Half-life period of a radioactive substance is `6 h`. After `24 h` activity is `0.01 muC`, what was the initial activity ?

Options

A. `0.04 muC`
B. `0.08 muC`
C. `0.24 muC`
D. `0.16 muC`

Answer

To find the initial activity of a radioactive substance given its half-life and activity after a certain time, we can follow these steps:

### Step 1: Understand the relationship between activity and time
The activity \( R \) at any time \( t \) is related to the initial activity \( R_0 \) by the equation:
\[ R = R_0 e^{-\lambda t} \]
where \( \lambda \) is the decay constant.

### Step 2: Calculate the decay constant \( \lambda \)
The decay constant \( \lambda \) is related to the half-life \( T_{1/2} \) by the formula:
\[ \lambda = \frac{\ln 2}{T_{1/2}} \]
Given that the half-life \( T_{1/2} = 6 \, \text{h} \):
\[ \lambda = \frac{\ln 2}{6 \, \text{h}} \]

### Step 3: Substitute the values into the activity equation
We know that after \( 24 \, \text{h} \), the activity \( R = 0.01 \, \mu C \). We can substitute \( t = 24 \, \text{h} \) into the activity equation:
\[ 0.01 = R_0 e^{-\lambda \cdot 24} \]

### Step 4: Calculate \( \lambda \cdot t \)
Substituting \( \lambda \) into the equation:
\[ -\lambda \cdot 24 = -\left(\frac{\ln 2}{6}\right) \cdot 24 = -4 \ln 2 \]
Thus, the equation becomes:
\[ 0.01 = R_0 e^{-4 \ln 2} \]

### Step 5: Simplify the equation
Using the property \( e^{\ln a} = a \):
\[ e^{-4 \ln 2} = (e^{\ln 2})^{-4} = 2^{-4} = \frac{1}{16} \]
So we can rewrite the equation as:
\[ 0.01 = R_0 \cdot \frac{1}{16} \]

### Step 6: Solve for \( R_0 \)
To find \( R_0 \):
\[ R_0 = 0.01 \cdot 16 = 0.16 \, \mu C \]

### Conclusion
The initial activity \( R_0 \) is \( 0.16 \, \mu C \).

### Final Answer
The initial activity was \( 0.16 \, \mu C \).

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