/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 Radioactive gold-198 is used in ... [FREE SOLUTION] | 91Ó°ÊÓ

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Radioactive gold-198 is used in the diagnosis of liver problems. \({ }^{198}\) Au decays in a first-order process, emitting a \(\beta\) particle (electron). The half-life of this isotope is 2.7 days. You begin with a 5.6 -mg sample of the isotope. Calculate how much gold-198 remains after 1.0 day.

Short Answer

Expert verified
After 1.0 day, approximately 4.33 mg of gold-198 remains.

Step by step solution

01

Understand the Decay Process

Gold-198 undergoes a first-order decay process characterized by a half-life, which means that the amount of the substance decays exponentially over time. The relationship for a first-order reaction is given by the equation \( N_t = N_0 e^{-kt} \), where: \( N_t \) is the amount remaining at time \( t \), \( N_0 \) is the initial amount, and \( k \) is the decay constant.
02

Determine the Decay Constant \( k \)

The decay constant \( k \) is related to the half-life \( t_{1/2} \) by the equation \( k = \frac{\ln(2)}{t_{1/2}} \). For this problem, the half-life \( t_{1/2} \) is 2.7 days. Therefore, \( k = \frac{\ln(2)}{2.7} \approx 0.2569 \, \text{days}^{-1} \).
03

Calculate the Remaining Gold-198 After 1 Day

Use the first-order decay equation \( N_t = N_0 e^{-kt} \). Plug in \( N_0 = 5.6 \, \text{mg} \), \( k = 0.2569 \, \text{days}^{-1} \), and \( t = 1.0 \, \text{day} \): \[ N_t = 5.6 \, \text{mg} \times e^{-0.2569 \times 1.0} \approx 5.6 \, \text{mg} \times 0.7731 \approx 4.33 \, \text{mg} \].
04

Verify Solution

Double-check the calculations: the exponential \( e^{-0.2569} \approx 0.7731 \), and multiplying it by the initial mass gives \( 4.33 \, \text{mg} \), confirming the calculations are correct.

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Key Concepts

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

First-Order Process
A first-order process in chemistry, especially in nuclear decay, is one where the rate of decay is directly proportional to the amount of substance present. This means that the decay does not happen at a constant rate, but rather at a rate that changes as the quantity of the radioactive substance decreases. In the case of gold-198, the first-order decay can be represented by the equation:\[ N_t = N_0 e^{-kt} \]Here:
  • \( N_t \) is the amount of substance remaining after time \( t \).
  • \( N_0 \) is the initial amount of the substance (5.6 mg in this example).
  • \( k \) is the decay constant, a crucial value determining the speed of the decay.
As the reaction proceeds, the amount of the substance decreases exponentially over time. This characteristic makes first-order processes unique, as they involve a proportionate and continuous decrease, not a linear one. The decay of gold-198 being a first-order process means it will grow slower over time as the available amount for decay diminishes.
Half-Life Calculation
The half-life (\( t_{1/2} \)) of a radioactive substance is the time required for half of the substance to decay. It is a key component in understanding radioactive decay processes. In the case of gold-198, which has a half-life of 2.7 days, this characteristic implies that after 2.7 days, only half of the initial amount remains.To find the decay constant (\( k \)), which is essential for the first-order decay equation, use the relationship:\[ k = \frac{\ln(2)}{t_{1/2}} \]For gold-198, given the half-life of 2.7 days, the decay constant is calculated as:\[ k \approx 0.2569 \, \text{days}^{-1} \]Knowing \( k \) allows the determination of how much of the substance is left after a certain amount of time. Understanding half-life helps predict how quickly a radioactive material will lose its activity, which is essential in fields like medicine and environmental science.
Beta Particle Emission
When gold-198 undergoes radioactive decay, it emits beta particles. A beta particle is essentially an electron that is emitted from the nucleus of an atom during a radioactive process. This occurs because a neutron in the gold-198 nucleus is transformed into a proton, emitting an electron and an antineutrino in the process. This transformation not only produces an electron but also alters the element itself because the number of protons in the nucleus changes. Thus, beta decay can convert one element into another - in the case of gold-198, it transforms into mercury-198 once it emits a beta particle. The emission of beta particles is crucial in various applications:
  • In medicine, it can be used for diagnosis and treatment, as beta emissions can target specific areas, like the liver in diagnostic studies involving gold-198.
  • Beta decay also plays a critical role in understanding nuclear reactions and the stability of isotopes.
Grasping the concept of beta particle emission is vital for comprehending how radioactive isotopes change and apply their properties in real-world scenarios.

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Most popular questions from this chapter

The rate of decay of a radioactive solid is independent of the temperature of that solid-at least for temperatures easily obtained in the laboratory. What does this observation imply about the activation energy for this process?

If a reaction has the experimental rate law, Rate \(=k[\mathrm{~A}]^{2}\), explain what happens to the rate when (a) the concentration of A is tripled. (b) the concentration of A is halved.

Suppose a chemical reaction has an activation energy of \(76 \mathrm{~kJ} / \mathrm{mol}\), as in the example in Figure \(11.12 .\) Calculate by what factor the rate of the reaction at \(50 .{ }^{\circ} \mathrm{C}\) is increased over its rate at \(25^{\circ} \mathrm{C}\).

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Make an Arrhenius plot and calculate the activation energy for the gas-phase reaction \(2 \mathrm{NOCl}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g})\) \begin{tabular}{lc} \hline\(T(\mathrm{~K})\) & Rate Constant \(\left(\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}\right)\) \\ \hline \(400 .\) & \(6.95 \times 10^{-4}\) \\ \(450 .\) & \(1.98 \times 10^{-2}\) \\ \(500 .\) & \(2.92 \times 10^{-1}\) \\ \(550 .\) & 2.60 \\ \(600 .\) & 16.3 \\ \hline \end{tabular}

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