Half lives of two radioactive nuclei A and B are 10 minutes and 20 minutes, respectively. If initial

Q: Why can't we just take the ratio of the undecayed nuclei?

Because the question specifically asks for the ratio of the *decayed* numbers of nuclei. Taking the ratio of undecayed nuclei is a very common trap set by examiners to test your attention to detail.

Q: Can I use the exponential decay formula $N = N_0 e^{-\lambda t}$ instead?

Yes, absolutely! However, since the total time is an exact multiple of the half-lives for both samples, using the $N = N_0 / 2^n$ formula is much faster and less prone to calculation errors.

Q: What if the initial number of nuclei for A and B were different?

If they were different, say $N_{0A}$ and $N_{0B}$, you would simply keep those variables in your final ratio expression. The problem states they are equal, which allows us to cancel $N_0$ out.

Q: Why isn't the ratio of decayed nuclei just the inverse of their half-lives?

Radioactive decay is an exponential process, not a linear one. The amount decayed depends on $2^n$, so the relationship between time and decayed amount is not a simple direct or inverse proportion.

Q: Is the number of decayed nuclei always greater than the undecayed nuclei?

Not always! It depends on the time elapsed. Before one half-life has passed, the undecayed nuclei are greater. Exactly at one half-life, they are equal. After one half-life, the decayed nuclei become greater.

Question

Half lives of two radioactive nuclei $A$ and $B$ are $10$ minutes and $20$ minutes, respectively. If initially a sample has equal number of nuclei, then after $60$ minutes, the ratio of decayed numbers of nuclei $A$ and $B$ will be

Options:

(A) $3 : 8$

(B) $1 : 8$

(C) $8 : 1$

(D) $9 : 8$

Sigma Percentile · Accepted Answer

### The Tale of Two Isotopes

Imagine you have two glowing boxes of radioactive materials, Sample A and Sample B. They both start with the exact same number of radioactive nuclei, which we will call $N_0$. However, they decay at different speeds. Sample A is the sprinter, with a half-life of just $10 	ext{ minutes}$. Sample B is the marathon runner, taking $20 	ext{ minutes}$ to halve its population.

Our mission is to find the ratio of their **decayed** nuclei after exactly $60 	ext{ minutes}$.

### The Master Equation

Before we can find out how many nuclei have decayed, we must first determine how many have *survived*. The fundamental law of radioactive decay tells us that the number of undecayed nuclei $N$ after $n$ half-lives is given by:

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

Here, $n$ is simply the total elapsed time divided by the half-life of the substance ($n = \frac{t}{T_{1/2}}$).

### Analyzing Sample A

Let's focus on our sprinter, Sample A. The total time is $60 	ext{ minutes}$, and its half-life is $10 	ext{ minutes}$.

The number of half-lives it undergoes is:

$$n_A = \frac{60}{10} = 6$$

Plugging this into our master equation, the number of **undecayed** nuclei left is:

$$N_A = \frac{N_0}{2^6} = \frac{N_0}{64}$$

But wait! This is where many students fall into a trap. The question asks for the **decayed** nuclei, not the survivors. To find the decayed amount, we subtract the survivors from the initial population:

$$N_{dA} = N_0 - \frac{N_0}{64} = \frac{63N_0}{64}$$

### Analyzing Sample B

Now, let's look at the marathon runner, Sample B. The total time is still $60 	ext{ minutes}$, but its half-life is $20 	ext{ minutes}$.

The number of half-lives it undergoes is:

$$n_B = \frac{60}{20} = 3$$

Using our equation, the number of **undecayed** nuclei left is:

$$N_B = \frac{N_0}{2^3} = \frac{N_0}{8}$$

Again, we must find the **decayed** amount by subtracting this from the initial population:

$$N_{dB} = N_0 - \frac{N_0}{8} = \frac{7N_0}{8}$$

### The Final Calculation

We have successfully navigated the traps and found the decayed amounts for both samples. The final step is to find their ratio:

$$	ext{Ratio} = \frac{N_{dA}}{N_{dB}} = \frac{\frac{63N_0}{64}}{\frac{7N_0}{8}}$$

The $N_0$ terms beautifully cancel out, leaving us with pure arithmetic:

$$	ext{Ratio} = \frac{63}{64} 	imes \frac{8}{7}$$

Simplifying the fractions ($63/7 = 9$ and $8/64 = 1/8$), we arrive at our final, elegant answer:

$$	ext{Ratio} = \frac{9}{8}$$

Always remember to read the question carefully. Distinguishing between "decayed" and "undecayed" is the key to conquering these problems!

Animated Solution for Physics - Atoms and Nuclei: Half lives of two radioactive nuclei $A$ and $B$ are $10$ minutes and $20$ minutes, respectively. If initially a sample has equal number of nuclei, then after $60$ minutes, the ratio of decayed numbers of nuclei $A$ and $B$ will be

Select Answer:

Visualized Solution

$Visualizing the Setup$

$The Decay Formula$

$Half-lives for Sample A$

$Decayed Nuclei of A$

$Half-lives for Sample B$

$Decayed Nuclei of B$

$Ratio of Decayed Nuclei$

$Conclusion$

The Sigma Insight: Radioactivity

The Tale of Two Isotopes

The Master Equation

Analyzing Sample A

Analyzing Sample B

The Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Decay Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Half-lives for Sample A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Decayed Nuclei of A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Half-lives for Sample B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Decayed Nuclei of B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio of Decayed Nuclei

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion