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Chapter 26: Problem 20

Sulfur-38 can be incorporated into proteins to follow certain aspects of protein metabolism. If a protein sample initially has an activity of 10000 disintegrations \(\cdot \min ^{-1}\) calculate the activity 6.00 h later. The half-life of sulfur- 38 is 2.84 h. Hint: Use the fact that the rate of decay is proportional to \(N(t)\) for a first-order process.

Short Answer

Expert verified
The activity 6 hours later is approximately 2310 disintegrations/min.

Step by step solution

01

Understand the decay formula

The decay of radioactive substances can be described using the first-order kinetic formula: \[ N(t) = N_0 e^{-kt} \] where \( N(t) \) is the remaining activity at time \( t \), \( N_0 \) is the initial activity, \( k \) is the decay constant, and \( t \) is the time elapsed.
02

Determine the decay constant

To find the decay constant \( k \), we use the relation that involves half-life \( \frac{1}{2} = e^{-k t_{1/2}} \). Knowing the half-life \( t_{1/2} = 2.84 \) hours, we can solve for \( k \):\[ k = \frac{\ln(2)}{t_{1/2}} = \frac{0.693}{2.84} \approx 0.244 \text{ h}^{-1} \].
03

Calculate the remaining activity

Substitute the initial activity \( N_0 = 10000 \) disintegrations/min, the decay constant \( k \approx 0.244 \text{ h}^{-1} \), and the time elapsed \( t = 6.00 \) h into the decay formula:\[ N(t) = 10000 \times e^{-0.244 \times 6.00} \].
04

Compute and interpret the result

Calculate \( N(t) \) to find the activity: \[ N(t) \approx 10000 \times e^{-1.464} \].Using a calculator, \( e^{-1.464} \approx 0.231 \), hence:\[ N(t) \approx 10000 \times 0.231 = 2310 \] disintegrations/min.So, the activity 6 hours later is approximately 2310 disintegrations/min.

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Key Concepts

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

Half-Life
In radioactive decay, the half-life is a fundamental concept. It represents the time taken for half of the radioactive nuclei in a sample to decay. During each half-life, the number of disintegrations reduces by 50%. This process continues regardless of the sample's size or activity level. For sulfur-38, the half-life is given as 2.84 hours. This means that every 2.84 hours, sulfur-38 will lose half of its radioactive activity. Understanding half-life helps us track the rate of decay and predict how long a sample will remain active. It is particularly useful when following metabolic processes like protein metabolism through radioactive tracing.
Decay Constant
The decay constant (\(k\)) is a key parameter in radioactive decay equations. This constant provides insight into the rate at which a radioactive substance decays over time. In first-order kinetics, the decay constant allows us to relate the half-life of a substance to its decay rate. For sulfur-38, we calculate the decay constant using the formula:\[ k = \frac{\ln(2)}{t_{1/2}} \]Here, \(t_{1/2}\) is the half-life. Substituting \(t_{1/2} = 2.84\) hours for sulfur-38, we find:\[ k \approx 0.244 \text{ h}^{-1} \]This means 24.4% of sulfur-38's activity decays every hour.
First-Order Kinetics
Radioactive decay follows first-order kinetics, where the rate of decay is proportional to the current amount of substance present. This can be mathematically represented by the equation:\[ N(t) = N_0 e^{-kt} \]Here, \(N(t)\) is the remaining activity at time \(t\), and \(N_0\) is the initial activity. The first-order nature simplifies calculations and is applicable to many different types of decay processes. With sulfur-38, knowing the initial activity and decay constant, we can predict future activity levels at specific times, such as 6 hours later. This property makes radioactive isotopes important tools in fields like medicine and biology.
Sulfur-38
Sulfur-38 is a radioactive isotope of sulfur used in various scientific applications, particularly in the study of protein metabolism. Its radioactive properties make it a useful tracer, allowing researchers to trace the path and rate of metabolic processes in organisms. Sulfur-38 decays according to its characteristic half-life and decay constant. Because it's incorporated into proteins, it provides a dynamic view of metabolic pathways. As it decays, the changing levels of radioactivity can be measured, offering insights into how quickly and efficiently proteins are synthesized and broken down within cells.
Protein Metabolism
Protein metabolism is the comprehensive set of metabolic processes that involve proteins, including their biosynthesis and degradation. Radioactive isotopes like sulfur-38 are integral in studying these complicated processes. By incorporating isotopes into proteins, scientists can monitor how proteins are constructed from amino acids, and how they are recycled within organisms. This gives detailed insights into cellular functions and how proteins maintain homeostasis. Tracking sulfur-38 as it decays helps in understanding specific aspects of protein metabolism, such as the speed of protein turnover and synthesis鈥攙ital for developments in health and disease research.

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Most popular questions from this chapter

The gas-phase rearrangement reaction vinyl allyl ether \(\longrightarrow\) allyl acetone has a rate constant of \(6.015 \times 10^{-5} \mathrm{s}^{-1}\) at \(420 \mathrm{K}\) and a rate constant of \(2.971 \times 10^{-3} \mathrm{s}^{-1}\) at \(470 \mathrm{K} .\) Calculate the values of the Arrhenius parameters \(A\) and \(E_{\mathrm{g}^{*}}\) Calculate the values of \(\Delta^{\dagger} H^{\circ}\) and \(\Delta^{\dagger} S^{\circ}\) at \(420 \mathrm{K} .\) (Assume ideal-gas behavior.)

