The activity ...

Question

The activity of a sample of radioactive material is $R_{1}$ at time $t_{1}$ and $R_{2}$ at time $t_{2} (t_{2} > t_{1})$ . Its mean life is T. Then

R_{1} t_{1} = R_{2} t_{2}

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

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

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

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Solution

The correct option is C $R_{2} = R_{1} e^{(t_{1} - t_{2}) / T}$
We have present activity $(N)$ given as $N = N_{0} e^{- t / T}$
where $N_{0}$ is initial activity.
Given that $R_{1} = N_{0} e^{- t_{1} / T}$ and $R_{2} = N_{0} e^{- t_{2} / T}$
Dividing both we get $\frac{R_{1}}{R_{2}} = e^{(t_{2} - t_{1}) / T}$
or $R_{2} = R_{1} e^{(t_{1} - t_{2}) / T}$
Concept used is $\frac{a^{b}}{a^{c}} = a^{b - c}$ and $1 / e^{a} = e^{- a}$
Option C is correct.

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