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A convenient source of gamma rays for radiation chemistry research is \(^{60}\) Co, which undergoes the following decay process: \(_{27}^{60} \mathrm{Co} \rightarrow_{28}^{60} \mathrm{Ni}+\beta^{-}+\gamma .\) The half-life of \(^{60} \mathrm{Co}\) is \(1.9 \times 10^{3}\) days a. What is the rate constant for the decay process? b. How long will it take for a sample of \(^{60}\) Co to decay to half of its original concentration?

Short Answer

Expert verified
The rate constant for the decay process of Cobalt-60 is approximately \(3.65 \times 10^{-4} \ \text{days}^{-1}\) and it will take \(1.9 \times 10^{3} \mathrm{days}\) for a sample of Co-60 to decay to half of its original concentration.

Step by step solution

01

1. Finding the rate constant for the decay process

Using the relationship between the half-life and the rate constant for first-order decay reactions: Half-life = \(t_{1/2}\) = \( \frac{ln(2)}{k} \) Where \(t_{1/2}\) is the half-life and k is the rate constant. We know that the half-life of Co-60 is 1.9 x 10鲁 days. We can now solve for the rate constant k: \(k = \frac{ln(2)}{t_{1/2}} \) \(k = \frac{ln(2)}{1.9 \times 10^3 \text{ days}} \)
02

2. Calculating the rate constant

Now, we can plug the half-life of Cobalt-60 into the equation to find the rate constant: \(k = \frac{ln(2)}{1.9 \times 10^3 \text{ days}} \approx 3.65 \times 10^{-4} \ \text{days}^{-1} \) The rate constant for the decay process of Cobalt-60 is approximately 3.65 x 10鈦烩伌 days鈦宦.
03

3. Finding the time it takes for the sample to decay to half of its original concentration

To find the time it takes for the sample to decay to half of its original concentration, we use the fact that the half-life is the time it takes for the sample to decay to half of its original concentration. In other words, the half-life is the answer to this part of the problem: Half-life of Cobalt-60 = \(1.9 \times 10^{3} \mathrm{days}\) Therefore, it will take 1.9 x 10鲁 days for a sample of Co-60 to decay to half of its original concentration.

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Key Concepts

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

Gamma rays
Gamma rays are high-energy electromagnetic waves that originate from the atomic nucleus. Unlike alpha or beta particles, gamma rays have no charge and are unimpeded by magnetic fields. This makes them highly penetrative and capable of passing through materials that would typically absorb other types of radiation. Due to these properties, gamma rays are extensively used in both medical applications and scientific research.

In radiation chemistry, gamma rays can be harnessed to study chemical processes that occur when materials interact with radiation. When cobalt-60 undergoes decay, it emits gamma rays as part of the decay process. This decay changes cobalt-60 into nickel-60, releasing a beta particle and a gamma ray as it transforms. The gamma rays emitted are useful in sterilizing medical equipment and in cancer therapy. However, due to their high-energy, appropriate safety measures must be observed to protect against exposure.
Half-life
The concept of "half-life" is crucial for understanding radioactive decay. The half-life of a radioactive isotope is the time required for half of the substance to decay. For cobalt-60, this half-life is 1.9 x 10鲁 days, or about 5.2 years.

This means that after 5.2 years, only half of the original amount of cobalt-60 will remain. After another 5.2 years, a further half of the remaining amount will have decayed, and so on. This predictable pattern allows scientists to accurately measure and predict the decay of isotopes over time, which is particularly important in fields like nuclear medicine and radiological dating.

Understanding half-lives also helps in determining how long a radioactive source will remain active and effective鈥攃ritical information for both diagnostic and therapeutic applications in medicine and industry.
Rate constant
The rate constant, often denoted as "k", is a parameter that provides insight into how quickly a radioactive material undergoes decay. In the context of radioactive decay, it helps describe the decay kinetics of an isotope, meaning how fast or slow this process occurs.

For first-order decay processes, such as that of cobalt-60, the relationship between the rate constant and the half-life is given by the equation:
\[ k = \frac{\ln(2)}{t_{1/2}} \]
Here, \(\ln(2)\) is the natural logarithm of 2, approximately equal to 0.693, and \(t_{1/2}\) is the half-life of the decay process. As shown in the solution, the rate constant for cobalt-60 is approximately \(3.65 \times 10^{-4} \, \text{days}^{-1}\).

