A radioactive sample is emitting 64 times radiations than non hazardous limit. if its half life is 2 hours, after what time it becomes non-hazardous:

Options

A. 16 hr
B. 12 hr
C. 8 hr
D. 4 hr

Answer

To solve the problem, we need to determine how long it takes for a radioactive sample emitting 64 times the non-hazardous limit of radiation to become non-hazardous again, given its half-life is 2 hours.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- The radioactive sample is currently emitting radiation at a level that is 64 times the non-hazardous limit. We can denote the non-hazardous limit as \( N_0 \). Therefore, the current activity \( N \) is:
\[ N = 64 \times N_0 \]

2. **Using the Half-Life Concept**:
- The half-life (\( t_{1/2} \)) of the sample is given as 2 hours. This means that every 2 hours, the amount of radioactive material (or its activity) is halved.

3. **Relating Current Activity to Non-Hazardous Activity**:
- The relationship between the initial activity \( N_0 \) and the current activity \( N \) after \( n \) half-lives can be expressed as:
\[ \frac{N_0}{N} = \left(\frac{1}{2}\right)^n \]
- Substituting the values we have:
\[ \frac{N_0}{64 \times N_0} = \left(\frac{1}{2}\right)^n \]
- Simplifying this gives:
\[ \frac{1}{64} = \left(\frac{1}{2}\right)^n \]

4. **Expressing 64 as a Power of 2**:
- We know that \( 64 = 2^6 \). Thus, we can rewrite the equation:
\[ \frac{1}{2^6} = \left(\frac{1}{2}\right)^n \]
- Since the bases are the same, we equate the exponents:
\[ n = 6 \]

5. **Calculating the Total Time**:
- The total time taken for the activity to reduce from 64 times the non-hazardous limit to the non-hazardous limit is given by:
\[ \text{Total Time} = n \times t_{1/2} \]
- Substituting the values we have:
\[ \text{Total Time} = 6 \times 2 \text{ hours} = 12 \text{ hours} \]

6. **Conclusion**:
- Therefore, the radioactive sample will become non-hazardous after **12 hours**.