Animated Solution for Physics - Atoms and Nuclei: The nuclear activity of a radioactive element becomes $(1/8)$ th of its initial value in $30\text{ yr}$. The half-life of radioactive element is ...... yr.

Imagine you are holding a piece of a radioactive element. It might look like an ordinary rock, but deep within its atomic structure, a silent, invisible countdown is taking place. Nuclei are spontaneously decaying, transforming into different elements, and releasing energy in the process. This is the fascinating quantum phenomenon of radioactivity.

One of the most profound aspects of radioactive decay is that it is entirely governed by probability. We cannot predict exactly when a single nucleus will decay, but when we have trillions of them, their collective behavior follows a beautifully precise mathematical law. This brings us to the concept of the half-life ($t_{1/2}$)—the time required for exactly half of the radioactive nuclei in a sample to decay.

A Race Against Time

In our problem, we are observing a radioactive element over a span of $30\text{ years}$. We are told that its nuclear activity—which is directly proportional to the number of undecayed nuclei—has plummeted to just $\frac{1}{8}\text{th}$ of its initial value.

Let's define our variables to make this concrete: - Let the initial activity at $t=0$ be $A_0$. - The final activity after $t=30\text{ years}$ is $A=\frac{A_0}{8}$.

Our mission is to uncover the hidden clockwork of this element: its half-life, $t_{1/2}$.

Half-Lives Explained

While the fundamental law of radioactive decay is an exponential function ($A=A_0{e}^{-\lambda t}$), there is a much more intuitive and rapid way to solve problems when the remaining fraction is a clean power of $\frac{1}{2}$.

Every time one half-life passes, the activity is multiplied by $\frac{1}{2}$. - After 1 half-life: Activity is $\frac{A_0}{2}$ - After 2 half-lives: Activity is $\frac{A_0}{4}$ - After $n$ half-lives: Activity is $A_0{\left(\frac{1}{2}\right)}^n$

This gives us our master equation:

Where $n$ represents the total number of half-lives that have elapsed. We can also relate $n$ to the total time $t$ and the half-life $t_{1/2}$ with a simple ratio:

Finding the Missing Piece

Now, let's substitute the information we were given into our master equation. We know the final activity $A$ is $\frac{A_0}{8}$.

The initial activity $A_0$ appears on both sides, so we can elegantly cancel it out. This mathematically proves that the half-life is completely independent of how much material you started with!

Now we ask ourselves: to what power must we raise $\frac{1}{2}$ to get $\frac{1}{8}$? Since $2\times 2\times 2=8$, we know that ${\left(\frac{1}{2}\right)}^3=\frac{1}{8}$.

Therefore, $n=3$.

We have discovered that exactly 3 half-lives have passed during the $30\text{-year}$ observation period.

Since the total time $t$ is the number of half-lives multiplied by the duration of one half-life, we can write:

Dividing both sides by 3, we arrive at our final answer:

The half-life of this mysterious radioactive element is exactly 10 years. This means every decade, half of it vanishes, ticking away like a perfect, atomic metronome.