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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?

Step by step solution

01

Determine the Relationship Between Constants

Since it is given that initially, at time \( t = 0 \), there is no \( ^{230}_{88}Th \), we have \( N_2(0) = 0 \). Substituting \( t = 0 \) in the trial solution \( N_2(t) = N_{10} [h_1 e^{-u_1 t} + h_2 e^{-u_2 t}] \) provides: \( N_2(0) = N_{10} [h_1 + h_2] = 0 \). Therefore, \( h_1 + h_2 = 0 \).

02

Calculate the Derivative of the Trial Solution

Taking the derivative of \( N_2(t) \) with respect to \( t \) yields \( \frac{dN_2(t)}{dt} = N_{10}[-h_1u_1 e^{-u_1 t} - h_2u_2 e^{-u_2 t}] \).

03

Substitute and Equate Coefficients

Substitute \( \frac{dN_2(t)}{dt} \) into the differential equation: \( \frac{dN_2(t)}{dt} = u_1 N_{01} e^{-u_1 t} - u_2 N_2(t) \). This gives \( N_{10}[-h_1u_1 e^{-u_1 t} - h_2u_2 e^{-u_2 t}] = u_1 N_{01} e^{-u_1 t} - u_2 N_{10} [h_1 e^{-u_1 t} + h_2 e^{-u_2 t}] \). Collect the coefficients of \( e^{-u_1 t} \) and \( e^{-u_2 t} \) and set them to zero: \(-h_1u_1 = u_1 - u_2 h_1\) and \(-h_2u_2 = -u_2 h_2\).

04

Solve for Constants

From the first equation, \(-h_1u_1 = u_1 - u_2 h_1\), rearrange to get \(h_1(u_1 - u_2) = u_1 \), so \( h_1 = \frac{u_1}{u_1 - u_2} \). Since \( h_1 + h_2 = 0 \), we find \( h_2 = -\frac{u_1}{u_1 - u_2} \).

05

Calculate the Mass of Thallium at a Given Time

Given \( t = 2.46 \times 10^5 \) years, calculate the number of remaining \( ^{234}_{92}U \) nuclei using \( N(t) = N_{10} e^{-u_1 t} \). With \( N_{10} = 30.0 \) g of \( ^{234}_{92}U \), convert this mass into mols (using its atomic mass) to find initial nuclei. Use \( N_2(t) = N_{10} [h_1 e^{-u_1 t} + h_2 e^{-u_2 t}] \) to find \( N_2(t) \), convert back to mass using \( ^{230}_{88}Th \)'s atomic mass.

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Key Concepts

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

Decay Constant

In the study of radioactive decay, we often encounter the term 'decay constant'. This is a measure of how quickly a particular isotope undergoes decay. It is represented by the Greek letter \( \lambda \). The decay constant is specific to each radioactive isotope and indicates the probability per unit time that a nucleus will decay.
For instance, in the decay process of \( ^{234}_{92}U \rightarrow ^{230}_{88}Th \), the decay constant \( \lambda_1 \) tells us the rate at which uranium-234 transforms into thorium-230. Each decay step in a sequence or chain has its own decay constant, reflecting how individual isotopes uniquely behave in their decay processes.

• The higher the decay constant, the faster the rate of decay.
• It is a crucial variable in calculating other important properties, like half-life.

Being able to determine the decay constant is essential for understanding the timescales over which radioactive materials change form.

Differential Equation

Differential equations play an important role in describing the process of radioactive decay. They allow us to express the changing number of radioactive nuclei mathematically.
In the context of our problem, the differential equation \( \frac{dN_2(t)}{dt} = \lambda_1 N_{10} e^{-\lambda_1 t} - \lambda_2 N_2(t) \) provides a dynamic view of how the quantity of thorium-230 changes over time. This equation accounts for both the production of thorium-230 (from uranium-234) and its further decay to radium-226.
Here are a few key points about differential equations in this context:

• They involve derivatives, which represent rates of change, such as how fast uranium-234 decays into thorium-230.
• The solution to these equations allows us to predict the quantity of a substance at any given time.

By solving these equations, we can gain insights into the nature of the decay process and evaluate parameters such as time-based changes in isotope amounts.

Half-Life

The concept of half-life is fundamental in nuclear physics when examining radioactive elements. It describes the time required for half of a radioactive substance to decay.
In our example, uranium-234 has a half-life of 2.46 x \( 10^5 \) years, meaning that it takes this period for half of any given amount of uranium-234 to transform into other substances via decay processes.
Half-life has several significant implications:

• It is a constant that is unique to each radioactive isotope.
• Understanding half-life helps determine how long a substance will remain active or hazardous.
• It is directly related to the decay constant by the formula \( T_{1/2} = \frac{0.693}{\lambda} \).

The half-life is a convenient measure to appreciate the longevity and persistence of radioactive materials in natural or engineered environments.

Nuclear Physics

Nuclear physics is a branch of physics that deals with the components and behavior of atomic nuclei. It encompasses phenomena such as nuclear decay, reactions, and the underlying forces.
In the context of our problem, nuclear physics allows us to analyze radioactive decay sequences such as \( ^{234}_{92}U \rightarrow ^{230}_{88}Th \rightarrow ^{226}_{84}Ra \).
Key aspects of nuclear physics include:

• The study of nuclear decay processes, which involves changes in nuclear structure and energy release.
• Understanding fundamental forces, like the weak nuclear force, which governs beta decay.
• Applications in sectors such as energy production through fission, medicine with radioactive tracers, and archaeology with radiocarbon dating.

Nuclear physics not only deepens our scientific knowledge but also has practical applications that impact diverse fields from healthcare to environmental science.