Suppose `50` bacterial are placed in a flask containing nutrients
for the bacteria so that they can multiply. A study at `35^(@)C` gave the following results:
Time: (minutes) `0, 15, 30, 45, 60`
Number of bacteria `100,, 200, 400, 800, 1600`
The `t_(1//2)` for the reaction is,

Options

A. `1.5` sec
B. `1.5` min
C. `150` min
D. `15` min

Explanation

To solve the problem of determining the half-life (t₁/₂) for the bacterial growth in the flask, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Growth Pattern:**
- We start with 50 bacteria, and the number of bacteria doubles every 15 minutes. The given data shows:
- At 0 minutes: 100 bacteria
- At 15 minutes: 200 bacteria
- At 30 minutes: 400 bacteria
- At 45 minutes: 800 bacteria
- At 60 minutes: 1600 bacteria

2. **Identifying the Doubling Time:**
- The time taken for the bacteria to double in number is 15 minutes. This is crucial for calculating the half-life.

3. **Relating Doubling Time to Half-Life:**
- For a first-order reaction, the relationship between the doubling time (t_d) and half-life (t₁/₂) can be expressed as:
\[ t_d = t_{1/2} \cdot \ln(2) \]
- Since we know that the doubling time is 15 minutes, we can rearrange this formula to find the half-life:
\[ t_{1/2} = \frac{t_d}{\ln(2)} \]

4. **Calculating Half-Life:**
- The natural logarithm of 2 (ln(2)) is approximately 0.693. Thus, we can substitute the values:
\[ t_{1/2} = \frac{15 \text{ minutes}}{0.693} \approx 21.65 \text{ minutes} \]

5. **Conclusion:**
- The half-life for the reaction, based on the doubling time of the bacteria, is approximately 21.65 minutes.

### Final Answer:
The half-life (t₁/₂) for the reaction is approximately **21.65 minutes**.

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