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 3rd and 7th half-lives, respectively, the ratio $\frac{A_1}{A_2}$ is _________.

Answer 1

The number of radioactive nuclei N(t) at a given time t is given by:
N(t) = N0 $\times$exp(-λt)
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 = -$\frac {dN}{dt}$ = λ$\times$ N(t)
Given that the activities of sources S1 and S2 are A1 and A2, respectively, we can write:
A1 = λ1 $\times$ N1(t)
A2 = λ2 $\times$ N2(t)
After the completion of the 3rd half-life for source S1, the remaining number of radioactive nuclei is:
N1(t) = N0 $\times$ ($\frac{1}{2}$)3 = N0 $\times$ $\frac {1}{8}$
Similarly, after the completion of the 7th half-life for source S2, the remaining number of radioactive nuclei is:
N2(t) = N0 $\times$ $[\frac{1}{2}]$7 = N0 $\times$ $\frac {1}{128}$
Now, let's substitute these values into the equations for activities A1 and A2 :
A1 = λ1 $\times$ N0 $\times$ ($\frac {1}{8}$)

A2 = λ2 $\times$ N0 $\times$ ($\frac {1}{128}$)

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

($\frac{A_1}{A_2}$) = ($\frac{\lambda_1}{\lambda_2}$) $\times$ ($\frac {1}{8}$ ) $\times$ ($\frac {128}{1}$)
Since both sources have the same activity at time t = 0, we can assume their decay constants are the same (λ1 = λ2). Therefore, the ratio simplifies to:
($\frac{A_1}{A_2}$) = $\frac {1}{8}$ $\times$ $\frac {128}{1}$ = $\frac {128}{8}$ = 16

Therefore, the ratio $\frac{A_1}{A_2}$ is 16.

So, the answer is $16$.