\(^{90} \mathrm{Kr}\left(t_{1 / 2}=32 \mathrm{~s}\right)\) is used to study respiration. How soon after being made must a sample be administered to the patient if the activity must be at least \(90 \%\) of the original activity?

Half-life is a core concept in the study of radioactive decay. It represents the time required for half of the radioactive nuclei in a sample to decay. In our exercise, the half-life of \( ^{90} \text{Kr} \) is given as 32 seconds. This means that every 32 seconds, the amount of \( ^{90} \text{Kr} \) in the sample is reduced by half. Understanding this helps us predict how quickly a substance will lose its radioactivity.
Knowing the half-life allows us to calculate how much of the original material remains after a given period. For example, after one half-life (32 seconds), 50% of the original \( ^{90} \text{Kr} \) would remain. After two half-lives (64 seconds), only 25% would remain.
This principle is crucial for calculating the administration time needed to ensure the sample retains sufficient activity when used in medical applications.

Exponential decay describes the process in which the quantity of a radioactive substance decreases exponentially over time. The formula used is: \( A(t) = A_0 e^{- \frac{t}{\tau}} \)
In this formula, \( A(t)\) is the activity at time \( t\), \( A_0 \) is the original activity, and \( \tau \) is the mean lifetime of the radioactive substance. The factor \( -\frac{t}{\tau} \) represents the fraction of the mean lifetime that has elapsed.
The exponential decay formula is essential because it provides a way to predict how the activity of a substance changes over time. This equation was used in the exercise to determine how long it takes for the activity to drop to 90% of its original value.

The decay constant, often denoted as \tau\, is a measure of the average lifetime or stability of a radioactive substance. It is calculated using the given half-life with the formula: \( τ = \frac{t_{1/2}}{\text{ln} 2} \)
For the given problem, where the half-life \( t_{1/2} \) is 32 seconds, we calculate \tau\ as follows: \( \tau = \frac{32 \text{s}}{0.693} \).
Using the value of \(\text{ln} 2 \) which is approximately 0.693, we find \( \tau \) to be about 46.14 seconds. This value was then used in the exponential decay formula to find the time needed for the activity to reduce to 90% of the original.

Activity retention refers to how much of the sample's activity remains after a certain period. In our exercise, we want to ensure that the sample retains at least 90% of its original activity when administered to the patient. To find out how long the sample can be allowed to decay while retaining at least 90% of its original activity, we use the relation:
\[ \frac{A(t)}{A_0} = e^{-\frac{t}{\tau}} = 0.9 \] Taking the natural logarithm of both sides, we have: \[ - \frac{t}{46.14 \text{s}} = \text{ln}(0.9) \] Knowing that \( \text{ln}(0.9) \) is approximately \( -0.1054 \), we solve for \( t \): \[ t = 46.14 \text{s} \times 0.1054 \] Which gives us \( t \) approximately 4.86 seconds.
Hence, the sample can be used for about 4.86 seconds after preparation to retain 90% of its original activity.