The activity of a sample of a radioactive meterial is `A_1` , at time `t_1` , and `A_2` at time `t_2(t_2 gt t_1)` . If its mean life T, then

Options

A. `A_1t_1=A_2t_2`
B. `(A_1-A_2)/(t_2-t_1) = "constant"`
C. `A_2=A_1e^(t_1-t_2//T)`
D. `A_2=A_1e^(t_1//Tt_2)`

Answer

To solve the problem, we need to establish the relationship between the activities \( A_1 \) and \( A_2 \) of a radioactive material at two different times \( t_1 \) and \( t_2 \), given that the mean life \( T \) is known.

### Step-by-Step Solution:

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

2. **Write the Expressions for \( A_1 \) and \( A_2 \)**:
At time \( t_1 \):
\[ A_1 = A_0 e^{-\lambda t_1} \quad \text{(Equation 1)} \]
At time \( t_2 \):
\[ A_2 = A_0 e^{-\lambda t_2} \quad \text{(Equation 2)} \]

3. **Divide Equation 2 by Equation 1**:
To find the ratio of \( A_2 \) to \( A_1 \):
\[ \frac{A_2}{A_1} = \frac{A_0 e^{-\lambda t_2}}{A_0 e^{-\lambda t_1}} = e^{-\lambda t_2} \cdot e^{\lambda t_1} = e^{-\lambda (t_2 - t_1)} \]

4. **Rearranging the Equation**:
From the above step, we can express \( A_2 \) in terms of \( A_1 \):
\[ A_2 = A_1 e^{-\lambda (t_2 - t_1)} \]

5. **Relate the Decay Constant \( \lambda \) to Mean Life \( T \)**:
The mean life \( T \) is related to the decay constant \( \lambda \) by:
\[ T = \frac{1}{\lambda} \quad \Rightarrow \quad \lambda = \frac{1}{T} \]

6. **Substituting \( \lambda \) into the Equation**:
Substitute \( \lambda \) back into the equation for \( A_2 \):
\[ A_2 = A_1 e^{-\frac{(t_2 - t_1)}{T}} \]

### Final Result:
Thus, the relation between \( A_2 \) and \( A_1 \) is:
\[ A_2 = A_1 e^{-\frac{(t_2 - t_1)}{T}} \]