Animated Solution for Physics - Atoms and Nuclei: A radioactive substance decays to $\frac{1}{16}$th of its initial activity in $80$ days. The half-life of the radioactive substance expressed in days is ...... .

Decoding Radioactive Decay

Imagine you are a scientist observing a glowing piece of radioactive material in a laboratory. Initially, it is highly active, emitting radiation at a certain rate. Let's call this initial activity $A_0$.

As time passes, the unstable nuclei within the sample decay, and the overall activity drops. The problem tells us that after exactly $80$ days, the activity has plummeted to just $\frac{1}{16}$th of its original value. Our mission is to find the half-life of this mysterious substance.

Radioactive decay is a game of halves. The half-life, denoted as $T_{1/2}$, is the precise amount of time it takes for the activity (or the number of active nuclei) to reduce to exactly half of its current value.

Because the final fraction given in the problem is $\frac{1}{16}$, and $16$ is a perfect power of $2$, we can solve this problem intuitively by tracing the decay chain step-by-step.

Let's walk through the timeline: - After the first half-life, the activity drops to $\frac{A_0}{2}$. - After the second half-life, it halves again, reaching $\frac{A_0}{4}$. - After the third half-life, it drops to $\frac{A_0}{8}$. - Finally, after the fourth half-life, it halves one more time, landing perfectly at $\frac{A_0}{16}$.

By simply counting the steps in our mental chain, we can see that it took exactly $4$ half-lives to reach the target activity of $\frac{A_0}{16}$.

We are given that this entire process took $80$ days. Therefore, we can set up a very simple linear equation relating the total time $t$, the number of half-lives $n$, and the duration of one half-life $T_{1/2}$:

Substituting our known values into the equation:

Now, we just divide both sides by $4$ to isolate the half-life:

And there we have it! The half-life of the radioactive substance is $20$ days.

While this intuitive chain method is incredibly fast for fractions that are powers of $2$, always remember the universal exponential decay law, $A=A_0{e}^{-\lambda t}$. If the problem had given us a fraction like $\frac{1}{10}$, we would have needed to rely on the master equation and logarithms to find the answer. Keep both tools sharp in your physics arsenal!