Using a nuclear counter, the count rate of emitted particles from a radioactive source is measured.

Q: Why didn't we use the formula $A = A_0 e^{-\lambda t}$?

You absolutely can! However, when the initial and final values are clean powers of 2 (like 1600 and 100), the logical chain method is much faster and less prone to algebraic mistakes.

Q: What if the time wasn't an exact multiple of the half-life?

If the time was, say, 5 seconds, you wouldn't land cleanly on a node in our chain. In that case, you *must* use the exponential formula $A = A_0 (1/2)^{t/T_{1/2}}$ to find the exact value.

Q: Does the count rate mean the same thing as activity?

Yes, in the context of these problems, the count rate measured by a detector is directly proportional to the actual activity (rate of decay) of the sample.

Q: Why does the activity halve exactly?

Because the number of active, undecayed nuclei halves during one half-life. Since activity is directly proportional to the number of active nuclei ($A = \lambda N$), the activity also halves.

Q: Is the half-life constant for a given radioactive material?

Yes, the half-life is a fundamental, unchanging property of a specific radioactive isotope, regardless of the amount of material present or external environmental conditions.

Question

Using a nuclear counter, the count rate of emitted particles from a radioactive source is measured. At $t =0$, it was $1600$ counts per second and $t =8	ext{ s}$, it was $100$ counts per second. The count rate observed as counts per second at $t =6	ext{ s}$ is close to

Options:

(A) 400

(B) 200

(C) 150

(D) 360

Sigma Percentile · Accepted Answer

### Visualizing the Radioactive Journey

Let's embark on a journey with our radioactive sample. Imagine you are observing a nuclear counter. We start our stopwatch at $t = 0$, and the counter is ticking rapidly at an initial activity of $A_0 = 1600 	ext{ s}^{-1}$.

We let the clock run, and fast forward to $t = 8 	ext{ s}$. The ticking has slowed down significantly, dropping all the way to $A = 100 	ext{ s}^{-1}$. Our mission is to figure out what the count rate was at exactly $t = 6 	ext{ s}$.

### The Power of the Half-Life

To solve this, we need to understand how a radioactive substance decays. It follows a beautiful, predictable pattern governed by its **half-life** ($T_{1/2}$). In every half-life duration, the activity becomes exactly half of what it was previously.

Instead of jumping straight into complex exponential formulas, let's trace this halving process logically. Starting from $1600$, one half-life brings it down to $800$. Another half-life takes it to $400$. A third one reduces it to $200$. And finally, a fourth half-life brings it down to $100$.

$$1600 \xrightarrow{T_{1/2}} 800 \xrightarrow{T_{1/2}} 400 \xrightarrow{T_{1/2}} 200 \xrightarrow{T_{1/2}} 100$$

By simply counting the steps, we can see that it took exactly **$4$ half-lives** to reach the state of $100 	ext{ counts/s}$.

### Calculating the Time

We know from the problem statement that this entire journey from $1600$ to $100$ took $8 	ext{ seconds}$. Since there are $4$ half-lives packed into this duration, we can easily find the length of a single half-life.

$$4 T_{1/2} = 8 	ext{ s}$$

Dividing both sides by $4$, we find that the half-life is:

$$T_{1/2} = 2 	ext{ s}$$

### Finding the Target Activity

The question asks for the count rate at $t = 6 	ext{ s}$. Since each half-life is $2 	ext{ seconds}$, $6 	ext{ seconds}$ corresponds to exactly $3$ half-lives ($6 / 2 = 3$).

Let's look back at our decay chain. After one, two, and three half-lives, where do we land? Following the third arrow in our sequence, we land squarely on **$200$**.

Therefore, the count rate at $6 	ext{ seconds}$ is **$200 	ext{ counts per second}$**. This logical chain method is incredibly fast and helps avoid the calculation errors that can sometimes happen with the exponential decay formula $A = A_0 e^{-\lambda t}$.

Select Answer:

Visualized Solution

Initial and Final States

The Half-Life Rule

Tracing the Decay Chain

Calculating Half-Life

State at $t = 6 s$

Final Answer

Alternative Method

The Sigma Insight: Radioactivity

Visualizing the Radioactive Journey

The Power of the Half-Life

Calculating the Time

Finding the Target Activity

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