Two radioactive substances X and Y originally have N1 and N2 nuclei, respectively. Half-life of X is

Q: Why do we use $N = N_0 (1/2)^n$ instead of the exponential formula $N = N_0 e^{-\lambda t}$?

Both formulas are perfectly valid and will give the same result! However, when the time elapsed is given as a clean integer multiple of the half-life, using $N = N_0 (1/2)^n$ is much faster and avoids dealing with natural logarithms and the decay constant $\lambda$.

Q: How did we figure out that $X$ went through 6 half-lives?

The total time elapsed is $3 T_Y$. Since the half-life of $X$ is half of $Y$ ($T_X = T_Y / 2$), we can substitute $T_Y = 2 T_X$ into the time equation. This gives $t = 3(2 T_X) = 6 T_X$. Therefore, $6$ half-lives of $X$ have passed.

Q: What does it mean physically that the ratio $N_1 / N_2$ is 8?

It means that because substance $X$ decays much faster than substance $Y$, you need to start with a significantly larger amount of $X$ (specifically, 8 times more) so that after the given time, both samples have dwindled down to the exact same number of surviving nuclei.

Q: Can we solve this using decay constants $\lambda$?

Yes! Since $T_X = T_Y / 2$, we know $\lambda_X = 2\lambda_Y$. You can set up the equation $N_1 e^{-\lambda_X t} = N_2 e^{-\lambda_Y t}$. Substitute $t = 3 \ln(2) / \lambda_Y$ and solve. You will get the exact same ratio of 8.

Q: What if the time given was not an integer multiple of the half-lives?

If the time was something like $2.5 T_Y$, you could still use $N = N_0 (1/2)^{2.5}$, but it's often safer and less confusing to switch to the exponential decay formula $N = N_0 e^{-\lambda t}$ to handle fractional powers smoothly.

Question

Two radioactive substances $X$ and $Y$ originally have $N_1$ and $N_2$ nuclei, respectively. Half-life of $X$ is half of the half-life of $Y$. After three half-lives of $Y$, number of nuclei of both are equal. The ratio $\frac{N_1}{N_2}$ will be equal to

Options:

(A) $\frac{1}{8}$

(B) $\frac{3}{1}$

(C) $\frac{8}{1}$

(D) $\frac{1}{3}$

Sigma Percentile · Accepted Answer

### The Race of Radioactive Decay

Imagine a race between two runners, $X$ and $Y$, but instead of running forward, they are shrinking away! This is the essence of radioactive decay. In this problem, we are given two radioactive substances, $X$ and $Y$, starting with initial populations of $N_1$ and $N_2$ nuclei, respectively.

The core of the problem lies in their decay rates. We are told that the half-life of $X$ is exactly half of the half-life of $Y$. Mathematically, we can write this as:

$$T_X = \frac{T_Y}{2}$$

This simple relation means that substance $X$ is decaying twice as fast as substance $Y$.

### The Time Factor

To find out how much of each substance is left, we need to know how much time has passed. The problem states that we are observing the samples after **three half-lives of $Y$**.

So, the total time elapsed is:

$$t = 3 T_Y$$

Now, we must determine how many half-lives each substance has experienced during this time $t$. For substance $Y$, it is straightforward—it has gone through exactly $3$ half-lives ($n_Y = 3$).

But what about substance $X$? Since its half-life is half as long, it will undergo twice as many half-lives in the same duration. Let's calculate it formally:

$$n_X = \frac{t}{T_X} = \frac{3 T_Y}{T_Y / 2} = 6$$

Substance $X$ has gone through a whopping $6$ half-lives!

### The Core Calculation

We know that after $n$ half-lives, the remaining amount of a radioactive substance is given by the formula:

$$N = \frac{N_0}{2^n}$$

Let's apply this to both substances. For substance $X$, the remaining nuclei will be:

$$N_X = \frac{N_1}{2^6} = \frac{N_1}{64}$$

For substance $Y$, the remaining nuclei will be:

$$N_Y = \frac{N_2}{2^3} = \frac{N_2}{8}$$

### The Grand Finale

The final piece of the puzzle is the condition that after this time, the number of nuclei of both substances is equal. This means we can equate our two expressions:

$$N_X = N_Y$$

$$\frac{N_1}{64} = \frac{N_2}{8}$$

To find the required ratio $\frac{N_1}{N_2}$, we simply rearrange the equation:

$$\frac{N_1}{N_2} = \frac{64}{8} = 8$$

And there we have it! Because substance $X$ decays so much faster, we needed $8$ times more of it initially just to tie with substance $Y$ at the end of the observation period. The correct option is **(c)**.

Select Answer:

Visualized Solution

Initial Setup

Half-life Relation

Time Elapsed

Number of Half-lives

Remaining Nuclei

Equating and Solving

Conclusion

The Sigma Insight: Radioactivity

The Race of Radioactive Decay

The Time Factor

The Core Calculation

The Grand Finale

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