The activity of a radioactive substance is `R_(1)` at time `t_(1)` and `R_(2)` at time `t_(2)(gt t_(1))`. Its decay cosntant is `lambda`. Then .

Options

A. `R_(1)t_(1)=R_(2)t_(2)`
B. `R_(1)=R_(1)e^(-(lambda^(t_(1)-t_(2))))`
C. `(R_(1)-R_(2))/(t_(2)-t_(1))`=constant
D. `R_(2)=R_(1)e^(lambda(t_(2)-t_(1))`

Answer

To solve the problem, we need to understand the relationship between the activity of a radioactive substance at two different times and its decay constant. Here are the steps to derive the required relationship:

### Step-by-Step Solution:

1. **Understanding Activity**: The activity \( R \) of a radioactive substance is defined as the number of decays per second. It can be expressed in terms of the decay constant \( \lambda \) and the number of radioactive nuclei \( N \) present at any time \( t \):
\[ R = \lambda N \]

2. **Activity at Two Different Times**: Given the activity at time \( t_1 \) is \( R_1 \) and at time \( t_2 \) is \( R_2 \):
\[ R_1 = \lambda N_1 \quad \text{and} \quad R_2 = \lambda N_2 \]

3. **Relating the Number of Nuclei**: The number of radioactive nuclei decreases over time according to the equation:
\[ N(t) = N_0 e^{-\lambda t} \]
where \( N_0 \) is the initial number of nuclei.

4. **Expressing \( N_1 \) and \( N_2 \)**:
- At time \( t_1 \):
\[ N_1 = N_0 e^{-\lambda t_1} \]
- At time \( t_2 \):
\[ N_2 = N_0 e^{-\lambda t_2} \]

5. **Substituting \( N_1 \) and \( N_2 \) into the Activity Equations**:
- For \( R_1 \):
\[ R_1 = \lambda N_0 e^{-\lambda t_1} \]
- For \( R_2 \):
\[ R_2 = \lambda N_0 e^{-\lambda t_2} \]

6. **Taking the Ratio of Activities**:
\[ \frac{R_1}{R_2} = \frac{\lambda N_0 e^{-\lambda t_1}}{\lambda N_0 e^{-\lambda t_2}} = \frac{e^{-\lambda t_1}}{e^{-\lambda t_2}} = e^{-\lambda (t_1 - t_2)} \]

7. **Rearranging the Equation**:
\[ R_1 = R_2 e^{-\lambda (t_1 - t_2)} \]

### Conclusion:
The relationship derived is:
\[ R_1 = R_2 e^{-\lambda (t_1 - t_2)} \]
This corresponds to option B.