JEE Advanced 2023 — Paper 2 — PHY-III solution

JEE Advanced2023Paper 2PHY-III

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 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 __________.

Your answer

Solution

For each source, the activity decays as $A (t) = A_{0} \cdot (\frac{1}{2})^{t / T_{1/2}}$ . The starting activities are equal: $A_{0}$ for both.

At the moment described, $S_{1}$ has completed $3$ half-lives so $A_{1} = A_{0} \cdot (\frac{1}{2})^{3}$ . Similarly $S_{2}$ has completed $7$ half-lives, so $A_{2} = A_{0} \cdot (\frac{1}{2})^{7}$ .

Therefore

$\frac{A _{1}}{A _{2}} = \frac{( 1/2 ) ^{3}}{( 1/2 ) ^{7}} = 2^{4} = 16$ .

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