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Beta Decay

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Question

The activity of a radioactive sample is $A_{1}$ at time $t_{1}$ and $A_{2}$ at time $t_{2}$ . If τ is average life of sample then the number of nuclei decayed in time $(t_{2} - t_{1})$ is

A

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

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B

$\frac{(A_{1} - A_{2})}{2} τ$

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C

$(A_{1} - A_{2}) (t_{2} - t_{1})$

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D

$(A_{1} - A_{2}) τ$

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Solution

The correct option is D
$(A_{1} - A_{2}) τ$

Let $N_{0}$ bet the initial number of nuclei, then
$N_{1} = N_{0} e^{- λ t_{1}}$

and $N_{2} = N_{0} e^{- λ t_{2}}$

$∴$ Number of nuclei decayed = $N_{1} - N_{2} = N_{0} (e^{- λ t_{1}} - e^{- λ t_{2}})$

$= \frac{A_{0}}{λ} (e^{- λ t_{1}} - e^{- λ t_{2}}) = \frac{A_{1} - A_{2}}{λ} = (A_{1} - A_{2}) τ .$

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