In a radioactive decay process, the activity is defined as where is the number of radioactive nuclei at time . Two radioactive sources, and , have the same activity at time . At a later time, the activities of and are and , respectively. When and

Question

In a radioactive decay process, the activity is defined as $A = - \frac{d N}{d t},$ where $N (t)$ is the number of radioactive nuclei at time $t$ . Two radioactive sources, $S_{1}$ and $S_{2}$ , have the 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 $\frac{A_{1}}{A_{2}}$ is _________. [2023]

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(16)

When radioactive sources just completed their $3^{rd}$ and $7^{th}$ half-lives, then the ratio

$\frac{A_{1}}{A_{2}} = \frac{A_{0} e^{- 3 \ln 2}}{A_{0} e^{- 7 \ln 2}} = \frac{2^{- 3}}{2^{- 7}} = \frac{1}{2^{- 4}} = 2^{4} = 16$

Solution

Ans.
(16)

When radioactive sources just completed their $3^{rd}$ and $7^{th}$ half-lives, then the ratio

$\frac{A_{1}}{A_{2}} = \frac{A_{0} e^{- 3 \ln 2}}{A_{0} e^{- 7 \ln 2}} = \frac{2^{- 3}}{2^{- 7}} = \frac{1}{2^{- 4}} = 2^{4} = 16$

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Solution

Ans. (16) When radioactive sources just completed their 3rd and 7th half-lives, then the ratio A1A2=A0e-3ln2A0e-7ln2=2-32-7=12-4=24=16

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Ans.
(16)

When radioactive sources just completed their $3^{rd}$ and $7^{th}$ half-lives, then the ratio

$\frac{A_{1}}{A_{2}} = \frac{A_{0} e^{- 3 \ln 2}}{A_{0} e^{- 7 \ln 2}} = \frac{2^{- 3}}{2^{- 7}} = \frac{1}{2^{- 4}} = 2^{4} = 16$