Activity of a radioactive substance is `A_(1)` at time `t_(1)` and `A_(2)` at time `t_(2)(t_(2) gt t_(1))` , then the ratio of f`(A_(2))/(A_(1))` is:

Options

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

Answer

To solve the problem of finding the ratio of the activity of a radioactive substance at two different times, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Activity**:
- The activity \( A \) of a radioactive substance is defined as the number of disintegrations per second. It is directly proportional to the number of undecayed nuclei \( n \).

2. **Establish the Relationship**:
- The activity can be expressed mathematically as:
\[ A = A_0 e^{-\lambda t} \]
where:
- \( A_0 \) is the initial activity,
- \( \lambda \) is the decay constant,
- \( t \) is the time elapsed.

3. **Write the Expressions for Activities at Two Times**:
- For time \( t_1 \):
\[ A_1 = A_0 e^{-\lambda t_1} \]
- For time \( t_2 \):
\[ A_2 = A_0 e^{-\lambda t_2} \]

4. **Find the Ratio of Activities**:
- To find the ratio \( \frac{A_2}{A_1} \):
\[ \frac{A_2}{A_1} = \frac{A_0 e^{-\lambda t_2}}{A_0 e^{-\lambda t_1}} \]
- The \( A_0 \) terms cancel out:
\[ \frac{A_2}{A_1} = \frac{e^{-\lambda t_2}}{e^{-\lambda t_1}} = e^{-\lambda t_2 + \lambda t_1} = e^{\lambda (t_1 - t_2)} \]

5. **Final Expression**:
- Thus, the ratio of the activities is:
\[ \frac{A_2}{A_1} = e^{\lambda (t_1 - t_2)} \]

### Conclusion:
The correct answer is \( e^{\lambda (t_1 - t_2)} \).