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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.

Short Answer

Expert verified
Approximately \(1.13 \times 10^{19}\) \(^{137}Cs\) atoms per kilogram of tissue are needed.

Step by step solution

01

Calculate the Energy Released per Decay

The total energy released per decay is given as 1.17 MeV. Converting this to joules, we know 1 MeV is equivalent to \(1.602 \times 10^{-13}\) J. Therefore, the energy released per decay in joules is \(1.17 \times 1.602 \times 10^{-13}\) J.
02

Calculate the Energy in Joules from One Decay

Using our previous conversion, \(E_{decay} = 1.17 \times 1.602 \times 10^{-13} = 1.87434 \times 10^{-13}\) J per decay.
03

Convert the Dose to Energy Deposited

Given that 3.5 Sv is the equivalent dose, which is \(3.5 \, \text{J/kg}\) for 1 kg of tissue. For a week, the energy deposited is \(3.5 \, \text{J/sec} \times 7 \times 24 \times 3600 \, \text{seconds}\). This equals 2.11 \(\times 10^6\) J.
04

Determine the Number of Decays Required

The total number of decay events needed to deposit 2.11 \(\times 10^6\) J on 1 kg of tissue is calculated as \(\frac{2.11 \times 10^6}{1.87434 \times 10^{-13}}\). This equals approximately \(1.13 \times 10^{19}\) decays.
05

Calculate the Number of \(^{137}Cs\) Atoms Needed

Each decay assumes one \(^{137}Cs\) atom, so approximately \(1.13 \times 10^{19}\) \(^{137}Cs\) atoms need to be present in the 1 kg of tissue.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radioactive Decay
Radioactive decay is a natural process where unstable atomic nuclei lose energy by emitting radiation. This process results in the transformation of an element into a different element or a different isotope. At the core of radioactive decay is the release of energy, often in the form of particles or electromagnetic waves. This decay can occur through different modes:
  • Alpha Decay: The nucleus emits an alpha particle (two protons and two neutrons). This results in the element changing into a different one with a mass number reduced by four and atomic number by two.
  • Beta Decay: A neutron in the nucleus is transformed into a proton and an electron, with the electron (beta particle) being emitted.
  • Gamma Decay: Following alpha or beta decay, the nucleus releases excess energy by emitting a gamma ray, which is a high-energy photon.
Understanding radioactive decay is crucial because it explains how certain elements change over time and how they emit energy, which can be both useful and dangerous.
Cs-137
Cs-137, or Cesium-137, is a radioactive isotope formed as a byproduct of nuclear fission, commonly found in nuclear reactors and weapon tests. In the Chernobyl disaster, a significant portion of the released radiation was due to Cs-137.
Cs-137 is particularly concerning due to its relatively long half-life and its ability to spread in the environment. It is absorbed by plants and enters the food chain, ultimately affecting humans and animals. This isotope emits beta particles during its decay, contributing to radiation exposure. Addressing the presence of Cs-137 in the environment is crucial to limit its long-term ecological and health impacts.
Radiation Dose
Radiation dose is the measure of radiation absorbed by an object or person. It is crucial in assessing potential harm and health risks associated with exposure. Measurement units include grays (Gy) and sieverts (Sv), with sieverts considering the biological effects of different radiation types.
For instance, a dose of 3.5 Sv in one week significantly affects human health, and managing the deposition of such doses is essential. The dose can derive from absorbing the energy of radioactive isotopes like Cs-137. Understanding radiation doses helps ensure safety standards in environments where radioactive materials exist.
Half-Life
The concept of half-life is a cornerstone of understanding radioactive decay processes. Half-life refers to the time required for half of a given amount of a radioactive element to decay into another element or isotope.
For Cs-137, the half-life is around 30.07 years, meaning it takes this period for half of its atoms to undergo decay. This long half-life signifies that Cs-137 remains in the environment for extended periods, posing ongoing risks after nuclear events. Studying half-lives allows scientists to predict decay patterns and manage radioactive elements safely.
Beta Decay
Beta decay is a mode of radioactive decay where an unstable atom emits a beta particle, which is identical to an electron, to achieve a more stable state. This process involves a neutron converting into a proton with the release of a beta particle (an electron) and an antineutrino.
Cs-137 undergoes beta decay, transforming into a stable isotope of barium. During its decay, Cs-137 releases energy, including gamma rays alongside the beta particles, contributing to radiation exposure concerns post-nuclear incidents like Chernobyl. Recognizing beta decay's role helps in understanding the behavior and eventual stabilization of radioactive materials.

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Most popular questions from this chapter

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?

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?

Thorium \(^{230}_{90}Th\) decays to radium \(^{226}_{88}Ra\) by a emission. The masses of the neutral atoms are 230.033134 u for \(^{230}_{90}Th\) and 226.025410 u for \(^{226}_{88}Ra\). If the parent thorium nucleus is at rest, what is the kinetic energy of the emitted \(\alpha\) particle? (Be sure to account for the recoil of the daughter nucleus.)

\(\textbf{Radiation Treatment of Prostate Cancer}\). In many cases, prostate cancer is treated by implanting 60 to 100 small seeds of radioactive material into the tumor. The energy released from the decays kills the tumor. One isotope that is used (there are others) is palladium (\(^1$$^0$$^3\)Pd), with a half-life of 17 days. If a typical grain contains 0.250 g of \(^1$$^0$$^3\)Pd, (a) what is its initial activity rate in Bq, and (b) what is the rate 68 days later?

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?

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