Animated Solution for Physics - Atoms and Nuclei: Two radioactive substances $A$ and $B$ have decay constants $5\lambda$ and $\lambda$, respectively. At $t=0$, a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become ${\left(\frac{1}{e}\right)}^2$ will be

Imagine a race, not of speed, but of disappearance. We have two radioactive substances, $A$ and $B$, starting with the exact same number of nuclei, $N_0$. However, they are decaying at vastly different rates. Substance $A$ is highly unstable, decaying with a constant of $5\lambda$, while substance $B$ is much more relaxed, decaying at a rate of just $\lambda$.

The question asks us to find the exact moment in time when the ratio of their remaining nuclei hits a specific target: ${\left(\frac{1}{e}\right)}^2$, or $e^{-2}$.

To track this race, we need our master tool: the radioactive decay law. The number of active nuclei $N$ remaining at any time $t$ is given by the exponential decay formula:

Let's apply this to our two contenders. For substance $A$, the decay constant is $5\lambda$, so its population over time is:

For substance $B$, the decay constant is simply $\lambda$, giving us:

If you were to graph these, you would see the curve for $N_A$ plummeting much steeper than the curve for $N_B$. Because $A$ decays faster, at any time $t>0$, there will always be fewer nuclei of $A$ left than $B$.

We are looking for the time when the ratio of their nuclei is $e^{-2}$. Since $e^{-2}$ is a fraction less than 1, and we know $N_A<N_B$, the ratio must logically be $\frac{N_A}{N_B}$. If we had set it up as $\frac{N_B}{N_A}$, the ratio would have to be greater than 1.

Let's set up our master equation:

Now, we substitute our decay expressions into the ratio:

The initial population $N_0$ cancels out beautifully, leaving us with a pure exponential equation. Using the laws of exponents, when dividing terms with the same base, we subtract the denominator's exponent from the numerator's exponent:

Since the bases on both sides of the equation are identical ($e$), their exponents must be equal for the statement to hold true. We can now drop the bases and solve the simple linear equation:

Dividing both sides by $-4\lambda$, we isolate $t$:

And there is our final answer. At exactly $t=\frac{1}{2\lambda }$, the fast-decaying substance $A$ will have dwindled down so much compared to $B$ that their ratio perfectly hits $e^{-2}$.