A substance has half of 3 min. After 6 min, the count rate was observed to be 400. What was its count rate at zero time?

1. Analyze the Question

We are dealing with a radioactive decay problem. We know the half-life ($T_{1/2}$) is 3 minutes, the elapsed time ($t$) is 6 minutes, and the final count rate ($N_t$) is 400. We need to find the initial count rate ($N_0$).

2. Explain the Concept

The radioactive decay formula relates the initial amount to the final amount over a period of time:

Where:

• $N_t$ = Final count rate
• $N_0$ = Initial count rate
• $n$ = Number of half-lives elapsed, calculated as $n = \frac{t}{T_{1/2}}$

3. Show the Solution

Step 1: Calculate the number of half-lives ($n$)

This means the substance has gone through 2 half-lives.

Step 2: Use the decay formula

Step 3: Solve for $N_0$ Multiply both sides by 4:

4. Identify the Answer

The initial count rate was 1600. This corresponds to option A.

5. Exam Tip

For simple half-life problems where the number of half-lives is a whole number, you don't always need the formula. Just work backward:

• At 6 mins: 400
• At 3 mins (one half-life earlier): $400 \times 2 = 800$
• At 0 mins (two half-lives earlier): $800 \times 2 = 1600$ This "doubling back" method is faster and helps you avoid calculation errors during the exam!