The decay constant of a radio isotope is $λ$ . If $A_{1}$ and $A_{2}$ are its activities at times $t_{1}$ and $t_{2}$ respectively , then the number of nuclei which have decayed during the time $(t_{1}^{2} - t_{2}) : -$

A_{1} - A_{2}

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(A_{1} - A_{2}) / λ

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λ (A_{1} - A_{2})

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A_{1} t_{1} - A_{2} t_{2}

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Solution

The correct option is A $A_{1} - A_{2}$
The decay constant of a ratio isotope is $λ$ if $a_{1}$ and $a_{2}$ are
its activates at times $t_{1}$ and $t_{2}$ respectively no of nuclei which have decayed during the time $(t_{1} - t_{2})$
So, $A_{1}$ =lambda $N_{1}$ at time $t_{1} = λ N_{1}$
$A_{2}$ =lambda $N_{2}$ at time $t_{2} = λ N_{2}$
therefore no of nuclei decayed during time interval $(t_{1} - t_{2})$
$\begin{matrix} λ N_{1} - λ N_{2} = A_{1} - A_{2} λ t a k i n g c o m m o n s o λ (N_{1} - N_{2}) = A_{1} - A_{2} ∴ N_{1} - N_{2} = A_{1} - A_{2} \end{matrix}$
Hence, Option $A$ is correct.

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