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Nuclei of radioactive element A are produced at rate $‘ t^{2} ’$ at an 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 $t = t_{0} .$ $\frac{d N}{d t}$ is minimum. Then the number of nuclei of element at A at time $t = t_{0}$ is

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$\frac{2 t_{0} - λ t_{0}^{2}}{λ^{2}}$

$\frac{t_{0} - λ t_{0}^{2}}{λ^{2}}$

$\frac{2 t_{0} - λ t_{0}^{2}}{λ}$

$\frac{t_{0} - λ t_{0}^{2}}{λ}$

answer is A.

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Detailed Solution

$\frac{d N}{d t} = t^{2} - λ N$ for $\frac{d N}{d t}$ to be minimum: $\frac{d^{2} N}{d t^{2}} = O$

$\Rightarrow \frac{d^{2} N}{d t^{2}} = 2 t - λ (t^{2} - λ N) = 0 O r N = \frac{2 t_{0} - λ t_{0}^{2}}{λ^{2}}$

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