In this problem, we will derive Equation 26.32 from the rate law (Equation 26.31 ) \\[ -\frac{d[\mathrm{A}]}{d t}=k[\mathrm{A}][\mathrm{B}] \\] Use the reaction stoichiometry of Equation 26.30 to show that \([\mathrm{B}]=[\mathrm{B}]_{0}-[\mathrm{A}]_{0}+[\mathrm{A}]\) Use this result to show that Equation 1 can be written as \\[ -\frac{d[\mathrm{A}]}{d t}=k[\mathrm{A}]\left|[\mathrm{B}]_{0}-[\mathrm{A}]_{0}+[\mathrm{A}]\right| \\] Now separate the variables and then integrate the resulting equation subject to its initial conditions to obtain the desired result, Equation 26.32 \\[ k t=\frac{1}{[\mathbf{A}]_{0}-[\mathbf{B}]_{0}} \ln \frac{[\mathbf{A}][\mathbf{B}]_{0}}{[\mathbf{B}][\mathbf{A}]_{0}} \\]

Show that if A reacts to form either \(\mathrm{B}\) or \(\mathrm{C}\) according to $$ \mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{B} \quad \text { or } \quad \mathrm{A} \stackrel{k_{2}}{\rightarrow} \mathrm{C} $$ then $$ [\mathrm{A}]=[\mathrm{A}]_{0} e^{-\left(k_{1}+k_{2}\right) t} $$ Now show that \(t_{1 / 2}\), the half-life of \(\mathrm{A}\), is given by $$ t_{1 / 2}=\frac{0.693}{k_{1}+k_{2}} $$ Show that \([\mathrm{B}] /[\mathrm{C}]=k_{1} / k_{2}\) for all times \(t\). For the set of initial conditions \([\mathrm{A}]=[\mathrm{A}]_{0}\), \([\mathrm{B}]_{0}=[\mathrm{C}]_{0}=0\), and \(k_{2}=4 k_{1}\), plot \([\mathrm{A}],[\mathrm{B}]\), and \([\mathrm{C}]\) as a function of time on the same graph. The following six problems deal with the decay of radioactive isotopes, which is a firstorder process. Therefore, if \(N(t)\) is the number of a radioactive isotope at time \(t\), then \(N(t)=N(0) e^{-k t}\), where \(N(0)\) is the corresponding number at \(t=0 .\) In dealing with radioactive decay, we use the half-life, \(t_{1 / 2}=0.693 / k\), almost exclusively to describe the rate of decay ( the kinetics of decay).

The kinetics of a chemical reaction can be followed by a variety of experimental techniques, including optical spectroscopy, NMR spectroscopy, conductivity, resistivity, pressure changes, and volume changes. When using these techniques, we do not measure the concentration itself but we know that the observed signal is proportional to the concentration; the exact proportionality constant depends on the experimental technique and the species present in the chemical system. Consider the general reaction given by $$ v_{\mathrm{A}} \mathrm{A}+v_{\mathrm{B}} \mathrm{B} \longrightarrow v_{\mathrm{Y}} \mathrm{Y}+v_{\mathrm{Z}} \mathrm{Z} $$ where we assume that \(\mathrm{A}\) is the limiting reagent so that \([\mathrm{A}] \rightarrow 0\) as \(t \rightarrow \infty\). Let \(p_{i}\) be the proportionality constant for the contribution of species \(i\) to \(S\), the measured signal from the instrument. Explain why at any time \(t\) during the reaction, \(S\) is given by $$ S(t)=p_{\mathrm{A}}[\mathrm{A}]+p_{\mathrm{B}}[\mathrm{B}]+p_{\mathrm{Y}}[\mathrm{Y}]+p_{\mathrm{Z}}[\mathrm{Z}] $$ Show that the initial and final readings from the instrument are given by $$ S(0)=p_{\mathrm{A}}[\mathrm{A}]_{0}+p_{\mathrm{B}}[\mathrm{B}]_{0}+p_{\mathrm{Y}}[\mathrm{Y}]_{0}+p_{\mathrm{Z}}[\mathrm{Z}]_{0} $$ tand $$ S(\infty)=p_{\mathrm{B}}\left([\mathrm{B}]_{0}-\frac{v_{\mathrm{B}}}{v_{\mathrm{A}}}[\mathrm{A}]_{0}\right)+p_{\mathrm{Y}}\left([\mathrm{Y}]_{0}+\frac{v_{\mathrm{Y}}}{v_{\mathrm{A}}}[\mathrm{A}]_{0}\right)+p_{\mathrm{Z}}\left([\mathrm{Z}]_{0}+\frac{v_{\mathrm{Z}}}{v_{\mathrm{A}}}[\mathrm{A}]_{0}\right) $$ Combine Equations 1 through 3 to show that $$ [\mathrm{A}]=[\mathrm{A}]_{0} \frac{S(t)-S(\infty)}{S(0)-S(\infty)} $$ (Hint: Be sure to express [B], [Y], and [Z] in terms of their initial values, [A] and \([\mathrm{A}]_{0}\).)

Copper-64 \(\left(t_{1 / 2}=12.8 \mathrm{h}\right)\) is used in brain scans for tumors and in studies of Wilson's disease (a genetic disorder characterized by the inability to metabolize copper). Calculate the number of days required for an administered dose of copper- 64 to drop to \(0.10 \%\) of the initial value injected. Assume no loss of copper- 64 except by radioactive decay.
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