine is produced using piston rings with a total mass of 100 g, which includes \(9.4 \mu Ci\) of \(^{59}Fe\) whose half-life is 45 days. The engine is test-run for 1000 hours, after which the oil is drained and its activity is measured. If the activity of the engine oil is 84 decays/s, how much mass was worn from the piston rings per hour of operation?

Many radioactive decays occur within a sequence of decays for example, \(^{234}_{92}U\) \(\rightarrow\) \(^{230}_{88}Th\) \(\rightarrow\) \(^{226}_{84}Ra\). The half-life for the \(^{234}_{92}U\) \(\rightarrow\) \(^{230}_{88}Th\) decay is \(2.46 \times 10^{5}\) y, and the half-life for the \(^{230}_{88}Th\) \(\rightarrow\) \(^{226}_{84}Ra\) decay is \(7.54 \times 10^{4}\) y. Let 1 refer to \(^{234}_{92}U\), 2 to \(^{230}_{88}Th\), and 3 to \(^{226}_{84}Ra\); let \(\lambda\)1 be the decay constant for the \(^{234}_{92}U\) \(\rightarrow\) \(^{230}_{88}Th\) decay and \(\lambda\)2 be the decay constant for the \(^{230}_{88}Th\) \(\rightarrow\) \(^{226}_{84}Ra\) decay. The amount of \(^{230}_{88}Th\) present at any time depends on the rate at which it is produced by the decay of \(^{234}_{92}U\) and the rate by which it is depleted by its decay to \(^{226}_{84}Ra\). Therefore, \(d$$N_{2}\)(\(t\))/\(d$$t\) = \(\lambda$$_1$$N$$_1\)(\(t\)) - \(\lambda$$_2$$N$$_2\)(\(t\)). If we start with a sample that contains \(N_{10}\) nuclei of \(^{234}_{92}U\) and nothing else, then \(N{(t)}\) = \(N_{01}\)e\(^{-\lambda_{1}t}\). Thus \(dN_{2}(t)/dt\) = \(\lambda\)1\(N_{01}\)e\(^{-\lambda_{1}t}\)- \(\lambda2N_{2}(t)\). This differential equation for \(N_{2}(t)\) can be solved as follows. Assume a trial solution of the form \(N_{2}(t)\) = \(N_{10} [h_{1}e^{-\lambda_1t}\) + \(h_{2}e^{-\lambda_2t}\)] , where \(h_{1}\) and \(h_{2}\) are constants. (a) Since \(N_{2}(0)\) = 0, what must be the relationship between \(h_{1}\) and \(h_{2}\)? (b) Use the trial solution to calculate \(dN_{2}(t)/dt\), and substitute that into the differential equation for \(N_{2}(t)\). Collect the coefficients of \(e^{-\lambda_{1}t}\) and \(e^{-\lambda_{2}t}\). Since the equation must hold at all t, each of these coefficients must be zero. Use this requirement to solve for \(h_{1}\) and thereby complete the determination of \(N_{2}(t)\). (c) At time \(t = 0\), you have a pure sample containing 30.0 g of \(^{234}_{92}U\) and nothing else. What mass of \(^{230}_{88}Th\) is present at time \(t = 2.46 \times 10^{5}\) y, the half-life for the \(^{234}_{92}U\) decay?

Which reaction produces \(^{131}\)Te in the nuclear reactor? (a) \(^{130}Te + n \rightarrow ^{131}Te\); (b) \(^{130}I + n \rightarrow ^{131}Te\); (c) \(^{132}Te + n \rightarrow ^{131}Te\); (d) \(^{132}I + n \rightarrow ^{131}Te\).