Question

In a radioactive material, the activity at time \( t_1 \) is \( A_1 \) and at a later time \( t_2 \), it is \( A_2 \). If the decay constant of the material is \( \lambda \), then:

• A. \( A_1 = A_2 e^{-\lambda(t_1 - t_2)} \)
• B. \( A_1 = A_2 e^{\lambda(t_1 - t_2)} \)
• C. \( A_1 = A_2 \frac{t_2}{t_1} \)
• D. \( A_1 = A_2 \)

Hint:

The activity of a radioactive material decays exponentially over time. To find the activity at one time from another, use the relationship \( A_1 = A_2 e^{-\lambda(t_1 - t_2)} \), where \( \lambda \) is the decay constant.

Answer 1

The Correct Option is A

Radioactive decay follows an exponential decay law, and the activity of a radioactive material at any time \( t \) is given by: \[ A(t) = A_0 e^{-\lambda t} \] where \( A_0 \) is the initial activity, \( \lambda \) is the decay constant, and \( t \) is the time elapsed. 
Step 1: Let \( A_1 \) be the activity at time \( t_1 \) and \( A_2 \) be the activity at time \( t_2 \). Using the decay law for both times: At time \( t_1 \): \[ A_1 = A_0 e^{-\lambda t_1} \] At time \( t_2 \): \[ A_2 = A_0 e^{-\lambda t_2} \] 
Step 2: To find the relationship between \( A_1 \) and \( A_2 \), divide the equation for \( A_1 \) by the equation for \( A_2 \): \[ \frac{A_1}{A_2} = \frac{A_0 e^{-\lambda t_1}}{A_0 e^{-\lambda t_2}} = e^{-\lambda(t_1 - t_2)} \] Thus, the relationship between the activities is: \[ A_1 = A_2 e^{-\lambda(t_1 - t_2)} \] Hence, the correct answer is \( \boxed{{A}} \).