Problem 14 - Precalculus

Question

In Exercises \(14-18,\) we list some radioactive isotopes and their associated half-lives. Assume that each decays according to the formula \(A(t)=A_{0} e^{k t}\) where \(A_{0}\) is the initial amount of the material and \(k\) is the decay constant. For each isotope: \- Find the decay constant \(k\). Round your answer to four decimal places. \- Find a function which gives the amount of isotope \(A\) which remains after time \(t\). (Keep the units of \(A\) and \(t\) the same as the given data.) \- Determine how long it takes for \(90 \%\) of the material to decay. Round your answer to two decimal places. (HINT: If \(90 \%\) of the material decays, how much is left?) Cobalt 60 , used in food irradiation, initial amount 50 grams, half-life of 5.27 years.

Step-by-Step Solution

Answer

The decay constant is approximately \(-0.1316\). The decay function is \(A(t) = 50e^{-0.1316t}\). It takes about 17.50 years for 90% to decay.

Step 1: Understanding the Half-life and Decay Constant Relation

The half-life of a substance is the time it takes for half of the initial amount to decay. For a radioactive isotope, the amount remaining is given by \(A(t) = A_0 e^{kt}\). When half the amount remains, \(A(t) = \frac{A_0}{2}\). Substitute the half-life \(t = 5.27\) years and \(A(t) = \frac{A_0}{2}\) into the equation to solve for \(k\): \[ \frac{A_0}{2} = A_0 e^{k \cdot 5.27} \] Divide both sides by \(A_0\): \[ \frac{1}{2} = e^{k \cdot 5.27} \] Take the natural logarithm of both sides: \[ \ln \left( \frac{1}{2} \right) = k \cdot 5.27 \] Solve for \(k\): \[ k = \frac{\ln \left( \frac{1}{2} \right)}{5.27} \] _(1)_

Step 2: Calculating the Decay Constant

Now compute \(k\) using the formula derived above: \[ k = \frac{\ln(0.5)}{5.27} \] Calculate \(k\): \(k \approx \frac{-0.6931}{5.27} \approx -0.1316\) Thus, the decay constant \(k\) is approximately \(-0.1316\) (rounded to four decimal places).

Step 3: Forming the Exponential Decay Function

Now that we have \(k\), we can construct the function to express the amount of cobalt-60 remaining after time \(t\). Use the general formula: \[ A(t) = A_0 e^{kt} \] Plug in \(A_0 = 50\) grams and \(k = -0.1316\): \[ A(t) = 50 e^{-0.1316t} \] This is the decay function for cobalt-60.

Step 4: Calculating Time for 90% Decay

To find the time when 90% of cobalt-60 has decayed, only 10% remains. Start with the equation: \[ A(t) = 0.1A_0 = 50 \cdot 0.1 = 5 \] Substitute into the decay formula: \[ 5 = 50 e^{-0.1316t} \] Solve for \(t\): \[ e^{-0.1316t} = \frac{5}{50} = 0.1 \] Take the natural logarithm: \[ -0.1316t = \ln(0.1) \] Solve for \(t\): \[ t = \frac{\ln(0.1)}{-0.1316} \] Calculate \(t\): \(t \approx \frac{-2.3026}{-0.1316} \approx 17.50\) Therefore, it takes approximately 17.50 years for 90% of the cobalt-60 to decay.

Key Concepts

Exponential decay

Exponential decay is a type of process where a quantity decreases at a rate proportional to its current value. This means that as time passes, the quantity diminishes exponentially, rather than linearly. A classic example of this is radioactive decay, where unstable isotopes lose particles, transforming into different, more stable elements.

We describe exponential decay using the mathematical formula: \[ A(t) = A_0 e^{kt} \] where:

• \(A(t)\) is the amount remaining at time \(t\).
• \(A_0\) is the initial amount present.
• \(e\) is the base of the natural logarithm (approximately 2.718).
• \(k\) is the decay constant, which is negative as the substance is decaying.

Understanding exponential decay helps us predict how quickly radioactive substances reduce over time. For Cobalt-60, our task relies on using this formula to calculate how much remains after a given period.

Half-life

The half-life of a radioactive isotope is the time it takes for half the amount of the substance to decay. It is a crucial concept because it allows us to determine how quickly a substance will lose its radioactivity.

By definition, we utilize the half-life to calculate the decay constant using the relationship:\[ k = \frac{\ln \left( \frac{1}{2} \right)}{t_{1/2}} \] where \(t_{1/2}\) is the half-life. This tells us how fast a certain quantity is reducing. Knowing this is important for applications like nuclear medicine, where precise dosages depend on the decay speed.

For instance, Cobalt-60 has a half-life of 5.27 years, meaning in 5.27 years, only half its initial radioactivity remains. This measurement helps in planning how often medical sources need refreshing.

Decay constant

The decay constant \(k\) is a fundamental component of the exponential decay formula and signifies the rate at which the substance decays. It is a negative number because it indicates a decrease over time.

We determine \(k\) using the equation:\[ k = \frac{\ln(0.5)}{t_{1/2}} \] For the cobalt-60 isotope, the decay constant \(k\) calculates as approximately -0.1316. This value is derived by inserting the half-life of cobalt-60, which is 5.27 years, into the equation.

Understanding \(k\) is vital for more than just nuclear physics. In fields such as archaeology, by identifying \(k\), we can determine the age of artifacts based on isotope decay rates, providing insights into historical timelines.

Cobalt-60 isotope

Cobalt-60 is a radioactive isotope used extensively in fields like medicine and industry. Its ability to emit gamma rays makes it useful for cancer treatment, sterilizing medical equipment, and even in food irradiation to eliminate pathogens.

The radioactive decay of cobalt-60 is well-documented by its exponential behavior, and understanding this is crucial for safe handling and efficient usage. Its half-life is 5.27 years, meaning it remains viable for certain applications over several years.

Calculating the decay constant and remaining amount of cobalt-60 can help facilities manage their resources better. For instance, after 17.5 years (three half-lives), over 90% of the initial cobalt-60 will have decayed, informing decisions about when to replace sources for consistent effectiveness in applications.