Question

Radioactive Decay The half-life of phosphorus- 32 is about 14 days. There are 6.6 grams present initially. (a) Express the amount of phosphorus-32 remaining as a function of time \(t\). (b) When will there be 1 gram remaining?

Short answer

The function representing the amount of phosphorus-32 remaining after time \(t\) is \(N(t) = 6.6 \times (1/2)^{t/14}\). Based on the calculations, there will be approximately 1 gram of phosphorus-32 remaining after \(\approx 52\) days (values depending on how you rounded).

Step-by-step solution

Create the Decay Function

The decay formula is given by \(N(t) = N_0 \times (1/2)^{t/T}\). Here \(N_0\) is the initial amount (6.6 g), \(t\) is the time in days, and \(T\) is the half-life (14 days). We can substitute these values to get the function for phosphorus-32.

Find the Expression for Amount Remaining

We substitute the initial amount \(N_0 = 6.6\) g and the half-life \(T = 14\) days into the decay equation. So the function becomes \(N(t) = 6.6 \times (1/2)^{t/14}\). Therefore, the amount of phosphorus-32 remaining as a function of time \(t\) is represented by this function.

Determine Time for 1 gram Remaining

To find when there will be 1 gram of phosphorus-32, set the function \(N(t) = 1\) and solve for \(t\). That is, solve the equation \(1 = 6.6 \times (1/2)^{t/14}\) for \(t\). Taking the natural logarithm on both sides simplifies the calculation. The final value of \(t\), rounded to the nearest whole number, gives the number of days when there will be 1 gram left.

Key concepts

Half-Life

The concept of half-life is central to the understanding of radioactive decay. It is defined as the time required for one half of the radioactive substance to decay. In simpler terms, after one half-life, only half of the original amount of the radioactive material will remain.

For example, if the half-life of a particular isotope is 14 days, this means that every 14 days, the quantity of that isotope will reduce to half of its initial amount. This process of decay continues, so after another 14 days, you'll have half of the remaining half, and so on. This is why half-life is used as a measure of the rate at which a radioactive substance decays.

The importance of understanding half-life lies in its diverse applications, which include dating archaeological finds, understanding medical treatments that involve radioactive substances, and managing nuclear waste.

Exponential Decay Function

Radioactive decay is a prime example of an exponential decay process. The exponential decay function represents how the quantity of a substance decreases over time at a rate that is proportional to its current value. Mathematically, for a radioactive element, this function is often represented as:
\( N(t) = N_0 \times (1/2)^{t/T} \).

In this function, \( N(t) \) is the amount of substance at time \( t \), \( N_0 \) is the initial amount of the substance, \( t \) is the elapsed time, and \( T \) is the half-life of the substance. The base of the exponential function is \( 1/2 \), reflecting the concept of half-life, as the substance is reduced by half over each period of time \( T \).

This type of function is crucial because it can be used to predict the future behavior of the decaying substance. It shows us that the rate of decay slows down over time. The substance never fully reaches zero because it's always decaying by halves, but it gets infinitely close as time progresses.

Logarithmic Calculations

Logarithmic calculations are essential when dealing with exponential functions, especially in the determination of time in radioactive decay problems. When a question asks for the time at which a certain amount of a radioactive substance remains, we often employ logarithms to solve for the variable.

Given the decay function \( N(t) = N_0 \times (1/2)^{t/T} \), to solve for \( t \) you can use logarithms to 'un-exponentiate' the equation. This process involves taking the log of both sides of the equation which then allows for the isolation and solving of the unknown variable \( t \).

For example, to find when there will be 1 gram of a substance left, you set the function equal to 1 gram and solve for \( t \). Using natural logs, this looks like \( \text{ln}(N(t)) = \text{ln}(N_0) + (t/T) \text{ln}(1/2) \), from which you can solve for \( t \) by isolating it on one side of the equation. Logarithms are a powerful mathematical tool that simplify these complex exponential equations and make it possible to calculate the duration of these decay processes in a practical manner.