Animated Solution for Physics - Atoms and Nuclei: In a radioactive material, fraction of active material remaining after time $t$ is $\frac{9}{16}$. The fraction that was remaining after time $t/2$ is

Welcome to a fascinating exploration of radioactive decay! This problem is a beautiful example of how the laws of physics, when expressed through the elegant language of mathematics, allow us to find answers with surprising simplicity. We don't need to crunch massive numbers or know the specific properties of the material; we just need to understand the shape of nature's decay curve.

At the heart of this problem is the Rutherford-Soddy law of radioactive decay. This fundamental principle states that the rate at which a radioactive substance decays is directly proportional to the number of active nuclei currently present.

Mathematically, this translates to an exponential function:

Here, $M(t)$ is the mass (or number of nuclei) remaining at time $t$, $M_0$ is the initial mass at $t=0$, and $\lambda$ is the decay constant, a unique fingerprint for every radioactive isotope.

If we rearrange this equation to find the fraction of material remaining, we get:

The problem gives us a very specific piece of information: after a certain time $t$, the fraction of active material remaining is $\frac{9}{16}$.

Let's plug this directly into our decay equation. This gives us our foundational anchor for the rest of the problem:

Notice that we don't know what $\lambda$ is, and we don't know what $t$ is. And the beautiful part? We don't need to.

The question asks for the fraction remaining at time $\frac{t}{2}$. Let's write out the expression for the fraction at this new time:

Now, we employ a classic rule of exponents. Remember that $x^{a\cdot b}=(x^a{)}^b$. We can rewrite our expression to isolate the term we already know:

This is where the magic happens. We already established that $e^{-\lambda t}=\frac{9}{16}$. We can simply substitute this value into our new expression:

Raising a fraction to the power of $\frac{1}{2}$ is exactly the same as taking its square root.

And there we have it! The fraction of active material remaining at time $\frac{t}{2}$ is $\frac{3}{4}$.

This problem perfectly illustrates the power of algebraic substitution in physics. By recognizing the structure of the exponential function, we bypassed tedious calculations and arrived at the solution with pure logic.