Animated Solution for Physics - Atoms and Nuclei: The half-life of ${}^{215}\text{At}$ is $100\mu \text{s}$. The time taken for the activity of a sample of ${}^{215}\text{At}$ to decay to $\frac{1}{16}\text{th}$ of its initial value is

Imagine you are holding a piece of a radioactive isotope, such as ${}^{215}\text{At}$. At the microscopic level, the nuclei of this isotope are inherently unstable. They are constantly undergoing a spontaneous transformation, emitting radiation to reach a more stable state. This process is entirely governed by the probabilistic laws of quantum mechanics. We cannot predict exactly when a single specific nucleus will decay, but when we look at a massive collection of these nuclei, a beautifully predictable statistical pattern emerges.

This pattern is known as the Law of Radioactive Decay. It states that the rate at which a radioactive sample decays—its activity—is directly proportional to the number of undecayed nuclei currently present. Mathematically, this leads to an exponential decay curve, meaning the activity drops off rapidly at first and then more slowly over time.

To make sense of this exponential decay, physicists use a highly intuitive concept called the half-life ($t_{1/2}$). The half-life is defined as the exact amount of time it takes for exactly half of the radioactive nuclei in a given sample to decay.

Because the decay is exponential, this time interval is constant. If you start with $1000$ nuclei, after one half-life, you will have $500$ left. After a second half-life, you won't have zero; you will have half of the $500$, which is $250$. The amount halves with every passing half-life.

This gives rise to a very powerful and simplified formula for calculating the remaining activity $R$ after a certain number of half-lives, $n$:

where $R_0$ is the initial activity of the sample.

Let's look at the specific problem at hand. We are given a sample of ${}^{215}\text{At}$ with a known half-life:

We are asked to find the total time $t$ it takes for the activity of this sample to drop to exactly $\frac{1}{16}\text{th}$ of its initial value. In mathematical terms, our final activity $R$ is:

We can directly apply our simplified half-life formula. By substituting the given final activity into the equation, we get:

The beauty of this equation is that the initial activity $R_0$ completely cancels out from both sides. This proves a fundamental principle: the time it takes to reach a certain fraction of the initial amount is completely independent of how much material you started with!

Now, we just need to solve for $n$. We recognize that $16$ is a perfect power of $2$. Specifically, $16=2^4$. Therefore, we can rewrite the equation as:

By comparing the exponents, it is immediately clear that:

This tells us that the sample must undergo exactly $4$ complete half-lives to reach $\frac{1}{16}\text{th}$ of its original activity.

Now that we know the number of half-lives, finding the total elapsed time is a simple multiplication. The total time $t$ is the number of half-lives $n$ multiplied by the duration of a single half-life $t_{1/2}$:

Substituting our values:

And there we have it! It takes $400\mu \text{s}$ for the sample's activity to decay to the specified level.

It is worth noting that this problem was particularly elegant because the remaining fraction, $\frac{1}{16}$, was a perfect power of $\frac{1}{2}$. What if the question had asked for the time to reach $\frac{1}{10}\text{th}$ of the initial activity?

In such cases, $n$ would not be a neat integer. We would have to rely on the general exponential decay equation:

where $\lambda$ is the decay constant, related to the half-life by $\lambda =\frac{\ln 2}{t_{1/2}}$. By taking the natural logarithm of both sides, we can solve for any arbitrary time $t$. However, for fractions that are powers of two, the $R=R_0(1/2{)}^n$ shortcut is a massive time-saver in competitive exams!