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Matter Waves

Activity of a...

Question

Activity of a radioactive substance is $R_{1}$ at time $t_{1}$ and $R_{2}$ at time $t_{2} (t_{2} > t_{1})$ . Then the ratio $\frac{R_{2}}{R_{1}}$ is:

t_{2} t_{1}

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e^{- λ (t_{1} + t_{2})}

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e (\frac{t_{1} + t_{2}}{λ})

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e^{λ (t_{1} - t_{2})}

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Solution

The correct option is D $e^{λ (t_{1} - t_{2})}$
$\begin{matrix} Given - & * Activity of t = t_{1} is R_{1} * Activity at t = t_{2} is R_{2} (t_{2} > t_{1}) \end{matrix}$

$\begin{matrix} we know that - Activity of a substance(R) at time(t) is given as R = R_{0} \cdot e^{λ t} where R_{0} = Initial activity λ = disintegration constant \end{matrix}$

$\begin{matrix} \Rightarrow R_{1} = R_{0} e^{- λ t_{1}} and R_{2} = R_{0} e^{- λ t_{2}} \Rightarrow \frac{R_{2}}{R_{1}} = \frac{e^{- λ t_{2}}}{e^{- λ t_{1}}} = e^{λ t_{1} - λ t_{2}} = e^{λ (t_{1} - t_{2})} Hence, option (d) is correct. \end{matrix}$

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