As accident in a nuclear laboratory resulting in deposition of a certain amount of radioactive material of half life `18 `days inside the laboratory Tests revealed that the radiation was `64` times more than the permissible level required for save operation of the laboratory what is the minimum number of days after which the laboratory can be considered safe for use?

Options

A. 64
B. 90
C. 108
D. 120

Answer

To solve the problem step by step, we will use the concept of half-life and the relationship between the initial activity of a radioactive substance and its activity after a certain number of half-lives.

### Step 1: Understand the Problem
We are given:
- Half-life of the radioactive material, \( t_{1/2} = 18 \) days.
- The radiation level is currently 64 times more than the permissible level.

### Step 2: Set Up the Equation
The activity \( A \) of a radioactive substance after \( n \) half-lives can be expressed as:
\[ A = A_0 \left(\frac{1}{2}\right)^n \]
Where \( A_0 \) is the initial activity. Since the current activity is 64 times the permissible level, we can set \( A_0 = 1 \) (for simplicity), and we need to find \( n \) such that:
\[ A = \frac{1}{64} \]

### Step 3: Relate the Current Activity to Half-Lives
From the equation:
\[ \frac{1}{64} = 1 \left(\frac{1}{2}\right)^n \]
This implies:
\[ \left(\frac{1}{2}\right)^n = \frac{1}{64} \]

### Step 4: Solve for \( n \)
We know that \( 64 = 2^6 \), thus:
\[ \frac{1}{64} = \frac{1}{2^6} \]
This gives us:
\[ n = 6 \]

### Step 5: Calculate the Time Required
Now, we can calculate the total time required for the activity to reduce to a safe level:
\[ \text{Total time} = n \times t_{1/2} = 6 \times 18 \text{ days} \]
Calculating this gives:
\[ \text{Total time} = 108 \text{ days} \]

### Conclusion
The minimum number of days after which the laboratory can be considered safe for use is **108 days**.

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