The half-life period of a first order reaction is 15 minutes. The amount of substance left after one hour will be:

Options

A. `(1)/(4)` of the original amount
B. `(1)/(8)` of the original amount
C. `(1)/(16)` of the original amount
D. `(1)/(32)` of the original amount

Explanation

To solve the problem, we need to determine the amount of substance left after one hour for a first-order reaction with a given half-life period.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Half-life (t₁/₂) = 15 minutes
- Total time (T) = 1 hour = 60 minutes

2. **Calculate the Number of Half-Lives (n):**
- The number of half-lives (n) can be calculated using the formula:
\[ n = \frac{T}{t_{1/2}} = \frac{60 \text{ minutes}}{15 \text{ minutes}} = 4 \]
- This means that in 60 minutes, 4 half-lives have passed.

3. **Determine the Remaining Amount of Substance:**
- For a first-order reaction, the relationship between the initial amount (A₀) and the amount remaining (A) after n half-lives is given by:
\[ \frac{A_0}{A} = 2^n \]
- Therefore, we can express A as:
\[ A = \frac{A_0}{2^n} \]
- Substituting n = 4:
\[ A = \frac{A_0}{2^4} = \frac{A_0}{16} \]

4. **Conclusion:**
- The amount of substance left after one hour is:
\[ A = \frac{A_0}{16} \]
- This means that if we started with an initial amount A₀, after one hour, the remaining amount will be \( \frac{1}{16} \) of the initial amount.

### Final Answer:
The amount of substance left after one hour will be \( \frac{A_0}{16} \).

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