In a radioactive decay process, the activity is defined as A = - d N d t , where N ( t ) is the number of radioactive nuclei at time t . Two radioactive sources, S 1 and S 2 have same activity at time

Question

In a radioactive decay process, the activity is defined as A = - d N d t , where N ( t ) is the number of radioactive nuclei at time t . Two radioactive sources, S 1 and S 2 have same activity at time t = 0 . At a later time, the activities of S 1 and S 2 are A 1 and A 2 , respectively. When S 1 and S 2 have just completed their 3 rd and 7 th half-lives, respectively, the ratio A 1 A 2 is _____.

TestHub · Accepted Answer

Activity for both the cases can be written as, $A_{1} = A_{0} e^{- λ_{1} t_{1}}$
also $A_{2} = A_{0} e^{- λ_{2} t_{2}}$ .
Half-life for the reaction can be written as, ${(t_{\frac{1}{2}})}_{1} = \frac{\ln 2}{λ_{1}}$ and hence
at $t_{1} = \frac{3 \ln 2}{λ_{1}}$ .
$A_{1} = A_{0} e^{- λ_{1} (\frac{3 \ln 2}{λ_{1}})}$
$\Rightarrow A_{1} = A_{0} e^{- 3 \ln 2} . . . (i)$
Similarly, for the second case, we can write
$t_{2} = 7 (\frac{\ln 2}{λ_{2}})$ and therefore,
$A_{2} = A_{0} e^{- λ_{2} (\frac{7 \ln 2}{λ_{2}})}$
$A_{2} = A_{0} e^{- 7 \ln 2} . . . (ii)$
From $(i)$ and $(ii)$ , we get
$\frac{A_{1}}{A_{2}} = \frac{A_{0} e^{- 3 \ln 2}}{A_{0} e^{- 7 \ln 2}} = \frac{A_{0} e^{\ln 2^{- 3}}}{A_{0} e^{\ln 2^{- 7}}} = \frac{2^{- 3}}{2^{- 7}} = 16$
Therefore, required ration would be $∴ \frac{A_{1}}{A_{2}} = 16$

Physics - Modern Physics Question with Solution | TestHub

Doubts & Discussion