After two hours, one-sixteenth of tge starting amount if a certain radioactive isotope remained undecayed. The half-life of the isotope is

Options

A. 15 min
B. 30 min
C. 45 min
D. 1 hour

Answer

To find the half-life of a certain radioactive isotope given that after two hours, one-sixteenth of the starting amount remains undecayed, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- After 2 hours, the remaining amount of the isotope is \( \frac{1}{16} \) of the initial amount.
- We denote the initial amount as \( N_0 \) and the remaining amount as \( N \).
- Therefore, \( N = \frac{N_0}{16} \).

2. **Use the Decay Formula**:
- The relationship between the remaining amount and the initial amount in terms of half-life is given by the formula:
\[ N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
- Here, \( T_{1/2} \) is the half-life we need to find, and \( t \) is the time elapsed (2 hours in this case).

3. **Substitute the Known Values**:
- We can substitute \( N \) and \( N_0 \) into the equation:
\[ \frac{N_0}{16} = N_0 \left( \frac{1}{2} \right)^{\frac{2}{T_{1/2}}} \]
- Dividing both sides by \( N_0 \) (assuming \( N_0 \neq 0 \)):
\[ \frac{1}{16} = \left( \frac{1}{2} \right)^{\frac{2}{T_{1/2}}} \]

4. **Express \( \frac{1}{16} \) in Terms of Powers of 2**:
- We know that \( \frac{1}{16} = \frac{1}{2^4} \).
- Therefore, we can rewrite the equation as:
\[ \frac{1}{2^4} = \left( \frac{1}{2} \right)^{\frac{2}{T_{1/2}}} \]

5. **Set the Exponents Equal**:
- Since the bases are the same, we can set the exponents equal to each other:
\[ 4 = \frac{2}{T_{1/2}} \]

6. **Solve for the Half-Life**:
- Rearranging the equation gives:
\[ T_{1/2} = \frac{2}{4} = \frac{1}{2} \text{ hours} \]
- Converting this to minutes, we find:
\[ T_{1/2} = 30 \text{ minutes} \]

### Final Answer:
The half-life of the isotope is **30 minutes**.