Question

In a radioactive decay process, the activity is defined as A = - \(\frac {dN}{dt}\), where 𝑁(𝑡) is the number of radioactive nuclei at time 𝑡. Two radioactive sources, 𝑆₁ and 𝑆₂ have same activity at time 𝑡 = 0. At a later time, the activities of 𝑆₁ and 𝑆₂ are 𝐴₁ and 𝐴₂, respectively. When 𝑆₁ and 𝑆₂ have just completed their 3^rd and 7^th half-lives, respectively, the ratio \(\frac{A_1}{A_2}\) is _____.

Answer 1

Correct Answer: 16

Radioactive Decay and Activity Calculation 

Given: The number of radioactive nuclei \( N(t) \) at a given time \( t \) is given by:

\(N(t) = N_0 \times \exp(-\lambda t)\)

The activity \( A \) of a radioactive source is defined as the rate of decay, which is the derivative of the number of radioactive nuclei with respect to time:

\(A = - \frac{dN}{dt} = \lambda \times N(t)\)

Given that the activities of sources \( S_1 \) and \( S_2 \) are \( A_1 \) and \( A_2 \), respectively, we can write:

\(A_1 = \lambda_1 \times N_1(t)\)

\(A_2 = \lambda_2 \times N_2(t)\)

After the completion of the 3rd half-life for source \( S_1 \), the remaining number of radioactive nuclei is:

\(N_1(t) = N_0 \times \left( \frac{1}{2} \right)^3 = N_0 \times \frac{1}{8}\)

Similarly, after the completion of the 7th half-life for source \( S_2 \), the remaining number of radioactive nuclei is:

\(N_2(t) = N_0 \times \left( \frac{1}{2} \right)^7 = N_0 \times \frac{1}{128}\)

Now, let's substitute these values into the equations for activities \( A_1 \) and \( A_2 \):

\(A_1 = \lambda_1 \times N_0 \times \left( \frac{1}{8} \right)\)

\(A_2 = \lambda_2 \times N_0 \times \left( \frac{1}{128} \right)\)

To find the ratio \( \frac{A_1}{A_2} \), we can simplify:

\(\frac{A_1}{A_2} = \frac{\lambda_1}{\lambda_2} \times \left( \frac{1}{8} \right) \times \left( \frac{128}{1} \right)\)

Since both sources have the same activity at time \( t = 0 \), we can assume their decay constants are the same (\( \lambda_1 = \lambda_2 \)). Therefore, the ratio simplifies to:

\(\frac{A_1}{A_2} = \frac{1}{8} \times \frac{128}{1} = \frac{128}{8} = 16\)

Thus, the ratio \( \frac{A_1}{A_2} \) is 16.