Animated Solution for Physics - Atoms and Nuclei: There are $10^{10}$ radioactive nuclei in a given radioactive element. Its half-life time is $1\text{ min}$. How many nuclei will remain after $30\text{ s}$? ($\sqrt{2}=1.414$)

Imagine you are observing a sample of a highly radioactive element. At the very beginning of our observation, which we call $t=0$, there are a massive $10^{10}$ radioactive nuclei present. This is our initial count, denoted as $N_0$.

We are given a crucial piece of information: the half-life of this element is $1\text{ min}$. The half-life, $t_{1/2}$, is the time it takes for exactly half of the radioactive nuclei in a sample to decay. Because we are asked to find the number of nuclei remaining after $30\text{ s}$, it is a smart move to convert our half-life into seconds to maintain consistent units.

So, $t_{1/2}=1\text{ min}=60\text{ s}$.

To find the number of undecayed nuclei $N(t)$ remaining after any time $t$, we use the fundamental law of radioactive decay expressed in terms of half-life:

This elegant equation tells us that for every half-life that passes, we multiply our initial amount by $\frac{1}{2}$. The exponent $\frac{t}{t_{1/2}}$ simply counts how many half-lives have elapsed.

Now, let's substitute our known values into the master equation. We want to find the remaining nuclei at $t=30\text{ s}$.

Look at the exponent: $\frac{30}{60}$. This simplifies beautifully to $\frac{1}{2}$. This makes perfect physical sense—$30\text{ s}$ is exactly half of a half-life!

Mathematically, raising a number to the power of $\frac{1}{2}$ is identical to taking its square root. Therefore, our expression becomes:

The problem kindly provides the approximation $\sqrt{2}\approx 1.414$. Let's plug that in:

If you recall your standard mathematical approximations, $\frac{1}{\sqrt{2}}\approx 0.707$.

To match the standard scientific notation of the given options, we shift the decimal point one place to the right, which decreases the exponent by one:

This perfectly matches option (b). The beauty of this problem lies in recognizing that $30\text{ s}$ is half of a half-life, leading directly to a factor of $\frac{1}{\sqrt{2}}$.