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According to current regulations, the maximum permissible dose of strontium-90 in the body of an adult is 1\(\mu \mathrm{Ci}\left(1 \times 10^{-6} \mathrm{Ci}\right) .\) Using the relationship rate \(=k N,\) calculate the number of atoms of strontium-90 to which this dose corresponds. To what mass of strontium-90 does this correspond? The half-life for strontium-90 is 28.8 yr.

Short Answer

Expert verified
The decay constant (\(\lambda\)) for strontium-90 can be calculated using its half-life (28.8 years) and the formula \(\lambda = \dfrac{\ln{2}}{{28.8}\,\text{yr}}\). The given maximum permissible dose of 1 \(\mu\text{Ci}\) can be converted to becquerels as \(3.7 \times 10^{4} \,\text{Bq}\). With the rate equation (rate \(= \lambda N\)), we can find the number of radioactive atoms \(N\). Finally, using the molar mass of strontium-90 (89.9079 g/mol) and Avogadro's number, we can calculate the mass of strontium-90 corresponding to the given dose.

Step by step solution

01

Calculate the decay constant

We know that the half-life of strontium-90 is 28.8 years. We can use the following formula to calculate the decay constant (\(\lambda\)): \(\lambda = \dfrac{\ln{2}}{T_{\frac{1}{2}}}\) where \(T_{\frac{1}{2}}\) is the half-life. Plugging in the values, we get: \(\lambda = \dfrac{\ln{2}}{{28.8}\,\text{yr}}\)
02

Calculate the activity

We are given the dose of strontium-90 in microcuries (\(\mu\text{Ci}\)) and need to convert it to activity (rate) in becquerels (\(\text{Bq}\)). There is a conversion factor: 1 \(\text{Ci}\) = \(3.7 \times 10^{10}\) Bq, so we have: 1 \(\mu\text{Ci}\) = \(1 \times 10^{-6} \,\text{Ci}\) = \(3.7 \times 10^{4} \,\text{Bq}\)
03

Calculate the number of radioactive atoms

We can now use the rate equation, rate \(= \lambda N\), to calculate the number of radioactive atoms: \(N = \dfrac{\text{rate}}{\lambda} = \dfrac{3.7 \times 10^{4} \, \text{Bq}}{\lambda}\) With the value of \(\lambda\) from Step 1, we can find the number of atoms \(N\).
04

Calculate the mass

We know the molar mass of strontium-90 is 89.9079 g/mol. 1 mole of strontium-90 has \(6.022 \times 10^{23}\) atoms. So, we can find the mass of our calculated number of atoms (\(N\)) using the following formula: \(\text{mass of strontium}=\dfrac{N\,\text{atoms}\times 89.9079\,\text{g/mol}}{6.022 \times 10^{23}\,\text{atoms/mol}}\) Now you can calculate the mass of strontium-90 corresponding to the given dose.

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Key Concepts

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

Strontium-90
Strontium-90 is a radioactive isotope that is commonly found as a byproduct in nuclear reactions and is a part of the radioactive fallout from nuclear tests. It is a beta-emitting radionuclide, meaning it emits beta particles as it decays. Strontium-90 is particularly concerning in environmental science because it substitutes calcium in bone tissues. Exposure to strontium-90 can therefore lead to harmful biological effects, primarily due to its long residence time in bones where it continues to emit radiation.

This element plays a significant role in nuclear waste management and in assessing radiation exposure to the public after nuclear activity. Understanding its behavior and decay is crucial for radiological protection.
  • Common Sources: Nuclear fallout, nuclear power plants.
  • Health Risks: Bone cancer, leukemia due to radiation emission.
  • Properties: Beta-emitter, with a physical half-life of 28.8 years.
Half-Life
The half-life of a radioactive substance is the time required for half of the radioactive atoms in a sample to decay. For strontium-90, this period is notably long, sitting at 28.8 years. This duration means that it remains active and potentially harmful for several decades after its release into the environment.

