Animated Solution for Physics - Atoms and Nuclei: A radioactive sample $S_1$ having an activity of $5\mu \text{Ci}$ has twice the number of nuclei as another sample $S_2$ which has an activity of $10\mu \text{Ci}$. The half lives of $S_1$ and $S_2$ can be

Imagine you are a nuclear physicist handed two mysterious radioactive samples, $S_1$ and $S_2$. You place them in your detectors and observe their activities. Sample $S_1$ is ticking away at $5\mu \text{Ci}$, while sample $S_2$ is twice as active, registering $10\mu \text{Ci}$. But here is the catch: your mass spectrometer reveals that sample $S_1$ actually contains twice as many radioactive nuclei as sample $S_2$.

How can a sample with more radioactive material be less active? The secret lies in the half-life. Let's unravel this beautiful relationship between activity, population size, and time.

To solve this mystery, we need to consult the fundamental law of radioactive decay. The activity $A$ of a sample—which is the number of decays per second—is directly proportional to the number of undecayed nuclei $N$ present in the sample. The constant of proportionality is the decay constant $\lambda$.

But what exactly is $\lambda$? It represents the probability of decay per unit time. We can relate it to a much more intuitive concept: the half-life $T_{1/2}$, which is the time required for half of the nuclei to decay. The relationship is given by:

Substituting this back into our activity equation, we get a master formula that connects all our variables:

We want to find the half-lives, so let's rearrange our master formula to solve for $T_{1/2}$:

Since $\ln 2$ is just a constant number, we can see a clear proportional relationship: the half-life is directly proportional to the number of nuclei $N$ and inversely proportional to the activity $A$.

This inverse relationship perfectly explains our initial puzzle! Sample $S_1$ has more nuclei but lower activity because its half-life must be significantly longer. It decays at a much more sluggish pace.

Let's set up a ratio to compare the two samples directly:

Notice how the activity ratio is inverted ($A_2/A_1$) because of the inverse proportionality.

Now, it is time for the grand finale. Let's plug in the numbers given to us in the problem. We know that $S_1$ has twice as many nuclei as $S_2$, so:

We also know their activities: $A_1=5\mu \text{Ci}$ and $A_2=10\mu \text{Ci}$. So the inverted activity ratio is:

Substituting these ratios back into our half-life equation:

This tells us that the half-life of sample $S_1$ is exactly four times the half-life of sample $S_2$.

Looking at our multiple-choice options, we need to find a pair of half-lives where the first is four times the second. - Option (a) offers $20\text{ yr}$ and $5\text{ yr}$. Since $20/5=4$, this is a perfect match! - Option (b) offers $20\text{ yr}$ and $10\text{ yr}$, which is a ratio of 2. - Options (c) and (d) offer equal half-lives, a ratio of 1.

Therefore, the correct answer is undeniably (a). The math elegantly confirms our physical intuition!