Question

A radioactive material, such as the isotope thorium- \(234,\) disintegrates at a rate proportional to the amount currently present. If \(Q(t)\) is the amount present at time \(t,\) then \(d Q / d t=-r Q\) where \(r>0\) is the decay rate. (a) If \(100 \mathrm{mg}\) of thorium- 234 decays to \(82.04 \mathrm{mg}\) in 1 week, determine the decay rate \(r\). (b) Find an expression for the amount of thorium- 234 present at any time \(t .\) (c) Find the time required for the thorium- 234 to decay to one-half its original amount.

Short answer

Answer: The decay rate of thorium-234 is approximately 0.194 per week. The expression for the remaining amount at any given time is Q(t) = 100e^(-0.194t). It takes approximately 3.57 weeks for the thorium-234 to decay to half its original amount.

Step-by-step solution

(Part a: Determine decay rate r)

First, we are given the information: Initial amount: \(Q_0 = 100\) mg Amount after 1 week: \(Q_1 = 82.04\) mg Time elapsed: \(t_1 = 1\) week We know that the equation governing radioactive decay is \(\frac{dQ}{dt} = -rQ\) To find the decay rate \(r\), we will first solve the differential equation: Separating variables, we get: \(\frac{dQ}{Q} = -r dt\) Integrating both sides, we have: \(\int \frac{dQ}{Q} = -r \int dt\) This yields: \(ln(Q) = -rt + C\) Now, we will plug in the initial conditions to find the constant \(C\): At \(t=0\), \(Q=Q_0=100\): \(ln(100) = -r(0) + C\) \(ln(100) = C\) Now we plug in the information for one week later to find the decay rate \(r\): At \(t=t_1=1\) week, \(Q=Q_1=82.04\): \(ln(82.04) = -r(1) + ln(100)\) Solving for \(r\): \(r = ln(100) - ln(82.04)\) \(r \approx 0.194\) So the decay rate of thorium-234 is approximately \(0.194\) per week.

(Part b: Expression for Q(t))

Using the equation we derived from the integration: \(ln(Q) = -rt + C\) We will now solve for Q(t): \(Q(t) = e^{-rt + C}\) Using the fact that \(C = ln(100)\), we have: \(Q(t) = e^{-rt + ln(100)}\) \(Q(t) = e^{ln(100) - rt}\) \(Q(t) = 100e^{-rt}\) The expression for the amount of thorium-234 present at any time t is: \(Q(t) = 100e^{-0.194t}\)

(Part c: Time for half-decay)

Now we need to find the time it takes for the thorium-234 to decay to half of its original amount: \(\frac{1}{2}Q_0 = 100e^{-0.194t}\) Dividing both sides by 100: \(\frac{1}{2} = e^{-0.194t}\) Taking the natural logarithm of both sides: \(ln \left(\frac{1}{2}\right) = -0.194t\) Solve for t: \(t = \frac{ln \left(\frac{1}{2}\right)}{-0.194}\) \(t \approx 3.57\) weeks Therefore, it takes approximately 3.57 weeks for the thorium-234 to decay to one-half its original amount.

Key concepts

Exponential Decay Rate

The exponential decay rate is a constant that describes how quickly a quantity decreases over time. In the context of radioactive materials, it is crucial to understand how they lose their activity and become less potent. The differential equation \(\frac{dQ}{dt} = -rQ\) represents the rate of decay, where \(Q(t)\) is the quantity of the substance at time \(t\), and \(r\) is the decay rate. The negative sign indicates a decrease in the quantity over time.

Solving for the constant \(r\) involves observing how much the substance has decayed over a given period, usually provided in an exercise, and applying mathematical methods to find its value. The decay rate helps us predict future behavior of the radioactive material, making it particularly important in fields such as pharmacology, nuclear physics, and environmental science.

An important point here is understanding that the decay rate is proportional to the current amount of the substance, meaning that as the substance decays and there is less of it, the rate at which it decays also slows down. This property leads us to the concept of half-life, which is a stable measure of the decay characteristic of a substance regardless of its initial amount.

Solving Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. In the case of radioactive decay, the equation \(\frac{dQ}{dt} = -rQ\) is a first-order linear differential equation and is key to modeling the decay process. Solving such an equation involves separating variables, integrating both sides, and then applying initial conditions to determine any constants of integration.

In the exercise example, the separation of variables and integration allowed us to derive the formula \(ln(Q) = -rt + C\), which we then converted to \(Q(t) = 100e^{-rt}\) after applying the initial conditions. This process is a staple in differential equations, where the general solution is found first, followed by the particular solution after considering the given data points, like initial amount and decay after a certain period.

One common improvement to understanding the steps involves a graphical representation, using software or calculators, where students can visualize how the exponential function decreases over time as \(t\) increases. Visual aids can build intuition in students for the relationship between the decay rate and the quantity of the radioactive material over time.

Half-Life Calculation

The half-life of a radioactive substance is the time it takes for half of the original amount of the substance to decay. It is a measure that's independent of the initial quantity, making it a characteristic property of the substance. In the example provided, the half-life calculation is derived by setting up an equation where the final quantity \(Q(t)\) is half of the initial amount \(\frac{1}{2}Q_0\), then solving for \(t\).

The calculation leads to the logarithmic equation \(ln\left(\frac{1}{2}\right) = -0.194t\), which isolates the time \(t\). The result is an estimation of the half-life period for the radioactive material based on the decay rate that was previously determined. In practice, half-life information is vital for tasks such as determining safe storage times for radioactive waste, dosage calculations for medical treatments, and dating archaeological discoveries.

Understanding and calculating half-life helps students grasp the concept of exponential decay in a tangible way, demonstrating how the theoretical equations relate to physical phenomena they might encounter in various scientific fields. It's also beneficial to reinforce the problem-solving process by encouraging students to learn about different methods for solving exponential equations, such as graphical representations or using logarithm properties.