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Q.

In a radioactive material the activity at time $t_{1}$ is $R_{1}$ and at a later time $t_{2},$ it is $R_{2} .$ If the decay constant of the material is $λ,$ then

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a

$R_{1} = R_{2} e^{λ (t_{1} - t_{2})}$

b

$R_{1} = R_{2} e^{(\frac{t_{2}}{t_{1}})}$

c

$R_{1} = R_{2}$

d

$R_{1} = R_{2} e^{- λ (t_{1} - t_{2})}$

answer is D.

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Detailed Solution

Let at time $t_{1} & t_{2},$ number of particles be $N_{1} & N_{2} .$ So,

$\begin{array}{l} R_{1} = - \frac{{dN}_{1}}{dt} = λ N_{1}; R_{2} = - \frac{{dN}_{2}}{dt} = λ N_{2} \\ \frac{R_{1}}{R_{2}} = \frac{λ N_{1}}{λ N_{2}} = \frac{N_{1}}{N_{1} e^{- λ (t_{2} - t_{1})}} = e^{λ (t_{2} - t_{1})} \\ R_{1} = R_{2} e^{λ (t_{2} - t_{1})} = R_{2} e^{- λ (t_{1} - t_{2})} \end{array}$

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