Animated Solution for Physics - Atoms and Nuclei: A radioactive nucleus $A$ with a half-life $T$, decays into a nucleus $B$. At $t=0$, there is no nucleus $B$. After sometime $t$, the ratio of the number of $B$ to that of $A$ is $0.3$. Then, $t$ is given by

Radioactive decay is one of the most elegant phenomena in physics. It is entirely governed by probability, yet it follows a strict, predictable mathematical law when dealing with large numbers of atoms. In this problem, we are tasked with finding the exact time $t$ when the ratio of daughter nuclei to parent nuclei reaches a specific value.

A Decaying into B

Imagine a sealed container. At time $t=0$, it is filled purely with radioactive nucleus $A$. There is absolutely no nucleus $B$ present. As the clock ticks, atoms of $A$ undergo radioactive decay and transform into stable atoms of $B$.

We are given a fascinating piece of information: at some specific time $t$, the ratio of the number of $B$ atoms to $A$ atoms is $0.3$. Mathematically, this is written as:

Instead of wrestling with abstract variables like $N_0$ right away, let's use a powerful pedagogical trick. The ratio $0.3$ is exactly the same as the fraction $\frac{30}{100}$.

Let's imagine that at this specific time $t$, we have exactly $100$ atoms of $A$ remaining. To satisfy the ratio, we must have exactly $30$ atoms of $B$ formed.

Now, think about the conservation of atoms. Every single atom of $B$ was originally an atom of $A$. Therefore, the initial number of $A$ atoms at $t=0$ must be the sum of the atoms that are still $A$ and the atoms that have turned into $B$:

So, we started with $130$ atoms, and at time $t$, we have $100$ atoms of $A$ left.

Now we bring in the heavy artillery: the radioactive decay law. This law states that the number of undecayed nuclei at any time $t$ is given by:

Let's substitute our assumed numbers into this master equation:

Dividing both sides by $130$, we get:

Which simplifies beautifully to:

To solve for $t$, we need to get it out of the exponent. First, let's take the reciprocal of both sides to eliminate the negative sign in the exponent. The reciprocal of $\frac{1}{1.3}$ is simply $1.3$, and the reciprocal of $e^{-\lambda t}$ is $e^{\lambda t}$:

Now, we take the natural logarithm ($\ln$ or ${\log }_e$) of both sides:

The problem gives us the half-life $T$, not the decay constant $\lambda$. However, we know the fundamental relationship between the two:

Substituting this into our equation gives:

Finally, we rearrange the terms to isolate $t$:

Since $\ln$ is just another notation for ${\log }_e$, we can write our final answer as:

This perfectly matches option (a). By assigning concrete numbers to a ratio, we bypassed messy algebraic fractions and arrived at the solution with absolute clarity!