Half-life is a fundamental concept in radiochemistry and affects how we handle radioactive substances:
  • Determines how long a substance remains hazardous.
  • Is crucial for calculations involving radioactive decay and substance elimination.
  • Helps in determining the timing for medical treatments using radiopharmaceuticals.
Recognizing the half-life aids in developing plans for storage, disposal, and environmental decontamination of radioactive materials.
Decay Constant
The decay constant, denoted as \( \lambda \), is a probability factor that characterizes the rate of radioactive decay for a given isotope. It can be calculated using the formula \( \lambda = \frac{\ln{2}}{T_{\frac{1}{2}}} \), where \( \ln{2} \) is the natural logarithm of 2, and \( T_{\frac{1}{2}} \) is the half-life of the isotope.

For strontium-90, given its half-life of 28.8 years, we can determine \( \lambda \) through substitution into the formula. This constant is critical for understanding radioactive decay processes:
  • Provides the proportion of atoms that decay per unit time.
  • Integral in calculations to predict activity over time.
  • Aids in determining the dose received from radiation over a period.
The decay constant helps in converting between half-life and activity, facilitating better planning for managing radioactive materials.
Activity
Activity, in the context of radioactivity, refers to the number of decays that occur per second from a radioactive source. It is measured in becquerels (Bq), where one becquerel corresponds to one decay per second. In the initial problem, we deal with activity in microcuries (µCi) and convert it to becquerels since 1 Ci = \(3.7 \times 10^{10}\) Bq.

Activity is crucial because it allows scientists and health professionals to understand how "active" a radioactive source is - or how quickly it is releasing energy through decay:
  • Determines the potential hazard of a radioactive material.
  • Used in calculating the exposure dose to individuals and the environment.
  • Help assess the safety of radioactive waste and clearance levels.
Understanding activity and converting between units ensures proper safety protocols can be enacted when handling radioactive substances.

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Most popular questions from this chapter

A wooden artifact from a Chinese temple has a \(^{14} \mathrm{C}\) cactivity of 38.0 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age. From the half-life for \(^{14} \mathrm{C}\) decay, 5715 yr, determine the age of the artifact.

(a) Which of the following are required characteristics of an isotope to be used as a fuel in a nuclear power reactor? (i) It must emit gamma radiation. (ii) On decay, it must release two or more neutrons. (iii) It must have a half-life less than one hour. (iv) It must undergo fission upon the absorption of a neutron. (b) What is the most common fissionable isotope in a commercial nuclear power reactor?

In 1930 the American physicist Ernest Lawrence designed the first cyclotron in Berkeley, California. In 1937 Lawrence bombarded a molybdenum target with deuterium ions, producing for the first time an element not found in nature. What was this element? Starting with molybdenum-96 as your reactant, write a nuclear equation to represent this process.

Nuclear scientists have synthesized approximately 1600 nuclei not known in nature. More might be discovered with heavy-ion bombardment using high-energy particle accelerators. Complete and balance the following reactions, which involve heavy-ion bombardments: \begin{equation} (a) \stackrel{6}{3} \mathrm{Li}+\stackrel{56}{28} \mathrm{Ni} \longrightarrow ? \end{equation}\begin{equation}(b) \stackrel{40}{20} \mathrm{Ca}+\stackrel {248}{96} \mathrm{Cm} \longrightarrow \stackrel{147}{62} \mathrm{Sm} + ? \end{equation}\begin{equation}(c) \stackrel{88}{38} \mathrm{Sr}+\stackrel{84}{36} \mathrm{Kr} \longrightarrow \stackrel{116}{46} \mathrm{Pd} + ?\end{equation}\begin{equation} (d)\stackrel{40}{20} \mathrm{Ca}+\stackrel{238}{92} \mathrm{U} \longrightarrow \stackrel{70}{30} \mathrm{Zn}+4 \stackrel{1}{0}\mathrm{n}+2 ?\end{equation}

Write balanced nuclear equations for the following transformations: \((\mathbf{a})\) bismuth-213 undergoes alpha decay; \((\mathbf{b})\) nitrogen-13 undergoes electron capture; \((\mathbf{c})\) technicium-98 undergoes electron capture; \((\mathbf{d})\) gold-188 decays by positron emission.

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