A human body required the 0.01 M activity of radioactive substance after 24 h. Half life of radioactive substance is 6h. Then injection of maximum activity of radioactie substance that can be injected will be

Options

A. 0.08
B. 0.04
C. 0.16
D. 0.32

Explanation

To solve the problem, we need to determine the maximum activity of a radioactive substance that can be injected into a human body so that after 24 hours, the activity is 0.01 M. Given that the half-life of the radioactive substance is 6 hours, we can follow these steps:

### Step 1: Calculate the number of half-lives in 24 hours
The half-life of the substance is given as 6 hours. To find out how many half-lives fit into 24 hours, we divide 24 by 6.

\[ \text{Number of half-lives} = \frac{24 \text{ hours}}{6 \text{ hours}} = 4 \]

### Step 2: Determine the final activity after 4 half-lives
We know that after 4 half-lives, the activity of the substance is reduced to 0.01 M.

### Step 3: Calculate the initial activity required
The activity of a radioactive substance halves with each half-life. Therefore, we can work backwards to find the initial activity (A0) that would result in 0.01 M after 4 half-lives.

Let’s denote the initial activity as \( A_0 \). After 1 half-life, the activity will be:

\[ A_1 = \frac{A_0}{2} \]

After 2 half-lives:

\[ A_2 = \frac{A_1}{2} = \frac{A_0}{4} \]

After 3 half-lives:

\[ A_3 = \frac{A_2}{2} = \frac{A_0}{8} \]

After 4 half-lives:

\[ A_4 = \frac{A_3}{2} = \frac{A_0}{16} \]

We know that \( A_4 = 0.01 \, \text{M} \). Thus, we can set up the equation:

\[ \frac{A_0}{16} = 0.01 \]

### Step 4: Solve for the initial activity \( A_0 \)
To find \( A_0 \), we multiply both sides of the equation by 16:

\[ A_0 = 0.01 \times 16 = 0.16 \, \text{M} \]

### Conclusion
The maximum activity of the radioactive substance that can be injected is **0.16 M**.

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