This value of "k" signifies that the substance is decaying at a certain rate every day, and it's a vital factor in predicting the behavior of radioactive substances in various chemical and physical processes.
Radiation chemistry
Radiation chemistry involves the study of chemical changes that occur when materials are exposed to radiation. It is a critical area of research that encompasses a variety of processes, from the decomposition of materials to the creation of new compounds.

When gamma rays, like those emitted from cobalt-60 decay, interact with a substance, they can excite or ionize its molecules. This interaction can lead to the breaking of chemical bonds, the formation of free radicals, and other chemical transformations.

Applications of radiation chemistry are vast and impactful. They include the modification of polymers, the sterilization of medical devices, and even cancer treatment through radiation therapy.
  • Studying radiation's effect on chemical compounds helps improve the safety and efficiency of these processes.
  • Innovations in this field can lead to enhanced material properties or the development of new substances.
The knowledge gained through radiation chemistry not only advances scientific understanding but also paves the way for practical applications that benefit society.

It's essential for scientists and engineers to understand the underlying principles and safety protocols associated with radiation exposure to harness the benefits while minimizing risks.

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Most popular questions from this chapter

Catalase is an enzyme that promotes the conversion of hydrogen peroxide \(\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)\) into water and oxygen. The diffusion constant and radius for catalase are \(6.0 \times 10^{-7} \mathrm{cm}^{2} \mathrm{s}^{-1}\) and 51.2 A. For hydrogen peroxide the corresponding values are \(1.5 \times 10^{-5} \mathrm{cm}^{2} \mathrm{s}^{-1}\) and \(r \sim 2.0\) A. The experimentally determined rate constant for the conversion of hydrogen peroxide by catalase is \(5.0 \times 10^{6} \mathrm{M}^{-1} \mathrm{s}^{-1} .\) Is this a diffusioncontrolled reaction?

Imidazole is a common molecular species in biological chemistry. For example, it constitutes the side chain of the amino acid histidine. Imidazole can be protonated in solution as follows: The rate constant for the protonation reaction is \(5.5 \times 10^{10} \mathrm{M}^{-1} \mathrm{s}^{-1} .\) Assuming that the reaction is diffusion controlled, estimate the diffusion coefficient of imidazole when \(D\left(\mathrm{H}^{+}\right)=9.31 \times 10^{-5} \mathrm{cm}^{2} \mathrm{s}^{-1}, r\left(\mathrm{H}^{+}\right) \sim 1.0 \AA\) and \(r\) (imidazole) \(=6.0\) A. Use this information to predict the rate of deprotonation of imidazole by \(\mathrm{OH}^{-}\left(D=5.30 \times 10^{-5} \mathrm{cm}^{2} \mathrm{s}^{-1}\right.\) and \(r=\sim 1.5 \AA)\)

An experiment is performed on the following parallel reaction: Two things are determined: (1) The yield for B at a given temperature is found to be 0.3 and (2) the rate constants are described well by an Arrhenius expression with the activation to \(\mathrm{B}\) and \(\mathrm{C}\) formation being 27 and \(34 \mathrm{kJ} \mathrm{mol}^{-1}\), respectively, and with identical preexponential factors. Demonstrate that these two statements are inconsistent with each other.

For a type II second-order reaction, the reaction is \(60 \%\) complete in 60 seconds when \([\mathrm{A}]_{0}=0.1 \mathrm{M}\) and \([\mathrm{B}]_{0}=0.5 \mathrm{M}\) a. What is the rate constant for this reaction? b. Will the time for the reaction to reach \(60 \%\) completion change if the initial reactant concentrations are decreased by a factor of \(2 ?\)

P35.32 The reaction of atomic chlorine with ozone is the first step in the catalytic decomposition of stratospheric ozone by \(\mathrm{Cl} \bullet\):$$\mathrm{Cl} \cdot(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{ClO} \cdot(g)+\mathrm{O}_{2}(g)$$ At \(298 \mathrm{K}\) the rate constant for this reaction is \(6.7 \times 10^{9} \mathrm{M}^{-1} \mathrm{s}^{-1}\) Experimentally, the Arrhenius pre-exponential factor was determined to be \(1.4 \times 10^{10} \mathrm{M}^{-1} \mathrm{s}^{-1} .\) Using this information determine the activation energy for this reaction.

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