Nuclei of radioactive element A are produced at rate ‘ $t^{2}$ ’ at any time t. The element A has decay constant $λ$ . Let N be the number of nuclei of element A at any time t. At time $t = t_{o}, \frac{d N}{d t}$ is minimum. Then the number of nuclei of element A at time t=t₀ is:

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$|\frac{t_{o} - λ t_{o}^{2}}{λ^{2}}|$

$|\frac{t_{o} - λ t_{o}^{2}}{λ}|$

$|\frac{2 t_{o} - λ t_{o}^{2}}{λ^{2}}|$

$|\frac{2 t_{o} - λ t_{o}^{2}}{λ}|$

answer is A.

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Detailed Solution

The rate of formation of radioactive nuclei is $t^{2}$
Rate of decay of radioactive nuclei is $λ N$
$∴ \frac{d N}{d t} = t^{2} - λ N$

$\begin{array}{l} For \frac{d N}{d t} to be minimum; \frac{d^{2} N}{d t^{2}} = 0 \\ \Rightarrow \frac{d^{2} N}{d t^{2}} = 2 t - λ \frac{d N}{d t} = 2 t - λ (t^{2} - λ N) = 0 \\ \Rightarrow N = \frac{2 t - λ t^{2}}{λ^{2}} \\ \Rightarrow N = \frac{2 t_{0} - λ t_{0}^{2}}{λ^{2}} (at t = t_{0}) \end{array}$