At time `t=0`, some radioactive gas is injected into a sealed vessel. At time `T`, some more of the same gas is injected into the same vessel. Which one of the following graphs best represents the variation of the logarithm of the activity `A` of the gas with time `t`?

Options

A.
B.
C.
D.

Answer

To solve the problem, we need to analyze the behavior of the activity \( A \) of a radioactive gas over time after two injections of gas into a sealed vessel. We will derive the relationship between the logarithm of the activity \( \ln A \) and time \( t \).

### Step-by-Step Solution:

1. **Understanding Radioactive Decay**:
The activity \( A \) of a radioactive substance is given by the equation:
\[ A(t) = A_0 e^{-\lambda t} \]
where \( A_0 \) is the initial activity, \( \lambda \) is the decay constant, and \( t \) is time.

2. **Taking the Natural Logarithm**:
To analyze the relationship in logarithmic terms, we take the natural logarithm of both sides:
\[ \ln A(t) = \ln A_0 - \lambda t \]
This equation represents a straight line with a negative slope.

3. **Behavior at Time \( t = 0 \)**:
At \( t = 0 \), the activity is at its maximum, \( A(0) = A_0 \). Thus, \( \ln A(0) = \ln A_0 \).

4. **Second Injection at Time \( T \)**:
At time \( T \), when more gas is injected, the total activity will increase due to the additional radioactive gas. Let’s denote the new activity after the second injection as \( A_1 \). The activity will still decay but will start from a new initial value:
\[ A_1(t) = A_0 e^{-\lambda (t-T)} + A_2 e^{-\lambda (t-T)} \]
where \( A_2 \) is the activity contributed by the newly injected gas.

5. **Logarithmic Representation After Second Injection**:
After the second injection, the logarithmic activity can be represented as:
\[ \ln A(t) = \ln(A_0 + A_2 e^{-\lambda (t-T)}) - \lambda t \]
This indicates that after time \( T \), the logarithm of the activity will show a different slope due to the contribution of the newly injected gas.

6. **Graphical Representation**:
The graph of \( \ln A \) versus \( t \) will show a linear decrease until time \( T \), and after \( T \), it will show a different linear decrease, indicating a change in slope. The first part of the graph will have a slope of \( -\lambda \), and after \( T \), it will have a different slope, reflecting the new activity level.

### Conclusion:
The correct graph will show a linear decrease in \( \ln A \) until \( t = T \) and then a different linear decrease after \( T \). The graph that best represents this behavior is option B.