Problem 8 - Essential Calculus Early Transcendentals

Question

Strontium-90 has a half-life of 28 days. (a) A sample has a mass of 50 \(\mathrm{mg}\) initially. Find a formula for the mass remaining after \(t\) days. (b) Find the mass remaining after 40 days. (c) How long does it take the sample to decay to a mass of 2 \(\mathrm{mg} ?\) (d) Sketch the graph of the mass function.

Step-by-Step Solution

Answer

(a) \(m(t) = 50 \times (1/2)^{t/28}\); (b) 35.35 mg; (c) 138.68 days.

Step 1: Understanding Half-life

The half-life of a substance is the time required for the amount of the substance to reduce to half its initial value. For Strontium-90, the half-life (\(t_{1/2}\)) is given as 28 days.

Step 2: Derive the Decay Formula

The formula for exponential decay can be obtained using the relation \(m(t) = m_0 \times (1/2)^{t/t_{1/2}}\), where \(m(t)\) is the mass at time \(t\), \(m_0\) is the initial mass, and \(t_{1/2}\) is the half-life. Plugging in the values, we get \(m(t) = 50 \times (1/2)^{t/28}\).

Step 3: Calculate Mass After 40 Days

Using the formula \(m(t) = 50 \times (1/2)^{t/28}\), substitute \(t = 40\) to find the mass. We calculate:\[m(40) = 50 \times (1/2)^{40/28} \approx 50 \times 0.707 \approx 35.35\, \text{mg}\]

Step 4: Solve for Time to Decay to 2 mg

We need to find \(t\) when \(m(t) = 2\, \text{mg}\). Set up the equation: \[2 = 50 \times (1/2)^{t/28}\] Divide both sides by 50: \[(1/2)^{t/28} = 0.04\] Take the logarithm of both sides: \[\log((1/2)^{t/28}) = \log(0.04)\]Use log properties: \[(t/28) \log(1/2) = \log(0.04)\]Solve for \(t\): \[t = 28 \times \frac{\log(0.04)}{\log(1/2)} \approx 138.68\, \text{days}\]

Step 5: Sketch the Graph of the Function

Sketch the graph of \(m(t) = 50 \times (1/2)^{t/28}\). The graph is an exponentially decreasing curve starting at 50 \(\text{mg}\) at \(t=0\), passing through approximately 35.35 \(\text{mg}\) when \(t=40\), and reaching close to 2 \(\text{mg}\) around \(t=138.68\, \text{days}\).

Key Concepts

Half-life

Half-life is a key concept in understanding radioactive decay. It represents the time it takes for half of a given quantity of a radioactive substance to decay. For example, Strontium-90, a radioactive isotope, has a half-life of 28 days. This means if you start with 50 mg of Strontium-90, after 28 days, only 25 mg will remain, as the other half will have decayed.
This pattern of decay continues over successive half-lives. For instance:

• After 56 days (two half-lives), the mass would reduce to 12.5 mg.
• After 84 days (three half-lives), it would further decline to 6.25 mg.

Understanding half-life is foundational in physics and chemistry to predict how a substance's amount decreases over time, especially for hazardous materials, which need careful monitoring.

Exponential Decay

Exponential decay is a mathematical process describing how quantities decrease rapidly at first, and then slowly over time. In the context of radioactive decay, this model fits perfectly, as it aligns with the half-life concept.
The decay formula in question, \[m(t) = m_0 \times (1/2)^{t/t_{1/2}}\] where:

• \(m(t)\) is the mass remaining at time \(t\).
• \(m_0\) is the initial mass.
• \(t_{1/2}\) is the half-life of the substance.

provides a way to calculate the remaining amount after a given period. By plugging different values of \(t\) into the formula, you can observe how the mass of the substance continuously decreases, showcasing exponential behavior. Exponential decay is pivotal in various fields, including finance, population studies, and engineering, wherever declines occur gradually over time.

Logarithmic Equations

Logarithmic equations are essential tools to solve exponential decay problems, especially when determining the time taken for a substance to reach a certain amount. To find out how long it will take for Strontium-90 to decay to a mass of 2 mg, we use logarithms.
The equation \[(1/2)^{t/28} = 0.04\] translates into a logarithmic form: \[(t/28) \log(1/2) = \log(0.04)\]This equation arises from the properties of logarithms, crucial for isolating the time variable \(t\) in decay problems.
The property \(\log(a^x) = x \log(a)\) is employed to solve for \(t\). By analyzing these logarithmic computations, students can grasp how logarithms allow for the manipulation and solving of exponential decay scenarios, making them versatile in various scientific and mathematical applications.