Starting with a sample of pure 66Cu, 7/8 of it decays into Zn in 15 min. The corresponding half-life

Q: Why do we use the remaining amount instead of the decayed amount in the formula?

The standard radioactive decay law $N = N_0 e^{-\lambda t}$ specifically calculates the number of nuclei that have *not yet decayed*. If you want to use the decayed amount, the formula becomes $N_{\text{decayed}} = N_0(1 - e^{-\lambda t})$, which is mathematically equivalent but often slightly more tedious to solve.

Q: How did $\ln 8$ become $3 \ln 2$?

This is a fundamental property of logarithms. Since $8 = 2^3$, we can write $\ln(8)$ as $\ln(2^3)$. The power rule of logarithms states that $\ln(a^b) = b \ln(a)$, so the $3$ comes down to the front, giving us $3 \ln 2$.

Q: Can I solve this without finding the decay constant $\lambda$?

Absolutely! In fact, it's highly recommended for competitive exams. Since the remaining fraction is $1/8$, and $1/8 = (1/2)^3$, we know exactly $3$ half-lives have passed. So, $3 \times t_{1/2} = 15 \text{ min}$, which instantly gives $t_{1/2} = 5 \text{ min}$.

Q: What if the remaining fraction wasn't a perfect power of $1/2$?

If the remaining fraction was something like $1/5$, you couldn't use the simple integer half-life method. You would be forced to use the logarithmic method: $\frac{1}{5} = e^{-\lambda t}$, find $\lambda$, and then calculate $t_{1/2} = \frac{\ln 2}{\lambda}$.

Q: Why is the half-life independent of the initial amount $N_0$?

Radioactive decay is a first-order kinetic process. This means the rate of decay is strictly proportional to the current amount present. Because of this exponential nature, the time it takes to halve the sample is a constant property of the isotope, regardless of whether you start with 1 gram or 1000 kilograms.

Question

Starting with a sample of pure $^{66}	ext{Cu}$, $7/8$ of it decays into Zn in $15	ext{ min}$. The corresponding half-life is

Options:

(A) $10	ext{ min}$

(B) $15	ext{ min}$

(C) $5	ext{ min}$

(D) $7\frac{1}{2}	ext{ min}$

Sigma Percentile · Accepted Answer

The phenomenon of radioactive decay is one of the most fascinating processes in quantum mechanics. It is entirely random for a single atom, yet perfectly predictable for a large sample.

In this problem, we are given a sample of Copper-66 ($^{66}	ext{Cu}$) and we need to determine its half-life based on how much of it decays over a specific period. Let's break down the physics and the math step-by-step.

### Visualizing the Decay Process

Imagine you have a pure block of Copper-66. At the very beginning, when our stopwatch reads $t = 0$, the entire block consists of undecayed Copper nuclei. We call this initial amount $N_0$.

The problem states that after $15 	ext{ min}$, exactly $\frac{7}{8}$ of this sample has decayed into Zinc ($	ext{Zn}$).

This is a crucial piece of information, but we must be careful. The standard radioactive decay formulas are designed to track what is **left over**, not what has decayed. So, our first logical step is to find the remaining amount of Copper.

If $\frac{7}{8}$ of the sample is gone, the remaining amount $N$ is simply the total minus the decayed portion:

$$N = N_0 - \frac{7}{8}N_0 = \frac{1}{8}N_0$$

This tells us that only one-eighth of our original Copper sample is still active.

### The Mathematical Rigor: Radioactive Decay Law

Now we bring in our heavy machinery: the **Radioactive Decay Law**. This law states that the number of undecayed nuclei decreases exponentially over time. The master equation is:

$$N = N_0 e^{-\lambda t}$$

Here, $\lambda$ is the decay constant, which dictates how fast the substance decays. Let's substitute the values we know into this equation. We know the remaining amount $N = \frac{1}{8}N_0$ and the time $t = 15 	ext{ min}$:

$$\frac{1}{8}N_0 = N_0 e^{-\lambda 	imes 15}$$

Notice how $N_0$ appears on both sides. This is beautiful because it means the initial mass doesn't matter! Whether we started with a gram or a ton, the fraction remaining is the same. Canceling $N_0$, we get:

$$\frac{1}{8} = e^{-15\lambda}$$

To make this easier to work with, let's take the reciprocal of both sides:

$$8 = e^{15\lambda}$$

### Solving for the Decay Constant

Our goal is to isolate $\lambda$. Since it is trapped in the exponent, we must use logarithms to bring it down. Taking the natural logarithm ($\ln$) of both sides yields:

$$\ln 8 = 15\lambda$$

We can simplify $\ln 8$ by recognizing that $8$ is $2^3$. Using the power rule of logarithms ($\ln(a^b) = b \ln a$), we can rewrite this as:

$$3 \ln 2 = 15\lambda$$

Now, isolating $\lambda$ is straightforward algebra:

$$\lambda = \frac{3 \ln 2}{15} = \frac{\ln 2}{5} 	ext{ min}^{-1}$$

We have successfully found the decay constant!

### The Final Half-Life Calculation

The question asks for the half-life ($t_{1/2}$), which is the time required for exactly half of the sample to decay. The relationship between half-life and the decay constant is given by:

$$t_{1/2} = \frac{\ln 2}{\lambda}$$

Let's substitute the $\lambda$ we just calculated into this formula:

$$t_{1/2} = \frac{\ln 2}{\frac{\ln 2}{5}}$$

The $\ln 2$ terms cancel out perfectly, flipping the $5$ to the numerator.

$$t_{1/2} = 5 	ext{ min}$$

And there we have it! The half-life of Copper-66 is exactly **$5 	ext{ min}$**.

### The Ninja Technique: Solving Intuitively

While the mathematical derivation is rigorous and foolproof, competitive exams like JEE demand speed. Let's look at an incredibly elegant, intuitive way to solve this without touching $e$ or natural logs.

We established early on that the remaining fraction of the sample is $\frac{1}{8}$.

Think about how half-lives work. After one half-life, $\frac{1}{2}$ remains. After two half-lives, $\frac{1}{4}$ remains. After three half-lives, $\frac{1}{8}$ remains!

Mathematically, we can express this as:

$$\frac{1}{8} = \left(\frac{1}{2}ight)^3$$

This instantly tells us that exactly **$3$ half-lives** have elapsed.

We are given that the total time elapsed is $15 	ext{ min}$. Therefore, three half-lives must equal $15 	ext{ min}$:

$$3 	imes t_{1/2} = 15 	ext{ min}$$

Dividing by $3$, we get:

$$t_{1/2} = 5 	ext{ min}$$

This intuitive method is lightning fast and perfectly highlights the physical meaning of half-life. Always keep an eye out for these clean fractions!

Animated Solution for Physics - Atoms and Nuclei: Starting with a sample of pure $^{66} Cu$ , $7/8$ of it decays into Zn in $15 min$ . The corresponding half-life is

Select Answer:

Visualized Solution

Initial Sample $N_{0}$

Decayed Amount

Remaining Amount $N$

Radioactive Decay Law

Simplifying the Equation

Solving for $λ$

Calculating Half-Life $t_{1/2}$

Alternative Intuitive Method

The Sigma Insight: Radioactivity

Visualizing the Decay Process

The Mathematical Rigor

Solving for the Decay Constant

The Final Half-Life Calculation

The Ninja Technique

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Animated Solution for Physics - Atoms and Nuclei: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Starting with a sample of pure 66Cu, 7/8 of it decays into Zn in 15 min. The corresponding half-life is