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 step by step, we will use the concept of half-life and the characteristics of radioactive decay.

### Step 1: Understand the Problem
We have a radioactive sample that emits radiation at a level 64 times higher than the non-hazardous limit. The half-life of this sample is given as 2 hours. We need to find out how long it will take for the radiation level to drop to the non-hazardous limit.

### Step 2: Define Variables
- Let \( N_0 \) be the initial radiation level.
- Let \( N_t \) be the radiation level after time \( t \).
- We know that \( N_t = \frac{N_0}{64} \) because the sample is initially 64 times more than the non-hazardous limit.

### Step 3: Use the First Order Kinetics Formula
For radioactive decay, we can use the formula:
\[ N_t = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \]
where \( t_{1/2} \) is the half-life of the substance.

### Step 4: Substitute Known Values
Substituting \( N_t \) and \( N_0 \) into the equation:
\[ \frac{N_0}{64} = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \]
We can cancel \( N_0 \) from both sides (assuming \( N_0 \neq 0 \)):
\[ \frac{1}{64} = \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \]

### Step 5: Express 64 as a Power of 2
We know that:
\[ 64 = 2^6 \]
Thus, we can rewrite the equation as:
\[ \frac{1}{2^6} = \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \]

### Step 6: Set the Exponents Equal
Since the bases are the same, we can set the exponents equal to each other:
\[ -6 = -\frac{t}{t_{1/2}} \]
This simplifies to:
\[ t = 6 \times t_{1/2} \]

### Step 7: Substitute the Half-Life Value
Given that the half-life \( t_{1/2} \) is 2 hours:
\[ t = 6 \times 2 \text{ hours} = 12 \text{ hours} \]

### Conclusion
Thus, the time it takes for the radioactive sample to become non-hazardous is **12 hours**.