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Law of Radioactivity

The activity ...

Question

The activity of a sample of radioactive element is $A_{1}$ at time $t_{1}$ and $A_{2}$ at time $t_{2} (t_{2} > t_{1})$ . The mean life of sample is $T,$ then :

A_{1} t_{1} = A_{2} t_{2}

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\frac{A_{1} - A_{2}}{t_{2} - t_{1}} = c o n s t a n t

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A_{2} = A_{1} e^{(t_{1} - t_{2}) / T}

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A_{2} = A_{1} e^{(t_{1} / T t_{2})}

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Solution

The correct option is D $A_{2} = A_{1} e^{(t_{1} - t_{2}) / T}$
As we know,
$t_{1} = \frac{1}{λ} ln \frac{A_{0}}{A_{1}}$

$t_{2} = \frac{1}{λ} ln \frac{A_{0}}{A_{2}}$

$t_{1} - t_{2} = \frac{1}{λ} ln \frac{A_{2}}{A_{1}}$

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

$= e^{(t_{1} - t_{2}) / T}$ (as $T = 1 / λ)$

Hence, option C is correct.

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The activity of a sample of radioactive element is A1 at time t1 and A2 at time t2(t2>t1). The mean life of sample is T, then :

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