Problem 15 - Precalculus

Question

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?) Phosphorus 32 , used in agriculture, initial amount 2 milligrams, half-life 14 days.

Step-by-Step Solution

Answer

Decay constant \(k\) is 0.0495 day\(^{-1}\). Function: \(A(t) = 2 e^{-0.0495t}\). 90% decays in 46.47 days.

Step 1: Understanding the Half-Life Equation

The half-life is the time it takes for half of a substance to decay. We can use the relationship between the decay constant and half-life to find \(k\) using the equation \(t_{1/2} = \frac{0.693}{k}\), where \(t_{1/2}\) is the half-life of the substance.

Step 2: Calculating the Decay Constant

For Phosphorus-32, the half-life \(t_{1/2} = 14\) days. Using the formula \(k = \frac{0.693}{t_{1/2}}\), we plug in the half-life value:\[ k = \frac{0.693}{14} \approx 0.0495 \] So, the decay constant \(k\) is approximately 0.0495 day\(^{-1}\).

Step 3: Finding the Decay Function

The amount of isotope that remains after time \(t\) is given by the equation \(A(t) = A_{0} e^{kt}\). Therefore, for Phosphorus-32:\[ A(t) = 2 e^{-0.0495t} \] This function represents the amount in milligrams remaining after \(t\) days.

Step 4: Calculating Time for 90% of the Material to Decay

If 90% of the material decays, 10% remains. We need to solve for \(t\) in the equation \(A(t) = 0.1 \times A_0\). For this, \(A(t) = 0.2\) since \(A_0 = 2\):\[ 0.2 = 2 e^{-0.0495t} \] Dividing both sides by 2, then taking the natural logarithm gives:\[ e^{-0.0495t} = 0.1 \]\[ -0.0495t = \ln(0.1) \]\[ t = \frac{\ln(0.1)}{-0.0495} \approx 46.47 \]Approximately, it takes 46.47 days for 90% of the material to decay.

Key Concepts

Half-Life

When learning about radioactive decay, the concept of half-life is crucial. It is defined as the time required for half of a given sample of a radioactive substance to decay and transform into another substance.
The half-life is a constant value that helps in understanding how quickly or slowly a substance decays. Scientists and engineers use this information to predict how long a substance will remain active or useful.
In the provided problem, Phosphorus-32 has a half-life of 14 days. This means that after 14 days, only half of the initial quantity of Phosphorus-32 remains. This characteristic helps in figuring out its decay process, guiding practical applications, such as its use in agriculture. Remember, different substances have varying half-lives, which can range from fractions of a second to millions of years.

Decay Constant

The decay constant, denoted by the symbol \(k\), is fundamental to understanding the rate at which a radioactive isotope decays. It relates directly to the half-life by the formula:

• \(k = \frac{0.693}{t_{1/2}}\)

Here, \(t_{1/2}\) represents the half-life. For Phosphorus-32, with a half-life of 14 days, we find the decay constant to be approximately 0.0495 day\(^{-1}\). This constant is specific to each isotope and changes with different elements or isotopes.
The decay constant allows us to calculate how fast or slow an isotope will decay, providing insight into its stability. As the decay constant increases, the isotope decays more rapidly. Thus, with a decay constant in hand, we can predict the time it takes for the isotope quantity to decrease to any desired level.

Exponential Decay Function

In radioactive decay, the remaining quantity of an isotope can be modeled using an exponential decay function. This function is typically expressed as:

• \(A(t) = A_0 e^{-kt}\)

Where:

• \(A(t)\) is the amount of the substance remaining at time \(t\)
• \(A_0\) is the initial amount of the substance
• \(e\) is the base of the natural logarithm (approximately 2.71828)
• \(k\) is the decay constant

For Phosphorus-32, given an initial amount of 2 milligrams and a decay constant of 0.0495 day\(^{-1}\), the decay function becomes \(A(t) = 2 e^{-0.0495t}\). This function provides precise calculations of how much Phosphorus-32 remains at any given time \(t\), helping to forecast when a specific amount will have decayed.

Phosphorus-32 Decay

Phosphorus-32 is a radioactive isotope commonly used in agriculture, particularly in fertilizer and plant studies. It has a half-life of 14 days, meaning that it undergoes relatively quick decay compared to isotopes with longer half-lives. Understanding the decay of Phosphorus-32 involves calculating how much of it will remain after a certain period or how long it takes for a significant portion to decay.
When determining the time for 90% of Phosphorus-32 to decay, only 10% of the initial amount would remain. Using the exponential decay formula, \(A(t) = 0.2\) (since the initial amount \(A_0 = 2\)), we solve the equation to find that it takes approximately 46.47 days for 90% of the substance to decay. This information is crucial for agricultural applications where timing is essential for maximizing effectiveness and safety.